From crose Tue Sep 9 09:58:17 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02112; Tue, 9 Sep 97 09:58:16 EDT Date: Tue, 9 Sep 97 09:58:16 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709091358.AA02112@MOGLI.rutgers.edu> To: 330_543 Subject: problem set 1 Status: RO Problem set 1 is available on the web. It is due 9/16/97 if you'd like the TA to look at it. No late submissions allowed. Cheers, Chris Rose From crose Thu Sep 11 21:17:16 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02680; Thu, 11 Sep 97 21:17:15 EDT Date: Thu, 11 Sep 97 21:17:15 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709120117.AA02680@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: mailing list Status: RO Hi Folks, This is the latest test of the mailing list! 1) 501 FOLK: Good class (for me... had fun). I"m going to have to do less proving and more lecturing... but I know I won't :) In any case, read up through the linear algebra stuff on matrix norms and onward to functions of matrices I'd like to get through chapter 1 by (AT THE LATEST) the end of next week. It's tough going, but it's useful as a thinking organizer 2) 543 FOLK: At Zhou Lin's suggestion I'm going to go over the TDM switch example (framining and "diagonal" discovery) again next class just ot make sure it stuck. Start reading chapter 3. I think we'll fly through it since most of the proofs are reasonably easy. Cheers, Chris Rose From crose@localhost.localdomain Mon Sep 15 22:02:09 1997 Return-Path: Received: from localhost.localdomain (ppp-13.ts-7.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA07497; Mon, 15 Sep 97 22:02:01 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id VAA29311 for 330_543@mogli.rutgers.edu; Mon, 15 Sep 1997 21:00:25 -0400 Date: Mon, 15 Sep 1997 21:00:25 -0400 From: Christopher Rose Message-Id: <199709160100.VAA29311@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: problem set 2 Status: RO Problem set 2 consists of Chapter 3: P1,P2,P5 & P8 Chapter 4: P1, P2, & P4 It is due on 9/23/97. Cheers Chris Rose From crose Tue Sep 16 16:18:32 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA08594; Tue, 16 Sep 97 16:18:31 EDT Date: Tue, 16 Sep 97 16:18:31 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709162018.AA08594@MOGLI.rutgers.edu> To: minfan@paul.rutgers.edu Subject: Re: Office hour for the TA Cc: 330_501, 330_543 Status: RO ************* Could you please tell me what is the office hour for the TA? I have lots of questions which I can figure out my self. Thanks. ********************* Hi, Have you tried posting your questions to the mailing list yet? That's the first thing to do. Also, to which course do you refer (501 or 543). Cheers, Chris Rose From crose Tue Sep 16 16:28:23 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA08641; Tue, 16 Sep 97 16:28:23 EDT Date: Tue, 16 Sep 97 16:28:23 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709162028.AA08641@MOGLI.rutgers.edu> To: 330_543 Subject: test Cc: crose@MOGLI.rutgers.edu Status: RO REMINDER: Test on Thursday 9/25 in class and period afterward as well NO CLASS Tuesday 9/30 CHeers, Chris Rose From minfan@paul.rutgers.edu Tue Sep 16 16:35:29 1997 Return-Path: Received: from paul.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA08661; Tue, 16 Sep 97 16:35:28 EDT Received: from localhost (minfan@localhost) by paul.rutgers.edu (8.8.5/8.8.5) with SMTP id PAA14395 for ; Tue, 16 Sep 1997 15:42:53 -0400 (EDT) Date: Tue, 16 Sep 1997 15:42:52 -0400 (EDT) From: Min Fan To: Christopher Rose Subject: Re: Office hour for the TA In-Reply-To: <9709162018.AA08594@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO > Hi, > > Have you tried posting your questions to the mailing list yet? > That's the first thing to do. Also, to which course do you refer > (501 or 543). I am attending 543. I think we have a FULL time TA, so at least FULL time TA should have office hour. I am in CS major. It looks I could not understand some of the assumption in the book. So that I have LOTS of questions. I think an interatvie discussion in person will help me. I look at your solutions for hw1. I still do not understand question 4 of chapter 2. So I do not think discuss it on line is a good way. Thanks. Min Fan > > Cheers, > > Chris Rose > From crose Tue Sep 16 16:43:29 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA08692; Tue, 16 Sep 97 16:43:28 EDT Date: Tue, 16 Sep 97 16:43:28 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709162043.AA08692@MOGLI.rutgers.edu> To: minfan@paul.rutgers.edu Subject: Re: Office hour for the TA Cc: 330_543 Status: RO thanks. I'll post the TA's office hours as soon as they are available. The TA by the way is Wenfeng Zhang and his email address is zwf@winlab.rutgers.edu so you can contact him directly. Cheers, CHris Rose From chriskly@er3.rutgers.edu Wed Sep 17 10:35:04 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA09625; Wed, 17 Sep 97 10:35:04 EDT Received: (from chriskly@localhost) by er3.rutgers.edu (8.8.5/8.8.5) id JAA00348; Wed, 17 Sep 1997 09:42:19 -0400 (EDT) Date: Wed, 17 Sep 97 9:42:19 EDT From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Cc: crose@mogli.rutgers.edu Subject: end of last class... Message-Id: Status: RO Professor Rose and anyone else who can answer- Thinking about the rules we went over at the end of class yesterday, wouldnt it be more exact to say that the max # of distinct entries per row is min(m1,r2)? Consider the case where m1=5 and r2=2. Then the max # entries is 2, not 5. Right? Or what am I missing? Christine From jpapa@mail.monmouth.com Wed Sep 17 12:06:10 1997 Return-Path: Received: from shell.monmouth.com by MOGLI.rutgers.edu (4.1/25-eef) id AA09733; Wed, 17 Sep 97 12:06:09 EDT Received: from rb-tc-ppp17.monmouth.com (rb-tc-ppp17.monmouth.com [208.7.184.101]) by shell.monmouth.com (8.8.5/8.7.3) with SMTP id LAA20120; Wed, 17 Sep 1997 11:10:05 -0400 (EDT) Received: by rb-tc-ppp17.monmouth.com with Microsoft Mail id <01BCC35A.5BDFD540@rb-tc-ppp17.monmouth.com>; Wed, 17 Sep 1997 11:10:51 -0400 Message-Id: <01BCC35A.5BDFD540@rb-tc-ppp17.monmouth.com> From: Joseph Papa To: "'Christine Kleiwerda'" Cc: "'330_543@mogli.rutgers.edu'" <330_543@mogli.rutgers.edu> Subject: RE: end of last class... Date: Wed, 17 Sep 1997 11:10:40 -0400 Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable Status: RO Yeah, you're exactly right (see Figure 6 in Chp.3). I think what Prof. = Rose was trying to say is that in practice r2 is usually greater than m1 = or n3, so the limiting factor is usually not r2 (IN PRACTICE).=20 Joe -----Original Message----- From: Christine Kleiwerda [SMTP:chriskly@eden.rutgers.edu] Sent: Wednesday, September 17, 1997 9:42 AM To: 330_543@MOGLI.rutgers.edu Cc: crose@MOGLI.rutgers.edu Subject: end of last class... Professor Rose and anyone else who can answer- Thinking about the rules we went over at the end of class yesterday, = wouldnt it be more exact to say that the max # of distinct entries per row is min(m1,r2)? =20 Consider the case where m1=3D5 and r2=3D2. Then the max # entries is 2, = not 5.=20 Right? Or what am I missing?=20 Christine From crose Wed Sep 17 13:00:57 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA09782; Wed, 17 Sep 97 13:00:56 EDT Date: Wed, 17 Sep 97 13:00:56 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709171700.AA09782@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, chriskly@eden.rutgers.edu Subject: Re: end of last class... Cc: crose Status: RO Yup! That would be the completely accurate statement and completely general (for clos point-2-point networks). The basic assumption in Joe's book however is that n1 >= m1 since the switch will be a blocking switch otherwise. Other comments welcome. Keep writing! CHeers, Chris Rose From elf_pub@email.rutgers.edu Wed Sep 17 23:01:50 1997 Return-Path: Received: from eden-backend.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA10870; Wed, 17 Sep 97 23:01:49 EDT Received: from chef24.rutgers.edu (chef24.rutgers.edu [165.230.129.153]) by eden-backend.rutgers.edu (8.8.5/8.8.5) with SMTP id WAA16101 for <330_543@mogli.rutgers.edu>; Wed, 17 Sep 1997 22:09:02 -0400 (EDT) Message-Id: <34208A7F.41B0@email.rutgers.edu> Date: Wed, 17 Sep 1997 21:57:19 -0400 From: Public ELF Workstation Reply-To: "no reply"@eden.rutgers.edu Organization: Rutgers University Libraries X-Mailer: Mozilla 3.01 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli.rutgers.edu Subject: distinct symbols Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Hi, At the end of last class, we was told that each raw of Paull's Matrix From vpopescu@caip.rutgers.edu Thu Sep 18 01:50:48 1997 Return-Path: Received: from caipfs.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA11072; Thu, 18 Sep 97 01:50:47 EDT Received: from tekka.rutgers.edu (tekka.rutgers.edu [128.6.43.14]) by caipfs.rutgers.edu (8.8.5/8.8.5) with ESMTP id AAA12145; Thu, 18 Sep 1997 00:57:45 -0400 (EDT) From: Viorel George Popescu Received: (vpopescu@localhost) by tekka.rutgers.edu (8.7.6/8.6.9) id AAA10333; Thu, 18 Sep 1997 00:57:44 -0400 (EDT) Date: Thu, 18 Sep 1997 00:57:44 -0400 (EDT) Message-Id: <199709180457.AAA10333@tekka.rutgers.edu> To: 330_543@mogli.rutgers.edu Cc: vpopescu@caip.rutgers.edu Status: RO They don't have to be distinct. Eg. assume m1=r2=n3; then you anyway don't have more than r2 symbols (second stage switches). Therefore they (collections of different symbols on row and collection of different symbols on column) overlap. But they don't have to be identical. If r2>m1 or r2>n3 then these collections have at least max(0, (min(m1,r2)+min(n3,r2)-r2) ) common symbols (you can find examples). My understanding of the problem is the following: 1. You get the maximum # of different symbols in a row if you consider the entire flow (traffic) going through only one switch in the first stage if r2>m1. The flow has to span to different second stage switches (no path overlaping). If r2m1=n3, by choosing to route the traffic through a single input switch and a single third stage switch; then you squeeze all symbols (m1=n3) in one element of the matrix. This is just an example. You cannot do that all the time because you have to solve conflicts in the switch (using input stage) and rearange the pachet order (third switch). For a non-blocking Clos network the proof of theorem (pp. 65) uses somehow the fact that you should have different symbols on row and columns, but I didn't get the meaning of that. My understanding of non-blocking for Clos network is as follows: you establish m1-1 connections (assume r2>m1, r2>n3) and you are trying to fit the last one. You want to be able to fit it anywhere: from any input to any output. To not disturb the already existent paths, you need n3 additional second stage switches, so that you can place any input in any output slot. If previous statements are not true, please send corrections and comments. Cheers, George > Hi , > > At the end of the class, we were told that each row of Paull's Matrix > can have at most m1 distinct symbols. Each columm can have at most n3 > distinct symbols. Here symbols are different from each other within a > row or a column. For instance, row a and column b may have m1+n3 > symbols, do they have to distinct each other? It depends on blocking of > the network? > if it is a nonblocking network? > > I will appreciate if some one gives the answer. Otherwise, it will be > covered next class. > From crose Thu Sep 18 02:21:55 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA11104; Thu, 18 Sep 97 02:21:54 EDT Date: Thu, 18 Sep 97 02:21:54 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709180621.AA11104@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, vpopescu@caip.rutgers.edu Subject: Paull's Matrix Status: RO Hi FOlks, The issue is one of point-2-point switching. If it's not point to point, then the distinctness constraint is gone. However, if you have point to point then each row can have a max of m1 symbols (all distinct... think about it... there are only m1 different switches you can get to at one time from a given input switch bank... you have to choose from r2>= m1 possible symbols, but you can only have m1 before you've used the switch up). Same argument for the columns (only now the limit it n3). By the way, a side issue which is kind of swept under the rug is that we do not consider networks which are blocking. We're only looking at rearrangeable networks and strict sense non-blocking networks. This is the reason that n1 >= m1 and m3 >= n3. Such networks are blocking --- can't have all inputs on a given input switch bank active at once (same for output switch banks) no matter what (can't sneak around the problem by rearrangement). Cheers, Chris Rose From crose Thu Sep 18 02:24:35 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA11115; Thu, 18 Sep 97 02:24:35 EDT Date: Thu, 18 Sep 97 02:24:35 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709180624.AA11115@MOGLI.rutgers.edu> To: 330_543 Subject: erratum Status: RO On my last mail message I said in the last sentence that such networks are blocking... what I meant was taht netorks with n1 < m1 and m3 < n3 are blocking. Sorry 'bout that.. From crose Thu Sep 18 02:26:37 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA11126; Thu, 18 Sep 97 02:26:37 EDT Date: Thu, 18 Sep 97 02:26:37 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709180626.AA11126@MOGLI.rutgers.edu> To: 330_543 Subject: by the way Status: RO Hey Folks, There is an archive of all email snt by the class on this address (330_543) linked to the web page. It's in text form (just a copy of my save file for class correspondence). Cheers (again) From chriskly@er3.rutgers.edu Wed Sep 17 10:35:04 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA09625; Wed, 17 Sep 97 10:35:04 EDT Received: (from chriskly@localhost) by er3.rutgers.edu (8.8.5/8.8.5) id JAA00348; Wed, 17 Sep 1997 09:42:19 -0400 (EDT) Date: Wed, 17 Sep 97 9:42:19 EDT From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Cc: crose@mogli.rutgers.edu Subject: end of last class... Message-Id: Status: RO Professor Rose and anyone else who can answer- Thinking about the rules we went over at the end of class yesterday, wouldnt it be more exact to say that the max # of distinct entries per row is min(m1,r2)? Consider the case where m1=5 and r2=2. Then the max # entries is 2, not 5. Right? Or what am I missing? Christine From jpapa@mail.monmouth.com Wed Sep 17 12:06:10 1997 Return-Path: Received: from shell.monmouth.com by MOGLI.rutgers.edu (4.1/25-eef) id AA09733; Wed, 17 Sep 97 12:06:09 EDT Received: from rb-tc-ppp17.monmouth.com (rb-tc-ppp17.monmouth.com [208.7.184.101]) by shell.monmouth.com (8.8.5/8.7.3) with SMTP id LAA20120; Wed, 17 Sep 1997 11:10:05 -0400 (EDT) Received: by rb-tc-ppp17.monmouth.com with Microsoft Mail id <01BCC35A.5BDFD540@rb-tc-ppp17.monmouth.com>; Wed, 17 Sep 1997 11:10:51 -0400 Message-Id: <01BCC35A.5BDFD540@rb-tc-ppp17.monmouth.com> From: Joseph Papa To: "'Christine Kleiwerda'" Cc: "'330_543@mogli.rutgers.edu'" <330_543@mogli.rutgers.edu> Subject: RE: end of last class... Date: Wed, 17 Sep 1997 11:10:40 -0400 Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable Status: RO Yeah, you're exactly right (see Figure 6 in Chp.3). I think what Prof. = Rose was trying to say is that in practice r2 is usually greater than m1 = or n3, so the limiting factor is usually not r2 (IN PRACTICE).=20 Joe -----Original Message----- From: Christine Kleiwerda [SMTP:chriskly@eden.rutgers.edu] Sent: Wednesday, September 17, 1997 9:42 AM To: 330_543@MOGLI.rutgers.edu Cc: crose@MOGLI.rutgers.edu Subject: end of last class... Professor Rose and anyone else who can answer- Thinking about the rules we went over at the end of class yesterday, = wouldnt it be more exact to say that the max # of distinct entries per row is min(m1,r2)? =20 Consider the case where m1=3D5 and r2=3D2. Then the max # entries is 2, = not 5.=20 Right? Or what am I missing?=20 Christine From crose Thu Sep 18 11:17:39 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA11785; Thu, 18 Sep 97 11:17:38 EDT Date: Thu, 18 Sep 97 11:17:38 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709181517.AA11785@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, chriskly@eden.rutgers.edu Subject: Re: end of last class... Cc: crose@mogli.rutgers.edu Status: RO Hi Christine, The issue (I think this has come up in a few emails before) is that if r2=2 and m1=5, the switch is automatically blocking... you can never use all m1 inputs at the same time. So m1 is always assumed to be less than or equal to r2. A similar argument applies to r2 and n3. Hope that helps.... Cheers Chris Rose From crose Thu Sep 18 11:19:23 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA11791; Thu, 18 Sep 97 11:19:22 EDT Date: Thu, 18 Sep 97 11:19:22 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709181519.AA11791@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, chriskly@eden.rutgers.edu Subject: Re: end of last class... Cc: crose@mogli.rutgers.eduret Status: RO Ignore my last mail since it was a rehash of a previously asked and answered question. (I had an open mail tool which I did not delete from yesterday). Cheers, Chris Rose From nirajp@er5.rutgers.edu Fri Sep 19 11:13:42 1997 Return-Path: Received: from eden-backend.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA00206; Fri, 19 Sep 97 11:10:36 EDT Received: from er5.rutgers.edu (nirajp@er5.rutgers.edu [165.230.180.133]) by eden-backend.rutgers.edu (8.8.5/8.8.5) with ESMTP id AAA05676 for <330_543@MOGLI.rutgers.edu>; Fri, 19 Sep 1997 00:12:43 -0400 (EDT) Received: from localhost (nirajp@localhost) by er5.rutgers.edu (8.8.5/8.8.5) with SMTP id AAA10453 for <330_543@MOGLI.rutgers.edu>; Fri, 19 Sep 1997 00:09:24 -0400 (EDT) Date: Fri, 19 Sep 1997 00:09:23 -0400 (EDT) From: Niraj Prabhavalkar To: 330_543@MOGLI.rutgers.edu Subject: vci Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO hi, i'm not very sure about the vci concept yet but this is what i think. its called virtual because it changes from time to time and is not dependent on the physical link but on the number of active users at a time. so the address changes accordingly. another thing is about todays class. i was slightly lost. can someone tell me about the second and third theorems please. waiting for a reply niraj. From crose Fri Sep 19 12:59:43 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00370; Fri, 19 Sep 97 12:59:42 EDT Date: Fri, 19 Sep 97 12:59:42 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709191659.AA00370@MOGLI.rutgers.edu> To: nirajp@eden.rutgers.edu Subject: Re: vci Cc: crose@MOGLI.rutgers.edu Status: RO Hi, I'll take a crack at your questions first: 1) A VIRTUAL ckt as opposed to a real circuit is a concept related to packetization of information. In the old days, a real circuit implied a real wire and a real hunk of bandwidth on various links that was used ALL THE TIME by the call in progress. And since we were usually talking TDM or FDM (frequency/time division multiplexing) that implied particular frequencies and times were assigned to carry a call on each leg of it's journey. In a VIRTUAL setting there are no explicit revervations of time slots and channels. There is just a guarantee of availability. SO, that means each packet has to have an identifier which states that it has a reservation for resources AND THIS MUST BE DONE FOR EACH LEG OF THE PACKET's TRIP! Thus, you have a VIRTUAL circuit IDENTIFIER for each packet and at each leg, you might have different ID numbers for the same conversation (VCI translation tables). 2) Not sure how to help here. What I'd suggest is follow the logic through (why must m1 < n1, and similarly for the output swtiches) and then play devil's advocate and find worst case scenarios for when you can't find paths without rearangement. For the rearrangment, draw another picture corresponding to the case where you need to do one rearrangment so you get a physical feel for what's happening. Then do two... I think the rearrangement proof will become clear at that point. For the double chain argument for maximum numbers of rearrangments, that's just a counting proof which is clear if you understand the rearrangeability proof I think. Hope that helps, but everyone chime in please as well! Cheers, Chris Rose From chriskly@er5.rutgers.edu Fri Sep 19 15:57:55 1997 Return-Path: Received: from er5.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA00729; Fri, 19 Sep 97 15:57:55 EDT Received: (from chriskly@localhost) by er5.rutgers.edu (8.8.5/8.8.5) id PAA22250; Fri, 19 Sep 1997 15:04:49 -0400 (EDT) Date: Fri, 19 Sep 97 15:04:49 EDT From: Christine Kleiwerda To: Niraj Prabhavalkar Cc: 330_543@MOGLI.rutgers.edu Subject: Re: vci In-Reply-To: Your message of Fri, 19 Sep 1997 00:09:23 -0400 (EDT) Message-Id: Status: RO > hi, > i'm not very sure about the vci concept yet but this is what i think. > its called virtual because it changes from time to time and is not > dependent on the physical link but on the number of active users at a > time. so the address changes accordingly. > another thing is about todays class. i was slightly lost. > can someone tell me about the second and third theorems please. > waiting for a reply > niraj. > > > i'm not sure which ones you are callining the 2nd and 3rd theorems, but i think you mean the rearrangable non blockign ones (RNB) ... 1. a 3 stage clos net is RNB if r2 > max (m1, n3). To me, this means that the number of middle switches must be greater than both the number of inputs and the number of outputs. Otherwise, say the number of middle switches is less than teh number of inputs, then it is possible that all the middle switches are being used when a new input is looking for a connection. Then this newest input is blocked (ther is no possible way for a connection to be made). Same argument for the output side. 2. an upper bound on the max # of rearrangements is min (r1,r3) - 1 Using Paull's matrix, each time you do a rearrangement, you are eliminating one row and one col from the "pool" where you can make the rearrangments. (I'm notquite sure what this translates to in the switch terminology... i think it's something like each time you do a rearrangement, you are changing which switch a path goes through and you can only make this rearrangement once??) Anyway, back to Paull's matrix.. .you can keep making rearrangements until you run out of rows or cols , whichever comes first. There are r1 rows and r3 cols total in the matrix. When you start, you immediately eliminate the row and col you are trying to connect, so you are left with r1-1 possible rows and r3-1 possible cols. The max # of rearrangements is the min of these, min(r1-1, r3-1) = min (r1, r3) - 1. Does this make sense? Is this what you were looking for? Christine From chriskly@er6.rutgers.edu Fri Sep 19 22:21:32 1997 Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01348; Fri, 19 Sep 97 22:21:29 EDT Received: (from chriskly@localhost) by er6.rutgers.edu (8.8.5/8.8.5) id VAA03619; Fri, 19 Sep 1997 21:28:22 -0400 (EDT) Date: Fri, 19 Sep 97 21:28:21 EDT From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu, linzhou@ece.rutgers.edu Subject: [Lin Zhou : Re: vci] Message-Id: Status: RO >From what I understood from class, here's the way to do it: We know there is at least one switch available to connect to a and another switch available to connect to b. Our problem is that these are not the same switches. Call a's free switch Sa and b's free switch Sb. Since Sa does not equal Sb, we know that there must be an entry in col b equal to Sa. We look in the row where Sa is for an entry of Sb. If Sb is not in that row, then we can change Sa to Sb and use Sa for the connection from a to b. If Sb is in that row, then we search the column where Sb is, looking for Sa. If there is no Sa, then we can change the Sb to Sa, and the Sa (from row b) to Sb and use Sa to complete the connection from a to b. And you can continue this process.... Basically, I think the concept is that we are going to use Sa to complete the connection. But, Sa is already used by b, so we need to replace that connection by Sb, which may already be used in that row, so we replace it with Sa, which may already be used in that col, so we replace it with Sb, which may already be used in that row, etc..... Eventually, we will reach a point where the switch we want to use is not already being used in that row/col and then we can go ahead and make all the rearrangements. Does this help at all? I'm trying to figure out why we are guaranteed "eventually, we will reach a point where the switch we want to use is not already being used in that row/col" and cant quite see the answer yet...... Anyone have an idea??? Somehow, it must be because we know Sa is available in row a and Sb is available in row b. Christine --------------- Received: from ece.rutgers.edu (ece.rutgers.edu [128.6.46.12]) by eden-backend.rutgers.edu (8.8.5/8.8.5) with SMTP id PAA02548 for ; Fri, 19 Sep 1997 15:44:01 -0400 (EDT) Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id PAA03785 for ; Fri, 19 Sep 1997 15:39:45 -0400 Date: Fri, 19 Sep 1997 15:39:45 -0400 (EDT) From: Lin Zhou To: Christine Kleiwerda Subject: Re: vci In-Reply-To: Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Hi Christine, The upbound of rearrangement nonblocking network is min(r1,r3)-1 with your explanation. How can you do the arrangement one by one after delete one row and one column? Though you can eat at most min(r1,r3) row or col finally, How is arrangement doing? The goal is to put a distinct entry to [a,b]. zhou lin From chriskly@er3.rutgers.edu Fri Sep 19 22:36:50 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01424; Fri, 19 Sep 97 22:36:49 EDT Received: (from chriskly@localhost) by er3.rutgers.edu (8.8.5/8.8.5) id VAA04197; Fri, 19 Sep 1997 21:43:42 -0400 (EDT) Date: Fri, 19 Sep 97 21:43:42 EDT From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu, crose@mogli.rutgers.edu Subject: exam Message-Id: Status: RO Professor Rose- Is the exam next thursday open book? What chapters will it cover? Christine From crose Fri Sep 19 22:45:50 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA01439; Fri, 19 Sep 97 22:45:49 EDT Date: Fri, 19 Sep 97 22:45:49 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709200245.AA01439@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, chriskly@eden.rutgers.edu, crose@mogli.rutgers.edu Subject: Re: exam Cc: crose@MOGLI.rutgers.edu Status: RO Hi Christine, 1) thanks for the various explanations: you're eventually guaranteed to stop because you'll eventually run out of rows and columns to find stuff in! If you're lucky and run out of things to find before then, then you count your blessings. 2) The test is NOT open book. However, take a peek at the info sheet and you'll see that you can have a single 2-sided sheet of 8.5x11 note paper with handwritten notes on it. That should help both as a study aid and security blanket :) The test will be on chapters 1-3 and possibly part of chapter 4. Cheers (all) Chris Rose From aonweller@maerskdata-usa.com Sat Sep 20 14:16:17 1997 Return-Path: Received: from msrvr.maerskdata-usa.com by MOGLI.rutgers.edu (4.1/25-eef) id AA02068; Sat, 20 Sep 97 14:16:16 EDT Received: from [10.1.25.41] ([207.242.160.66]) by msrvr.maerskdata-usa.com (Netscape Mail Server v2.02) with SMTP id AAA96 for ; Sat, 20 Sep 1997 17:26:04 +0000 Message-Id: <34240673.4BA7@maerskdata-usa.com> Date: Sat, 20 Sep 1997 13:23:00 -0400 From: aonweller@maerskdata-usa.com (Allen Onweller) X-Mailer: Mozilla 3.01-C-MACOS8 (Macintosh; I; PPC) Mime-Version: 1.0 To: crose@mogli.rutgers.edu Subject: Web Site not accessable Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Hi Prof. Rose. I seem to be having problems accessing any sites in the ece.rutgers.edu domain (I get a connection refused error.) I can get to the main Rutgers site (www.rutgers.edu) just fine though. The problem is that I haven't downloaded problem set two yet and I was planning on working on it today and Sunday. I was wondering if it wouldn't be too much trouble for you to e-mail the pproblem set to the e-mail list. (I'm assuming that there are others in need also.) Regards, Allen From crose Sat Sep 20 19:48:08 1997 To: cripop@ece.rutgers.edu Subject: Re: mailing list error Cc: s corresp/543 Status: O Glad to repost. The machines were down on thursday... Cheers, From cripop@ece.rutgers.edu Sat Sep 20 17:49:33 1997 Return-Path: Received: from zen.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02185; Sat, 20 Sep 97 17:49:32 EDT Received: from chess.rutgers.edu (chess.rutgers.edu [128.6.46.113]) by zen.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id QAA02602 for ; Sat, 20 Sep 1997 16:52:18 -0400 From: Dimitrie Popescu Message-Id: <199709202052.QAA02602@zen.rutgers.edu> Subject: mailing list error To: crose@ece.rutgers.edu Date: Sat, 20 Sep 1997 16:46:48 -0400 (EDT) Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit Content-Length: 2992 Status: RO Dear dr. Rose, I have tried to send the following message to the 330_501 mailing list (330_501@mogli.rutgers.edu) but it has been returned twice with the same error message "User unknown". Did I do something wrong? Also, maybe you can post the following message to the mailing list. Otilia Popescu PS. I receive messages from 330_501@mogli.rutgers.edu, but I haven't been able to send any messages. >From Postmaster Fri Sep 19 12:43:15 1997 Received: from localhost (localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with internal id MAA10603; Fri, 19 Sep 1997 12:43:15 -0400 Date: Fri, 19 Sep 1997 12:43:15 -0400 From: Mail Delivery Subsystem Subject: Returned mail: User unknown Message-Id: <199709191643.MAA10603@ece.rutgers.edu> To: cripop MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="MAA10603.874687395/ece.rutgers.edu" Status: RO This is a MIME-encapsulated message --MAA10603.874687395/ece.rutgers.edu The original message was received at Fri, 19 Sep 1997 12:43:13 -0400 from cripop@localhost ----- The following addresses had delivery problems ----- 330_501@mogli.rutgers.edu (unrecoverable error) ----- Transcript of session follows ----- ... while talking to mogli.rutgers.edu.: >>> RCPT To:<330_501@mogli.rutgers.edu> <<< 550 /usr/crose/system/501: line 2: utgers.edu... User unknown 550 330_501@mogli.rutgers.edu... User unknown ----- Original message follows ----- --MAA10603.874687395/ece.rutgers.edu Content-Type: message/rfc822 Return-Path: cripop Received: (from cripop@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) id MAA10602 for 330_501@mogli.rutgers.edu; Fri, 19 Sep 1997 12:43:13 -0400 Date: Fri, 19 Sep 1997 12:43:13 -0400 From: Dimitrie Popescu Message-Id: <199709191643.MAA10602@ece.rutgers.edu> To: 330_501@mogli.rutgers.edu Subject: Cayley-Hamilton Theorem Content-Length: 1048 This is a proof of the Cayley-Hamilton Theorem in the general case. No assumption is made about matrix A; just that is a square matrix of dimension nxn. Let us denote the characteristic polynomial of A: Phi(s)=det(sI-A)=s^n+a_1s^(n-1)+...+a_n Let us also take the inverse of sI-A: (sI-A)^(-1)=(sI-A)*/det(sI-A)=B(s)/Phi(s) where (sI-A)*=B(s) is the adjoint matrix of sI-A. Rewrite the above relation as: Phi(s)I_n=(sI-A)B(s) because the highest power of s in Phi(s) is n, then the highest power of s in B(s) will be n-1. B(s) can be written as: B(s)=B_0s^(n-1)+B_1s(n-2)+...+B_(n-1) After performing the multiplication and identifying coefficients we get: s^n: I=B_0 s^(n-1): a_1I=B_1-AB_0 s^(n-2): a_2I=B_2-AB_1 ........ s^1: a_(n-1)I=B_(n-1)-AB_(n-2) s^0: a_nI= -AB_(n-1) These relations are equivalent with: B_0=I B_1=A+a_1I B_2=A^2+Aa_1+a_2I ......... B_(n-1)=A^(n-1)+A^(n-2)a_1+...+a_(n-1)I 0 =A^n+A^(n-1)a_1+...+Aa_(n-1)+a_nI The last relation is in fact Phi(A)=0 and the proof is completed. Otilia Popescu --MAA10603.874687395/ece.rutgers.edu-- From crose Sat Sep 20 19:53:28 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02296; Sat, 20 Sep 97 19:53:28 EDT Date: Sat, 20 Sep 97 19:53:28 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709202353.AA02296@MOGLI.rutgers.edu> To: 330_543 Subject: problem set Cc: crose@MOGLI.rutgers.edu Status: RO Problem set 2 consists of Chapter 3: P1,P2,P5 & P8 Chapter 4: P1, P2, & P4 It is due on 9/23/97. Cheers Chris Rose From crose Sun Sep 21 10:57:53 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03071; Sun, 21 Sep 97 10:57:45 EDT Date: Sun, 21 Sep 97 10:57:45 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709211457.AA03071@MOGLI.rutgers.edu> To: chriskly@eden.rutgers.edu, nirajp@eden.rutgers.edu Subject: Re: [Lin Zhou : Re: vci] Cc: 330_543@MOGLI.rutgers.edu, linzhou@ece.rutgers.edu Status: RO That's exactly right Niraj. If you run out of Cs and Ds then you do the rearrangement. The proofs are just for upper bonds (worst cases). Cheers, Chris Rose From nirajp@er6.rutgers.edu Sun Sep 21 01:38:57 1997 Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02704; Sun, 21 Sep 97 01:38:34 EDT Received: from localhost (nirajp@localhost) by er6.rutgers.edu (8.8.5/8.8.5) with SMTP id AAA09282; Sun, 21 Sep 1997 00:45:14 -0400 (EDT) Date: Sun, 21 Sep 1997 00:45:14 -0400 (EDT) From: Niraj Prabhavalkar To: Christine Kleiwerda Cc: 330_543@MOGLI.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: [Lin Zhou : Re: vci] In-Reply-To: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO hi about the rearrangement thing, i have one serious doubt. in the book they say that we keep the chain going and keep finding c's and d's to replace each other till we reach the end of the chain. however is it possible that we run out of c's and d's before we run out of the rows and columns??? in that case the upperbound will depend on the no of c's/d's that are available and not necessarily on the min(r1,r3)-1. anyone agrees or disagrees? niraj. $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ NIRAJ PRABHAVALKAR:- US HOUSING US MAILING ADD. PERMANENT ADDRESS BUELL 385 B.P.O 23952 44, POORNANAND BUSCH CAMPUS PO BOX 1119 WALKESHWAR RUTGERS UNIVERSITY PISCATAWAY BOMBAY 400006 NEW BRUNSWICK NJ 08855-1119 INDIA. NEW JERSEY 08855 USA USA. TEL: 732-8781648 TEL: 022-3681733/3623561 EMAIL: nirajp@eden.rutgers.edu EMAIL:nirajp@giasbma.vsnl.net.in nirajp@er5.rutgers.edu $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ From crose Sun Sep 21 11:14:48 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03104; Sun, 21 Sep 97 11:14:21 EDT Date: Sun, 21 Sep 97 11:14:21 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709211514.AA03104@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: ece machine Cc: crose@MOGLI.rutgers.edu Status: RO Hi Folks, The ECE machine seems to be fried. Hopefully it will be fixed by monday. In the mean time, even I cannot get access to the web page (both front AND back doors). I'll make copies of the PS solutions and distribute them in class. I'll also look into moving the web page to a more reliable machine. For some reason ece is flaky. Cheers, sort of, Chris ROse From jsucec@ece.rutgers.edu Mon Sep 22 02:22:14 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03910; Mon, 22 Sep 97 02:22:00 EDT Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA13344; Mon, 22 Sep 1997 01:24:15 -0400 Date: Mon, 22 Sep 1997 01:24:15 -0400 (EDT) From: John Sucec To: Niraj Prabhavalkar Cc: 330_543@MOGLI.rutgers.edu Subject: Re: vci In-Reply-To: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO This message is in response to an E-Mail sent below, on the subject of Virtual Circuit Identifier (VCI)... "Virtual" in the case of VCI, refers to the "circuit" part of VCI, not the "identifier" part. This is because in ATM, unlike TDM where there is an actual time slot allocated to a particular connection, communication needs are satisfied on a best effort basis by the network. If the ATM network is healthy, however, end users will not be able to tell the difference between the best effort cell transport mechanism of ATM from an actual TDM circuit. Hence, the ATM connectivity between end users is referred to as a virtual circuit. As far as the VCI changing from time to time, I don't think this is true, at least not in a dynamic sense. Once an ATM connection is established, network switches use cut-through switching to send cells out on the appropriate port based on the VCI field of each cell header. When the switch sends a cell out on the appropriate port, it will write a new value into the VCI field of the cell header. This new value, however, comes from a "look-up" table that functions as a static mapping of the VCI for the incoming cells of each virtual circuit to the appropriate destination port and a VCI value for outgoing cells of each virtual circuit. The VCI values for cells belonging to a particular virtual circuit, therefore, are maintained statically in the RAM of network switches that compose the path of the virtual circuit. Anyway, I believe that is how the VCI field is typically used. I hope this helps. ...John On Fri, 19 Sep 1997, Niraj Prabhavalkar wrote: > hi, > i'm not very sure about the vci concept yet but this is what i think. > its called virtual because it changes from time to time and is not > dependent on the physical link but on the number of active users at a > time. so the address changes accordingly. > another thing is about todays class. i was slightly lost. > can someone tell me about the second and third theorems please. > waiting for a reply > niraj. > > > From jsucec@ece.rutgers.edu Mon Sep 22 02:58:04 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03929; Mon, 22 Sep 97 02:57:51 EDT Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA13604 for <330_543@MOGLI.rutgers.edu>; Mon, 22 Sep 1997 02:00:07 -0400 Date: Mon, 22 Sep 1997 02:00:06 -0400 (EDT) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Re: vci (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Oops, I said that the VCI values for an ATM connection "are stored in the RAM of network switches." I did not mean to imply, however, that the switching of cells is done by looking up VCIs in RAM... The switching done by ATM nodes has to be at the hardware level in order to switch cells at rates required by typical ATM networks. Therefore, once an ATM connection is established (which is CPU and RAM intensive), the cut through switching of cells is performed 'with hardware'. Unfortunately, I do not understand very well the design of ATM switches at the hardware level, so I can't comment much further on how the VCI-in to VCI-out static mapping is maintained and accessed at the 'hardware level' within the switch. ...John ---------- Forwarded message ---------- Date: Mon, 22 Sep 1997 01:24:15 -0400 (EDT) From: John Sucec To: Niraj Prabhavalkar Cc: 330_543@MOGLI.rutgers.edu Subject: Re: vci This message is in response to an E-Mail sent below, on the subject of Virtual Circuit Identifier (VCI)... "Virtual" in the case of VCI, refers to the "circuit" part of VCI, not the "identifier" part. This is because in ATM, unlike TDM where there is an actual time slot allocated to a particular connection, communication needs are satisfied on a best effort basis by the network. If the ATM network is healthy, however, end users will not be able to tell the difference between the best effort cell transport mechanism of ATM from an actual TDM circuit. Hence, the ATM connectivity between end users is referred to as a virtual circuit. As far as the VCI changing from time to time, I don't think this is true, at least not in a dynamic sense. Once an ATM connection is established, network switches use cut-through switching to send cells out on the appropriate port based on the VCI field of each cell header. When the switch sends a cell out on the appropriate port, it will write a new value into the VCI field of the cell header. This new value, however, comes from a "look-up" table that functions as a static mapping of the VCI for the incoming cells of each virtual circuit to the appropriate destination port and a VCI value for outgoing cells of each virtual circuit. The VCI values for cells belonging to a particular virtual circuit, therefore, are maintained statically in the RAM of network switches that compose the path of the virtual circuit. Anyway, I believe that is how the VCI field is typically used. I hope this helps. ...John On Fri, 19 Sep 1997, Niraj Prabhavalkar wrote: > hi, > i'm not very sure about the vci concept yet but this is what i think. > its called virtual because it changes from time to time and is not > dependent on the physical link but on the number of active users at a > time. so the address changes accordingly. > another thing is about todays class. i was slightly lost. > can someone tell me about the second and third theorems please. > waiting for a reply > niraj. > > > From nirajp@er6.rutgers.edu Sun Sep 21 01:38:57 1997 Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02704; Sun, 21 Sep 97 01:38:34 EDT Received: from localhost (nirajp@localhost) by er6.rutgers.edu (8.8.5/8.8.5) with SMTP id AAA09282; Sun, 21 Sep 1997 00:45:14 -0400 (EDT) Date: Sun, 21 Sep 1997 00:45:14 -0400 (EDT) From: Niraj Prabhavalkar To: Christine Kleiwerda Cc: 330_543@MOGLI.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: [Lin Zhou : Re: vci] In-Reply-To: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO hi about the rearrangement thing, i have one serious doubt. in the book they say that we keep the chain going and keep finding c's and d's to replace each other till we reach the end of the chain. however is it possible that we run out of c's and d's before we run out of the rows and columns??? in that case the upperbound will depend on the no of c's/d's that are available and not necessarily on the min(r1,r3)-1. anyone agrees or disagrees? niraj. $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ NIRAJ PRABHAVALKAR:- US HOUSING US MAILING ADD. PERMANENT ADDRESS BUELL 385 B.P.O 23952 44, POORNANAND BUSCH CAMPUS PO BOX 1119 WALKESHWAR RUTGERS UNIVERSITY PISCATAWAY BOMBAY 400006 NEW BRUNSWICK NJ 08855-1119 INDIA. NEW JERSEY 08855 USA USA. TEL: 732-8781648 TEL: 022-3681733/3623561 EMAIL: nirajp@eden.rutgers.edu EMAIL:nirajp@giasbma.vsnl.net.in nirajp@er5.rutgers.edu $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ From crose Mon Sep 22 12:03:29 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA04263; Mon, 22 Sep 97 12:01:29 EDT Date: Mon, 22 Sep 97 12:01:29 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709221601.AA04263@MOGLI.rutgers.edu> To: jsucec@ece.rutgers.edu, nirajp@eden.rutgers.edu Subject: Re: vci Cc: 330_543@MOGLI.rutgers.edu Status: RO Nice explanation John! From shahid@er7.rutgers.edu Mon Sep 22 15:18:00 1997 Return-Path: Received: from er7.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA04449; Mon, 22 Sep 97 15:16:34 EDT Received: from localhost (shahid@localhost) by er7.rutgers.edu (8.8.5/8.8.5) with SMTP id OAA17663 for <330_543@MOGLI.rutgers.edu>; Mon, 22 Sep 1997 14:23:00 -0400 (EDT) Date: Mon, 22 Sep 1997 14:22:59 -0400 (EDT) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Subject: CHAPTER :3 : Page 74 & Crosspoint count + Crosspoint complexity Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Professor Rose, I have doubts regarding the following : 1.On page 74 of text book under the topic of Recursive Construction of Switching Networks there is a calculation for crosspoint complexity.How does 6^l appear in the expression. 2.I have a slight confusion regarding calculation of crosspoint count and crosspoint complexity.If you could explain taking example of a specific network that would be better. Waiting for your quick reply, Thank You. Shahid From crose@localhost.localdomain Mon Sep 22 17:03:13 1997 Return-Path: Received: from localhost.localdomain (ppp-72.ts-11.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA04624; Mon, 22 Sep 97 16:58:57 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id PAA06872; Mon, 22 Sep 1997 15:55:43 -0400 Date: Mon, 22 Sep 1997 15:55:43 -0400 From: Christopher Rose Message-Id: <199709221955.PAA06872@localhost.localdomain> To: 330_543@MOGLI.rutgers.edu, shahid@eden.rutgers.edu Subject: Re: CHAPTER :3 : Page 74 & Crosspoint count + Crosspoint complexity Status: RO Hi, You'll need to state your question (for part 2). As for the 6^l, think about where the recursion has to stop in going from equation 3.7.3 to 3.7.4 (look at the F() term argument). But in general, you'll be able to help everyone if you spell out your question in gory detail. Sometimes you'll spell it out wrong and taht will be an immediately recognizable source of the confusion which as a group we can address and from which we can benefit. Cheers, Chris Rose From shahid@er3.rutgers.edu Tue Sep 23 15:54:26 1997 X-UIDL: e4fc0560f1da072cf4cdc25bb60f5f3a Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06087; Tue, 23 Sep 97 15:50:57 EDT Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id OAA21898 for <330_543@MOGLI.rutgers.edu>; Tue, 23 Sep 1997 14:57:12 -0400 (EDT) Date: Tue, 23 Sep 1997 14:57:11 -0400 (EDT) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Subject: Calculation of Crosspoint Complexity Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Professor Rose, I wish to know how to calculate crosspoint complexity of a given network like : Point to Point, 3-Stage Clos network :a) Strict Sense Non Blocking and b) Rearrangably non blocking. I thought it was dependent upon the number of switching stages, number of switches in each stage and number of inputs and outputs. But I cannot find a proper answer . Thank You, Shahid. From crose Tue Sep 23 17:09:04 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06246; Tue, 23 Sep 97 17:06:40 EDT Date: Tue, 23 Sep 97 17:06:40 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709232106.AA06246@MOGLI.rutgers.edu> To: shahid@eden.rutgers.edu Subject: Re: Calculation of Crosspoint Complexity Cc: 330_543 Status: RO It is! BUt it's also dependent upon how you factor the network. That is, I don't believe anyone knows EXACTLY the minimum number of crosspoints one must use except for the most basic networks (or regular ones like Clos, etc). Cheers, Chris ROse From crose Tue Sep 23 21:40:15 1997 X-UIDL: 493e75b84842bad940e63569aa3dec0a Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06623; Tue, 23 Sep 97 21:37:29 EDT Date: Tue, 23 Sep 97 21:37:29 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709240137.AA06623@MOGLI.rutgers.edu> To: 330_543 Subject: web stuff Status: RO Hi Folks, I've posted the solutions to problem set 2 on the web (hooray for Wenfeng!) I've also made a duplicate of the web page on a winlab machine so that just in case ece goes down, you can still get at the stuff http:/winwww.rutgers.edu/pub/about/people/crose/winlab543.html is the URL. Not everything works (like the signup for instance), but the PS files are there and readable as well as the email archive (which will be updated with your mailings in my absence so keep an eye out for it). See you when I get back (10/2). I should have the exams graded about a week from then is not sooner. Remember, there's no class on 9/30, unless you'd just like to watch the other students in 501 suffer! :) Cheers Chris Rose PS: Anybody know how to get from the airport in Budapest to local hotels? I've not a clue! From crose Tue Sep 23 22:50:27 1997 X-UIDL: f5ce9d185a9515d6d0764f4d1f5b2189 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06815; Tue, 23 Sep 97 22:43:19 EDT Date: Tue, 23 Sep 97 22:43:19 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709240243.AA06815@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: signup Status: RO Even the signup now works on winlab. So everything should be ok! Cheers, and good luck on the exams! Chris Rose From nirajp@er4.rutgers.edu Wed Sep 24 10:09:35 1997 Return-Path: Received: from eden-backend.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA00198; Wed, 24 Sep 97 10:09:29 EDT Received: from er4.rutgers.edu (nirajp@er4.rutgers.edu [165.230.180.142]) by eden-backend.rutgers.edu (8.8.5/8.8.5) with ESMTP id AAA03049 for ; Wed, 24 Sep 1997 00:32:37 -0400 (EDT) Received: from localhost (nirajp@localhost) by er4.rutgers.edu (8.8.5/8.8.5) with SMTP id AAA03910 for ; Wed, 24 Sep 1997 00:29:18 -0400 (EDT) Date: Wed, 24 Sep 1997 00:29:18 -0400 (EDT) From: Niraj Prabhavalkar To: Christopher Rose Subject: Re: ch 3 In-Reply-To: <9709240137.AA06623@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO hi i have a few questions and doubts. 1. in recursive factorisation something is said about factoring N into prime factors for optimization in case of RNB. could someone give me an ex. of factoring say N=100 or 200 into prime factors. 2. In todays class chris said that The max no of stages in RNB is k-1. i feel actually the max # of successive factorizations are k-1 and the no. of stages are in fact 2logN-1 viz. 2k-1. 3. in recursive factorization for ssnb the initial math is easy to follow but i didnt get the last step. >From F(N) = (2^n)(6^l)F(2) how do we get the last step. basically how does 6^l become logN^(log6) i would appreciate if someone helped soon. niraj. From aonweller@maerskdata-usa.com Wed Sep 24 10:12:56 1997 Return-Path: Received: from msrvr.maerskdata-usa.com by MOGLI.rutgers.edu (4.1/25-eef) id AA00212; Wed, 24 Sep 97 10:12:53 EDT Received: from [10.1.25.41] ([207.242.160.66]) by msrvr.maerskdata-usa.com (Netscape Mail Server v2.02) with SMTP id AAA40; Wed, 24 Sep 1997 11:30:45 +0000 Message-Id: <3428F94C.605C@maerskdata-usa.com> Date: Wed, 24 Sep 1997 07:28:13 -0400 From: aonweller@maerskdata-usa.com (Allen Onweller) X-Mailer: Mozilla 3.01-C-MACOS8 (Macintosh; I; PPC) Mime-Version: 1.0 To: Christopher Rose Cc: 330_501@MOGLI.rutgers.edu, 330_543@MOGLI.rutgers.edu Subject: Problems with alternate 543 Web Site? References: <9709240243.AA06815@MOGLI.rutgers.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO I think that the correct URL is: http://winwww.rutgers.edu/pub/about/people/staff/crose/winlab543.html From aonweller@maerskdata-usa.com Wed Sep 24 10:19:54 1997 Return-Path: Received: from msrvr.maerskdata-usa.com by MOGLI.rutgers.edu (4.1/25-eef) id AA00231; Wed, 24 Sep 97 10:19:46 EDT Received: from mxwnetf3 ([207.242.160.66]) by msrvr.maerskdata-usa.com (Netscape Mail Server v2.02) with SMTP id AAA138; Wed, 24 Sep 1997 13:29:17 +0000 Message-Id: <34291457.4E40@maerskdata-usa.com> Date: Wed, 24 Sep 1997 09:23:35 -0400 From: aonweller@maerskdata-usa.com (Allen Onweller) X-Mailer: Mozilla 3.01 (WinNT; U) Mime-Version: 1.0 To: crose@MOGLI.rutgers.edu Cc: 330_543@MOGLI.rutgers.edu Subject: Bad URL for alternate web site? Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO I think that the correct URL for the alternat 330:543 site is: http://winwww.rutgers.edu/pub/about/people/staff/crose/winlab543.html Regards, Allen From crose Wed Sep 24 11:20:06 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00362; Wed, 24 Sep 97 11:18:46 EDT Date: Wed, 24 Sep 97 11:18:46 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709241518.AA00362@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, nirajp@eden.rutgers.edu Subject: Re: ch 3 Cc: 330_543 Status: RO on point 2 ... I don't THINK I said the number of stages is k-1. The number of successive factorizations is k-1... which leads to the 2k-1 stages. If you check your notes, I think you'll see we indeed ended up with 2k-1 stages with N/2 switches in each. 6^l = (2^log6)^l = 2^(l log 6) but N = 2^k = 2^(2^l) log N = 2^l so 6^l = (2^l)^(log 6) = (logN)^(log6) I fyou remember, I left the exact relationship open by using an alpha in the exponenet for the logN. Hope that helps. Somebody out there should provide a factorization for your first question Cheers Chris Rose From SIZE=433@att.com Wed Sep 24 11:24:17 1997 Return-Path: Received: from att.com (cagw1.att.com) by MOGLI.rutgers.edu (4.1/25-eef) id AA00383; Wed, 24 Sep 97 11:24:15 EDT Received: by cagw1.att.com; Wed Sep 24 10:25 EDT 1997 Received: from localhost (localhost) by caig1.att.att.com (AT&T/GW-1.0) with internal id KAA16226; Wed, 24 Sep 1997 10:21:58 -0400 (EDT) Date: Wed, 24 Sep 1997 10:21:58 -0400 (EDT) From: Mail Delivery Subsystem Message-Id: <199709241421.KAA16226@caig1.att.att.com> To: crose@MOGLI.rutgers.edu Mime-Version: 1.0 Content-Type: multipart/report; report-type=delivery-status; boundary="KAA16226.875110918/caig1.att.att.com" Subject: Returned mail: Service unavailable Auto-Submitted: auto-generated (failure) Status: RO This is a MIME-encapsulated message --KAA16226.875110918/caig1.att.att.com The original message was received at Wed, 24 Sep 1997 10:21:36 -0400 (EDT) from nuucp@localhost ----- The following addresses had permanent fatal errors ----- ustad.att.com!rice ustad.att.com!jdodley ----- Transcript of session follows ----- ... while talking to ustad.ho.att.com.: >>> MAIL From: SIZE=333 <<< 554 Cannot bind to domain ustaddomain: can't communicate with ypbind 554 ustad.att.com!rice,ustad.att.com!jdodley... Service unavailable --KAA16226.875110918/caig1.att.att.com Content-Type: message/delivery-status Reporting-MTA: dns; caig1.att.att.com Arrival-Date: Wed, 24 Sep 1997 10:21:36 -0400 (EDT) Final-Recipient: RFC822; rice@ustad.ho.att.com Action: failed Status: 5.0.0 Remote-MTA: DNS; ustad.ho.att.com Diagnostic-Code: SMTP; 554 Cannot bind to domain ustaddomain: can't communicate with ypbind Last-Attempt-Date: Wed, 24 Sep 1997 10:21:57 -0400 (EDT) Final-Recipient: RFC822; jdodley@ustad.ho.att.com Action: failed Status: 5.0.0 Remote-MTA: DNS; ustad.ho.att.com Diagnostic-Code: SMTP; 554 Cannot bind to domain ustaddomain: can't communicate with ypbind Last-Attempt-Date: Wed, 24 Sep 1997 10:21:57 -0400 (EDT) --KAA16226.875110918/caig1.att.att.com Content-Type: message/rfc822 Return-Path: Received: (from nuucp@localhost) by caig1.att.att.com (AT&T/GW-1.0) id KAA16163; Wed, 24 Sep 1997 10:21:36 -0400 (EDT) >Received: by cagw1.att.com; Wed Sep 24 10:24 EDT 1997 Received: by cagw1.att.com; Wed Sep 24 10:24 EDT 1997 Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00375; Wed, 24 Sep 97 11:21:54 EDT Date: Wed, 24 Sep 97 11:21:54 EDT From: Christopher Rose Message-Id: <9709241521.AA00375@MOGLI.rutgers.edu> To: aonweller@maerskdata-usa.com Subject: Re: Problems with alternate 543 Web Site? Cc: 330_501@MOGLI.rutgers.edu, 330_543@MOGLI.rutgers.edu Content-Type: text What URL did I provide? Regardless, what you have looks right. You can also reach it by going to my home page http:/winwww.rutgers.edu/pub/about/people/staff/crose/crose.html and following the links at the bottom of the page Cheers, ALL --KAA16226.875110918/caig1.att.att.com-- From crose Wed Sep 24 11:24:50 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00375; Wed, 24 Sep 97 11:21:54 EDT Date: Wed, 24 Sep 97 11:21:54 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709241521.AA00375@MOGLI.rutgers.edu> To: aonweller@maerskdata-usa.com Subject: Re: Problems with alternate 543 Web Site? Cc: 330_501, 330_543 Status: RO What URL did I provide? Regardless, what you have looks right. You can also reach it by going to my home page http:/winwww.rutgers.edu/pub/about/people/staff/crose/crose.html and following the links at the bottom of the page Cheers, ALL From SIZE=996@att.com Wed Sep 24 11:33:31 1997 Return-Path: Received: from att.com (cagw1.att.com) by MOGLI.rutgers.edu (4.1/25-eef) id AA00424; Wed, 24 Sep 97 11:33:31 EDT Received: by cagw1.att.com; Wed Sep 24 10:35 EDT 1997 Received: from localhost (localhost) by caig1.att.att.com (AT&T/GW-1.0) with internal id KAA19966; Wed, 24 Sep 1997 10:31:26 -0400 (EDT) Date: Wed, 24 Sep 1997 10:31:26 -0400 (EDT) From: Mail Delivery Subsystem Message-Id: <199709241431.KAA19966@caig1.att.att.com> To: crose@MOGLI.rutgers.edu Mime-Version: 1.0 Content-Type: multipart/report; report-type=delivery-status; boundary="KAA19966.875111486/caig1.att.att.com" Subject: Returned mail: Service unavailable Auto-Submitted: auto-generated (failure) Status: RO This is a MIME-encapsulated message --KAA19966.875111486/caig1.att.att.com The original message was received at Wed, 24 Sep 1997 10:31:18 -0400 (EDT) from nuucp@localhost ----- The following addresses had permanent fatal errors ----- ustad.att.com!rice ustad.att.com!jdodley ----- Transcript of session follows ----- ... while talking to ustad.ho.att.com.: >>> MAIL From: SIZE=896 <<< 554 Cannot bind to domain ustaddomain: can't communicate with ypbind 554 ustad.att.com!rice,ustad.att.com!jdodley... Service unavailable --KAA19966.875111486/caig1.att.att.com Content-Type: message/delivery-status Reporting-MTA: dns; caig1.att.att.com Arrival-Date: Wed, 24 Sep 1997 10:31:18 -0400 (EDT) Final-Recipient: RFC822; rice@ustad.ho.att.com Action: failed Status: 5.0.0 Remote-MTA: DNS; ustad.ho.att.com Diagnostic-Code: SMTP; 554 Cannot bind to domain ustaddomain: can't communicate with ypbind Last-Attempt-Date: Wed, 24 Sep 1997 10:31:26 -0400 (EDT) Final-Recipient: RFC822; jdodley@ustad.ho.att.com Action: failed Status: 5.0.0 Remote-MTA: DNS; ustad.ho.att.com Diagnostic-Code: SMTP; 554 Cannot bind to domain ustaddomain: can't communicate with ypbind Last-Attempt-Date: Wed, 24 Sep 1997 10:31:26 -0400 (EDT) --KAA19966.875111486/caig1.att.att.com Content-Type: message/rfc822 Return-Path: Received: (from nuucp@localhost) by caig1.att.att.com (AT&T/GW-1.0) id KAA19903; Wed, 24 Sep 1997 10:31:18 -0400 (EDT) >Received: by cagw1.att.com; Wed Sep 24 10:34 EDT 1997 Received: by cagw1.att.com; Wed Sep 24 10:34 EDT 1997 Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00403; Wed, 24 Sep 97 11:31:28 EDT Date: Wed, 24 Sep 97 11:31:28 EDT From: Christopher Rose Message-Id: <9709241531.AA00403@MOGLI.rutgers.edu> To: 330_501@MOGLI.rutgers.edu, 330_543@MOGLI.rutgers.edu Subject: ARRRRRRRGGGGGGGGGGHHHHHHH!!! Cc: crose@MOGLI.rutgers.edu Content-Type: text Hi Folks, I left out a protion of the path for the alternate site url It should have read: http:/winwww.rutgers.edu/pub/about/people/staff/crose/winlabXXX.html where XXX is either 501 or 543. Sorry bout that.... Allen Onweller bailed me out on that one. Thanks Also, for 501 students. GIF files for the old midterm and solutions don't exist! Sorry for the inconvenience. You can always print them out at RU. But a better idea is to follow the links provided to get some ghostview software which you can load on your PC to view postscript files. I've never tried it, but also there's adobe acrobat which I would think should be able to view Postscript somehow (same company which puts out adobe postscript, no?) Cheers ALL I leave today. Let's hope I can avoid a flight 800 sort of deal :) --KAA19966.875111486/caig1.att.att.com-- From crose Wed Sep 24 11:33:42 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00403; Wed, 24 Sep 97 11:31:28 EDT Date: Wed, 24 Sep 97 11:31:28 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9709241531.AA00403@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: ARRRRRRRGGGGGGGGGGHHHHHHH!!! Cc: crose@MOGLI.rutgers.edu Status: RO Hi Folks, I left out a protion of the path for the alternate site url It should have read: http:/winwww.rutgers.edu/pub/about/people/staff/crose/winlabXXX.html where XXX is either 501 or 543. Sorry bout that.... Allen Onweller bailed me out on that one. Thanks Also, for 501 students. GIF files for the old midterm and solutions don't exist! Sorry for the inconvenience. You can always print them out at RU. But a better idea is to follow the links provided to get some ghostview software which you can load on your PC to view postscript files. I've never tried it, but also there's adobe acrobat which I would think should be able to view Postscript somehow (same company which puts out adobe postscript, no?) Cheers ALL I leave today. Let's hope I can avoid a flight 800 sort of deal :) From andyy@ece.rutgers.edu Tue Oct 7 16:37:09 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA16198; Tue, 7 Oct 97 16:37:08 EDT Received: (from andyy@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) id PAA27048 for crose@mogli.rutgers.edu; Tue, 7 Oct 1997 15:36:47 -0400 Date: Tue, 7 Oct 1997 15:36:47 -0400 From: Wingho Yuen Message-Id: <199710071936.PAA27048@ece.rutgers.edu> To: crose@mogli.rutgers.edu Subject: ee 543 Content-Length: 143 Status: RO Hello Prof Rose Would the Cantor Network be covered? I could not follow the proof of the SSNB property of Cantor Network. Thanks Andy Yuen From crose Tue Oct 7 20:38:08 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA16406; Tue, 7 Oct 97 20:37:59 EDT Date: Tue, 7 Oct 97 20:37:59 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710080037.AA16406@MOGLI.rutgers.edu> To: andyy@ece.rutgers.edu Subject: Re: ee 543 Cc: 330_543 Status: RO Nope, don't worry about Cantor. It's much ado about not too much :) Cheers, Chris Rose From linzhou@ece.rutgers.edu Sun Oct 12 16:39:34 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA22433; Sun, 12 Oct 97 16:39:12 EDT Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id PAA06216; Sun, 12 Oct 1997 15:37:55 -0400 Date: Sun, 12 Oct 1997 15:37:55 -0400 (EDT) From: Lin Zhou To: Christopher Rose Cc: andyy@ece.rutgers.edu, 330_543@MOGLI.rutgers.edu Subject: Re: ee 543 In-Reply-To: <9710080037.AA16406@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Dr. Rose, In last class, you said the power of inverse Banyan network which forms compact superconcentrator was (n/2)*(logn)*2. Since the total number of delta switch is (n/2)*(logn), connection pattern is exp2((n/2)*logn). Why the power is doubled to (n/2)*(log)*2 ? zhou lin From crose Sun Oct 12 17:52:36 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA22488; Sun, 12 Oct 97 17:51:54 EDT Date: Sun, 12 Oct 97 17:51:54 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710122151.AA22488@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: ee 543 Cc: 330_543@MOGLI.rutgers.edu, andyy@ece.rutgers.edu Status: RO Hi, Just after the class was over, I saw that I'd counted switch states in once case and delta switches in the other. Each delta switch has two states (cross and straight). The number 2N log N for multicast is the number of SWITCHES but the number of cross points (which we kind of us analogously to switch states) is 4N log N (just like the number of switch states for a Benes network is on the order of 2N log N (while the number of switches is of order N log N). Cheers Chris Rose From crose Sun Oct 12 17:52:36 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA22488; Sun, 12 Oct 97 17:51:54 EDT Date: Sun, 12 Oct 97 17:51:54 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710122151.AA22488@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: ee 543 Cc: 330_543@MOGLI.rutgers.edu, andyy@ece.rutgers.edu Status: RO Hi, Just after the class was over, I saw that I'd counted switch states in once case and delta switches in the other. Each delta switch has two states (cross and straight). The number 2N log N for multicast is the number of SWITCHES but the number of cross points (which we kind of us analogously to switch states) is 4N log N (just like the number of switch states for a Benes network is on the order of 2N log N (while the number of switches is of order N log N). Cheers Chris Rose From linzhou@ece.rutgers.edu Sun Oct 12 18:32:47 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA22520; Sun, 12 Oct 97 18:32:15 EDT Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id RAA10124; Sun, 12 Oct 1997 17:30:59 -0400 Date: Sun, 12 Oct 1997 17:30:58 -0400 (EDT) From: Lin Zhou To: Christopher Rose Cc: 330_543@MOGLI.rutgers.edu, andyy@ece.rutgers.edu Subject: Re: ee 543 In-Reply-To: <9710122151.AA22488@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Dear Dr. Rose, Thank you for your explanation on the number of switches and the number of states. So the number of connection pattern is the number of states, while the power of connection pattern is the number of delta swiches. You said the number of crosspoints is "analogously the number of states. Is this the actualy number of crosspoints? For instance, P2P delta switch has 2 states, only two crosspoints is enough? As I understand, four crosspoints is used for 2x2 switch, but two states exists.(one control bit). Please give me the answer. Thanks a lot. Best regards, zhou lin On Sun, 12 Oct 1997, Christopher Rose wrote: > Hi, > > Just after the class was over, I saw that I'd counted > switch states in once case and delta switches in the other. > Each delta switch has two states (cross and straight). > > The number 2N log N for multicast is the number of SWITCHES > but the number of cross points (which we kind of us analogously > to switch states) is 4N log N (just like the number of switch states > for a Benes network is on the order of 2N log N (while the number > of switches is of order N log N). > > Cheers > > Chris Rose > From crose Sun Oct 12 23:06:16 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA22763; Sun, 12 Oct 97 23:06:00 EDT Date: Sun, 12 Oct 97 23:06:00 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710130306.AA22763@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: ee 543 Cc: 330_543@MOGLI.rutgers.edu, andyy@ece.rutgers.edu Status: RO My Goodness, I think I must have been asleep. I just reread what I wrote. Here's the low down... Each connection pattern corresponds to (at least one) switch state. The power of a switch is the log number of connection patterns. You need at least log_2( switch power) crosspoints to realize a switch with a given number of connection patterns. Or another way to think about it is that you need at least that many delta switches (since like crosspoints, they have only two states). Hope that clarifies my ambiguous/wrong previous statement! Cheers, Chris Rose PS: What I did in class at the end (the board to the far left) was DEAD WRONG. You cannot take each delta switch and multiply by two to get the switch states. Not sure what I was thinking... AAAAAARRRRRRRRGGGGGGGGGG! Why? Because the number of switch STATES is 2^(number of binary decision elements) in the switch. Just to be careful, remember that the number of switch states is not necessarily the number of switching patterns! That's what I get for rushing at the end of class! Thanks for persisting in your question. From crose@localhost.localdomain Mon Oct 13 10:12:41 1997 Return-Path: Received: from localhost.localdomain (ppp-72.ts-4.nyc.idt.net) by boom.rutgers.edu (4.1/SMI-4.1) id AA01950; Mon, 13 Oct 97 10:12:39 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id KAA00761; Mon, 13 Oct 1997 10:00:22 -0400 Date: Mon, 13 Oct 1997 10:00:22 -0400 From: Christopher Rose Message-Id: <199710131400.KAA00761@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: grading Cc: crose@boom.rutgers.edu Status: RO Hi Folks, I"m done the grading for the 543 exam. The TA (Wenfeng) will collate the results and we'll return them on Tuesday. In all, I'm pretty pleased, except for the fact that most of you know very little probability... but that's correctable :) Cheers, Chris Rose From crose@localhost.localdomain Mon Oct 13 11:07:50 1997 Return-Path: Received: from localhost.localdomain (ppp-72.ts-4.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA23751; Mon, 13 Oct 97 11:07:13 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id KAA00761; Mon, 13 Oct 1997 10:00:22 -0400 Date: Mon, 13 Oct 1997 10:00:22 -0400 From: Christopher Rose Message-Id: <199710131400.KAA00761@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: grading Cc: crose@MOGLI.rutgers.edu Status: RO Hi Folks, I"m done the grading for the 543 exam. The TA (Wenfeng) will collate the results and we'll return them on Tuesday. In all, I'm pretty pleased, except for the fact that most of you know very little probability... but that's correctable :) Cheers, Chris Rose From crose Wed Oct 15 11:21:04 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA26625; Wed, 15 Oct 97 11:20:43 EDT Date: Wed, 15 Oct 97 11:20:43 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710151520.AA26625@MOGLI.rutgers.edu> To: 330_543 Subject: "proof" Cc: crose@MOGLI.rutgers.edu Status: RO Hi Folks, I hate my "proof" or ssnb for multipoint networks when viewed in the light of day. I'm convinced we need no more than $N$ centerstage switches (because the overlap argument does not hold water... if n_3-1 inputs are busy in the output stage switch and N-1 outputs are busy in the input stage switch, then THERE MUST BE OVERLAP (Salim Manji, I need to see your exam again). That is, those $n_3-1$ lines MUST BE completely included in the paths radiating from the first stage switch. But having opened that door, we need to figure out a tight lower bound (for the box in which we've placed ourselves). That is, is $N$ too many? Since I don't know the answer, let's open a competition. In a non-collaborative way I'd like arguments from each of you for why the number of center stage switches in a multipoint network must be ____. I'd like your answers by friday via email. You can make up a postscript file and mail it or just plain text if possible. This will be worth 10 points on your first quiz grade (which is why I don't want collaboration). Send your submissions to me (crose@mogli.rutgers.edu or crose@ece.rutgers.edu). GOOD LUCK! Cheers, Chris Rose From crose Wed Oct 15 11:23:59 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA26632; Wed, 15 Oct 97 11:23:46 EDT Date: Wed, 15 Oct 97 11:23:46 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710151523.AA26632@MOGLI.rutgers.edu> To: 330_543 Subject: ALSO Status: RO I'll defer posting solutions to that problem (well yes I DO know the answer now, but it's nice to generate excitement, no?) until friday. The other solutions will be available on the web today. Cheers From crose Wed Oct 15 12:15:41 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA26760; Wed, 15 Oct 97 12:11:40 EDT Date: Wed, 15 Oct 97 12:11:40 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710151611.AA26760@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: regrade policy Status: RO Hi Folks, Now that the solutions are out, you have a week to send me your complaints IN WRITING along with your exam. I'll regrade and that grade will be final. Don't hesitate to ask questions throug the regrade process. I'm not an ogre and will probably not dock yoy points on other sections unless I determine I was asleep when I graded them :) Cheers, Chris Rose From alap@winlab.rutgers.edu Wed Oct 15 22:24:32 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA27451; Wed, 15 Oct 97 22:24:31 EDT Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id VAA07261; Wed, 15 Oct 1997 21:27:12 -0400 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id VAA14270; Wed, 15 Oct 1997 21:27:12 -0400 Date: Wed, 15 Oct 1997 21:27:12 -0400 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199710160127.VAA14270@gsm.winlab.rutgers.edu> To: crose@mogli Subject: Exam problem Status: RO Prof Rose, I have a question about the exam problem: multicast is only one to many and never many to one (?) I understand that all switches must be capable of multicast, not only the switches in the first layer. Is that right ? Thanks a lot. Ana. From crose@localhost.localdomain Wed Oct 15 22:55:37 1997 Return-Path: Received: from localhost.localdomain (ppp-6.ts-9.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA27470; Wed, 15 Oct 97 22:55:17 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id VAA03443 for 330_543@mogli.rutgers.edu; Wed, 15 Oct 1997 21:47:54 -0400 Date: Wed, 15 Oct 1997 21:47:54 -0400 From: Christopher Rose Message-Id: <199710160147.VAA03443@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: problem sets Status: RO Hi Folks, Here's a gift from me to you! Problem set 3 (due 10/23/97) Chapter 5 problems 1,2,4 and 5 Problem set 4 (due 10/30/97) Chapter 6 problems 2,3,5,7 and 10 The test is on November 4th (tuesday) in class and beyond (the next period too). Remember to wait for 20minutes tomorrow (thursday) just in case I'm delayed by traffic (or the baby arrives by appointment -- we're hoping hard for the latter, otherwise I've got to repeat this routine next week too :( ) Cheers all Your sleepless professor soon to become sleepless^2 PS: I'm looking to wrap up the descriptive switching stuff pretty quickly so we can get to performance issues. By the way, might you like to read about how a 5ESS works (the AT&T old wonderswitch)? From crose Wed Oct 15 22:58:33 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA27487; Wed, 15 Oct 97 22:58:22 EDT Date: Wed, 15 Oct 97 22:58:22 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710160258.AA27487@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Exam problem Cc: 330_543 Status: RO Yup, all the switches are capable of multicast. However, you can make up your own version of multicast if you like (i.e. you can specify your own box as long as it is SSNB and can provide any multicast pattern AND IS A CLOS network :) You might also consider the issue of wide sense non-blocking where you establish rules for nonblocking behavior (this is harder and i'll double the credit if you can solve this one). Cheers, Chris Rose ******************* Prof Rose, I have a question about the exam problem: multicast is only one to many and never many to one (?) I understand that all switches must be capable of multicast, not only the switches in the first layer. Is that right ? Thanks a lot. Ana. From crose Wed Oct 15 23:02:05 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA27499; Wed, 15 Oct 97 23:01:14 EDT Date: Wed, 15 Oct 97 23:01:14 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710160301.AA27499@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Exam problem Cc: 330_543 Status: RO OH and yes we cannot have many to one (we'll save that loveliness for packet switching) From crose Thu Oct 16 12:40:11 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28263; Thu, 16 Oct 97 12:39:50 EDT Date: Thu, 16 Oct 97 12:39:50 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710161639.AA28263@MOGLI.rutgers.edu> To: jasonjen@ece.rutgers.edu Subject: Re: Question: NxN multipoint network?! Cc: 330_543 Status: RO The box i'm putting you in is tat any call can use any resources is so desires, even it it's inefficient. Think about what that means on the input side for a multicast switch. From alap@winlab.rutgers.edu Thu Oct 16 16:41:41 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA28553; Thu, 16 Oct 97 16:41:41 EDT Received: from cell.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id PAA18228; Thu, 16 Oct 1997 15:44:14 -0400 Received: by cell.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id PAA19823; Thu, 16 Oct 1997 15:44:13 -0400 Date: Thu, 16 Oct 1997 15:44:13 -0400 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199710161944.PAA19823@cell.winlab.rutgers.edu> To: crose@MOGLI Subject: Still question... Cc: alap@winlab.rutgers.edu Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Content-Md5: r3eUzTo3PzhGmc6zLqJXPg== Status: RO > > The box i'm putting you in is tat any call can use > any resources is so desires, even it it's inefficient. > So, if I have an empty input I should be able to send to "as many as I want (FREE) outputs". It does not matter HOW. The problem then only changes because now I have the possibility to send either one-to-one and/or one-to-many. That means that if I am braodcasting to many outputs that are in the same switch is the same problem as one-to-one, cause I just get in the switch input with one line and broadcast internally in the switch. Is that correct ? > > You might also consider the issue of wide sense non-blocking where > you establish rules for nonblocking behavior (this is harder and > i'll double the credit if you can solve this one). > Can you explain better what is "wide-sense non blocking" ? Is this the case where I follow some algorithm when I assign paths ? Thanks again. Ana. From crose Thu Oct 16 21:00:46 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28730; Thu, 16 Oct 97 21:00:45 EDT Date: Thu, 16 Oct 97 21:00:45 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710170100.AA28730@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Still question... Cc: crose@MOGLI.rutgers.edu Status: RO Wide sense non-blocking is a combination of switching hardware and route selection algorithms that render a switch SSNB eventhough it might be only RNB under an uncontrolled routing protocol. Cheers, Chris Rose From crose Thu Oct 16 21:01:34 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28735; Thu, 16 Oct 97 21:01:11 EDT Date: Thu, 16 Oct 97 21:01:11 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710170101.AA28735@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Still question... Cc: 330_543 Status: RO Wide sense non-blocking is a combination of switching hardware and route selection algorithms that render a switch SSNB eventhough it might be only RNB under an uncontrolled routing protocol. Cheers, Chris Rose From jasonjen@ece.rutgers.edu Thu Oct 16 21:35:40 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28765; Thu, 16 Oct 97 21:35:39 EDT Received: (from jasonjen@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) id UAA29436 for crose; Thu, 16 Oct 1997 20:33:39 -0400 Date: Thu, 16 Oct 97 20:33:38 EDT From: Jason Jen To: crose@ece.rutgers.edu Subject: More question about NxN ssnb clos network Message-Id: Status: RO Dr. Rose, Hi, is it okay to have only some inputs (say less than N inputs) to perform multicasting so that all N output terminals are active? In this case all idle input terminals cannot make any connection and it's not considered as "blocking". Thanks, Jason Jen From crose Thu Oct 16 22:07:57 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28807; Thu, 16 Oct 97 22:07:38 EDT Date: Thu, 16 Oct 97 22:07:38 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710170207.AA28807@MOGLI.rutgers.edu> To: jasonjen@ece.rutgers.edu Subject: Re: More question about NxN ssnb clos network Cc: 330_543 Status: RO Dr. Rose, Hi, is it okay to have only some inputs (say less than N inputs) to perform multicasting so that all N output terminals are active? In this case all idle input terminals cannot make any connection and it's not considered as "blocking". Thanks, Well, think about that one. If you're doing multicasting and don't allow an input to be inactive, then you'll end up with more than N connection requests (a no no). Chers, Chris Rose From crose Thu Oct 16 22:09:44 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28817; Thu, 16 Oct 97 22:09:26 EDT Date: Thu, 16 Oct 97 22:09:26 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710170209.AA28817@MOGLI.rutgers.edu> To: 330_543 Subject: competition Cc: crose@MOGLI.rutgers.edu Status: RO Hi Folks, the doors close on clos-multipoint entries tomorrow (friday) at 12 midnight! It's worth 10 whole points on your first quiz grade! Cheers, Chris Rose From shahid@er4.rutgers.edu Fri Oct 17 12:39:31 1997 X-UIDL: 8f931a26aa4b6cb9a6b9fe5942bc101f Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA29733; Fri, 17 Oct 97 12:39:30 EDT Received: from er4.rutgers.edu (shahid@er4.rutgers.edu [165.230.180.142]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id LAA07925 for ; Fri, 17 Oct 1997 11:37:22 -0400 Received: from localhost (shahid@localhost) by er4.rutgers.edu (8.8.5/8.8.5) with SMTP id LAA06159 for ; Fri, 17 Oct 1997 11:41:52 -0400 (EDT) Date: Fri, 17 Oct 1997 11:41:52 -0400 (EDT) From: Shahid Kagal To: crose@ece.rutgers.edu Subject: Multipoint NxN network Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Respected Prof. Rose, I have a doubt regarding the question you have put forward. You have written in the mail than there are n-1 inputs active at the output switch and N-1 outputs active at the input switch. But if there are total N inputs than how can we have N-1 outputs from a single switch active or for that matter do you mean outputs from all the first stage switches. If it has something to do with multicasting please explain. Also if I go beyond this point than the N-1 term in the expression (N-1 + n3-1) could according to the proof of SSNB networks mean inputs active at a time in a single first stage switch. Please also correct this interpretation if wrong. ( or I have confused the symbols) Thank You, Shahid. From crose Fri Oct 17 13:26:29 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA29806; Fri, 17 Oct 97 13:26:11 EDT Date: Fri, 17 Oct 97 13:26:11 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710171726.AA29806@MOGLI.rutgers.edu> To: shahid@eden.rutgers.edu Subject: Re: Multipoint NxN network Cc: 330_543 Status: RO YOU WROTE ************** I have a doubt regarding the question you have put forward. You have written in the mail than there are n-1 inputs active at the output switch and N-1 outputs active at the input switch. But if there are total N inputs than how can we have N-1 outputs from a single switch active or for that matter do you mean outputs from all the first stage switches. If it has something to do with multicasting please explain. Also if I go beyond this point than the N-1 term in the expression (N-1 + n3-1) could according to the proof of SSNB networks mean inputs active at a time in a single first stage switch. Please also correct this interpretation if wrong. ( or I have confused the symbols) ************************ COnsider this: What if the first input wants to sent info to N-1 outputs in a broadcast fashion. Can he/she set up N-1 paths through the first stage switch? (yes). Now, just to throw a monkey wrench in things. Suppose that you DON'T allow first stage multicast and only permit it at the second stage and later. How might this change the problem. Is the switch now blocking!?!?!? I want you all to think carefullly about your assumptions (many of you already have... thanks) and provide an answer! :) Cheers, From aonweller@maerskdata-usa.com Fri Oct 17 14:57:16 1997 Return-Path: Received: from msrvr.maerskdata-usa.com by MOGLI.rutgers.edu (4.1/25-eef) id AA00180; Fri, 17 Oct 97 14:57:16 EDT Received: from mxwnetf3 ([207.242.160.66]) by msrvr.maerskdata-usa.com (Netscape Mail Server v2.02) with SMTP id AAA141 for ; Fri, 17 Oct 1997 18:02:59 +0000 Message-Id: <3447A6DB.DCC@maerskdata-usa.com> Date: Fri, 17 Oct 1997 13:56:43 -0400 From: aonweller@maerskdata-usa.com (Allen Onweller) X-Mailer: Mozilla 3.01 (WinNT; U) Mime-Version: 1.0 To: Christopher Rose Subject: Re: Multipoint NxN network References: <9710171726.AA29806@MOGLI.rutgers.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Christopher Rose wrote: > > COnsider this: > > What if the first input wants to sent info to N-1 outputs > in a broadcast fashion. Can he/she set up N-1 paths > through the first stage switch? (yes). >.... Hello Professor Rose, I am sticking to my original submission for the extra credit, but I have a few comments with respect to your last e-mail: I'm not sure if I'm missing something obvious, but: If you mean a total of N-1 paths through the network I agree. But I tend to disagree that an single input can setup N-1 paths through the first stage in general. I would argue that it can set up at most n_1 paths where n_1 is the number of outputs from a first stage switch. After that, the second and/or third stages are needed to connect the remaining paths. Of course if n_1 = N, then your statement holds. Maybe this was implied in you discussion. Something that I need to look into further is: Ignoring fan out problems, if n_1 = N, do we really need a third stage? Or for that matter a second stage? Obviously it wouldn't be a Clos net without those stages. (Perhaps by tying corresponding outputs from the first stage together so that each of the N outputs is linked to r_1 lines from the r_1 switches forming a sort of output bus. Of couse bus arbitration would be needed.) Maybe I'll have time to look into this later next week. Regards, Allen From shahid@er4.rutgers.edu Fri Oct 17 15:47:41 1997 Return-Path: Received: from er4.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA00276; Fri, 17 Oct 97 15:47:40 EDT Received: from localhost (shahid@localhost) by er4.rutgers.edu (8.8.5/8.8.5) with SMTP id OAA28199 for ; Fri, 17 Oct 1997 14:50:03 -0400 (EDT) Date: Fri, 17 Oct 1997 14:50:03 -0400 (EDT) From: Shahid Kagal To: Christopher Rose Subject: Re: Multipoint NxN network In-Reply-To: <9710171726.AA29806@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Respected Prof. Rose, I am still a little bit confused. It is correct to suppose that a particular input if wants to broadcast to N-1 outputs will require N-1 paths thru the first stage switch. But does it mean a single switch or all the switches and if the latter than how I cannot figure out. Thank You, Shahid. On Fri, 17 Oct 1997, Christopher Rose wrote: > YOU WROTE > ************** > I have a doubt regarding the question you have put forward. You have > written in the mail than there are n-1 inputs active at the output switch > and N-1 outputs active at the input switch. But if there are total N > inputs than how can we have N-1 outputs from a single switch active or > for that matter do you mean outputs from all the first stage switches. If > it has something to do with multicasting please explain. Also if I go > beyond this point than the N-1 term in the expression (N-1 + n3-1) could > according to the proof of SSNB networks mean inputs active at a time in a > single first stage switch. Please also correct this interpretation if > wrong. ( or I have confused the symbols) > > ************************ > > COnsider this: > > What if the first input wants to sent info to N-1 outputs > in a broadcast fashion. Can he/she set up N-1 paths > through the first stage switch? (yes). > > Now, just to throw a monkey wrench in things. Suppose > that you DON'T allow first stage multicast and only permit it at > the second stage and later. How might this change the problem. > Is the switch now blocking!?!?!? > > I want you all to think carefullly about your assumptions (many of you > already have... thanks) and provide an answer! :) > > Cheers, > From crose@localhost.localdomain Sat Oct 18 03:40:11 1997 Return-Path: Received: from localhost.localdomain (ppp-20.ts-9.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA01037; Sat, 18 Oct 97 03:39:58 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id CAA00920 for 330_543@mogli.rutgers.edu; Sat, 18 Oct 1997 02:32:09 -0400 Date: Sat, 18 Oct 1997 02:32:09 -0400 From: Christopher Rose Message-Id: <199710180632.CAA00920@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: submissions Status: RO Submissions are formally closed as of now (2:45AM Saturday 10/18) for the clos multicast competition. I will print the results soon and evaluate your opuses :) From crose@localhost.localdomain Sat Oct 18 16:36:55 1997 Return-Path: Received: from localhost.localdomain (ppp-18.ts-1.nyc.idt.net) by boom.rutgers.edu (4.1/SMI-4.1) id AA02497; Sat, 18 Oct 97 16:36:52 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id QAA01368; Sat, 18 Oct 1997 16:24:17 -0400 Date: Sat, 18 Oct 1997 16:24:17 -0400 From: Christopher Rose Message-Id: <199710182024.QAA01368@localhost.localdomain> To: 330_543@localhost.localdomain Subject: averages Cc: crose@boom.rutgers.edu Status: RO Hi Folks, The class average was 127 +- 23 s.d. The high was 170 and the low was 50. All in all I was pretty pleased. Class average would probably fall somewhere around a low B to a solid B if I had to hand out grades today. Good job for the most part folks. Just have to teach you a bit of probability. Cheers, Chris Rose From crose@localhost.localdomain Sat Oct 18 17:33:16 1997 Return-Path: Received: from localhost.localdomain (ppp-18.ts-1.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA01613; Sat, 18 Oct 97 17:32:33 EDT Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id QAA01373 for 330_543@mogli.rutgers.edu; Sat, 18 Oct 1997 16:24:38 -0400 Date: Sat, 18 Oct 1997 16:24:38 -0400 From: Christopher Rose Message-Id: <199710182024.QAA01373@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: averages Status: RO Hi Folks, The class average was 127 +- 23 s.d. The high was 170 and the low was 50. All in all I was pretty pleased. Class average would probably fall somewhere around a low B to a solid B if I had to hand out grades today. Good job for the most part folks. Just have to teach you a bit of probability. Cheers, Chris Rose From xuyang@rutcor.rutgers.edu Fri Oct 24 00:13:11 1997 Return-Path: Received: from bartok.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA08757; Fri, 24 Oct 97 00:12:56 EDT Received: (from xuyang@localhost) by bartok.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.12) id XAA25168 for 330_543@mogli; Thu, 23 Oct 1997 23:14:13 -0400 Date: Thu, 23 Oct 97 23:14:12 EDT From: Yang Xu To: 330_543@mogli Subject: complexity definition? Message-Id: Status: RO Dr. Rose, The complexity is defined on what kind of operations? +, -, * counted as one single operation? How about assignment such as $a[1]\leftarrow 1$? If the assigment is counted as one single operation, then as I remember bucket sort can be completed in linear time when the inputs are some close set of integers. Thanks a lot, yang From alap@winlab.rutgers.edu Thu Oct 23 23:35:55 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA08667; Thu, 23 Oct 97 23:35:34 EDT Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id WAA05490; Thu, 23 Oct 1997 22:36:53 -0400 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id WAA12123; Thu, 23 Oct 1997 22:36:52 -0400 Date: Thu, 23 Oct 1997 22:36:52 -0400 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199710240236.WAA12123@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Sort Complexity Status: RO Prof. Rose, Isn't the complexity of "divide-sort-merge" algorithm NlogN - NC(1) as opposed to N(logN-1) - NC(1) given in class ?? Thanks a lot, Ana. From crose Fri Oct 24 07:25:01 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA08985; Fri, 24 Oct 97 07:24:10 EDT Date: Fri, 24 Oct 97 07:24:10 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710241124.AA08985@MOGLI.rutgers.edu> To: 330_543, xuyang@rutcor.rutgers.edu Subject: Re: complexity definition? Cc: 330_501 Status: RO Hi Yang, Good question! For the merge sort algorithm, the limiting step is the merge since at the bottom of the recursion you can parallelize the comparion operations. The merge takes two ordered lists, a1,a2,a3... and b1,b2,b3 and does the following (for non-decreasing sort) Compare a1 and b1 --> place smaller (if tied take a1) in merged list hold on to larger (call it R) Compare R and next element from the OTHER list -> place smaller in the merged list etc. As you can see, for a list of size N you end up with N comparison operations and THIS is the complexity we're talking about. The factor of logN comes in from the fact that we split the list logN times (recursion) and are therefore doing this merge on N elements logN times. There're also stores going on, but we ignore those or just lump them into the compare. Now as for the actual time complexity in general, that's a little more complex. For example, the time complexity of the algorithm in the book (concatenation of bitonic sorters) is of order (logN)^2 (no factor of N at all if you consider that each stage of sorting (using the 2x2 sorters) takes one unit of time. However, the overall COMPLEXITY is still N (logN)^2 since you have of order N sorters in each stage (N/2 to be exact). So the complexity issue is not just about time. Sorting requires at least N log N complexity no matter how you cut it. If someone says different, then they are misguided or lying to you :) TO answer your question about bucket sort, I'd need to remember bucket sort... BUT I DON'T :) Could you give a brief description to refresh my memory? Thanks. Oh, one last thing, I'm avoiding the guts of comparison at the bit level. That is, you can tell half the time (random integers of fixed maximum size) whether one number is bigger by looking at the first bit, 1/4 of the time you need to look at the second bit also, etc so that the average random comparison takes the examination of only 2 bits. Of course, as your list gets sorted better and better (nonrandom) you've got to compare more bits. Well not quite one last thing: if you want to go deeper into the theory, one could pose the sorting problem as a "twenty questions" game. That is, what is the minimum number of yes no questions you need to ask (binary comparisons you need to perform) to sort a list. Well, if any ordering is equally likely, you've got a uniform distribution on N! possibilities. Identifying which possibility is essentially equivalent to sorting the list. The entropy of a Uniform distribution with N! elements is log(N!) which again is of order N log N. You just can't escape from N log N can you !?!?!?! Cheers, Chris Rose From crose Fri Oct 24 07:26:04 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA08991; Fri, 24 Oct 97 07:25:55 EDT Date: Fri, 24 Oct 97 07:25:55 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710241125.AA08991@MOGLI.rutgers.edu> To: 330_543 Subject: whoops Status: RO Posted this to the wrong group! ******************* Hi Ana, Yup, I made a calculation error in counting the stages. The ORDER of the complexity does not change, but the number of stages of reduction is logN as you pointed out. Thanks for catching that error. Cheers, Chris Rose From xuyang@rutcor.rutgers.edu Fri Oct 24 12:37:17 1997 Return-Path: Received: from rutcor.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA09309; Fri, 24 Oct 97 12:37:17 EDT Received: (from xuyang@localhost) by rutcor.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.12) id LAA18519 for Christopher Rose ; Fri, 24 Oct 1997 11:38:28 -0400 Date: Fri, 24 Oct 97 11:38:28 EDT From: Yang Xu To: Christopher Rose Subject: Re: complexity definition? In-Reply-To: Your message of Fri, 24 Oct 97 07:24:10 EDT Message-Id: Status: RO Dr. Rose, Thanks for your exclusive answer! Yes, it has been shown that comparison based sorting has lower bound of NlogN. Bucket sorting is not comparison based sorting, as I understand. The idea is simple. Suppose we want to sort N numbers(must be integers!) in a list S. Creating an array A with size of O(N), and go over N numbers in list S and assign each $i\in S$ to A[i], i.e. $A[i]\leftarrow i, \forall i\in S$. Then, numbers are sorted in array A. If S is a set(no duplicate), then in each bucket of A there is at most 1 number. If S is not a set, there are maybe several nubmers in one bucket. But it doen't matter too much( we can use a counter for each bucket to record the number of element in each bucket). If we say assigning one number takes logN (actually it is the time to allocate the space A[i], given i), then we end up with magic number NlogN. If we assume finding A[i], given i, takes O(1) time, it is linear algorithm. I am having an algorithm book(by Cormen, et. al) in front of me now! I am sorry that I made a mistake. The algorithm I just described above is actually called counting sort in the book. The book says it uses operations other than comparison and rans in linear time. I think in our switch context, inputs are integers and counting sort is better than quick sort in performance and implementation. Cheers?! yang From linzhou@ece.rutgers.edu Sat Oct 25 14:02:44 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA10416; Sat, 25 Oct 97 14:02:43 EDT Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id MAA02089 for ; Sat, 25 Oct 1997 12:59:07 -0400 Date: Sat, 25 Oct 1997 12:59:07 -0400 (EDT) From: Lin Zhou To: Christopher Rose Subject: proof In-Reply-To: <9710241125.AA08991@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Dear Dr. Rose, When you proved the therom last class, you give the restrictions of input compact and output monotone. For both conditions, it allows increase in cycle? Why you restain the output as d(i+1)>=mod(d(i+1) Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA10760; Sat, 25 Oct 97 22:10:31 EDT Date: Sat, 25 Oct 97 22:10:31 EDT From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710260210.AA10760@MOGLI.rutgers.edu> To: linzhou@ece.rutgers.edu Subject: Re: proof Cc: 330_543 Status: RO That was the issue. The mod condition does not work. It has to be strictly monotone and less then N (remove the mod symbol like I did after Salim made his observation). Cheers, Chris ROse From alap@winlab.rutgers.edu Mon Oct 27 13:10:48 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA12617; Mon, 27 Oct 97 13:10:48 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id MAA25327; Mon, 27 Oct 1997 12:11:29 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id MAA17373; Mon, 27 Oct 1997 12:11:29 -0500 Date: Mon, 27 Oct 1997 12:11:29 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199710271711.MAA17373@gsm.winlab.rutgers.edu> To: crose@mogli Subject: Exam Problem Status: RO Prof. Rose, I would like to know if you graded the exam problem we sent via e-mail. In class you said the answer was N (for SSNB), but what about RNB ? Thnak you very much. Ana. From crose Mon Oct 27 15:45:13 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA12764; Mon, 27 Oct 97 15:45:12 EST Date: Mon, 27 Oct 97 15:45:12 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710272045.AA12764@MOGLI.rutgers.edu> To: tonguc@av.rutgers.edu Subject: Re: abstract Cc: crose@MOGLI.rutgers.edu Status: RO Hi, Thanks! Cheers, Chris PS: not sure what happened before... might I have sent it to Marios only? Who knows... From crose Mon Oct 27 15:48:41 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA12781; Mon, 27 Oct 97 15:48:41 EST Date: Mon, 27 Oct 97 15:48:41 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710272048.AA12781@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Exam Problem Cc: crose@MOGLI.rutgers.edu Status: RO Nope, not yet. I think at this point I'll just add the score on this to the score on your next aexam... no time right now.. Cheers, Chris Rose From crose Mon Oct 27 23:49:32 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA13340; Mon, 27 Oct 97 23:47:12 EST Date: Mon, 27 Oct 97 23:47:12 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710280447.AA13340@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: class on tuesday Status: RO Hi Folks, There is a serious chance that I'll have to cancel class on tuesday (10/28/97). I'll send email as early as possible (probably in the early PM). I'll be at the maternity hospital (in NY) and will laptop-email from there if at all possible. 501 STUDENTS: About the test: it is commulative (covers EVERYTHING from the begnning of the term until what we've done so far and maybe a little more besides). However, it will be heavily skewed towward the systems stuff we've been doing of late with a fair amount of physical modeling and solutions. Don't worry TOO much about the Lagrangian. Anything I give you on that will be heavily guided. But you'll need to know all about linearization, laplace transforms transfer functions (that was tomorrow's lecture among other things) and everything else we've done so far. 543 STUDENTS: Everything up through and including chapter 6 is fair game. I won't have your solutions ready (probably) for the answers some of you handed in via email, but will return them with your graded quiz II at the latest. You should know all about switching and sorting and packet networks etc. I'll also almost certainly spring some probability on you! But I will carefully guide you this time. EVERYONE: if we miss tomorrow we'll have to make up the class. The only day which seems feasible is on a friday late afternoon. We'll cross that bridge (schedule a class) when we come to it if necessary. It's also possible that I might try to give you a web lecture (but don't count on it because I dont' think the technology is quite there yet.... CHEERS ALL! Chris Rose From jasonjen@ece.rutgers.edu Tue Oct 28 00:55:34 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA13473; Tue, 28 Oct 97 00:55:34 EST Received: from localhost (jasonjen@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id XAA13241 for ; Mon, 27 Oct 1997 23:51:31 -0500 Date: Mon, 27 Oct 1997 23:51:30 -0500 (EST) From: Jason Jen To: Chris Rose Subject: homework solution? Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Dr. Rose, Hi, will the homework solutions be posted sometime this week? Thanks, Jason J a s o n J e n ------------------------------ Electrical Engineering Office: EE 214 Tel: (732) 445-5385 ------------------------------ From coslit@ece.rutgers.edu Tue Oct 28 13:10:24 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA13998; Tue, 28 Oct 97 13:09:19 EST Received: (from coslit@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) id MAA10153; Tue, 28 Oct 1997 12:05:11 -0500 Date: Tue, 28 Oct 1997 12:05:11 -0500 From: Dianne Coslit Message-Id: <199710281705.MAA10153@ece.rutgers.edu> To: 330_543@mogli.rutgers.edu, 330_501@mogli.rutgers.edu Status: RO CLASSES ARE CANCELLED FOR TONIGHT. D. COSLIT ECE From crose@localhost.localdomain Tue Oct 28 23:55:32 1997 Return-Path: Received: from localhost.localdomain (ppp-11.ts-3.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA14540; Tue, 28 Oct 97 23:55:08 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id MAA00683; Tue, 28 Oct 1997 12:39:47 -0500 Date: Tue, 28 Oct 1997 12:39:47 -0500 From: Christopher Rose Message-Id: <199710281739.MAA00683@localhost.localdomain> To: 330_501@mogli.rutgers.edu, 330_543@mogli.rutgers.edu Subject: class cancelled for tuesday 10/28 Cc: crose@MOGLI.rutgers.edu Status: RO Hi FOlks, Class is cancelled tonite. Baby seems to be making an entrance. Will contact 501 folks tonite or tomorrow about exam on thursday and further details. We'll need to reschedule class ... probably for a friday later in the term. Cheers, Chris Rose From crose Tue Oct 28 23:57:42 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA14561; Tue, 28 Oct 97 23:57:41 EST Date: Tue, 28 Oct 97 23:57:41 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710290457.AA14561@MOGLI.rutgers.edu> To: jasonjen@ece.rutgers.edu Subject: Re: homework solution? Cc: crose@MOGLI.rutgers.edu Status: RO yes, they will.. Cheers From crose Wed Oct 29 00:22:38 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA14649; Wed, 29 Oct 97 00:21:18 EST Date: Wed, 29 Oct 97 00:21:18 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710290521.AA14649@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: Baby Cc: crose@MOGLI.rutgers.edu Status: RO Hi FOlks, Our baby boy was born today at about 4:40 pm. Mother and baby are resting comfortably. Thanks for your forebearance (sorry I had to miss class). FOr 501 folks, exam is still on for thursday. FOr 543 solks, I'll see if we can get the solutions posted to the PS's by this weekend at the latest (friday is what I mean). Gotta go now... I'm almost asleep as you can probably tell my my typing. Cheers, Chris ROse From zwf@winlab.rutgers.edu Tue Oct 28 23:23:04 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA00351; Tue, 28 Oct 97 23:22:36 EST Received: from dilbert.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id UAA24756; Wed, 29 Oct 1997 20:23:01 -0500 Received: from winlab.rutgers.edu by dilbert.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id UAA01264; Wed, 29 Oct 1997 20:23:00 -0500 Sender: zwf@winlab.rutgers.edu Message-Id: <3457E172.555649F0@winlab.rutgers.edu> Date: Wed, 29 Oct 1997 20:22:58 -0500 From: Wenfeng Zhang X-Mailer: Mozilla 4.03 [en] (X11; I; SunOS 5.6 sun4u) Mime-Version: 1.0 To: "330_543@MOGLI" <330_543@MOGLI> Subject: TA announcement Content-Type: multipart/alternative; boundary="------------B3409D54AA002EB2F04BCB73" Status: RO --------------B3409D54AA002EB2F04BCB73 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Hello, This is TA announcement. As the midterm exam for 330:543 is coming, some of you may want to discuss with me about what you learned. I change my office hour of next Tuesday to Monday morning. Only for next week! If you want to meet with me at that time, you are highly desirable to mail me in advance with your problems. Your cooperation will save time of all of us. My email is zwf@winlab.rutgers.edu See you on next Monday. Regards, Wenfeng -- Wenfeng Zhang Post: BPO 23888, PO BOX 1119, Piscataway, NJ 08855 Phone: (908)463-1964 (H) http://www.eden.rutgers.edu/~wfzhang --------------B3409D54AA002EB2F04BCB73 Content-Type: text/html; charset=us-ascii Content-Transfer-Encoding: 7bit Hello,

This is TA announcement.
As the midterm exam for 330:543 is coming, some of you may want to discuss with me about what you learned.  I change my office hour of next Tuesday to Monday morning. Only for next week!  If you want to meet with me at that time, you are highly desirable to mail me in advance with your problems. Your cooperation will save time of all of us.
My email is zwf@winlab.rutgers.edu
See you on next Monday.

Regards,
Wenfeng 

-- 
Wenfeng Zhang
Post: BPO 23888, PO BOX 1119, Piscataway, NJ 08855
Phone: (908)463-1964 (H)
http://www.eden.rutgers.edu/~wfzhang
  --------------B3409D54AA002EB2F04BCB73-- From crose Wed Oct 29 03:15:37 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00687; Wed, 29 Oct 97 03:15:36 EST Date: Wed, 29 Oct 97 03:15:36 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710290815.AA00687@MOGLI.rutgers.edu> To: xuyang@rutcor.rutgers.edu Subject: Re: 2nd midterm Cc: crose@MOGLI.rutgers.edu Status: RO Thanks Yang! I'll figure something out. WHat I think we'll do if possible is to start you at 3pm somewhere and then we'll migrate over to SEC117 for the second half. I'll get back to you wth details soon. However, if for some reason we've not connected by Sunday, please email me a reminder. At worst you can sit in my lab for the first half. Cheers, Chris Rose From crose Wed Oct 29 03:15:37 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00687; Wed, 29 Oct 97 03:15:36 EST Date: Wed, 29 Oct 97 03:15:36 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710290815.AA00687@MOGLI.rutgers.edu> To: xuyang@rutcor.rutgers.edu Subject: Re: 2nd midterm Cc: crose@MOGLI.rutgers.edu Status: RO Thanks Yang! I'll figure something out. WHat I think we'll do if possible is to start you at 3pm somewhere and then we'll migrate over to SEC117 for the second half. I'll get back to you wth details soon. However, if for some reason we've not connected by Sunday, please email me a reminder. At worst you can sit in my lab for the first half. Cheers, Chris Rose From crose Wed Oct 29 04:07:42 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00864; Wed, 29 Oct 97 04:06:05 EST Date: Wed, 29 Oct 97 04:06:05 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710290906.AA00864@MOGLI.rutgers.edu> To: 330_501, 330_543 Subject: mail problems Cc: crose@MOGLI.rutgers.edu Status: RO hi Folks, I've been having some mailer problems which may have sent some confusing mail. 1) there is NO CLASS for 330_543 on this coming thursday. The exam is on tuesday and I'll have solutions to the last two PS out on the web by friday (actually Wenfeng is working on this now). 2) there is NO CLASS for 330_501 on tuesday next. Thanks for all the well wishes on the new baby. Steph and I really appreciate it. Cheers, Chris Rose From shahid@er3.rutgers.edu Wed Oct 29 12:55:22 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01753; Wed, 29 Oct 97 12:55:21 EST Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id JAA26813; Thu, 30 Oct 1997 09:55:37 -0500 (EST) Date: Thu, 30 Oct 1997 09:55:37 -0500 (EST) From: Shahid Kagal To: crose@ece.rutgers.edu Cc: crose@MOGLI.rutgers.edu Subject: Some problems in Chap 4 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Respected Prof. Rose, Following are some problems faced in Chap. 4: 1. Page 96 : For compact superconcentrators there is a case discussed when each second stage superconcentrator has more than one output.How do we deduce that there is more than one row at the output of second stage. 2. Prove that the banyan network is the mirror image of the inverse banyan network : Sir, I have a real problem flipping the network. 3.Many to one concentration page 103-104 : Second portion of assigning for each active input i an output Oi in order to satisfy the monotone condition. In that how is the boundary for the interval decided, i.e. Oi = [ Sum over i'< i (ni'), Sum over i'<=i (ni'-1) ] 4. Are we suppose to prepare the Pippenger network. Thank You, Shahid. From crose Wed Oct 29 14:31:35 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA01857; Wed, 29 Oct 97 14:29:45 EST Date: Wed, 29 Oct 97 14:29:45 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710291929.AA01857@MOGLI.rutgers.edu> To: shahid@eden.rutgers.edu Subject: Re: Some problems in Chap 4 Cc: 330_543 Status: RO Hi Shahid, Thanks for the questions. I'm reposting them to the group for comment. Anybody care to take a crack at Shahid's questions. let's get a discussion going so I know you're alive out there :) Cheers Chrs Rose ***************** Following are some problems faced in Chap. 4: 1. Page 96 : For compact superconcentrators there is a case discussed when each second stage superconcentrator has more than one output.How do we deduce that there is more than one row at the output of second stage. 2. Prove that the banyan network is the mirror image of the inverse banyan network : Sir, I have a real problem flipping the network. 3.Many to one concentration page 103-104 : Second portion of assigning for each active input i an output Oi in order to satisfy the monotone condition. In that how is the boundary for the interval decided, i.e. Oi = [ Sum over i'< i (ni'), Sum over i'<=i (ni'-1) ] 4. Are we suppose to prepare the Pippenger network. From crose Wed Oct 29 14:33:05 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA01860; Wed, 29 Oct 97 14:31:23 EST Date: Wed, 29 Oct 97 14:31:23 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710291931.AA01860@MOGLI.rutgers.edu> To: shahid@eden.rutgers.edu Subject: Re: Some problems in Chap 4 Cc: 330_543 Status: RO Oh and yes I'd like yoyu to know about the Pippinger network. We need to know just abot everything up through and including chapter 6. I'll exclude the appendix to chapter 6 which is amore of a culture piece than course material. Cheers Chris Rose From zwf@winlab.rutgers.edu Wed Oct 29 14:39:09 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA01873; Wed, 29 Oct 97 14:37:34 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id LAA01277; Thu, 30 Oct 1997 11:37:52 -0500 Received: from winlab.rutgers.edu by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id LAA24513; Thu, 30 Oct 1997 11:37:34 -0500 Sender: zwf@winlab.rutgers.edu Message-Id: <3458B7CC.C8C8C1FA@winlab.rutgers.edu> Date: Thu, 30 Oct 1997 11:37:32 -0500 From: Wenfeng Zhang X-Mailer: Mozilla 4.03 [en] (X11; I; SunOS 5.5.1 sun4u) Mime-Version: 1.0 To: "330_543@MOGLI" <330_543@MOGLI> Subject: Solution for Chapter 6 Content-Type: multipart/alternative; boundary="------------40A15F97CE14DDF5418B1317" Status: RO --------------40A15F97CE14DDF5418B1317 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Hi, everybody The copies of the problem solution for chapter 6 will be available after 4:30PM. You can come to my office (EE 218) and get them. I will put the copies in a box near the door. The solution covers all of the problems after Chapter 6. Sorry for that I am too busy to give you a PS version on the net by next Tuesday. Further, I just find I cannot access the solution for any previous chapters from the homepage of 330:543. I donot know whether you suffer the same thing. To give you a quick reference, I will make a mirror site for all the solutions on my homepage. It is available soon via http://www.eden.rutgers.edu/~wfzhang Regards, Wenfeng -- Wenfeng Zhang Post: BPO 23888, PO BOX 1119, Piscataway, NJ 08855 Phone: (908)463-1964 (H) http://www.eden.rutgers.edu/~wfzhang --------------40A15F97CE14DDF5418B1317 Content-Type: text/html; charset=us-ascii Content-Transfer-Encoding: 7bit Hi, everybody

The copies of the problem solution for chapter 6 will be available after 4:30PM. You can come to my office
(EE 218) and get them.  I will put the copies in a box near the door. The solution covers all of the problems after Chapter 6.   Sorry for that I am too busy to give you a PS version on the net by next Tuesday.

Further, I just find I cannot access the solution for any previous chapters from the homepage of 330:543. I donot know whether you suffer the same thing.  To give you a quick reference, I will make a mirror site for all the solutions on my homepage.  It is available soon via http://www.eden.rutgers.edu/~wfzhang
Regards,
Wenfeng 

-- 
Wenfeng Zhang
Post: BPO 23888, PO BOX 1119, Piscataway, NJ 08855
Phone: (908)463-1964 (H)
http://www.eden.rutgers.edu/~wfzhang
  --------------40A15F97CE14DDF5418B1317-- From crose Wed Oct 29 14:42:57 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA01888; Wed, 29 Oct 97 14:41:16 EST Date: Wed, 29 Oct 97 14:41:16 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710291941.AA01888@MOGLI.rutgers.edu> To: 330_543, zwf@winlab.rutgers.edu Subject: Re: Solution for Chapter 6 Status: RO WAY TO GO WENFENG!!!!!!! I'll check out the ece problems. it's NOTORIOUSLY FLAKY. I've been having mailer problems as well. From minfan@paul.rutgers.edu Thu Oct 30 13:07:19 1997 Return-Path: Received: from paul.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01292; Thu, 30 Oct 97 13:07:04 EST Received: from localhost (minfan@localhost) by paul.rutgers.edu (8.8.5/8.8.5) with SMTP id KAA12034; Fri, 31 Oct 1997 10:07:12 -0500 (EST) Date: Fri, 31 Oct 1997 10:07:12 -0500 (EST) From: Min Fan To: Christopher Rose Cc: shahid@eden.rutgers.edu, 330_543@MOGLI.rutgers.edu Subject: What will be covered in the midterm II In-Reply-To: <9710291929.AA01857@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO We did not go through chapter 6. Will we be tested on chapter 6? Thanks. ------------------------------------------------ Min Fan Office : Hill 355 Tel:(732)445-3213 ext. 20 Email: minfan@paul.rutgers.edu ------------------------------------------------ From shahid@er5.rutgers.edu Thu Oct 30 16:00:07 1997 Return-Path: Received: from er5.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01467; Thu, 30 Oct 97 16:00:07 EST Received: from localhost (shahid@localhost) by er5.rutgers.edu (8.8.5/8.8.5) with SMTP id NAA12051 for ; Fri, 31 Oct 1997 13:00:11 -0500 (EST) Date: Fri, 31 Oct 1997 13:00:09 -0500 (EST) From: Shahid Kagal To: Christopher Rose Subject: Re: Some problems in Chap 4 In-Reply-To: <9710291929.AA01857@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Respected Prof. Rose, I think I haven't got any answers from our team. Shall I post the questions once again or wait for sometime. Yours, Shahid. From alap@winlab.rutgers.edu Thu Oct 30 23:21:58 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA01818; Thu, 30 Oct 97 23:21:46 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id UAA19484; Fri, 31 Oct 1997 20:21:50 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id UAA29824; Fri, 31 Oct 1997 20:21:50 -0500 Date: Fri, 31 Oct 1997 20:21:50 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711010121.UAA29824@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: Some Problems in Chapter 4 Status: RO Shahid - 1. Page 96 : For compact superconcentrators there is a case discussed when each second stage superconcentrator has more than one output.How do we deduce that there is more than one row at the output of second stage. 2. Prove that the banyan network is the mirror image of the inverse banyan network : Sir, I have a real problem flipping the network. ------- 1. I believe this is up to you, I mean, it depends what kind of switchings you have available to use. 2. to see that banyan is the mirror of inverse banyan, just pick the inverse banyan and change input and output. You will see you got the banyan. I think this is what the book means by "mirror". I hope it helps (and I hope it's right :) ) -Ana. From alap@winlab.rutgers.edu Fri Oct 31 00:50:47 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA01949; Fri, 31 Oct 97 00:50:34 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id VAA19679; Fri, 31 Oct 1997 21:50:37 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id VAA29882; Fri, 31 Oct 1997 21:50:37 -0500 Date: Fri, 31 Oct 1997 21:50:37 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711010250.VAA29882@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: Some Problems in Ch. 4 Status: RO Shahid 3.Many to one concentration page 103-104 : Second portion of assigning for each active input i an output Oi in order to satisfy the monotone condition. In that how is the boundary for the interval decided, i.e. Oi = [ Sum over i'< i (ni'), Sum over i'<=i (ni'-1) ] --- In this case we know that it does not matter WHERE the copies will appear, we are just interested in having n_i copies from input i. So we just send copies from input 1 to the first n_1 outputs, copies from input 2 to the next n_2 outputs, and so on. I think if you work out a very simple example it will be clear. For example: n_1 = 3 n_2 = 2 n_3 = 5 then O_1 = [0,2] O_2 = [3,4] O_3 = [4,9] I hope it helps. Regards, Ana. From crose Fri Oct 31 01:38:03 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02000; Fri, 31 Oct 97 01:37:47 EST Date: Fri, 31 Oct 97 01:37:47 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710310637.AA02000@MOGLI.rutgers.edu> To: minfan@paul.rutgers.edu Subject: Re: What will be covered in the midterm II Cc: 330_543 Status: RO yes we went through chapter six and it's possible to cover material in a grad course which was not specifically discussed in class. Cheers, Chris Rose From crose Fri Oct 31 02:06:17 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02083; Fri, 31 Oct 97 02:04:37 EST Date: Fri, 31 Oct 97 02:04:37 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710310704.AA02083@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Some Problems in Chapter 4 Status: RO Hi Ana, I agree with your second comment, but don't understand your first comment about the compact SC which is ok because I did not really understand the questoin either! Care to elaborate for everyone (Shahid AND Ana) ? Cheers, Chris Rose From crose Fri Oct 31 02:06:18 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02089; Fri, 31 Oct 97 02:06:05 EST Date: Fri, 31 Oct 97 02:06:05 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9710310706.AA02089@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Some Problems in Ch. 4 Status: RO Thanks again Ana! From shahid@er3.rutgers.edu Wed Oct 29 12:55:22 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01754; Wed, 29 Oct 97 12:55:21 EST Received: from er3.rutgers.edu (shahid@er3.rutgers.edu [165.230.180.141]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id JAA04170 for ; Thu, 30 Oct 1997 09:50:57 -0500 Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id JAA26813; Thu, 30 Oct 1997 09:55:37 -0500 (EST) Date: Thu, 30 Oct 1997 09:55:37 -0500 (EST) From: Shahid Kagal To: crose@ece.rutgers.edu Cc: crose@mogli.rutgers.edu Subject: Some problems in Chap 4 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Respected Prof. Rose, Following are some problems faced in Chap. 4: 1. Page 96 : For compact superconcentrators there is a case discussed when each second stage superconcentrator has more than one output.How do we deduce that there is more than one row at the output of second stage. 2. Prove that the banyan network is the mirror image of the inverse banyan network : Sir, I have a real problem flipping the network. 3.Many to one concentration page 103-104 : Second portion of assigning for each active input i an output Oi in order to satisfy the monotone condition. In that how is the boundary for the interval decided, i.e. Oi = [ Sum over i'< i (ni'), Sum over i'<=i (ni'-1) ] 4. Are we suppose to prepare the Pippenger network. Thank You, Shahid. From shahid@er5.rutgers.edu Fri Oct 31 14:01:38 1997 Return-Path: Received: from er5.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02721; Fri, 31 Oct 97 14:01:26 EST Received: from localhost (shahid@localhost) by er5.rutgers.edu (8.8.5/8.8.5) with SMTP id LAA12779; Sat, 1 Nov 1997 11:01:22 -0500 (EST) Date: Sat, 1 Nov 1997 11:01:21 -0500 (EST) From: Shahid Kagal To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Some Problems in Ch. 4 In-Reply-To: <199711010250.VAA29882@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Thanks, Ana That will be quite helpful. Shahid. From jsucec@ece.rutgers.edu Fri Oct 31 17:04:03 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02848; Fri, 31 Oct 97 17:04:02 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id NAA04286; Sat, 1 Nov 1997 13:59:16 -0500 Date: Sat, 1 Nov 1997 13:59:15 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu, crose@MOGLI.rutgers.edu, zwf@liman.Rutgers.EDU Subject: Question on 0-1 Principle Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO TO ANYONE IN 543 WHO CAN HELP ME HERE... I don't understand very well the 0-1 principle as discussed in Ch6 P2 of the homework 4 set (I didn't get a chance to come to campus until late Friday night, and by that time the EE building doors were locked and I have not been able to pick up a hard copy of the homework set 4 solutions.) I understand the 0-1 principle well enough to realize that it allows a divide and conquer approach to sorting. That is, given a set of N unsorted objects, we segregate the set into 2 subsets... One subset consists of members greater than the median of the set and the other subset consists of members whose value is less than the median. This dividing approach is performed recursively on the resulting subsets until we have an ordered list of N subsets consisting of 1 member each. Hence, the original set of N objects has been sorted. Now, Ch6 P2 is seemingly asking us to compare the complexity of sorting zeros and ones with the complexity of sorting a permutation of distinct objects using the 0-1 principle as the basis of the sorting algorithm. Is this assumption correct? If so, it seems that it can be shown that these two activities are equivalent, and therefore, have the same order of complexity. Is this correct? Since I don't have a copy of the solutions yet here is my take on Ch6 P2a... Obviously a sorting network with N objects, 0 to N-1, has N! possible input patterns. Now, the way I read P2a, is that we are asked to find the number of 0-1 input patterns that exist if we represent the N distinct objects, 0 to N-1, by zeros and ones. Is this a correct interpretation? Based on this interpretation, it seems to me that if N=2^n, then we need n bits (0-1s) to represent each of the N distinct objects. To represent all N inputs, therefore, we need N*n=N*logN 0-1s. The number of permutations of a list of N*logN bits (0-1s) is 2^(N*logN). When comparing this with Stirling's approximation of N!, it can then be shown that 2^(N*logN) permutations is on the same order of magnitude as N! permutations. Sorting complexity of zeros and ones, therefore, is of the same order of magnitude of sorting a permutation of 0 to N-1. Does this answer appear to correctly address problem Ch6 P2a? I think if I fully understand Ch6 P2a I should be able to figure out parts b and c. So, if anyone can answer any of the questions I have posed in this E-Mail, then your answers may help me move forward also. However, if anyone wants to add comments on parts b and c, or on Ch6 P1 for that matter, I will appreciate these remarks, too. Thanks. ...John From alap@winlab.rutgers.edu Fri Oct 31 20:05:57 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA02993; Fri, 31 Oct 97 20:05:40 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id RAA24276; Sat, 1 Nov 1997 17:05:35 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id RAA00607; Sat, 1 Nov 1997 17:05:35 -0500 Date: Sat, 1 Nov 1997 17:05:35 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711012205.RAA00607@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Compact Superconcentrators Status: RO Hello, I am confused about superconcentrators. I understand we pick a set (A,B) where A is subset of inputs, B is subset of outputs and A and B are the same size. Then in section (4.3) a crossbar switch MxN is considered and it shows that if we number correctly the active lines, a two stage factorization produces a compact superconcentrator. OK> But then in the example (inverse banyan) he considers M=N. Does that mean that he is only picking those active inputs ?? What happens with the other inputs ?? Or better: once A and B chosen they will be fixed "forever" ?? If someone can help, I would appreciate. Thanks. Ana. From alap@winlab.rutgers.edu Fri Oct 31 18:59:29 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA02959; Fri, 31 Oct 97 18:59:07 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id PAA24087; Sat, 1 Nov 1997 15:59:03 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id PAA00550; Sat, 1 Nov 1997 15:59:02 -0500 Date: Sat, 1 Nov 1997 15:59:02 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711012059.PAA00550@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Self Routing Status: RO If someone can help me: Prof Rose said in class that the Banyan Net is not self routing. The book says it is. I believe it is if we number the outputs correctly. Any idea ? Thanks. Ana. From jsucec@ece.rutgers.edu Fri Oct 31 20:09:28 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02998; Fri, 31 Oct 97 20:09:17 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id RAA06390; Sat, 1 Nov 1997 17:04:30 -0500 Date: Sat, 1 Nov 1997 17:04:29 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Self Routing In-Reply-To: <199711012059.PAA00550@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, I believe what the book says (or at least implies) is that the INVERSE Banyan network is self-routing (Ch5 P4, page 135). I believe Prof. Rose`s class statement that the Banyan network is not self-routing is correct (Actually, I recall contradicting him on this statement in class, and was subsequently shot down when a simple example showed that I was wrong... I too, was thinking at the time of the book's statement on the Inverse Banyan network and had confused it with the plain old Banyan network case which Prof. Rose was discussing.) Further, I believe you are correct in saying that the Banyan network can be modified to be self-routing, provide we relabel the inputs and outputs appropriately. The exact relabelling scheme escapes now, but I recall that we did an example of this in class for an 8x8 Banyan network. Anyway, I hope this helps. ...John On Sat, 1 Nov 1997, Ana Lucia Pinheiro wrote: > > If someone can help me: > Prof Rose said in class that the Banyan Net is not self routing. The book > says it is. I believe it is if we number the outputs correctly. > Any idea ? > Thanks. > Ana. > From jsucec@ece.rutgers.edu Fri Oct 31 20:09:28 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02998; Fri, 31 Oct 97 20:09:17 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id RAA06390; Sat, 1 Nov 1997 17:04:30 -0500 Date: Sat, 1 Nov 1997 17:04:29 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Self Routing In-Reply-To: <199711012059.PAA00550@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, I believe what the book says (or at least implies) is that the INVERSE Banyan network is self-routing (Ch5 P4, page 135). I believe Prof. Rose`s class statement that the Banyan network is not self-routing is correct (Actually, I recall contradicting him on this statement in class, and was subsequently shot down when a simple example showed that I was wrong... I too, was thinking at the time of the book's statement on the Inverse Banyan network and had confused it with the plain old Banyan network case which Prof. Rose was discussing.) Further, I believe you are correct in saying that the Banyan network can be modified to be self-routing, provide we relabel the inputs and outputs appropriately. The exact relabelling scheme escapes now, but I recall that we did an example of this in class for an 8x8 Banyan network. Anyway, I hope this helps. ...John On Sat, 1 Nov 1997, Ana Lucia Pinheiro wrote: > > If someone can help me: > Prof Rose said in class that the Banyan Net is not self routing. The book > says it is. I believe it is if we number the outputs correctly. > Any idea ? > Thanks. > Ana. > From alap@winlab.rutgers.edu Fri Oct 31 20:26:06 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA03008; Fri, 31 Oct 97 20:25:54 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id RAA24312; Sat, 1 Nov 1997 17:25:49 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id RAA00618; Sat, 1 Nov 1997 17:25:49 -0500 Date: Sat, 1 Nov 1997 17:25:49 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711012225.RAA00618@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Banyan Status: RO John, thanks a lot for the reply. Yes, I agree with you that the inverse banyan is self routing and agree also that if we top-down number the outputs the banyan is NOT self routing. But I do not agree that the book inplies that: in page 130 it is written: "we may also top-down numbering for nodes and links for the other networks shown in figure 10". Figure 10 has the plain banyan (not inverse). That's why I got so confused... Anyway, once more: thank you. Regards, Ana. From jsucec@ece.rutgers.edu Fri Oct 31 20:42:56 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03027; Fri, 31 Oct 97 20:42:44 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id RAA06767; Sat, 1 Nov 1997 17:37:57 -0500 Date: Sat, 1 Nov 1997 17:37:56 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Compact Superconcentrators In-Reply-To: <199711012205.RAA00607@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, I agree with you that some clarification of the example given by Figure 7 (page 98) is required. My take on this matter of the book describing the Inverse Banyan network as 2-stage compact superconcentrator (or visa-versa, depending on how you look at it), is that the Inverse Banyan network is being used as an example of how 2-stage compact superconcentration might be implemented. That is, we have an example of its implementation without using a brute force, and wasteful, MxN crossbar switch. As you point out, the book's Inverse Banyan network example considers M=N. Now, I believe that the Inverse Banyan network is still a useful structure even for the case when the size (=K) of the sets (A,B) is less than M=N. For example, suppose we have an 8x8 Inverse Banyan network with input leads 0, 2, 5 and 7 being active (4 out of 8 possible inputs). In this case, K=4. Although, I have not done the analysis myself, I believe that the Inverse Banyan network will then concentrate these 4 disperse inputs to outputs 0, 1, 2 and 3, in a monotonically increasing order. That is, the input at stage 0 input lead 0 goes to goes stage 2 output lead 0, the input at stage 0 input lead 2 goes to stage 2 output lead 1, and etc.. Now, how the Inverse Banyan network actually routes the active inputs to the correct compact output is a little fuzzy in my mind. But I believe that, in this case, the self-routing properties of the Inverse Banyan are NOT used. I believe, instead, that the Row Major Assignment algorithm is employed to achieve the compact superconcentration. I don't know whether my understanding of compact superconcentration is 100% correct, but I this is what I have been thinking to explain in my mind some of the issues you have raised. Maybe someone else can confirm or provide some additional insight. Thanks. ...John On Sat, 1 Nov 1997, Ana Lucia Pinheiro wrote: > > Hello, > > I am confused about superconcentrators. I understand we pick a set (A,B) where > A is subset of inputs, B is subset of outputs and A and B are the same size. > Then in section (4.3) a crossbar switch MxN is considered and it shows that > if we number correctly the active lines, a two stage factorization produces > a compact superconcentrator. OK> But then in the example (inverse banyan) > he considers M=N. Does that mean that he is only picking those active inputs ?? > What happens with the other inputs ?? Or better: once A and B chosen they will > be fixed "forever" ?? > > If someone can help, I would appreciate. > > Thanks. > Ana. > > From jsucec@ece.rutgers.edu Fri Oct 31 21:06:35 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03043; Fri, 31 Oct 97 21:06:23 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id SAA07049; Sat, 1 Nov 1997 18:01:35 -0500 Date: Sat, 1 Nov 1997 18:01:35 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Banyan In-Reply-To: <199711012225.RAA00618@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Good point about that reference on page 130. I had forgotton about that confusing comment in our book. The only thing I can think of which might justify this terribly confusing statement, is that perhaps the "top-down numbering for nodes and links" which the author describes on the previous page is in fact valid even for the plain vanilla Banyan network. I have not done the analysis, however, to verify this. I'm just taking it on faith for now. I am fairly convinced, however, by the homework and class examples that all of these "Banyan-like" networks are self-routing (provided we label the inputs and outputs correctly, as you mentioned). The bottom line, however, is that this statement on page 130 is very confusing to me also because I typically don't think of how to number the actual nodes and links that are INTERNAL to the network! (When a network diagram example appears on a homework problem, the first thing I do is label the input and output leads of the network from 00...0 to 11...1) So when I first read the statement on page 130, it took me a while to realize that it was actually referring to the derivation of the self-routing scheme employed for the Shuffle Exchange network describe on page 129 which discusses edges and switches internal to the network. ...John On Sat, 1 Nov 1997, Ana Lucia Pinheiro wrote: > > John, thanks a lot for the reply. Yes, I agree with you that the inverse banyan is > self routing and agree also that if we top-down number the outputs the banyan is > NOT self routing. But I do not agree that the book inplies that: in page 130 it > is written: "we may also top-down numbering for nodes and links for the other > networks shown in figure 10". Figure 10 has the plain banyan (not inverse). > That's why I got so confused... > Anyway, once more: thank you. > Regards, > Ana. > From jsucec@ece.rutgers.edu Fri Oct 31 21:33:27 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03052; Fri, 31 Oct 97 21:33:17 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id SAA07367 for <330_543@MOGLI.rutgers.edu>; Sat, 1 Nov 1997 18:28:29 -0500 Date: Sat, 1 Nov 1997 18:28:29 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Re: Some Problems in Chapter 4 (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Shahid, I agree with Ana's explaination of both (1) and (2) given below. Concerning (1), here is an example I worked out this afternoon while studying this material... M=16, N=8 * Pick p=4 * Pick q=N/2=M/p=4 --> Stage 1 consists of 4 4x4 (i.e., p M/p x q) compact superconcentrators --> Stage 2 consists of 4 4x2 (i.e., q p x N/q) compact superconcentrators Since, N/q for this example is 2, this represents a case where we have more than 1 row at the output of the 2nd stage. Anyway, this example clarified things in my mind. I hope, however, it does not confuse anyone who may have contrived other examples. ...John ---------- Forwarded message ---------- Date: Fri, 31 Oct 1997 20:21:50 -0500 From: Ana Lucia Pinheiro To: 330_543@mogli.rutgers.edu Subject: Re: Some Problems in Chapter 4 Shahid - 1. Page 96 : For compact superconcentrators there is a case discussed when each second stage superconcentrator has more than one output.How do we deduce that there is more than one row at the output of second stage. 2. Prove that the banyan network is the mirror image of the inverse banyan network : Sir, I have a real problem flipping the network. ------- 1. I believe this is up to you, I mean, it depends what kind of switchings you have available to use. 2. to see that banyan is the mirror of inverse banyan, just pick the inverse banyan and change input and output. You will see you got the banyan. I think this is what the book means by "mirror". I hope it helps (and I hope it's right :) ) -Ana. From crose Fri Oct 31 23:00:17 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03148; Fri, 31 Oct 97 23:00:17 EST Date: Fri, 31 Oct 97 23:00:17 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711010400.AA03148@MOGLI.rutgers.edu> To: linzhou@ece.rutgers.edu Subject: Re: Problems in Ch. 6 Cc: crose@MOGLI.rutgers.edu Status: RO Hi, I can't answer your question right now, bu i'm going to post it to the group for commment. Cheers, Chris Rose From crose Fri Oct 31 23:04:31 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03156; Fri, 31 Oct 97 23:04:12 EST Date: Fri, 31 Oct 97 23:04:12 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711010404.AA03156@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Self Routing Cc: crose@MOGLI.rutgers.edu Status: RO Hi Ana, This is probably already answered but I'll tell you anyway. The Banyan net as shown in the book IS NOT self routing since it would natrually be assumed that the output ordering is from top to bottom (try it). However, if you renumber the outputs (I think we did this in class) then all is ok. Think abot how the outputs could be reordered. Folks KEEP THE QUESTIONS COMING TO THE GROUP! This is the best way to explore and answer questions. Cheers, Chris Rose PS: I won't be able to answer all questions immediately, so don't dispair ifyou dont' hear from me quickly. From crose Fri Oct 31 23:07:12 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03170; Fri, 31 Oct 97 23:07:02 EST Date: Fri, 31 Oct 97 23:07:02 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711010407.AA03170@MOGLI.rutgers.edu> To: 330_543 Subject: self-routing Status: RO ] RIGHT ON, John! The only thing I disagree with is the renumbering of the INPUTS. if a switch is self-routing then renumbering the inputs won't matter at all. it's the outpus which kill you if not numbered correctly. Cheers, Chris Rose From alap@winlab.rutgers.edu Fri Oct 31 23:16:46 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA03183; Fri, 31 Oct 97 23:16:36 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id UAA24689; Sat, 1 Nov 1997 20:16:30 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id UAA01060; Sat, 1 Nov 1997 20:16:30 -0500 Date: Sat, 1 Nov 1997 20:16:30 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711020116.UAA01060@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Problem 5.1 Status: RO Hello, Related to Problem 5.1 , why is the probability that it requires r searches the sum form r to infinity ? How can we search in more that r paths if there are only r paths to search? Am I missing something? Also, even if I accept that, I believe the equation for the average number of search has a typo, , the first sum should have "i" instead of "n". And also , I do not see how I can write the sum as the differentiation. Besides that I undersatnd everything (I mean the very simple differentiation :) ) I appreciate if someone gives me a hint. Thank you. Ana. From crose Fri Oct 31 23:26:45 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03194; Fri, 31 Oct 97 23:26:34 EST Date: Fri, 31 Oct 97 23:26:34 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711010426.AA03194@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Problem 5.1 Cc: crose@MOGLI.rutgers.edu Status: RO Mean of a geometric distribution. i've not looked at the solutoins but you get the expression fot eh mean of a geometric by differentiating the expression for the sum of p^k from k to r. I know this sounds cryptic, but I'm in a hurry this weekend trying to get the house ready for arrival of wife (baby will remain in hospital for some while... and am rnning around making sure taht goes well too). Sorry to be the absent prof. Oh and Lin Zhou, I'm not ignoring you, just could not read all your questions right away.... I'll get to them. Chers, Chris ROse From crose Fri Oct 31 23:27:06 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03203; Fri, 31 Oct 97 23:26:54 EST Date: Fri, 31 Oct 97 23:26:54 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711010426.AA03203@MOGLI.rutgers.edu> To: 330_543 Status: RO >From linzhou@ece.rutgers.edu Fri Oct 31 15:43:44 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA02791; Fri, 31 Oct 97 15:43:43 EST Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id MAA03542 for ; Sat, 1 Nov 1997 12:38:58 -0500 Date: Sat, 1 Nov 1997 12:38:58 -0500 (EST) From: Lin Zhou To: Christopher Rose Subject: Problems in Ch. 6 In-Reply-To: <9710310706.AA02089@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: R Dear Prof Rose, In problem set 5 of chapter 6, N elements by PxQ are sorted. P is the number of raw while q is the number of column. The sorter of each raw should be qxq since each raw has q elements. Thus, there are p qxq sorter in horizontal planes. But the solution gives q pxp sorter which I think p q should be swapped. Am I right? Best regards, lin zhou ************************** Lin Zhou Department of Electrical Engineering Rutgers University LPO 16389 P.O.Box 5064 New Brunswick NJ 08903 USA Tel: 732-699-0182 Email: linzhou@ece.rutgers.edu From crose Fri Oct 31 23:47:18 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03304; Fri, 31 Oct 97 23:47:06 EST Date: Fri, 31 Oct 97 23:47:06 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711010447.AA03304@MOGLI.rutgers.edu> To: 330_543 Subject: just posted Cc: crose@MOGLI.rutgers.edu Status: RO Hi FOlks, I just posted a question about sorting. I'd like one of you (students) to take a look and comment. Why? Because you've got a test coming up and any critical thinking will help on this sort of problem (this one is simple, though). I've got aabout 5 minutes to spare this evening soe let me take the time to try to re-answer Ana's question about the mean number of paths. You find a path in the first try with prob 1-q Second try -> q(1-q) kth try -> q^{k-1} (1-q) The prob you don't find a path at all is q^r let's make sure this is a prob dist: \sum_{k=0}^{r-1} (1-q)q^k + q^r = via the identity on the summation of a geometric series (1-q) [(1-q^r)/(1-q)] + q^r = 1 (checks out ok) Remembering that you search r paths both if you find a free path on the last try AND when you don't find a free path at all we have the mean number of paths searched as \sum_{k=1}^{r} k prob(k) = \sum_{k=0}^{r-1} (k+1) (1-q)q^k + rq^r The problem is the k term in the sum. If you don't remember the identity for the mean of a geometric dist, you can derive it from differentiating the expression for the sum of the geometric series (notice how the k in the exponent pops down when you differentiate with respect to q ?) Hope that helps, Chris Rose (in a hurry, used fifteen minutes as opposed to 5 egad!) From jsucec@ece.rutgers.edu Sat Nov 1 02:51:32 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03533; Sat, 1 Nov 97 02:51:31 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id XAA11009; Sat, 1 Nov 1997 23:46:40 -0500 Date: Sat, 1 Nov 1997 23:46:40 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu, crose@MOGLI.rutgers.edu, zwf@liman.Rutgers.EDU Subject: Question on 0-1 Principle (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO ALL, I am no longer in dire need to have my questions concerning Ch6 P2 answered... I was lucky enough to meet a WINLAB student this evening who let me browse his copy of the Ch6 homework solutions (thanks Hua). I don't mean to sound overly critical and negative, but I think many will agree with me that the wording of this problem, as stated in the text, is somewhat ambiguous (at least for part a). That aside, I think I can add some additional comments to the solution handout for this problem. First of all, I agree with the solution statement that a list of N 0-1s can be sorted in O(N) time. This is done by the well known Counting Sort algorithm which uses the a priori information about the range of objects to be sorted to reduce the sorting time from order N*logN (lower bound complexity for when no a priori information available about the objects to be sorted) to order N. This algorithm, I believe, requires its computation to be done by a single centralized processor and thus does not lend itself adequately to the distributed nature of a Sorting network. My additional thoughts about this are as follows: Why can't we use the Counting Sort algorithm to sort N objects which are known to be in the range [0,N-1]? Thus, we can sort the N! permutations in O(N) time just as the list of N 0-1 elements is sorted in O(N) time. I believe this is a valid approach to sorting the N objects from the N! possible input patterns because the only assumption of the Counting Sort algorithm is "that each of the N elements is an element in the range 1 to K" (from Introduction to Algorithms, by Cormen, Leiserson and Rivest). I still don't have my own copy of the Ch6 solutions, so I'm going from memory here, but I seem to recall that the solution statement says we require O(N*logN) effort to sort the N objects in range [0,N-1]. Based on my algorithm reference book, however, it seems that we can sort this in O(N) time just as we can for the 0-1 sequence, ASSUMING WE HAVE A SINGLE CPU TO PERFORM ALL OF THE SEQUENTIAL ALGORITHM WORK. This brings me to comment on the whole point of this particular problem. I think the author's objective in putting this problem in the book was to demonstrate that we have to deal with some inherent inefficiencies of the distributed nature of the Sorting network. That is, although distributed processing provides considerable speedup for certain applications (e.g. the Bitonic Sorting algorithm lends itself well to parallel speedup, in theory at least), there are other applications that are of mostly sequential nature (e.g. the Counting Sort algorithm), which should not be used by a Sorting Network / distributed processing architecture. Like I said, I still don't have my own copy of the solution handout, but I think I have recalled correctly what is stated on the handout. Also, I believe I have correctly applied the Counting Sort algorithm to N! permutations of the N objects... PROVIDED these objects are in the range [0,N-1] and we have a sequential processor doing the number crunching. If I have stated something incorrectly, please feel free to steer me right! Anyway, I better start looking at some of the other Ch6 problems! ...John ---------- Forwarded message ---------- Date: Sat, 1 Nov 1997 13:59:15 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu, crose@MOGLI.rutgers.edu, zwf@winlab.rutgers.edu Subject: Question on 0-1 Principle TO ANYONE IN 543 WHO CAN HELP ME HERE... I don't understand very well the 0-1 principle as discussed in Ch6 P2 of the homework 4 set (I didn't get a chance to come to campus until late Friday night, and by that time the EE building doors were locked and I have not been able to pick up a hard copy of the homework set 4 solutions.) I understand the 0-1 principle well enough to realize that it allows a divide and conquer approach to sorting. That is, given a set of N unsorted objects, we segregate the set into 2 subsets... One subset consists of members greater than the median of the set and the other subset consists of members whose value is less than the median. This dividing approach is performed recursively on the resulting subsets until we have an ordered list of N subsets consisting of 1 member each. Hence, the original set of N objects has been sorted. Now, Ch6 P2 is seemingly asking us to compare the complexity of sorting zeros and ones with the complexity of sorting a permutation of distinct objects using the 0-1 principle as the basis of the sorting algorithm. Is this assumption correct? If so, it seems that it can be shown that these two activities are equivalent, and therefore, have the same order of complexity. Is this correct? Since I don't have a copy of the solutions yet here is my take on Ch6 P2a... Obviously a sorting network with N objects, 0 to N-1, has N! possible input patterns. Now, the way I read P2a, is that we are asked to find the number of 0-1 input patterns that exist if we represent the N distinct objects, 0 to N-1, by zeros and ones. Is this a correct interpretation? Based on this interpretation, it seems to me that if N=2^n, then we need n bits (0-1s) to represent each of the N distinct objects. To represent all N inputs, therefore, we need N*n=N*logN 0-1s. The number of permutations of a list of N*logN bits (0-1s) is 2^(N*logN). When comparing this with Stirling's approximation of N!, it can then be shown that 2^(N*logN) permutations is on the same order of magnitude as N! permutations. Sorting complexity of zeros and ones, therefore, is of the same order of magnitude of sorting a permutation of 0 to N-1. Does this answer appear to correctly address problem Ch6 P2a? I think if I fully understand Ch6 P2a I should be able to figure out parts b and c. So, if anyone can answer any of the questions I have posed in this E-Mail, then your answers may help me move forward also. However, if anyone wants to add comments on parts b and c, or on Ch6 P1 for that matter, I will appreciate these remarks, too. Thanks. ...John From jsucec@ece.rutgers.edu Sat Nov 1 03:18:19 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03565; Sat, 1 Nov 97 03:18:05 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id AAA11212 for <330_543@MOGLI.rutgers.edu>; Sun, 2 Nov 1997 00:13:14 -0500 Date: Sun, 2 Nov 1997 00:13:14 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Correction: Compact Superconcentrators (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Oops, in the E-Mail I sent yesterday afternoon, I incorrectly described an example of compact superconcentration via Inverse Banyan network by saying that a "stage 0" input is routed to a particular "stage 2" output... This wording implies that there are 3 stages which is incorrect because we're talking about 2 STAGE network. I should have stated that the network inputs enter stage 1 and exit stage 2. Sorry if this error caused any confusion. ...John ---------- Forwarded message ---------- Date: Sat, 1 Nov 1997 17:37:56 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Compact Superconcentrators Ana, I agree with you that some clarification of the example given by Figure 7 (page 98) is required. My take on this matter of the book describing the Inverse Banyan network as 2-stage compact superconcentrator (or visa-versa, depending on how you look at it), is that the Inverse Banyan network is being used as an example of how 2-stage compact superconcentration might be implemented. That is, we have an example of its implementation without using a brute force, and wasteful, MxN crossbar switch. As you point out, the book's Inverse Banyan network example considers M=N. Now, I believe that the Inverse Banyan network is still a useful structure even for the case when the size (=K) of the sets (A,B) is less than M=N. For example, suppose we have an 8x8 Inverse Banyan network with input leads 0, 2, 5 and 7 being active (4 out of 8 possible inputs). In this case, K=4. Although, I have not done the analysis myself, I believe that the Inverse Banyan network will then concentrate these 4 disperse inputs to outputs 0, 1, 2 and 3, in a monotonically increasing order. That is, the input at stage 0 input lead 0 goes to goes stage 2 output lead 0, the input at stage 0 input lead 2 goes to stage 2 output lead 1, and etc.. Now, how the Inverse Banyan network actually routes the active inputs to the correct compact output is a little fuzzy in my mind. But I believe that, in this case, the self-routing properties of the Inverse Banyan are NOT used. I believe, instead, that the Row Major Assignment algorithm is employed to achieve the compact superconcentration. I don't know whether my understanding of compact superconcentration is 100% correct, but I this is what I have been thinking to explain in my mind some of the issues you have raised. Maybe someone else can confirm or provide some additional insight. Thanks. ...John On Sat, 1 Nov 1997, Ana Lucia Pinheiro wrote: > > Hello, > > I am confused about superconcentrators. I understand we pick a set (A,B) where > A is subset of inputs, B is subset of outputs and A and B are the same size. > Then in section (4.3) a crossbar switch MxN is considered and it shows that > if we number correctly the active lines, a two stage factorization produces > a compact superconcentrator. OK> But then in the example (inverse banyan) > he considers M=N. Does that mean that he is only picking those active inputs ?? > What happens with the other inputs ?? Or better: once A and B chosen they will > be fixed "forever" ?? > > If someone can help, I would appreciate. > > Thanks. > Ana. > > From chriskly@er3.rutgers.edu Sat Nov 1 03:26:28 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03570; Sat, 1 Nov 97 03:26:17 EST Received: (from chriskly@localhost) by er3.rutgers.edu (8.8.5/8.8.5) id AAA26727; Sun, 2 Nov 1997 00:26:09 -0500 (EST) Date: Sun, 2 Nov 97 0:26:09 EST From: Christine Kleiwerda To: Christopher Rose Cc: 330_543@MOGLI.rutgers.edu In-Reply-To: Your message of Fri, 31 Oct 97 23:26:54 EST Message-Id: Status: RO > >From linzhou@ece.rutgers.edu Fri Oct 31 15:43:44 1997 > Return-Path: > Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) > id AA02791; Fri, 31 Oct 97 15:43:43 EST > Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id MAA03542 for ; Sat, 1 Nov 1997 12:38:58 -0500 > Date: Sat, 1 Nov 1997 12:38:58 -0500 (EST) > From: Lin Zhou > To: Christopher Rose > Subject: Problems in Ch. 6 > In-Reply-To: <9710310706.AA02089@MOGLI.rutgers.edu> > Message-Id: > Mime-Version: 1.0 > Content-Type: TEXT/PLAIN; charset=US-ASCII > Status: R > > Dear Prof Rose, > > In problem set 5 of chapter 6, N elements by PxQ are sorted. P is the > number of raw while q is the number of column. The sorter of each raw > should be qxq since each raw has q elements. Thus, there are p qxq sorter > in horizontal planes. But the solution gives q pxp sorter which I think p > q should be swapped. Am I right? > > > Best regards, > > lin zhou > > > > ************************** > Lin Zhou > Department of Electrical Engineering > Rutgers University > LPO 16389 P.O.Box 5064 > New Brunswick NJ 08903 > USA > > Tel: 732-699-0182 > Email: linzhou@ece.rutgers.edu > > > > I'm not sure that I exactly understand the question, but here is my take on the problem: the sorting network corresponding to shear sort is a series of q pxp sorters, followed by p qxq sorters, followed by q pxp sorters, etc... In total, there are 2*log(p) stages of sorters here. Each stage of q pxp sorters correspond to step 1 of the shear sort algorithm and the p qxq sorters correspond to step 2. Finally, there is one last stage for step 3. This is a stage of q pxp sorters. To answer the question that I think was asked, the answer is that there are both p qxq sorters and q pxp sorters in the representation. Hope this helps. :) If not, maybe you could rephrase the question, or else someone else could try to answer. Christine From chriskly@er3.rutgers.edu Sat Nov 1 03:32:04 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA03580; Sat, 1 Nov 97 03:31:53 EST Received: (from chriskly@localhost) by er3.rutgers.edu (8.8.5/8.8.5) id AAA26985 for 330_543@mogli.rutgers.edu; Sun, 2 Nov 1997 00:31:46 -0500 (EST) Date: Sun, 2 Nov 97 0:31:46 EST From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Subject: answers to chpt 5 probs??? Message-Id: Status: RO Were answers to chpt 5 problems given out? Are they on the web? christine From crose@localhost.localdomain Sat Nov 1 03:43:37 1997 Return-Path: Received: from localhost.localdomain (ppp-34.ts-7.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA03596; Sat, 1 Nov 97 03:41:28 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id BAA01990 for 330_501@mogli.rutgers.edu; Sun, 2 Nov 1997 01:29:28 -0500 Date: Sun, 2 Nov 1997 01:29:28 -0500 From: Christopher Rose Message-Id: <199711020629.BAA01990@localhost.localdomain> To: 330_501@mogli.rutgers.edu Subject: solutions Status: RO hi Christine, The solutions are avaialble from Wenfeng (outside his office) in hardcopy. He's been unable to translate them to PS for web distribution. From alap@winlab.rutgers.edu Sat Nov 1 04:34:09 1997 X-UIDL: d54494fcd41a7a7793ec73b38e97e501 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA03678; Sat, 1 Nov 97 04:33:51 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA25003; Sun, 2 Nov 1997 01:33:43 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA01489; Sun, 2 Nov 1997 01:33:42 -0500 Date: Sun, 2 Nov 1997 01:33:42 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711020633.BAA01489@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: answers to chpt 5 probs??? Status: RO Christine , yes, the answers are on the TA web page. He actually sent an e-mail with his web page address (I can't find right now... :( I hope you can get there). Regards, Ana. From crose Sat Nov 1 05:03:47 1997 X-UIDL: d9157c35f10246dae7921bcbda796b2c Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA03736; Sat, 1 Nov 97 05:03:36 EST Date: Sat, 1 Nov 97 05:03:36 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711011003.AA03736@MOGLI.rutgers.edu> To: 330_543 Subject: ps 3 Cc: crose@MOGLI.rutgers.edu Status: RO hi Folks, I've put ps3soln.ps up on the course web page. you can also find it on Wenfeng's web page. I've not proofed them so consume at own risk. Don't stress TOO much about the quiz. The stuf we've done since the last quiz is pretty straightforward. And don't worry too much about Shear Sort, that's for culture. You can also forget the appendix in ch 6. Seems that this text has a few problems eh? After the quiz, i'd like to record complaints if possible. maybe we can convince Joe to do a revised edition? Cheers, Chris Rose From linzhou@ece.rutgers.edu Sat Nov 1 14:37:55 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA04169; Sat, 1 Nov 97 14:37:38 EST Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id LAA02572; Sun, 2 Nov 1997 11:32:43 -0500 Date: Sun, 2 Nov 1997 11:32:43 -0500 (EST) From: Lin Zhou To: Christopher Rose Cc: 330_543@MOGLI.rutgers.edu Subject: Re: self-routing In-Reply-To: <9711010407.AA03170@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Hi, For self routing network, if the renembering of the input won't matter at all, it will not be the same as suffle network. Therefore, it could be blocked even compact input and monotone out put are set. lin zhou On Fri, 31 Oct 1997, Christopher Rose wrote: > ] > RIGHT ON, John! The only thing I disagree with is the renumbering > of the INPUTS. if a switch is self-routing then renumbering the > inputs won't matter at all. it's the outpus which kill you if not > numbered correctly. > > Cheers, > > Chris Rose > From alap@winlab.rutgers.edu Sat Nov 1 19:49:39 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA04346; Sat, 1 Nov 97 19:49:20 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA26841; Sun, 2 Nov 1997 16:49:05 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id QAA01862; Sun, 2 Nov 1997 16:49:04 -0500 Date: Sun, 2 Nov 1997 16:49:04 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711022149.QAA01862@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Ch 6: Three-phase algorithm Status: RO Could someone give me a help in Figure 14 (page 158) , I have problems understanding the PHASE II part. He explains that in the last paragraph in page 156, I don't see what is k , i and j_i in this case. I appreciate any help. Thank you very much. Ana. From shahid@er5.rutgers.edu Sat Nov 1 20:19:44 1997 Return-Path: Received: from er5.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA04378; Sat, 1 Nov 97 20:19:27 EST Received: from localhost (shahid@localhost) by er5.rutgers.edu (8.8.5/8.8.5) with SMTP id RAA25740 for <330_543@MOGLI.rutgers.edu>; Sun, 2 Nov 1997 17:19:10 -0500 (EST) Date: Sun, 2 Nov 1997 17:19:08 -0500 (EST) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Subject: Staggering the Phases Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO I seem to figure out no idea of how to stagger the phases of the 3 phase algorithm in order to save the overhead time of request and acknowledgement. Could anybody who has been able to solve the problem help me . Thanks. Shahid. From jsucec@ece.rutgers.edu Sun Nov 2 01:24:07 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA04674; Sun, 2 Nov 97 01:22:11 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id WAA14249; Sun, 2 Nov 1997 22:17:11 -0500 Date: Sun, 2 Nov 1997 22:17:10 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Ch 6: Three-phase algorithm In-Reply-To: <199711022149.QAA01862@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, this is my take on PHASE II of Figure 14... * k represents the output of the Sort network and in the input to the parallel Acknowlegment Sort network (used for the Acknowledgement packets). * i represents the source of the overall network and the destination of the Acknowledgement Sort network. The address of i must be included in the header of the request packet in order for the Acknowledgement packet to be properly addressed. * j_i represents the destination of the overall network requested by input i. I hope this helps. Actually, I too have a question on this same figure. What is the purpose of the "feedback-like" lines drawn between the Sorting network and Banyan network of the PHASE II parallel Acknowledgement Sort-Banyan network? As far as I know, no feedback or acknowledgement is required because the Acknowledgement packets are guarenteed to be non-blocking due to the purging algorithm which takes place at the end of PHASE I. Anyone else have thoughts on this ??? ...John On Sun, 2 Nov 1997, Ana Lucia Pinheiro wrote: > > Could someone give me a help in Figure 14 (page 158) , I have problems > understanding the PHASE II part. He explains that in the last paragraph > in page 156, I don't see what is k , i and j_i in this case. > > I appreciate any help. > Thank you very much. > Ana. > From jsucec@ece.rutgers.edu Sun Nov 2 01:36:56 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA04688; Sun, 2 Nov 97 01:36:38 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id WAA14569 for <330_543@MOGLI.rutgers.edu>; Sun, 2 Nov 1997 22:31:38 -0500 Date: Sun, 2 Nov 1997 22:31:37 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Many to 1! How is this possible? Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO On pages 102-103 and in Figure 11, the concept of Many to 1 concentration is discussed. I have not been able to justify this my mind... I thought that when 2 or more distinct inputs requested the same output, then this constituted blocking! Since Figure 11 shows 2 distinct input contending for a single output of a compact superconcentrator, one of these inputs will be blocked. Does anyone understand this figure? If so, please help me out here with what you know about it. Thanks! ...John From crose@localhost.localdomain Sun Nov 2 02:21:07 1997 Return-Path: Received: from localhost.localdomain (ppp-58.ts-7.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA04731; Sun, 2 Nov 97 02:20:40 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id AAA03008; Mon, 3 Nov 1997 00:09:42 -0500 Date: Mon, 3 Nov 1997 00:09:42 -0500 From: Christopher Rose Message-Id: <199711030509.AAA03008@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: An excellent Idea! Cc: 330_501@mogli.rutgers.edu Status: RO Hi Folks, A student in my linear systems class brought to my attention that I sometimes use relatively sophisticated English idioms when phrasing problems. It was felt that having a dictionary handy for translation from different languages would be helpful. SO, for those of you whose mother tongue is not English, I invite you to bring a dictionary for translation to the test tomorrow just in case you run across words which you do not understand. For those of you who speak and write English better than I do I apologize in advance for the seeming condescension. That is not my intention. I simply want everyone on equal technical footing. The nice thing about technical courses is that they are, more than many other courses, language-independent. That is, as mathematicians we speak a lovely higher form of language which travels well around the world. It therefore seems particularly distasteful to raise the spectre of plebian language difficulties in a technical course examination when it can be easily avoided. Cheers, Chris Rose PS: For 501 folks, the same rule will apply on the final. PPS: I know I DON'T HAVE TO MENTION THIS TO YOU FOLKS, but unfortunately my avatar of ECE Professor requires it: don't abuse this priviledge by bringing a reference book disguised as a dictionary. Sorrow will be your first, last and middle name :) Think of it this way. Your career is a lovely butterfly emerging from its crysalis, spreading its wings slowly, gently raising and lowering them, readying for flight in a sunlit meadow filled with flowers. A disguised reference book is the same butterfly emerging from its crysalis on a miserable rainy day in the left lane of the NJ Turnpike during rush hour :) From crose Sun Nov 2 02:26:41 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA04753; Sun, 2 Nov 97 02:26:17 EST Date: Sun, 2 Nov 97 02:26:17 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711020726.AA04753@MOGLI.rutgers.edu> To: 330_543@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Many to 1! How is this possible? Status: RO This is quick so I can answer it in the 2 minutes I have right now. many to 1 is just Joe's way of prepping you for the knockout switch or any other output queued switch. It does not really make sense otherwise. Oh and thatnks for your comments on phasing. Don't have the figure in question in front of me right now so I can't comment without confusing the issue (I could tell you what Phasing IS, but it might not be the same pedagogical path Joe was taking.... basic idea is to stagger timing so you can use parallel hardware to speed things up... like pipelining in computer architecture... yes your quetsion had to do with the correctness of the posed protcol, but I daren't go there withtout the figure in front of me :) Cheers, Chris From shahid@er3.rutgers.edu Sun Nov 2 14:31:08 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA05467; Sun, 2 Nov 97 14:30:50 EST Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id LAA17225 for <330_543@MOGLI.rutgers.edu>; Mon, 3 Nov 1997 11:30:27 -0500 (EST) Date: Mon, 3 Nov 1997 11:30:27 -0500 (EST) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Subject: Problems: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Hi everyone. I have doubts regarding the following question: Why isn't a two stage superconcentrator a superconcentrator? I started on the track that in case of a superconcentrator the outputs are neither compact nor row majored. Hence any subset of the outputs can appear at the output of the first stage. But now I am not able to think about it further. Any suggestions altogether different will be welcome. Shahid From shahid@er3.rutgers.edu Sun Nov 2 16:18:28 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA05607; Sun, 2 Nov 97 16:18:27 EST Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id NAA08644; Mon, 3 Nov 1997 13:17:59 -0500 (EST) Date: Mon, 3 Nov 1997 13:17:59 -0500 (EST) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Cc: crose@MOGLI.rutgers.edu Subject: Exercises: Chapter 5 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Dear all, I have problems solving Exercises 6 and 7 of chapter 5. In exercise 6 I have been able to deduce some facts of the hypercube network with N=8 but then I cannot generalize the facts. As an example for N=8 the fanout will be 2 . Now if I increase N then the first question that comes to my mind is that since it is hypercube what should I select my next value of N. Say if I take N=12 then the question is which side of the hypercube I should add the four nodes. If anybody could help solving it I will appreciate. In exercise 7 I cannot find solid reasons (qualitative) to prove the given facts. Thanks, Shahid. From shahid@er3.rutgers.edu Sun Nov 2 16:19:12 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA05612; Sun, 2 Nov 97 16:19:11 EST Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id NAA08889 for ; Mon, 3 Nov 1997 13:18:48 -0500 (EST) Date: Mon, 3 Nov 1997 13:18:47 -0500 (EST) From: Shahid Kagal To: crose@MOGLI.rutgers.edu Subject: Problems: (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO ---------- Forwarded message ---------- Date: Mon, 3 Nov 1997 11:30:27 -0500 (EST) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Subject: Problems: Hi everyone. I have doubts regarding the following question: Why isn't a two stage superconcentrator a superconcentrator? I started on the track that in case of a superconcentrator the outputs are neither compact nor row majored. Hence any subset of the outputs can appear at the output of the first stage. But now I am not able to think about it further. Any suggestions altogether different will be welcome. Shahid From shahid@er3.rutgers.edu Sun Nov 2 16:27:42 1997 Return-Path: Received: from er3.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA05616; Sun, 2 Nov 97 16:27:41 EST Received: from localhost (shahid@localhost) by er3.rutgers.edu (8.8.5/8.8.5) with SMTP id NAA10572; Mon, 3 Nov 1997 13:27:17 -0500 (EST) Date: Mon, 3 Nov 1997 13:27:17 -0500 (EST) From: Shahid Kagal To: 330_543@MOGLI.rutgers.edu Cc: crose@MOGLI.rutgers.edu Subject: Chapter 4: Complexity of NxN multipoint Pippenger Network. Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Page 107: There are 4 banyan networks each with.... How do we deduce ( and I think that has been done by induction) that the total complexity is 1/2(N)(logN)^2 from the previous argument . Page :118 Paull's connection matrix . In case of multirate connections the entries of Paul Matrix are fractional as long as the row and column sum is equal to one. How? Can anyone clarify because I cannot even get started on this statement. From alap@winlab.rutgers.edu Sun Nov 2 19:15:49 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA05850; Sun, 2 Nov 97 19:15:35 EST Received: from monk-a-asy-5.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA10148; Mon, 3 Nov 1997 16:15:07 -0500 Message-Id: <345E660E.6EE8@winlab.rutgers.edu> Date: Mon, 03 Nov 1997 16:02:22 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: John Sucec Cc: 330_543@mogli Subject: Three-phase algorithm and staggering References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO I think I have a better idea now how the phasing works. Please reffer to Figure 14 in chapter 6. Phase 1 is easy , just purge lower Phase 2: use feedback lines (that are fixed - from k to k) to send packets with the addresses of inputs (i) that will be able to transmit (in this case i= 2,3,4). The inputs that receive these packets (in this case we have k = 1,2 and 4) will send an ack packet (forward) addressed to 2,3 and 4. At this point we have the connection: (k,i) = (1,2), (2,4) and (4,3). AT THIS POINT I BELIEVE THERE IS SOMETHING MISSING BECAUSE THE INPUT OF SORTING NET SHOULD BE ABLE TO LISTEN TO THE OUTPUT OF BANYAN. I think there should be a fixed connection between them also. Actually in the Figure it is written: "input paired". With the fixed conections the inputs can have listen to the ack without the back propagation. After that comes phase 3, wich is just to send the data. I believe this is a very complicated algorithm, and the reason we want to use (in opposed to the simple back propagation of acknoledgement in page 157 - fig 13) is because we are able to stagger the phases (see problem 8 in this chapter). John Sucec wrote: > > Ana, this is my take on PHASE II of Figure 14... > > * k represents the output of the Sort network and in the input to the > parallel Acknowlegment Sort network (used for the Acknowledgement > packets). > > * i represents the source of the overall network and the destination of > the Acknowledgement Sort network. The address of i must be included in > the header of the request packet in order for the Acknowledgement packet > to be properly addressed. > > * j_i represents the destination of the overall network requested by > input i. > > I hope this helps. Actually, I too have a question on this same figure. > What is the purpose of the "feedback-like" lines drawn between the Sorting > network and Banyan network of the PHASE II parallel Acknowledgement > Sort-Banyan network? As far as I know, no feedback or acknowledgement is > required because the Acknowledgement packets are guarenteed to be > non-blocking due to the purging algorithm which takes place at the end of > PHASE I. Anyone else have thoughts on this ??? > > ...John > From alap@winlab.rutgers.edu Sun Nov 2 19:15:51 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA05853; Sun, 2 Nov 97 19:15:38 EST Received: from monk-a-asy-5.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA10157; Mon, 3 Nov 1997 16:15:11 -0500 Message-Id: <345E6836.1676@winlab.rutgers.edu> Date: Mon, 03 Nov 1997 16:11:34 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli Subject: Question in last exam Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Prof Rose (or who can help): I am trying to understand Problem 2 in last exam (blocking in TDM) and I don't undersatnd that formula. I accept the binomial expression and I also accept that there should have a term to compensate for the repetitions. But I tried the very simple case where C=1, N=2, p=1/2 and the formula gives Pb= 1/32. But if I number the outputs as "1" and "0" I will have Pb = P(two packets for "1") + P(two packets for "0") and this is 1/2*1/2 + 1/2*1/2 = 1/8. Am I missing something? Thank you very much. Ana. From alap@winlab.rutgers.edu Sun Nov 2 19:44:22 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA05874; Sun, 2 Nov 97 19:44:01 EST Received: from monk-a-asy-5.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA10774; Mon, 3 Nov 1997 16:43:33 -0500 Message-Id: <345E6F5F.130C@winlab.rutgers.edu> Date: Mon, 03 Nov 1997 16:42:07 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: Shahid Kagal Cc: 330_543@mogli Subject: Re: Staggering the Phases References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Shahid Kagal wrote: > > I seem to figure out no idea of how to stagger the phases of the 3 phase > algorithm in order to save the overhead time of request and > acknowledgement. Could anybody who has been able to solve the problem help > me . Thanks. > > Shahid. Shahid - Take a look in Problem 8 of chapter 6 , I think that will answer your question. Regards, Ana. From jsucec@ece.rutgers.edu Sun Nov 2 22:02:38 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06014; Sun, 2 Nov 97 22:02:27 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id SAA05892; Mon, 3 Nov 1997 18:57:17 -0500 Date: Mon, 3 Nov 1997 18:57:16 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Three-phase algorithm and staggering In-Reply-To: <345E660E.6EE8@winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, I think I agree with you... Instead of using a totally seperate Sort-Banyon network return acknowledgement packets, we use the same Sort-Banyan network for a 3 phases (not two physically seperate parallel networks as I had assumed). This makes makes the diagram perfectly clear in my mind now. Thanks. ...John On Mon, 3 Nov 1997, Ana Lucia Pinheiro wrote: > I think I have a better idea now how the phasing works. > Please reffer to Figure 14 in chapter 6. > > Phase 1 is easy , just purge lower > > Phase 2: use feedback lines (that are fixed - from k to k) to send packets with the > addresses of inputs (i) that will be able to transmit (in this case i= 2,3,4). > The inputs that receive these packets (in this case we have k = 1,2 and 4) > will send an ack packet (forward) addressed to 2,3 and 4. At this point we have the > connection: (k,i) = (1,2), (2,4) and (4,3). > AT THIS POINT I BELIEVE THERE IS SOMETHING MISSING BECAUSE THE INPUT OF SORTING NET > SHOULD BE ABLE TO LISTEN TO THE OUTPUT OF BANYAN. I think there should be a fixed > connection between them also. Actually in the Figure it is written: "input paired". > With the fixed conections the inputs can have listen to the ack without the back > propagation. > > After that comes phase 3, wich is just to send the data. > > I believe this is a very complicated algorithm, and the reason we want to use > (in opposed to the simple back propagation of acknoledgement in page 157 - fig 13) > is because we are able to stagger the phases (see problem 8 in this chapter) From crose Sun Nov 2 22:10:42 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06018; Sun, 2 Nov 97 22:10:41 EST Date: Sun, 2 Nov 97 22:10:41 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711030310.AA06018@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Question in last exam Cc: crose@MOGLI.rutgers.edu Status: RO HUH? I get 1/8 just like you do..... cof C-1 \frac{(1-\frac{p}{N})^N + p - 1}{p} = 1/8 when p 1/2 and N = 2 The result is the same when you evaluate the expression for general C as well. Cheers, From alap@winlab.rutgers.edu Mon Nov 3 20:30:19 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06100; Mon, 3 Nov 97 20:30:07 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id UAA12975; Mon, 3 Nov 1997 20:30:37 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id UAA04210; Mon, 3 Nov 1997 20:30:37 -0500 Date: Mon, 3 Nov 1997 20:30:37 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040130.UAA04210@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Still question in last exam Status: RO Prof Rose, thanks for the reply. What I have for N=2, p=1/2 and C=1 is: the sum is going to have only one term (m=2) , then: Pb= (1/2) (p/N)^2 = (1/2) (1/4)^2 = (1/2)(1/16) = 1/32 (??) In the case I considered myself in last e-mail (Pb=1/8) I had a typo, because I would have: Prob (two "0") + Prob(two "1") = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 1/2 So I am comparing my result (1/2) with your result (1/32) I may be doing something wrong, but I can't see... :( I really appreciate if someone could help me. Thanks, Ana. From crose Mon Nov 3 22:07:30 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06193; Mon, 3 Nov 97 22:07:09 EST Date: Mon, 3 Nov 97 22:07:09 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711040307.AA06193@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Still question in last exam Cc: crose@MOGLI.rutgers.edu Status: RO Ana, What expression are you using? I think you're MULTIPLYING by p instead of DIVIDING by p (mean number of blocks over mean number of arrivals per output rail (Cp)). Cheers, Chris Rose From crose Mon Nov 3 22:12:46 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06213; Mon, 3 Nov 97 22:12:22 EST Date: Mon, 3 Nov 97 22:12:22 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711040312.AA06213@MOGLI.rutgers.edu> To: 330_543 Subject: HEY!!! Cc: crose@MOGLI.rutgers.edu Status: RO Why do I keep seeing the same old names on all the mailings!!!!!!!!!!!! Folks, let's get the mail group cooking! I've only seen three or four regular contributors! (John, Ana and Shahid... with a sometimes blibbit from George, Christine and I seem to remember either Joe Papa or Eldar). From alap@winlab.rutgers.edu Mon Nov 3 22:22:42 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06222; Mon, 3 Nov 97 22:22:33 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id WAA13525; Mon, 3 Nov 1997 22:23:03 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id WAA04358; Mon, 3 Nov 1997 22:23:02 -0500 Date: Mon, 3 Nov 1997 22:23:02 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040323.WAA04358@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: Still question in last exam Status: RO Prof Rose, I am using the formula that is the Exam solution in the course web page for question 2, item (a), where p = prob of receiveing a packet , N is the number of inputs/outputs and C in this case (item a) is 1. The formula is a sum from m to N, in our case (N=2, p=1/2) the sum is only one term with m=2. I am sorry for asking about this problem so many times but I really would like to understand where is my mistake. Thanks again! Ana. From alap@winlab.rutgers.edu Mon Nov 3 22:53:34 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06241; Mon, 3 Nov 97 22:53:18 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id WAA13944; Mon, 3 Nov 1997 22:53:47 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id WAA04391; Mon, 3 Nov 1997 22:53:47 -0500 Date: Mon, 3 Nov 1997 22:53:47 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040353.WAA04391@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Shahid Questions Status: RO Shahid: About Pippenger, I am also confused. We will use a banyan as a N/2xN/2 concentrator, and I think in this case we need 2, one for the top and another for the botton. How come 4, 8, etc appear there ?? At leats do you understand this part ?? Page 118 Paull's Matrix: The fractional part is due to the fact that you are working with frames. Before you had only ONE link from each first stage switch to each second stage switch. NOW we still have ONLY one link, but each link can carry , say, N slots. so you can have connections to at most N differnt outputs using the same link (you are using Time Divisiion Multiplex). That's why the fractions take place. (for example the middle switch A could appear N times as A/N). Hope this helps. :) Could someone give a help in the Pippenger Net (page 107) ? THANKS! Regards, Ana. From crose Mon Nov 3 23:21:58 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06277; Mon, 3 Nov 97 23:21:38 EST Date: Mon, 3 Nov 97 23:21:38 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711040421.AA06277@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Re: Still question in last exam Cc: 330_543 Status: RO (1/p)(p/N)^2 (1-p/N)^0 = (2)(1/4)^2 (1) = 2*(1/16) = 1/8 From jsucec@ece.rutgers.edu Mon Nov 3 23:23:14 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06283; Mon, 3 Nov 97 23:23:04 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id XAA20265; Mon, 3 Nov 1997 23:18:49 -0500 Date: Mon, 3 Nov 1997 23:18:49 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Still question in last exam In-Reply-To: <199711040130.UAA04210@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, I think I see where you may be getting a little confused. Unfortunately, I can not extend your approach to yield Pb=1/32, only to Pb=1/16. In the example you provide below where you obtain a Pb=1/2, you have assumed that both of the N=2 inputs are active. Given an a priori of 2 active inputs, then yes, the probability of ONE of these 2 packets being blocked is 1/2. But, the probability that input i (i=1 or 2) is blocked is (m-1)/m = 1/2 times the probability of blocking --> This means probability input i being blocked GIVEN m=2, is 1/4 = p^2. Further, the problem statement has no a priori. Therefore, in order to use your approach for calculating Pb with N=2, we must first calculate the probability that we have 2 simultaneous active inputs... This is p^2. Multiplying the probibility of input i being blocked given m=2 by the probability of m=2, we get p^2*p^2 = p^4 = 1/16. How we get 1/32 from this point, as the exam formula states, escapes me! Maybe there is something wrong with the exam solution formula. More likely, though, I'm just getting sleepy and missing some basic probability principle here in my analysis. Perhaps someone else can identify another probability principle accounts for the missing 1/2 factor. ...John On Mon, 3 Nov 1997, Ana Lucia Pinheiro wrote: > > Prof Rose, thanks for the reply. > What I have for N=2, p=1/2 and C=1 is: > the sum is going to have only one term (m=2) , then: > Pb= (1/2) (p/N)^2 = (1/2) (1/4)^2 = (1/2)(1/16) = 1/32 (??) > > In the case I considered myself in last e-mail (Pb=1/8) > I had a typo, because I would have: > Prob (two "0") + Prob(two "1") = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 1/2 > > So I am comparing my result (1/2) with your result (1/32) > I may be doing something wrong, but I can't see... :( > > I really appreciate if someone could help me. > Thanks, > Ana. > From jasonjen@ece.rutgers.edu Tue Nov 4 01:04:33 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06404; Tue, 4 Nov 97 01:04:15 EST Received: from localhost (jasonjen@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA26810 for <330_543@mogli.rutgers.edu>; Tue, 4 Nov 1997 01:00:01 -0500 Date: Tue, 4 Nov 1997 01:00:00 -0500 (EST) From: Jason Jen To: 330_543@mogli.rutgers.edu Subject: pippenger network In-Reply-To: <199711040353.WAA04391@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO For pippenger network, I think the N/2*N/2 network is for the first stage concentrator. In this case it requires 4 N/2*N/2 banyan networks. Now, if recursively break down the one of the sencod stage multi-pt switches (N/2*N/2) to 2 more N/2*N/4 concentrators then each N/2*N/4 concentartor can be factor into another 2 N/4*N/4 banyan networks..and so on... so, the complexity would be 4*(N/4)log(N/2)+8*(N/8)log(N/4)+16*(N/16)log(N/8)+.....up to the last 2x2 banyan network. I didn't have the time to find out the exact value of this summation but I believe the sum should be very close to 1/2*N(logN)^2! Hope my explaination about the pippenger network is correct! J a s o n J e n ------------------------------ Electrical Engineering Office: EE 214 Tel: (732) 445-5385 ------------------------------ On Mon, 3 Nov 1997, Ana Lucia Pinheiro wrote: > > Shahid: > > About Pippenger, I am also confused. We will use a banyan as a N/2xN/2 > concentrator, and I think in this case we need 2, one for the top and another > for the botton. How come 4, 8, etc appear there ?? At leats do you understand > this part ?? > > Page 118 Paull's Matrix: > The fractional part is due to the fact that you are working with frames. > Before you had only ONE link from each first stage switch to each second > stage switch. NOW we still have ONLY one link, but each link can carry , say, > N slots. so you can have connections to at most N differnt outputs using > the same link (you are using Time Divisiion Multiplex). That's why the fractions > take place. (for example the middle switch A could appear N times as A/N). > Hope this helps. :) > > Could someone give a help in the Pippenger Net (page 107) ? THANKS! > > Regards, > Ana. > From jsucec@ece.rutgers.edu Tue Nov 4 01:17:29 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06413; Tue, 4 Nov 97 01:17:21 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA27916 for <330_543@MOGLI.rutgers.edu>; Tue, 4 Nov 1997 01:13:07 -0500 Date: Tue, 4 Nov 1997 01:13:07 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Re: Still question in last exam (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Actually, I have convinced myself that the probability of blocking (Pb) for a given input i, with p=1/2 and N=2, is in fact 1/16 as described below. A simple coin flipping experiment with 2 pairs of coins (i.e. 16 total states) can be used to verify this. I go through the details of this coin flipping experiment, below (in case anyone really cares), but the point I would like to make is that I think the expression given as the solution for problem 2a of EXAM 1 actually represents the probability that a packet at input i desiring to reach a specific output j, is blocked. That is, it represents the probability of a specific input-->output PAIR. This particular condition, however, occurs only probability 1/N... Since there are N possible outputs, the probability that input i wants to reach output j is only 1/N. Therefore, the solution given for EXAM 1 P2a must be multiplied by N to obtain actual probability that a packet at input rail i is blocked. COIN FLIPPING EXAMPLE: * 2 pairs of coins (4 coins total) * Pair 1 (2 quarters say) represents desired I/O pattern at input rail 1 * Pair 2 (2 nickels say) represents desired I/O pattern at input rail 2 * Coin pair (I/O) states: - Pair state TT-->input i to output 1 - Pair state HH-->input i to output 2 - HT ~ input i inactive - TH ~input i inactive. STATE COIN PAIR 1|COIN PAIR 2 ----- ----------- ----------- 1 TT | TT <-- Blocking State (both inputs request output 1) 2 TT | TH 3 TT | HT 4 TT | HH 5 TH | TT 6 TH | TH 7 TH | HT 8 TH | HH 9 HT | TT 10 HT | TH 11 HT | HT 12 HT | HH 13 HH | TT 14 HH | TH 15 HH | HT 16 HH | HH <-- Blocking State (both inputs request output 2) Clearly, there are 2 blocking states out of a total of 16 possible coin (i.e. switch) states. Therefore, assuming random purge probabilty, a packet at input rail i, has probability of 1/16 of being blocked. This is twice the value indicated by the solution of Exam 1 P2a. I think the solution statement should be ammended to indicate more clearly that it represents only the probability of blocking event of a single input-->output PAIR. The actual total probability of blocking for a given input packet is "N times" this. ...John ---------- Forwarded message ---------- Date: Mon, 3 Nov 1997 23:18:49 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Still question in last exam Ana, I think I see where you may be getting a little confused. Unfortunately, I can not extend your approach to yield Pb=1/32, only to Pb=1/16. In the example you provide below where you obtain a Pb=1/2, you have assumed that both of the N=2 inputs are active. Given an a priori of 2 active inputs, then yes, the probability of ONE of these 2 packets being blocked is 1/2. But, the probability that input i (i=1 or 2) is blocked is (m-1)/m = 1/2 times the probability of blocking --> This means probability input i being blocked GIVEN m=2, is 1/4 = p^2. Further, the problem statement has no a priori. Therefore, in order to use your approach for calculating Pb with N=2, we must first calculate the probability that we have 2 simultaneous active inputs... This is p^2. Multiplying the probibility of input i being blocked given m=2 by the probability of m=2, we get p^2*p^2 = p^4 = 1/16. How we get 1/32 from this point, as the exam formula states, escapes me! Maybe there is something wrong with the exam solution formula. More likely, though, I'm just getting sleepy and missing some basic probability principle here in my analysis. Perhaps someone else can identify another probability principle accounts for the missing 1/2 factor. ...John On Mon, 3 Nov 1997, Ana Lucia Pinheiro wrote: > > Prof Rose, thanks for the reply. > What I have for N=2, p=1/2 and C=1 is: > the sum is going to have only one term (m=2) , then: > Pb= (1/2) (p/N)^2 = (1/2) (1/4)^2 = (1/2)(1/16) = 1/32 (??) > > In the case I considered myself in last e-mail (Pb=1/8) > I had a typo, because I would have: > Prob (two "0") + Prob(two "1") = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 1/2 > > So I am comparing my result (1/2) with your result (1/32) > I may be doing something wrong, but I can't see... :( > > I really appreciate if someone could help me. > Thanks, > Ana. > From alap@winlab.rutgers.edu Tue Nov 4 01:19:17 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06418; Tue, 4 Nov 97 01:19:06 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA14699; Tue, 4 Nov 1997 01:19:34 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA04700; Tue, 4 Nov 1997 01:19:34 -0500 Date: Tue, 4 Nov 1997 01:19:34 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040619.BAA04700@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Pippenger Net Status: RO Jason - As given in class and also from the book (page 98 - Figure 7) a Banyan net NxN can be factored in (N/2)logN switches (2x2 switches) The complexity is given by the number of 2x2 switches. If we have 4 banyans, isn't the complexity just 4(N/2)logN ?? I mean, why do we NEED the recursive formula if we already know the answer for 1 single banyan ?? Maybe I am missing something... By the way: THANKS for the reply, specilly given the time !!!! Ana. From alap@winlab.rutgers.edu Tue Nov 4 01:33:18 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06422; Tue, 4 Nov 97 01:33:06 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA14724; Tue, 4 Nov 1997 01:33:34 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA04708; Tue, 4 Nov 1997 01:33:34 -0500 Date: Tue, 4 Nov 1997 01:33:34 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040633.BAA04708@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Coins... Status: RO John, thanks. I agree with your example , but isn't 2 blocking states out of 16 possible states going to give 2/16=1/8 probability of blocking ??? If so, then it agrees with my thinking that: P(blocking) = P(two destinations = "0") + P(two dest = "1") = [P(active)*P("0")]^2 + [P(active)*P("1")]^2 = (1/4)^2 + (1/4)^2 = 1/8 which would match with your 2/16 (if that was the case and not 1/16 in your last mail. Thanks again!! -Ana. From jasonjen@ece.rutgers.edu Tue Nov 4 01:33:50 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06427; Tue, 4 Nov 97 01:33:37 EST Received: from localhost (jasonjen@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA28322; Tue, 4 Nov 1997 01:29:22 -0500 Date: Tue, 4 Nov 1997 01:29:21 -0500 (EST) From: Jason Jen To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Pippenger Net In-Reply-To: <199711040619.BAA04700@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Let take a look at NxN/2 concentrator. Since we know it can be constructed with 2 N/2*N/2 banyan networks (pg 106 last paragraph). And, each N/2*N/2 banyan network has log(N/2) stages plus each stage has (N/2)/2 delta switches. So the number of delta switches should be (N/4)log(N/2) for one N/2*N/2 banyan network. In first stage we have two N*N/2 concentrators so it can be constructed with 4 banyan (N/2*N/2) networks. So the total number of delta switches should be 4*(N/4)log(N/2). J a s o n J e n ------------------------------ Electrical Engineering Office: EE 214 Tel: (732) 445-5385 ------------------------------ On Tue, 4 Nov 1997, Ana Lucia Pinheiro wrote: > > Jason - > As given in class and also from the book (page 98 - Figure 7) > a Banyan net NxN can be factored in (N/2)logN switches (2x2 switches) > The complexity is given by the number of 2x2 switches. > If we have 4 banyans, isn't the complexity just 4(N/2)logN ?? > I mean, why do we NEED the recursive formula if we already know the answer > for 1 single banyan ?? Maybe I am missing something... > > By the way: THANKS for the reply, specilly given the time !!!! > Ana. > From alap@winlab.rutgers.edu Tue Nov 4 01:40:42 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06441; Tue, 4 Nov 97 01:40:30 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA14765; Tue, 4 Nov 1997 01:40:58 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA04713; Tue, 4 Nov 1997 01:40:58 -0500 Date: Tue, 4 Nov 1997 01:40:58 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040640.BAA04713@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Still Pippenger Status: RO Jason, I agree with you about the first stage (actually I saw that before , just wrote it wrong in the last e-mail). WHAT I DO NOT SEE is where does 8, 16, etc comes from (the recursion). They come from the second stage ?? Thanks again. Ana. From jasonjen@ece.rutgers.edu Tue Nov 4 01:46:41 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA06470; Tue, 4 Nov 97 01:46:32 EST Received: from localhost (jasonjen@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA28550; Tue, 4 Nov 1997 01:42:17 -0500 Date: Tue, 4 Nov 1997 01:42:16 -0500 (EST) From: Jason Jen To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Still Pippenger In-Reply-To: <199711040640.BAA04713@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO At second stage, the N/2*N/2 multi-pt switch can be further break down to 1x2 demuxs and N/2*N/4 concentrators. Right behind each concentrator is a N/4*N/4 banyan network. Hope this will help! J a s o n J e n ------------------------------ Electrical Engineering Office: EE 214 Tel: (732) 445-5385 ------------------------------ On Tue, 4 Nov 1997, Ana Lucia Pinheiro wrote: > > Jason, I agree with you about the first stage (actually I saw that > before , just wrote it wrong in the last e-mail). WHAT I DO NOT SEE > is where does 8, 16, etc comes from (the recursion). They come from the > second stage ?? > Thanks again. > Ana. > From alap@winlab.rutgers.edu Tue Nov 4 01:55:59 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA06475; Tue, 4 Nov 97 01:55:47 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA14796; Tue, 4 Nov 1997 01:56:14 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA04721; Tue, 4 Nov 1997 01:56:14 -0500 Date: Tue, 4 Nov 1997 01:56:14 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711040656.BAA04721@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Pippenger Status: RO Jason - that helped a lot !!!!! I see now !!! Thanks once more. -Ana. From jpapa@mail.monmouth.com Tue Nov 4 09:55:05 1997 Return-Path: Received: from shell.monmouth.com by MOGLI.rutgers.edu (4.1/25-eef) id AA06831; Tue, 4 Nov 97 09:54:43 EST Received: from rb-tc-ppp58.monmouth.com (rb-tc-ppp58.monmouth.com [208.7.184.142]) by shell.monmouth.com (8.8.5/8.7.3) with SMTP id JAA17906; Tue, 4 Nov 1997 09:50:52 -0500 (EST) Received: by rb-tc-ppp58.monmouth.com with Microsoft Mail id <01BCE8FF.97FBD720@rb-tc-ppp58.monmouth.com>; Tue, 4 Nov 1997 08:56:52 -0500 Message-Id: <01BCE8FF.97FBD720@rb-tc-ppp58.monmouth.com> From: Joseph Papa To: "'Christopher Rose'" , "330_543@MOGLI.rutgers.edu" <330_543@MOGLI.rutgers.edu> Subject: Notes?? Date: Tue, 4 Nov 1997 08:56:49 -0500 Return-Receipt-To: Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Status: RO What's the policy on bringing sheets of notes?? 1 or 2 sheets? From crose Tue Nov 4 10:07:19 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA06856; Tue, 4 Nov 97 10:07:00 EST Date: Tue, 4 Nov 97 10:07:00 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711041507.AA06856@MOGLI.rutgers.edu> To: 330_543@MOGLI.rutgers.edu, crose@MOGLI.rutgers.edu, jpapa@mail.monmouth.com Subject: Re: Notes?? Status: RO TWO GLORIOUS SHEETS!!!!!!!!!!!!!!!!!!!!!!!!!!! BOTH SIDES!!!!!!!! From jsucec@ece.rutgers.edu Tue Nov 4 13:23:45 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA07277; Tue, 4 Nov 97 13:22:21 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id NAA10834; Tue, 4 Nov 1997 13:17:59 -0500 Date: Tue, 4 Nov 1997 13:17:59 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Coins... In-Reply-To: <199711040633.BAA04708@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, 1/8 represents the probability of blocking of either of the two input packets. 1/16 represents the probability that a particular packet is blocked (because even in the event that blocking occurs in our 2x2 network, only 1 of the 2 packets will be blocked). I'm glad someone actually bothered to read my note! Thanks. ...John On Tue, 4 Nov 1997, Ana Lucia Pinheiro wrote: > > John, thanks. > I agree with your example , but isn't 2 blocking states out of 16 possible states > going to give 2/16=1/8 probability of blocking ??? > > If so, then it agrees with my thinking that: > P(blocking) = P(two destinations = "0") + P(two dest = "1") > = [P(active)*P("0")]^2 + [P(active)*P("1")]^2 = (1/4)^2 + (1/4)^2 = 1/8 > > which would match with your 2/16 (if that was the case and not 1/16 in your last > mail. > > Thanks again!! > -Ana. > > From chriskly@er5.rutgers.edu Tue Nov 4 15:18:47 1997 Return-Path: Received: from er5.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA07372; Tue, 4 Nov 97 15:18:39 EST Received: (from chriskly@localhost) by er5.rutgers.edu (8.8.5/8.8.5) id PAA21025; Tue, 4 Nov 1997 15:18:59 -0500 (EST) Date: Tue, 4 Nov 97 15:18:58 EST From: Christine Kleiwerda To: John Sucec Cc: Ana Lucia Pinheiro , 330_543@MOGLI.rutgers.edu Subject: Re: Coins... In-Reply-To: Your message of Tue, 4 Nov 1997 13:17:59 -0500 (EST) Message-Id: Status: RO > Ana, 1/8 represents the probability of blocking of either of the two input > packets. 1/16 represents the probability that a particular packet is > blocked (because even in the event that blocking occurs in our 2x2 > network, only 1 of the 2 packets will be blocked). > > I'm glad someone actually bothered to read my note! Thanks. > > ...John > > On Tue, 4 Nov 1997, Ana Lucia Pinheiro wrote: > > > > > John, thanks. > > I agree with your example , but isn't 2 blocking states out of 16 possible states > > going to give 2/16=1/8 probability of blocking ??? > > > > If so, then it agrees with my thinking that: > > P(blocking) = P(two destinations = "0") + P(two dest = "1") > > = [P(active)*P("0")]^2 + [P(active)*P("1")]^2 = (1/4)^2 + (1/4)^2 = 1/8 > > > > which would match with your 2/16 (if that was the case and not 1/16 in your last > > mail. > > > > Thanks again!! > > -Ana. > > > > > > John- I am not sure if this si what you are saying, but I wanted to say for clarification on the blocking problem, that I do not believe that you can say that if hte blocking on one output is p and there are N outputs, then the total blockign is Np. I tried this on teh exam, and later was convinced that my argument was incorrect. The reason goes something like: you cannot assume that the blockign on all links is independent. What happens on one link could in fact affect the others. We must look a thte long run averages of the total number of packets that get blocked adn divide by the total number of packets that arrive. Hope this helped and made sense (and is correct)...... Christine From crose Tue Nov 4 16:34:31 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA07460; Tue, 4 Nov 97 16:34:09 EST Date: Tue, 4 Nov 97 16:34:09 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711042134.AA07460@MOGLI.rutgers.edu> To: chriskly@eden.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Coins... Cc: 330_543@MOGLI.rutgers.edu, alap@liman.Rutgers.EDU Status: RO Excellent argument Christine and completely correct. From jsucec@ece.rutgers.edu Tue Nov 4 22:42:26 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA07878; Tue, 4 Nov 97 22:42:12 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id WAA28134; Tue, 4 Nov 1997 22:37:44 -0500 Date: Tue, 4 Nov 1997 22:37:44 -0500 (EST) From: John Sucec To: Christine Kleiwerda Cc: Ana Lucia Pinheiro , 330_543@MOGLI.rutgers.edu Subject: Re: Coins... In-Reply-To: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Christine, what you say makes sense. I still think, however, that based on the coin tossing example I outlined, that the Exam 1 P2a solution statement should be either modified or ammended because the solution does not appear to yield the proper result for the N=2, p=1/2, C=1 network Ana proposed. As the coin tossing example "shows", 2 out of the 16 states which are possible in Ana's network are blocking, for an overall network blocking event probability of 1/8. Further, since the distribution of the blocking event is divided with equal weight between the 2 possible input packets, the probability that a particular packet gets blocked is 1/16. Neither of these appear to be predicted by the Exam 1 P2a solution as it is currently worded. Anyway, maybe I'm wrong here, but the exam solution really does not appear to satisfy the network Ana proposed. Perhaps a correction notice for the P2a solution was made already by Dr. Rose during the exam review session we had in class. I'll check my notes. ...John On Tue, 4 Nov 1997, Christine Kleiwerda wrote: > > Ana, 1/8 represents the probability of blocking of either of the two input > > packets. 1/16 represents the probability that a particular packet is > > blocked (because even in the event that blocking occurs in our 2x2 > > network, only 1 of the 2 packets will be blocked). > > > > I'm glad someone actually bothered to read my note! Thanks. > > > > ...John > > > > On Tue, 4 Nov 1997, Ana Lucia Pinheiro wrote: > > > > > > > > John, thanks. > > > I agree with your example , but isn't 2 blocking states out of 16 possible states > > > going to give 2/16=1/8 probability of blocking ??? > > > > > > If so, then it agrees with my thinking that: > > > P(blocking) = P(two destinations = "0") + P(two dest = "1") > > > = [P(active)*P("0")]^2 + [P(active)*P("1")]^2 = (1/4)^2 + (1/4)^2 = 1/8 > > > > > > which would match with your 2/16 (if that was the case and not 1/16 in your last > > > mail. > > > > > > Thanks again!! > > > -Ana. > > > > > > > > > > > > > John- > > I am not sure if this si what you are saying, but I wanted to say for > clarification on the blocking problem, that I do not believe that you can say > that if hte blocking on one output is p and there are N outputs, then the > total blockign is Np. I tried this on teh exam, and later was convinced that > my argument was incorrect. The reason goes something like: you cannot > assume that the blockign on all links is independent. What happens on one > link could in fact affect the others. We must look a thte long run averages > of the total number of packets that get blocked adn divide by the total > number of packets that arrive. > > Hope this helped and made sense (and is correct)...... > > Christine > From crose Tue Nov 4 23:10:33 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA07889; Tue, 4 Nov 97 23:10:11 EST Date: Tue, 4 Nov 97 23:10:11 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711050410.AA07889@MOGLI.rutgers.edu> To: chriskly@eden.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Coins... Cc: 330_543@MOGLI.rutgers.edu, alap@liman.Rutgers.EDUr330_543@mogli.rutgers.edu Status: RO HI Folks, ABOUT THE BLOCKING PROBABILITY: There's an easy way for you to see if the solution for the first quiz (problem 2) is correct (it is) by simply writing a simulation and checking the statistical results agains the equations. However, from an analytical basis: The probability of blocking given you're arriving on a particular rail and leaving on a particular rail is the same as the overall blocking probability for uniform dispersal (1/N prob of exiting on a given rail, indep for all packets) and uniform arrivals to all rails (prob p of arrival in each slot indep of which rail). Let pb(i,j) be the probability of blocking for a particular input i and output j pair. The overall blocking probability is Pb = sum_{ij} P(i,j) pb(i,j) But the probaiblity of blocking is the same for any i,j pair so pb(i,j) = q. THerefore Pb= q But the question now is "what is q" Prob there is another arrival to MY output rail (I'm a packet which is trying to get out) is (1/2 --- arrival prob on other input rail)(1/2 prob other packet goes to MY output rail). Then there's prob 1/2 that I AM THE ONE that's blocked ---> 1/8. Now, what about the other guy in this particular example? Pb is defined per packet. It is not defined for an ensemble of packets in the same frame. That is Pb for a particular packet GIVEN THAT ANOTHER PACKET WAS BLOCKED OR GOT THROUGH in the same frame is not the probability of blocking. Hope this lays the problem to rest. The answer 1/8 is obtained no matter how you look at the problem for p = 1/2, C = 1 and N=2. That is the expressions in the solutions evaluate to 1/8 AND 1/8 is the correct answer. Cheers, Chris Rose PS: How did you all like the test! I tried to make it interesting an fun. This was probably harder than the last exam though :( From jsucec@ece.rutgers.edu Wed Nov 5 22:12:21 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA09216; Wed, 5 Nov 97 22:12:17 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id WAA28135; Wed, 5 Nov 1997 22:07:42 -0500 Date: Wed, 5 Nov 1997 22:07:42 -0500 (EST) From: John Sucec To: Christopher Rose Cc: alap@liman.rutgers.edu, chriskly@eden.rutgers.edu Subject: Re: Coins... In-Reply-To: <9711050410.AA07889@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, thanks for taking time out to resolve this issue. Thanks also to Ana and Christine for their insightful comments. (Since the 3 of you have resolved this problem with some creative thinking, and my contribution has been simply to poke holes in the solution posted on the web page, I'll do the rest of the class a favor by keeping this E-mail confined to the 4 of us.) As I recalled last night, you did correct the solution for this problem during the review session of the exam you held in class back on 16-Oct. I'm checking my notes, right now, and I see that you wrote on the board during this review, the following: "Pb = (mean # packets blocked)/(mean # arrivals)" This makes sense to me and does yield the correct blocking probabilities for the N=2, p=1/2 and C=1 network. Anyway, my seemingly excessive E-Mails were written because the equation provided in the web page solution statement for P2a does not represent the probability of blocking, but rather, only the mean number of packets blocked. As you pointed out in class, this term must be divided by the mean number of arrivals in order to obtain Pb. Now, remember when you and Ana went through the 2x2 network example the first time around? We plugged N=2 and p=1/2 into the web page solution equation and got 1/32! This is because the web page equation for Pb actually represents the mean number of packets blocked. Dividing 1/32 by the mean number of arrivals at an output rail then correctly yields the probability of a given packet being blocked in Ana's network. Anyway, I am sending this E-Mail primarily to assure you that I am singing from the same hymn notes as the rest of the class. Although the web page solution statement for P2a caused considerable confusion for me, your explaination given in class was more than adequate (I should have referred to it all along). Also, I appreciate you going through the detail of the 2x2 example below. I was there myself on Monday night, with my nickels and quarters simulation experiment! ...John P.S.: If you are not satisfied with the interpretation provided in this E-Mail, perhaps we can spend 2 or 3 minutes between classes tomorrow and attempt to clear any differences in intepretation of the web page solution for P2a. If this doesn't resolve any remaining issue, we can then just agree to disagree and move forward with other class topics! On Tue, 4 Nov 1997, Christopher Rose wrote: > HI Folks, > > ABOUT THE BLOCKING PROBABILITY: > > There's an easy way for you to see if the solution for the first quiz > (problem 2) is correct (it is) by simply writing a simulation and > checking the statistical results agains the equations. > > However, from an analytical basis: > > The probability of blocking given you're arriving on a particular > rail and leaving on a particular rail is the same as the overall > blocking probability for uniform dispersal (1/N prob of exiting > on a given rail, indep for all packets) and uniform arrivals > to all rails (prob p of arrival in each slot indep of which rail). > > Let pb(i,j) be the probability of blocking for a particular input i > and output j pair. The overall blocking probability is > > Pb = sum_{ij} P(i,j) pb(i,j) > > But the probaiblity of blocking is the same for any i,j pair > so pb(i,j) = q. THerefore > > Pb= q > > But the question now is "what is q" > > Prob there is another arrival to MY output rail (I'm a packet > which is trying to get out) is (1/2 --- arrival prob on other > input rail)(1/2 prob other packet goes to MY output rail). > Then there's prob 1/2 that I AM THE ONE that's blocked ---> 1/8. > > Now, what about the other guy in this particular example? Pb > is defined per packet. It is not defined for an ensemble > of packets in the same frame. That is Pb for a particular packet > GIVEN THAT ANOTHER PACKET WAS BLOCKED OR GOT THROUGH in the same > frame is not the probability of blocking. > > Hope this lays the problem to rest. The answer 1/8 is obtained > no matter how you look at the problem for p = 1/2, C = 1 and > N=2. That is the expressions in the solutions evaluate to 1/8 > AND 1/8 is the correct answer. > > > > Cheers, > > Chris Rose > > > PS: How did you all like the test! I tried to make it interesting > an fun. This was probably harder than the last exam though :( > > From JOHNSUCEC@delphi.com Thu Nov 6 02:56:57 1997 Return-Path: Received: from bos1d.delphi.com by MOGLI.rutgers.edu (4.1/25-eef) id AA09445; Thu, 6 Nov 97 02:56:56 EST Received: from delphi.com by delphi.com (PMDF V5.1-8 #23839) id <01IPO868JU6O8WWAQM@delphi.com> for crose@MOGLI.rutgers.edu; Thu, 6 Nov 1997 02:05:14 EST Date: Thu, 06 Nov 1997 02:05:13 -0500 (EST) From: JOHNSUCEC@delphi.com Subject: Possible alternate solution to Exam 1 P2a To: crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Message-Id: <01IPO868K3TU8WWAQM@delphi.com> X-Vms-To: INTERNET"crose@MOGLI.rutgers.edu" X-Vms-Cc: INTERNET"jsucec@ece.rutgers.edu" Mime-Version: 1.0 Content-Type: TEXT/PLAIN; CHARSET=US-ASCII Status: RO Prof. Rose, I a have tried to come up with my own formula for Exam 1 P2a. Following your explaination of the problem you gave in class: Pb = E[# of packets blocked] / E[# of packets arrived] I have gotten as far as deriving the expression for E[# of packets blocked]: Let: * E[ ] = Expected value operator * E[B] = Expected number of packets blocked * E[A] = Expected number of packets arriving at an output * C(N,m) = Combination operator, that is, "N choose m" * sum(m=i,j){ } = Summation over m = i to j of the expression { } Now, using the probability distribution, p_M(m), provided in the web page solution statement to represent the probability of m packets arriving at a given output we have: * p_M(m) = C(N,m)*(p/N)^m*(1-p/N)^(N-m) I agree with this distribution. Further, I agree with your expression for E[# packets blocked] = E[B]: * E[B] = sum(m=2,N){(m-1)*C(N,m)*(p/N)^m*(1-p/N)^(N-m) However, according to the statement made in class, Pb = E[B]/E[A], I don't think I can agree any further with the way the solution on the web page is worded. For the mean number of arrivals at an output, I get: * E[A] = p*N/N = p This yields the following expression for Pb: * Pb = E[B]/E[A] = (1/p)*sum(m=2,N){(m-1)*C(N,m)*(p/N)^m*(1-p/N)^(N-m) This is different from the solution given on the page, and it satifies the network N=2, p=1/2, C=1. ACTUALLY, THE E-MAIL YOU SENT OUT TO ANA LATE MONDAY NIGHT, APPEARS TO MATCH THIS EQUATION! Now, I am not yet myself convinced that this is correct... I tried plugging another simple 2x2 network into this equation (N=2, p=1/3, C=1). Same as Ana's example, except p=1/3, and I do not get the results expected from brute force analysis of the network... I get Pb=1/12 rather than Pb=1/18. However, the web page solution gets this example wrong also (web page solution yields 1/72). Maybe my brute force solution with p=1/3, is wrong. Actually, for whatever reason, if we disregard E[A] in our calculation of Pb, and simply multiply E[B] by N, we get the correct probability of blocking value for the 2 networks I have discussed (N=1,p=1/2,C=1) and (N=2,p=1/3,C=1). This would yield: Pb = N*sum(m=2,N){(m-1)*C(N,m)*(p/N)^m*(1-p/N)^(N-m). Unfortunately, I don't believe N fits in to the process, at this point. Honestly, I would never have spent so much time belaboring this problem if somehow the web page equation yielded the correct probability of blocking value for the network Ana devised. Unfortunately, the web page equation yields 1/32, and consequently, I have been banging my head with this problem. If it I can be shown to me somehow that the web page yields 1/8 for Ana's network, I probably would be a believer. ...John From crose Thu Nov 6 09:08:30 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA09657; Thu, 6 Nov 97 09:08:29 EST Date: Thu, 6 Nov 97 09:08:29 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711061408.AA09657@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Coins... Cc: alap@liman.rutgers.edu, chriskly@eden.rutgers.eduret Status: RO HI All, NO problem at all! HOpefully my email did not sound testy since it was not meant to be. I just did not want you guys barking up the wrong tree for the rest of the term :) Let's take a few minutes AFTER class tonite to chat about this. I'm still confused since when I looked at the web page, I saw exactly the mean blocks/mean arrivals solution. I don't remember changing it mid stream.... See ya tonite! CHeers, Chris Rose From crose Thu Nov 6 09:08:32 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AB09659; Thu, 6 Nov 97 09:08:29 EST Date: Thu, 6 Nov 97 09:08:29 EST From: Mail Delivery Subsystem Full-Name: Mail Delivery Subsystem Message-Id: <9711061408.AB09659@MOGLI.rutgers.edu> Subject: Returned mail: User unknown To: crose Status: RO ----- Transcript of session follows ----- 550 chriskly@eden.rutgers.eduret... Host unknown 550 log... User unknown ----- Unsent message follows ----- Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA09657; Thu, 6 Nov 97 09:08:29 EST Date: Thu, 6 Nov 97 09:08:29 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711061408.AA09657@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Coins... Cc: alap@liman.rutgers.edu, chriskly@eden.rutgers.eduret HI All, NO problem at all! HOpefully my email did not sound testy since it was not meant to be. I just did not want you guys barking up the wrong tree for the rest of the term :) Let's take a few minutes AFTER class tonite to chat about this. I'm still confused since when I looked at the web page, I saw exactly the mean blocks/mean arrivals solution. I don't remember changing it mid stream.... See ya tonite! CHeers, Chris Rose From crose Thu Nov 6 09:13:31 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA09675; Thu, 6 Nov 97 09:13:11 EST Date: Thu, 6 Nov 97 09:13:11 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711061413.AA09675@MOGLI.rutgers.edu> To: JOHNSUCEC@delphi.com, crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Possible alternate solution to Exam 1 P2a Cc: crose@MOGLI.rutgers.edu Status: RO HI John, I'm COMPLETELY confused. I looked at the web page when this problem arose and I swore I saw the 1/p out front. I'll look again,,, or maybe WE BOTH need to look using your hardware. This is wierd!!!!!!! :) Cheers, CHris Rose From crose Thu Nov 6 09:22:30 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA09692; Thu, 6 Nov 97 09:22:12 EST Date: Thu, 6 Nov 97 09:22:12 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711061422.AA09692@MOGLI.rutgers.edu> To: 330_543 Subject: wild error Cc: crose@MOGLI.rutgers.edu Status: RO It seems that in illstrating a common error on te quiz 1 solutions I somehow forgot to include the correct solution on the copy I posted!!!!! We've been having a colloquy about this solution with me looking at what I have at home and you looking at what was on the web page (WHICH IS WRONG DAMMIT!). I'm reposting the solution and completely expunging the older solution. Criminy! I'm SO sorry for the confusion. I THOUGHT I'd checked the web page, but I could not have given this glaring error! Cheers, Chris Rose From jsucec@ece.rutgers.edu Sun Nov 9 18:41:51 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA13322; Sun, 9 Nov 97 18:41:12 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id SAA09560 for <330_543@MOGLI.rutgers.edu>; Sun, 9 Nov 1997 18:35:54 -0500 Date: Sun, 9 Nov 1997 18:35:54 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Questions on Section 7.4 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO To anyone who can help, I have a couple of questions on the material covered in section 7.4 of our text... Page 188, Equation 7.4.4: The expression for P_w(ja) contains an "O(1/K)" term. From where does this term come? I don't recall it being shown in the derivation of the Poisson distribution that was done in class, and I don't see how it gets added when going from equation 7.4.3 to 7.4.4. Page 191: In going equation from 7.4.6 to equation 7.4.7, our text uses an intermediary of step of [1 - t/d_i]^(d_i*(1/d_i)) in approximating (1 - t/d_i). What is the purpose of this step? Don't we already have (1 - t/d_i) ~= e^(-t/d_i), provided we ignore the 2nd order and above terms of the series representation of e^(-t/d_i) and assume t/d_i to be small? Anyway, section 7.4 is easy to understand compared to the rest of the chapter, but I would like to at least nail down the easy stuff. So, if I am missing something here, I'd appreciate it if someone could point it out to me. Thanks. ...John From alap@winlab.rutgers.edu Mon Nov 10 16:34:15 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA14678; Mon, 10 Nov 97 16:33:54 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA29477; Mon, 10 Nov 1997 16:33:14 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id QAA04307; Mon, 10 Nov 1997 16:33:14 -0500 Date: Mon, 10 Nov 1997 16:33:14 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711102133.QAA04307@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: Questions on Section 7.4 Status: RO John - Page 188, eq 7.4.4: O(1/K) This term O(1/K) comes from the product (1-1/K)(1-2/K)...(1-(j-1)/K) because if we multiply the terms we get = 1 + function of 1/K and O(1/K) is to represent this function of 1/K. In class we just said that those terms go to zero as K goes to infinity (which he also says in the book). I think he is just trying to show that those terms are there, but they will go to zero in the limit as K goes to infinity. About Page 191, I agree with you, that the expansion already gives e^(-t/d_i) = 1 - t/d_i and I don't understand the purpose of that step either! REgards, Ana. From jsucec@ece.rutgers.edu Tue Nov 11 08:37:48 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA15438; Tue, 11 Nov 97 08:36:22 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id IAA01356; Tue, 11 Nov 1997 08:30:47 -0500 Date: Tue, 11 Nov 1997 08:30:46 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Questions on Section 7.4 In-Reply-To: <199711102133.QAA04307@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, thanks. Looking again at page 191, I see that the expression "(1 - t/d_k)^(d_k*(1/d_k))" matches the the (1 - rho/K)^K ~= e^(-rho) approximation used on page 188 and by Prof. Rose in class. This technique must be a superior approximation to e^x than simply ignoring the 2nd order and above terms of series expansion of e^x. Concerning the work done in going from equation 7.4.3 to 7.4.4 on page 188, here is how I account for the terms in 7.4.3: * 1/j! --> No change, appears as is in 7.4.4 * (1 - 1/K), (1 - 2/K), ..., etc. --> For values of j< rho^j * (1-p)^(K-j) ~= (1-p)^K = (1-rho/K)^K ~= e^(-rho) The above 4 bullets appear to account for all of the terms in equation 7.4.3. I guess I will go with your explaination and simply assume that the author felt like including the O(1/K) approximation of the "artifacts" generated by the product of the (1-1/K)*(1-2/K)*...*(1-(j-1)/K). It just seems strange that in one sentence the author would say that each of these factors is effectively 1, and then in the next statement show the order of error made in this approximation. Again, thanks. ...John On Mon, 10 Nov 1997, Ana Lucia Pinheiro wrote: > > John - > > Page 188, eq 7.4.4: O(1/K) > > This term O(1/K) comes from the product (1-1/K)(1-2/K)...(1-(j-1)/K) > because if we multiply the terms we get = 1 + function of 1/K > and O(1/K) is to represent this function of 1/K. > In class we just said that those terms go to zero as K goes to infinity > (which he also says in the book). I think he is just trying to show that > those terms are there, but they will go to zero in the limit as K goes to infinity. > > About Page 191, I agree with you, that the expansion already gives > e^(-t/d_i) = 1 - t/d_i > and I don't understand the purpose of that step either! > > REgards, > Ana. > > From elf_pub@email.rutgers.edu Tue Nov 11 19:42:20 1997 Return-Path: Received: from eden-backend.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA16206; Tue, 11 Nov 97 19:41:38 EST Received: from alef06.rutgers.edu (alef06.rutgers.edu [165.230.226.142]) by eden-backend.rutgers.edu (8.8.5/8.8.5) with SMTP id TAA23726 for <330_543@mogli.rutgers.edu>; Tue, 11 Nov 1997 19:40:44 -0500 (EST) Message-Id: <3468EB77.67A4@email.rutgers.edu> Date: Tue, 11 Nov 1997 19:34:16 -0400 From: Public ELF Workstation Reply-To: "no reply"@eden.rutgers.edu Organization: Rutgers University Libraries X-Mailer: Mozilla 3.04 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli.rutgers.edu Subject: Questions Ch7 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Dear Sir, I have questions in Chapter Seven, if you would please answer some of these very trivial questions. Page 188 equation 7.3.1, 7.3.2 & 7.3.2 how did we arrive at these equations? Also it says as k-> infinity keep P'=kp constant, but if p=P'/k since k->infinity then p->0 I am confused please explain this to me. Statement " A Semi Markov Process is a Discrete Time Markov process if we ignore time durations between transitions" Please explain. What does " Sojourn Time " & Renewal Process in Alternating State Renewal Process mean? Also what is the difference between a Poisson Distribution & a Poisson Process? Are many poisson distributions considered together a Poisson Process What is an Embedded Markov process? Why Embedded? From crose Tue Nov 11 19:51:29 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA16217; Tue, 11 Nov 97 19:50:40 EST Date: Tue, 11 Nov 97 19:50:40 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711120050.AA16217@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, no@email.rutgers.edu, reply@eden.rutgers.edu Subject: Re: Questions Ch7 Cc: crose@MOGLI.rutgers.edu Status: RO Hi! Anybody care to take a shot at at the questions just raised! Ana, John, Allen, Salim???? Cheers, Chris Rose From alap@winlab.rutgers.edu Tue Nov 11 22:22:13 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA16379; Tue, 11 Nov 97 22:21:58 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id WAA28392; Tue, 11 Nov 1997 22:21:05 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id WAA07559; Tue, 11 Nov 1997 22:21:05 -0500 Date: Tue, 11 Nov 1997 22:21:05 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711120321.WAA07559@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: Questions Ch7 Status: RO Here are my comments related to that e-mail: 1) I do not have equations 7.3.2 and 7.3.2 in page 188. Maybe the page or the eq. number is not correct (?) Please resend. 2) if k goes to infinity, ro=kp can be constant if we let p be very small. 3) " A Semi Markov Process is a Discrete Time Markov process if we ignore time durations between transitions" : As Prof Rose explained today in class, the semi-markov process is when we consider ONLY the transitions probabilities. It does not matter WHEN you go from one state to another, what you are interested in is the Transition probability between states. If we see the 3 states semi-markov process , for example, we are only interested in probability of going to state i given you were in state j, and i is always different than j (there is no such an event "stay in the same state") because what you are looking at is "given you will change state, TO WHICH ONE are you going to". So, if you pick the discrete markov process and look only when there is a transition (ignoring the time durations between transitions), you get the same idea! 4) sojourn time=time you stay in a state (what Prof. called in class d_i|j) 5) Renewal Process : I've seen this to represent number of times some equipament fails (number of renewals that have to be made). I think in this case the author wants to represent each transition as a failure and then each transition back as a renewal. (?) It does not apply to the generalized case where n>2. So, I am not sure either. 6) A poisson distribution is the function that represents the probability distribution of a random variable. The poisson process is described in page 188 of book. They are related somehow: the number of arrivals in a Poisson Process is a random variable that obeys the Poisson distribution. On the other hand, the time between arrival is a random variable that obeys the exponential distribution. So we could say that the Poisson Process constists of arrivals that are represented by two random variables: t: time between arrivals: exponential random variable n: number of arrivals in an interval: poisson random variable. I hope that helps. I also hope someone will read and correct any mistakes I may have. Regards, Ana. From alap@winlab.rutgers.edu Tue Nov 11 22:25:56 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA16388; Tue, 11 Nov 97 22:25:41 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id WAA28442; Tue, 11 Nov 1997 22:24:49 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id WAA07563; Tue, 11 Nov 1997 22:24:49 -0500 Date: Tue, 11 Nov 1997 22:24:49 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711120324.WAA07563@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Problems in mailing list ?? Status: RO Prof Rose: here is the message I got from last e-mail: ----- Transcript of session follows ----- Connected to eden-backend.rutgers.edu: >>> RCPT To: <<< 550 ... User unknown 550 sgujar@eden.rutgers.edu... User unknown >>> RCPT To: <<< 550 ... User unknown 550 abhishek@eden.rutgers.edu... User unknown 550 log... User unknown Is there a problem with some students? Thanks. Ana. From alap@winlab.rutgers.edu Wed Nov 12 01:05:19 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA16553; Wed, 12 Nov 97 01:05:03 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA29428; Wed, 12 Nov 1997 01:04:09 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA07588; Wed, 12 Nov 1997 01:04:09 -0500 Date: Wed, 12 Nov 1997 01:04:09 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711120604.BAA07588@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Memoryless of exponential interarrivals Status: RO Hello! I would like to make a comment about the discussion we had in class today. Trying to figure out what is the residual waiting time distribution: p_t'|t_1(t'|t_1) for t' > t_1 = p_t',t_1(t',t_1) / p_t_1(t_1) [Bayes Rule] But the hard part was to find the numerator. p_t',t_1(t',t_1) = p_t'(t') = m exp(-mt') Is this a valid density function ? integral over (t',t_1) of m exp(-mt') = = integral (from t_1 = 0 to infinity) . integral (t' = t_1 to infinity) the first integral in t' gives exp(mt_1) integral (t_1 = 0 to infinity) of exp(mt_1) is not 1 !!! So I really believe that that approach was not correct , at least not precise. The answer is correct, but I think we should derive that using CDF. This is in Papoulis, I just copy here: F(t'|t>t_1) = P(t <= t' , t > t_1) / P(t>t_1) = {F(t') - F(t_1)} / {1-F(t_1)} differentianting with respect to t': f(t'|t>t_1) = f(t')/{1-F(t_1)} where F(x) = P(t<=x) and t is our random variable. This approach yields to the memorylessness of interarrivals. Hope that will help! Regards, Ana. From alap@winlab.rutgers.edu Wed Nov 12 01:28:12 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA16566; Wed, 12 Nov 97 01:27:42 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA29463; Wed, 12 Nov 1997 01:26:49 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA07597; Wed, 12 Nov 1997 01:26:48 -0500 Date: Wed, 12 Nov 1997 01:26:48 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711120626.BAA07597@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: A paradox? Status: RO Dave, I found your mail very interesting! I think I always have problems when I try to bring the memoryless idea to the real world. Anyway, I am not sure, I guess what is wrong in the approach is that we should not add both averages. You are saying that you have two random variables, say t_1 and t_2, and they are "time I waited" and "time I will wait". Both have mean 1/x. And then you say that interarrival time is t_1 + t_2 , which is a sum of 2 rv's, it will have mean = 2/x. For sure t_1 + t_2 is expo(x/2), but I think that intearrival time cannot be represented as this sum. Why ?? I don't really know... I hope other people will add more comments to this problem. Regards, Ana. From crose Wed Nov 12 02:16:26 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA16644; Wed, 12 Nov 97 02:16:14 EST Date: Wed, 12 Nov 97 02:16:14 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711120716.AA16644@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Memoryless of exponential interarrivals Cc: crose@MOGLI.rutgers.edu Status: RO THANKS ANA! From crose Wed Nov 12 02:30:23 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA16655; Wed, 12 Nov 97 02:30:11 EST Date: Wed, 12 Nov 97 02:30:11 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711120730.AA16655@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: A paradox? Status: RO Hi Folks, Keep the "paradox" talk going! GREAT OBSERVATION! I'll let you stew a little more on it and see if you resolve it (there IS a resolution) but you'll have to think carefully about conditional probs. I think Ana's formulation is the best approach since she's defined random variables. Now you need to define conditionals based on what is known. Also, perhaps more important, think carefully about Dave's problem as posed. There's a bit of random incidence going on there (remember the time you plunk down at is chosen at RANDOM). So why not figure out the mean length of an interval you plunk down into! I'll stop now before I give it all away (if I've not done so already :) tata, Chris Rose PS: Folks sorry about the interruption in class today (cellular phone). I'll have to beg your indulgence since until my son (Jason) makes it home from the hospital I need to be in touch with family and physicians. Jason needed surgery shortly after birth and is still recuperating. The surgeons and ICU folks were EXCELLENT, but as he's gotten better the care provided has been a little uneven and requires our constant attention (I just now came back from the hospital after terrorizing the night nurses :). From crose Wed Nov 12 02:30:23 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA16655; Wed, 12 Nov 97 02:30:11 EST Date: Wed, 12 Nov 97 02:30:11 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711120730.AA16655@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: A paradox? Status: RO Hi Folks, Keep the "paradox" talk going! GREAT OBSERVATION! I'll let you stew a little more on it and see if you resolve it (there IS a resolution) but you'll have to think carefully about conditional probs. I think Ana's formulation is the best approach since she's defined random variables. Now you need to define conditionals based on what is known. Also, perhaps more important, think carefully about Dave's problem as posed. There's a bit of random incidence going on there (remember the time you plunk down at is chosen at RANDOM). So why not figure out the mean length of an interval you plunk down into! I'll stop now before I give it all away (if I've not done so already :) tata, Chris Rose PS: Folks sorry about the interruption in class today (cellular phone). I'll have to beg your indulgence since until my son (Jason) makes it home from the hospital I need to be in touch with family and physicians. Jason needed surgery shortly after birth and is still recuperating. The surgeons and ICU folks were EXCELLENT, but as he's gotten better the care provided has been a little uneven and requires our constant attention (I just now came back from the hospital after terrorizing the night nurses :). From alap@winlab.rutgers.edu Wed Nov 12 03:27:56 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA16773; Wed, 12 Nov 97 03:27:33 EST Received: from monk-a-asy-1.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id DAA29645; Wed, 12 Nov 1997 03:26:36 -0500 Message-Id: <3469921B.7ECD@winlab.rutgers.edu> Date: Wed, 12 Nov 1997 03:25:15 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli Subject: paradox Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Dave, here is another approach: Consider t_1 the time of first arrival and t_2 the time of next arrival. Then (t_2 - t_1) is a random variable expo(x), where t_1 is given. The randomness is in t_2 (t_1 is fixed) t_1 t_2 |------------|---> t Now consider your problem: you still have t_1 and t_2 and now call T the time you arrive in the system t_1 T t_2 |-----|-------|---> t When you arrive BOTH t_1 and t_2 are unknonw and we only know T. The random variable (T-t_1) represents the time you have waited and is expo(X). The randomness is in t_1. The random variable (t_2-T) represents the time you will wait and is expo(X). The randomness is in t_2 The random variable (T-t_1) + (t_2-T) = (t_2 - t_1) is expo (X/2) but DOES NOT represent the interarrival time since BOTH t_1 and t_2 are unknow. Both random variables (T-t_1) and (t_2-T) represent the interarrival time, just looking from a different point of view. Imagine the following: You decide to tape the arrivals process. You arrive in an arbitrary time and start taping until the next arrival. The process will last (on average) 1/x. Now imagine you were able to tape backwards in time. It would take you (on av.) 1/x to get to the last arrival. But the film you are taping is actually the same, and both situations represent the same thing, just looking from a different perspective. Hope this is helpfull :) Ana. From linzhou@ece.rutgers.edu Wed Nov 12 11:18:55 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA17070; Wed, 12 Nov 97 11:17:15 EST Received: from localhost (linzhou@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id LAA04953; Wed, 12 Nov 1997 11:11:29 -0500 Date: Wed, 12 Nov 1997 11:11:28 -0500 (EST) From: Lin Zhou To: crose@ece.rutgers.edu Cc: 330_543@mogli.rutgers.edu Subject: sojourn time d_i|j In-Reply-To: <199711120321.WAA07559@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO prof. Rose, To calculate the steady state probability(Pss_i), why we don't use the difference equation and matrix Q? You introduced mean sojourn time d_i|j which is the holding time at state j before transtion to state i. you gave the formula P(ss)_i=sum_j(qi|j*pj*di|j)/sum_i()sum_j() if it applied to alternative states process, I get p(ss)_0=d_0|1/(d_0|1+d_1|0). The numerator is d_0|1 (mean holding time at state 1 before it goes to state 0). I guess it should be d_1|0. Do you think we can change the d_i|j in the formula to d_j|i? I am not sure. Regards, lin zhou From jsucec@ece.rutgers.edu Wed Nov 12 11:22:32 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA17082; Wed, 12 Nov 97 11:20:50 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id LAA05148 for <330_543@MOGLI.rutgers.edu>; Wed, 12 Nov 1997 11:15:04 -0500 Date: Wed, 12 Nov 1997 11:15:03 -0500 (EST) From: John Sucec To: 330_543@MOGLI.rutgers.edu Subject: Paradox (my interpretation) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Here's my take on how we might interpret the "paradox" of the interarrival process... * X = Duration between arrival events * X = 0 time at which at which last arrival occurred * P(X=x) = probability density function for time to next arrival = a*e^(-a*x) * E[X] = 1/a Given snapshot at time x=t: * Let A = state where arrival has occurred * Let B = state where arrival has not occurred * Prob(Arrival has occurred) = P(A) = 1 - e^(-a*t) * Prob(Arrival has NOT occurred) = P(B) = e^(-a*t) * Let Y = Arrival duration given snapshot at x=t * E[Y] = E[X | Arrival has occurred] + E[X | Arrival has not occurred] = E[X | A] + E[X | B] = E[X]*P(A) + E[X]*P(B) = E[X] = 1/a {since P(A) + P(B) = 1} ...and grinding through the arithmetic, we get... = (1/a)*(1 - e^(-a*t)) + (1/a)*(e^(-a*t)) = 1/a + (1/a)*(-e^(-a*t)) + (1/a)*(e^(-a*t)) = 1/a = E[X] As shown above, we get 1/a as the expected waiting time for the next arrival event which is consistent with the exponenential distribution model. The fact that we take a snapshot of the system at an arbitrary time, t, will not disrupt the statistical properties of the system (i.e., event Y is statistically the same as event X). Anyway, this analysis cleared up the "paradox" issue in my mind. I have not read Ana's E-Mail thoroughly, but it looks like she has already provided considerable detail. I hopes this helps. ...John From crose@MOGLI.rutgers.edu Wed Nov 12 12:30:20 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA17214; Wed, 12 Nov 97 12:30:18 EST Full-Name: Christopher Rose Received: from MOGLI.rutgers.edu (mogli.rutgers.edu [128.6.46.55]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id MAA08343; Wed, 12 Nov 1997 12:24:28 -0500 Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA17211; Wed, 12 Nov 97 12:30:13 EST Date: Wed, 12 Nov 97 12:30:13 EST From: Christopher Rose Message-Id: <9711121730.AA17211@MOGLI.rutgers.edu> To: crose@ece.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: sojourn time d_i|j Cc: 330_543@MOGLI.rutgers.edu Status: RO Hi, I might have reversed the subscripts in my haste at the board. However, it should be P_i^{ss} = [\sum_j (q_{j|i} p_i^* d_{j|i})] / [ \sum_{ij} (above) ] Please note that my notation is different than the book. BLAH_{i|j} means you start in state j going to state i. In the book the notation is BLAH_{ji} for the same thing. By the way folks, I put the above exptression in latex since you'll need latex to do your theses or write reports. Never too soon to get started. In fact, here's a better version of the same thing using the frac (fraction) function. The $$ are equation display delimiters. $$ P_i^{ss} = \frac{\sum_j q_{j|i} p_i^* d_{j|i}}{ \sum_{ij} \sum_j q_{j|i} p_i^* d_{j|i}} $$ From crose Wed Nov 12 12:31:49 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA17211; Wed, 12 Nov 97 12:30:13 EST Date: Wed, 12 Nov 97 12:30:13 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711121730.AA17211@MOGLI.rutgers.edu> To: crose@ece.rutgers.edu, linzhou@ece.rutgers.edu Subject: Re: sojourn time d_i|j Cc: 330_543@mogli.rutgers.edu Status: RO Hi, I might have reversed the subscripts in my haste at the board. However, it should be P_i^{ss} = [\sum_j (q_{j|i} p_i^* d_{j|i})] / [ \sum_{ij} (above) ] Please note that my notation is different than the book. BLAH_{i|j} means you start in state j going to state i. In the book the notation is BLAH_{ji} for the same thing. By the way folks, I put the above exptression in latex since you'll need latex to do your theses or write reports. Never too soon to get started. In fact, here's a better version of the same thing using the frac (fraction) function. The $$ are equation display delimiters. $$ P_i^{ss} = \frac{\sum_j q_{j|i} p_i^* d_{j|i}}{ \sum_{ij} \sum_j q_{j|i} p_i^* d_{j|i}} $$ From chriskly@er6.rutgers.edu Wed Nov 12 12:34:44 1997 Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA17229; Wed, 12 Nov 97 12:32:56 EST Received: (from chriskly@localhost) by er6.rutgers.edu (8.8.5/8.8.5) id MAA14363 for 330_543@mogli.rutgers.edu; Wed, 12 Nov 1997 12:31:57 -0500 (EST) Date: Wed, 12 Nov 97 12:31:57 EST From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Subject: THE PARADOX Message-Id: Status: RO I'd like to add my own 2 cents into the discussion about memoryless exponential. First, I believe that it is NOT true that you should add the two means together to find the total interarrival time. The reason is that the memoryless property says that the time you wait for a new arrival is independant of the time you have already waited. Trying to apply this to the RU screw bus system means that when I get to the busstop (just after the bus has left) I will have to wait an exponential amount of time (w/ mean 1/a) for the next bus to come. If Ana arrives sometime after I do, then she will also wait an exponential amt of time (with mean 1/a), regardless of how long I have been there. On AVERAGE, I will wait 1/a and on AVERAGE Ana will wait 1/a. The key here is that 1/a is the average waiting time, although in most cases the waiting time will either be longer or shorter. The key is in the distribution of hte exponential. There certainly are other distribution with mean 1/a where the memoryless property does not apply. Second, here is my favorite proof to show the memoryless-ness of the exponential. (Maybe someone else will also like this one...) We are trying to show that the time until the next event occurs does not depend on how long ago the previous even occurs. Let t be current time and t + s be some time after that. (ie, if I start waiting for ht ebus at time 0 and Ana arrives at time t. I am still waiting, so there has been no arrival in (0,t) We want to show that hte distribtution of the amount of time Ana waits does not depend on how long I have been waiting. Suppose the bus arrives at time t+s. We need to show that the time Ana waits, which is s, does not depend on the fact that I already waited t time units before she arrived, so I waited a total of t+s time units.) Pr(X > t+s | X > t) = Pr(X > t+s, X > t) / Pr (X > t) *** By Bayes Thm = Pr(X > t+s) / Pr(X>t) *** True for any distribution Now, if X is exponential with mean 1/a, then Pr(X > x) = e^(-ax) So we have Pr(X > t+s) | X > t) = e^(-a(t+s)) / e^(-at) = e^(-as) = Pr (X > s) ** independnat of t Hope this makes sense and I apoligize for the spelling misstakes (I am trying to get this done during my lunch break and still have time to eat! :) (and Ana, I hopeyou dont mind me using yoru name throughout my example!) Christine From crose Wed Nov 12 12:42:55 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA17256; Wed, 12 Nov 97 12:42:33 EST Date: Wed, 12 Nov 97 12:42:33 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711121742.AA17256@MOGLI.rutgers.edu> To: 330_543@mogli.rutgers.edu, chriskly@eden.rutgers.edu Subject: Re: THE PARADOX Status: RO Keep it up folks!!!!! I'm still wating for someone to calculate the average interval size into which a given random observation falls. That is I inspect the system at t_1. What is the mean duration of the interval? (the interval is the duration between the arrival occurring before and after my observation). From alap@winlab.rutgers.edu Wed Nov 12 13:11:09 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA17353; Wed, 12 Nov 97 13:10:23 EST Received: from monk-a-asy-3.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id NAA02758; Wed, 12 Nov 1997 13:09:21 -0500 Message-Id: <346A1AA9.5868@winlab.rutgers.edu> Date: Wed, 12 Nov 1997 13:07:53 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli Subject: Re: paradox References: <3469921B.7ECD@winlab.rutgers.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Prof. Rose: >From my Fig. 2 below, if we assume that interarrival time is expo(X), then the average size into which my observation (T) fails is (1/2x). *** But this is true because we assume the intearrival time exponential. *** The problem Dave had was different: he wants to use the information we have at time T to derive the exponential interarrival time. Which, as I try to show below, it's not possible. Of course we could just add (1/2x) (twice) and get the correct answer. But I think what we are trying is to see the mistake in his reasoning. Isn't that right? Am I missing something ? Ana Lucia Pinheiro wrote: > > Dave, here is another approach: > > Consider t_1 the time of first arrival and t_2 the time of next arrival. > Then (t_2 - t_1) is a random variable expo(x), where t_1 is given. > The randomness is in t_2 (t_1 is fixed) > > t_1 t_2 > |------------|---> t Fig.1 > > Now consider your problem: you still have t_1 and t_2 and now call T the > time you arrive in the system > > t_1 T t_2 > |-----|-------|---> t Fig.2 > > When you arrive BOTH t_1 and t_2 are unknonw and we only know T. > > The random variable (T-t_1) represents the time you have waited and is expo(X). > The randomness is in t_1. > > The random variable (t_2-T) represents the time you will wait and is expo(X). > The randomness is in t_2 > > The random variable (T-t_1) + (t_2-T) = (t_2 - t_1) is expo (X/2) > but DOES NOT represent the interarrival time since BOTH t_1 and t_2 > are unknow. > > Both random variables (T-t_1) and (t_2-T) represent the interarrival time, just > looking from a different point of view. > > Imagine the following: You decide to tape the arrivals process. You arrive > in an arbitrary time and start taping until the next arrival. The process > will last (on average) 1/x. Now imagine you were able to tape backwards in > time. It would take you (on av.) 1/x to get to the last arrival. But the film you > are taping is actually the same, and both situations represent the same thing, > just looking from a different perspective. > > Hope this is helpfull :) > Ana. From fam@winlab.rutgers.edu Wed Nov 12 16:00:42 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA17584; Wed, 12 Nov 97 16:00:15 EST Received: from winlab3.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id PAA05719; Wed, 12 Nov 1997 15:59:14 -0500 Received: by winlab3.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id PAA01363; Wed, 12 Nov 1997 15:59:12 -0500 Date: Wed, 12 Nov 1997 15:59:12 -0500 From: fam@winlab.rutgers.edu (David Famolari) Message-Id: <199711122059.PAA01363@winlab3.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: THE PARADOX Cc: fam@winlab.rutgers.edu X-Sun-Charset: US-ASCII Status: RO Hi, Thanks for all the responses about the paradox. I'm glad it generated a lot of discussion. To address Christine's point, I agree that you shouldn't just add the two means together to get the average interval length. However, it is true that you should add the average time that since the last arrival and the average time until the next arrival to get the average interval length that you happen upon. So to calculate the average length of the interval of time when I get to teh bus stop, I can add the time from the last bus to the expected time of the next bus, and that's completly valid. In the Possion process it turns out that the residual time and the age both have the same mean namely 1/a. But I think the paradox can be solved by realizing that the interval that we pick our random point in is NOT a typical interval and will NOT be distributed like the others (not exponetnial). Let's say that the interarrival time, T, is distributed exponentially with parameter a, that is the average interval length of time between intervals is 1/a. Now we pick a point in time. This point will fall within some interval marked by two arrivals. The key is, that point in time is more likely to fall within a large interval. Let's call this interval X. Now the distribution of X is not the same as T!. To show this we use the fact that the probability that the random point will fall in a longer interval is greater then the probability it will fall in an interval of smaller size. SO there is a proportionality to interval size. So let the probability that the chosen interval length, X, is between some x and x+dx be P[xFrom <@MOGLI:chriskly@er6.rutgers.edu> Wed Nov 12 12:32:00 1997 Date: Wed, 12 Nov 97 12:31:57 EST From: Christine Kleiwerda To: 330_543@mogli Subject: THE PARADOX I'd like to add my own 2 cents into the discussion about memoryless exponential. First, I believe that it is NOT true that you should add the two means together to find the total interarrival time. The reason is that the memoryless property says that the time you wait for a new arrival is independant of the time you have already waited. Trying to apply this to the RU screw bus system means that when I get to the busstop (just after the bus has left) I will have to wait an exponential amount of time (w/ mean 1/a) for the next bus to come. If Ana arrives sometime after I do, then she will also wait an exponential amt of time (with mean 1/a), regardless of how long I have been there. On AVERAGE, I will wait 1/a and on AVERAGE Ana will wait 1/a. The key here is that 1/a is the average waiting time, although in most cases the waiting time will either be longer or shorter. The key is in the distribution of hte exponential. There certainly are other distribution with mean 1/a where the memoryless property does not apply. Second, here is my favorite proof to show the memoryless-ness of the exponential. (Maybe someone else will also like this one...) We are trying to show that the time until the next event occurs does not depend on how long ago the previous even occurs. Let t be current time and t + s be some time after that. (ie, if I start waiting for ht ebus at time 0 and Ana arrives at time t. I am still waiting, so there has been no arrival in (0,t) We want to show that hte distribtution of the amount of time Ana waits does not depend on how long I have been waiting. Suppose the bus arrives at time t+s. We need to show that the time Ana waits, which is s, does not depend on the fact that I already waited t time units before she arrived, so I waited a total of t+s time units.) Pr(X > t+s | X > t) = Pr(X > t+s, X > t) / Pr (X > t) *** By Bayes Thm = Pr(X > t+s) / Pr(X>t) *** True for any distribution Now, if X is exponential with mean 1/a, then Pr(X > x) = e^(-ax) So we have Pr(X > t+s) | X > t) = e^(-a(t+s)) / e^(-at) = e^(-as) = Pr (X > s) ** independnat of t Hope this makes sense and I apoligize for the spelling misstakes (I am trying to get this done during my lunch break and still have time to eat! :) (and Ana, I hopeyou dont mind me using yoru name throughout my example!) Christine ----- End Included Message ----- ----- End Included Message ----- From jsucec@ming.rutgers.edu Wed Nov 12 17:00:28 1997 Return-Path: Received: from zen.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA17629; Wed, 12 Nov 97 17:00:27 EST Received: from ming.rutgers.edu.rutgers.edu (ming.rutgers.edu [128.6.46.125]) by zen.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id QAA03439; Wed, 12 Nov 1997 16:54:55 -0500 Received: from localhost by ming.rutgers.edu.rutgers.edu (SMI-8.6/SMI-SVR4) id QAA12375; Wed, 12 Nov 1997 16:58:37 -0500 Date: Wed, 12 Nov 1997 16:58:36 -0500 (EST) From: John Sucec To: Christopher Rose Cc: 330_543@MOGLI.rutgers.edu, chriskly@eden.rutgers.edu Subject: Re: THE PARADOX In-Reply-To: <9711121742.AA17256@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, as I mentioned earlier, an observation (or snapshot) of a process, does not change the statistical properties of that process. So, if the interarrivals of a bus is modelled as exponential with mean 1/a, then the expected duration between arrivals will be 1/a. Therefore, the interval into which a random observation/snapshot falls, will also be 1/a. Am I missing something here? Maybe I don't understand the problem statement completely. Well, if I am understanding the problem correctly then the mean value of the interval into which the observation falls will simply be the mean of the exponential process modelling the interarrival duration. Anyway, please let me know if I have misunderstood the problem statement. Thanks. ...John On Wed, 12 Nov 1997, Christopher Rose wrote: > Keep it up folks!!!!! I'm still wating for someone to calculate > the average interval size into which a given random observation falls. > That is I inspect the system at t_1. What is the mean duration of the > interval? > > (the interval is the duration between the arrival occurring before and after my observation). > > From jsucec@ece.rutgers.edu Wed Nov 12 17:55:33 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA17656; Wed, 12 Nov 97 17:55:06 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id RAA20051; Wed, 12 Nov 1997 17:49:14 -0500 Date: Wed, 12 Nov 1997 17:49:13 -0500 (EST) From: John Sucec To: "no reply"@eden.rutgers.edu Cc: 330_543@MOGLI.rutgers.edu Subject: Re: Questions Ch7 In-Reply-To: <3468EB77.67A4@email.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO I presume that the question below, "Page 188 equation...", actually refers to equation 7.4.1 on page 187, and equations 7.4.2 and 7.4.3 on page 188? Assuming this is correct, I'll share my interpretation of these equations. Firstly, the process W represents the aggregate traffic of K terminals. Each terminal can exist in a number of transmitting or non-transmitting states. I liken this to a bus network such as an Ethernet segment. Equation 7.4.1 is a probability mass function (pmf) describing the event of aggregate network traffic from K terminals. P_W(W=w), therefore, is the probability that the aggregate traffic is W=w. The assumption here is that each terminal has finite number of "traffic states" and associate with each terminal state there is a single traffic rate requirement. Therefore, if there are say, P permutations of the states of the K terminals whose traffic requirements sum to w, then probability that the network exists in one of these permutations will be the product of the probabilities of the K terminal states. This product for each of the P permutations summing to w. Equation 7.4.2 is devived from 7.4.1 when we add the restriction that each terminal has only 1 state for which the state's traffic requirement a_k is non-zero, and restrict this traffic requirement to be the same (that is, a) for all K terminals. The aggregate traffic process, W, can then be represented by the Bernoulli distribution that is equation 7.4.2. Equation 7.4.3 is derived from 7.4.2 when we take lim(K --> infinity). Factor out the j! of C(K,j) brings the 1/j! term out in front in 7.4.3.. Then K^j is factored out of the K!/(K-j)! term and stuck along with p^j in 7.4.2 to yield (K*p)^j in 7.4.3. This factoring also leaves us with K!/((K^j)*(K-j)! which (I guess) yields the (1-1/K)*(1-2/K)*... terms in equation 7.4.3. You don't inquire about it, but if you are interested in the step of going from equation 7.4.3 to equation 7.4.4, there were a few E-Mails earlier this week discussing that step. That's my take on these equations. I hope these were the equations about which you were inquiring and I hope this explaination makes sense. Thanks. ...John On Tue, 11 Nov 1997, Public ELF Workstation wrote: > Dear Sir, > I have questions in Chapter Seven, if you would please answer some of > these very trivial questions. > > Page 188 equation 7.3.1, 7.3.2 & 7.3.2 how did we arrive at these > equations? > Also it says as k-> infinity keep P'=kp constant, > but if p=P'/k since k->infinity then p->0 > I am confused please explain this to me. > > Statement " A Semi Markov Process is a Discrete Time Markov process if > we ignore time durations between transitions" Please explain. > > What does " Sojourn Time " & Renewal Process in Alternating State > Renewal Process mean? > > Also what is the difference between a Poisson Distribution & a Poisson > Process? > Are many poisson distributions considered together a Poisson Process > > What is an Embedded Markov process? Why Embedded? > From aonweller@maerskdata-usa.com Wed Nov 12 21:40:22 1997 Return-Path: Received: from msrvr.maerskdata-usa.com by MOGLI.rutgers.edu (4.1/25-eef) id AA17783; Wed, 12 Nov 97 21:40:21 EST Received: from [10.1.25.41] ([207.242.160.66]) by msrvr.maerskdata-usa.com (Netscape Mail Server v2.02) with SMTP id AAA145 for ; Thu, 13 Nov 1997 02:42:20 +0000 Message-Id: <346A6831.2B8C@maerskdata-usa.com> Date: Wed, 12 Nov 1997 21:38:48 -0500 From: aonweller@maerskdata-usa.com (Allen Onweller) X-Mailer: Mozilla 3.01-C-MACOS8 (Macintosh; I; PPC) Mime-Version: 1.0 To: Christopher Rose Subject: Re: THE PARADOX References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Hi Prof. Rose, Here's my take on the paradox, but I'm afraid that I haven't been able to keep up with the mail, so I don't what to confuse anyone. If this isn't too far out there, please feel free to repost: I haven't had a chance to read all of the paradox e-mail, so please excuse me if this has allready been stated or if I'm off the mark. Let t = the time of the obersvation and s = a time after t (just before the arrival) s-t = the time between the arrival and the observation. Prob{no obersvation by time t} = 1 - integral from 0 to t {a exp(-a(v))dv} = exp(-a(t)) Prob{no observation during the interval s-t} = 1 - integral from 0 to s-t {a exp(-a(v))dv) = exp(-a(s-t)) Prob{ no observation until s} = Prob{no observation by time t} and Prob{no observation during s-t} =exp(-a(t))exp(-(s-t)) =exp(-a(s)) This is exactly the same probability that the arrival did not occurr by time s. Basically this means that the observation that the arrival did not come by time t did not affect the probability of the arrival occuring by time s. True the expected arrival time changes to t + 1/a, but the probability that the arrival occurs by time s does not change. So to sum it up, the memoryless property makes no claims as to the affect of observations on expectation. It merely states that past observations are no guaratee of future performance. (Much like investment brokers.) Regards, Allen From crose Wed Nov 12 22:42:37 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA17871; Wed, 12 Nov 97 22:42:14 EST Date: Wed, 12 Nov 97 22:42:14 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711130342.AA17871@MOGLI.rutgers.edu> To: 330_543, fam@winlab.rutgers.edu Subject: Re: THE PARADOX Status: RO BRAVO DAVE! THAT"S EXACTLY THE RIGHT ANSWER!!!!!!!! Random incidence generates Atypical intervals for observation, skewed toward the longer intervals because you're more likely to have your randomly selected observation time fall into one of those intervals!!!! Cheers, Chris Rose From crose Wed Nov 12 22:44:55 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA17901; Wed, 12 Nov 97 22:44:14 EST Date: Wed, 12 Nov 97 22:44:14 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711130344.AA17901@MOGLI.rutgers.edu> To: jsucec@ming.rutgers.edu Subject: Re: THE PARADOX Cc: 330_543 Status: RO Hi John, See Dave Famolari's most recent explanation. Not observation does not change the statistics of the process per se (like in quantum mechanics), but what you observe depends upon how you make your observations! Cheers, Chris Rose From crose@localhost.localdomain Mon Nov 17 22:45:21 1997 Return-Path: Received: from localhost.localdomain (ppp-38.ts-10.nyc.idt.net) by boom.rutgers.edu (4.1/SMI-4.1) id AA02196; Mon, 17 Nov 97 22:45:17 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id XAA03363; Mon, 17 Nov 1997 23:31:03 -0500 Date: Mon, 17 Nov 1997 23:31:03 -0500 From: Christopher Rose Message-Id: <199711180431.XAA03363@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: grades! Cc: crose@boom.rutgers.edu Status: RO Hi Folks, I've finished grading. here they are. Please check the list against your previous quiz grade so we can nip errors in the bud. SSN Q1 Q2 7476 150 X 1850 145 115 3431 145 110 2692 160 110 1171 134 46 7837 95 70 4926 145 87 1403 135 61 1154 135 82 6974 100 79 7058 95 105 8296 160 110 0831 125 51 7793 145 95 8674 145 87 8441 135 97 2674 125 117 1930 120 47 2108 50 45 2111 115 60 2552 120 74 8931 135 143 1653 105 64 0505 115 85 0486 140 117 1368 130 80 mean: 85 sd: 30 hi: 143 lo: 45 Solutions out shortly! From crose@localhost.localdomain Mon Nov 17 22:46:00 1997 Return-Path: Received: from localhost.localdomain (ppp-38.ts-10.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA23834; Mon, 17 Nov 97 22:44:30 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id XAA03363; Mon, 17 Nov 1997 23:31:03 -0500 Date: Mon, 17 Nov 1997 23:31:03 -0500 From: Christopher Rose Message-Id: <199711180431.XAA03363@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: grades! Cc: crose@MOGLI.rutgers.edu Status: RO Hi Folks, I've finished grading. here they are. Please check the list against your previous quiz grade so we can nip errors in the bud. SSN Q1 Q2 7476 150 X 1850 145 115 3431 145 110 2692 160 110 1171 134 46 7837 95 70 4926 145 87 1403 135 61 1154 135 82 6974 100 79 7058 95 105 8296 160 110 0831 125 51 7793 145 95 8674 145 87 8441 135 97 2674 125 117 1930 120 47 2108 50 45 2111 115 60 2552 120 74 8931 135 143 1653 105 64 0505 115 85 0486 140 117 1368 130 80 mean: 85 sd: 30 hi: 143 lo: 45 Solutions out shortly! From crose@localhost.localdomain Mon Nov 17 22:55:26 1997 Return-Path: Received: from localhost.localdomain (ppp-38.ts-10.nyc.idt.net) by boom.rutgers.edu (4.1/SMI-4.1) id AA02201; Mon, 17 Nov 97 22:55:24 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id XAA03397; Mon, 17 Nov 1997 23:41:41 -0500 Date: Mon, 17 Nov 1997 23:41:41 -0500 From: Christopher Rose Message-Id: <199711180441.XAA03397@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: erratum Cc: crose@boom.rutgers.edu Status: RO The hi was 145 not 143 (grading error) From crose@localhost.localdomain Mon Nov 17 22:55:07 1997 Return-Path: Received: from localhost.localdomain (ppp-38.ts-10.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA23839; Mon, 17 Nov 97 22:54:46 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id XAA03397; Mon, 17 Nov 1997 23:41:41 -0500 Date: Mon, 17 Nov 1997 23:41:41 -0500 From: Christopher Rose Message-Id: <199711180441.XAA03397@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: erratum Cc: crose@MOGLI.rutgers.edu Status: RO The hi was 145 not 143 (grading error) From crose Mon Nov 17 23:18:58 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA23885; Mon, 17 Nov 97 23:18:44 EST Date: Mon, 17 Nov 97 23:18:44 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711180418.AA23885@MOGLI.rutgers.edu> To: 330_543 Subject: solutions Status: RO The solutions for q2 are on the web page. From alap@winlab.rutgers.edu Tue Nov 18 21:08:08 1997 X-UIDL: 16928e0b89a38ef75c998d009a264747 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA24993; Tue, 18 Nov 97 21:05:54 EST Received: from monk-a-asy-10.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id VAA04087; Tue, 18 Nov 1997 21:03:48 -0500 Message-Id: <347272DC.21E2@winlab.rutgers.edu> Date: Tue, 18 Nov 1997 21:02:20 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli Subject: Probabilistic Sorting Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Hello all, I would like to make some comments/questions about Problem 2 of exam (probabilistic sorting): We had 3 cases: N real numbers, N integer numbers, N real numbers between 0 and 1. Consider the second case (N integers numbers from 1 to N): I can see 2 situations here: 1) N numbers without repetition, 2) N numbers with repetition (in the range 1 to N) Professor Rose's solution assume case (1), which I believe is because it's a "set" of integers (no repetitions). But I think that if you know N and know that the numbers are from 1 to N (and no repetitions), then you actually need no inspections to write down your answer (just write the numbers in order). Actually he is using this same argument when he says that you need only (N-1) inspections. He assumes that if you inspect (N-1) numbers, you know the last. But I think if you inspect (N-2) then you also know the 2 last, and so on. So you need no inspections. (?) For case (2) you could use counting sort, which would take N inspections (which I think it's a more real situation). Consider now the third case (uniform[0,1]): I do not see how you can sort real numbers with complexity less than NlogN. I think on average you could have something better than NlogN, but this will not be the worst case, so that this will not be the complexity. I believe the complexity to sort N real numbers is always NlogN, no matter the prob. distribution. You may have something better on average (according to the prob dist), but is this the complexity ?? Hope someone can reply with some comments. Thanks! Regards, Ana. From chriskly@er7.rutgers.edu Tue Nov 18 23:16:19 1997 X-UIDL: 8e93c87a29355b6fada650df74fc5065 Return-Path: Received: from er7.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25078; Tue, 18 Nov 97 23:15:49 EST Received: (from chriskly@localhost) by er7.rutgers.edu (8.8.5/8.8.5) id XAA10308; Tue, 18 Nov 1997 23:13:42 -0500 (EST) Date: Tue, 18 Nov 97 23:13:41 EST From: Christine Kleiwerda To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Probabilistic Sorting In-Reply-To: Your message of Tue, 18 Nov 1997 21:02:20 -0800 Message-Id: Status: RO Ana - In the exam, I also had some problems with #2. I interpretted the problem to be the second situation you describe. > Consider the second case (N integers numbers from 1 to N): > > I can see 2 situations here: > 1) N numbers without repetition, > 2) N numbers with repetition (in the range 1 to N) > > Professor Rose's solution assume case (1), which I believe is because it's a "set" > of integers (no repetitions). But I think that if you know N and know that > the numbers are from 1 to N (and no repetitions), then you actually need no > inspections to write down your answer (just write the numbers in order). > Actually he is using this same argument when he says that you need > only (N-1) inspections. He assumes that if you inspect (N-1) numbers, you know the last. > But I think if you inspect (N-2) then you also know the 2 last, and so on. > So you need no inspections. (?) > I think your problem lies in the statement above "just write the numbers in order" You are given an unordered list and must put it in order. You look at one number and then know which position it must go in. (number k goes in position k). When you have filled in N-1 positions in the ordered list, the final number must go in the last position, so you only need N-1 inspections. After N-2 inspections, you have 2 numbers left to put in the ordered list and you will not know which of the 2 open spaces to put them in, unles you look at one of them. "Writing the numbers in order" means that you have to look at the numbers, thereby "inspecting" them. > For case (2) you could use counting sort, which would take N inspections (which I > think it's a more real situation). > I'm not familiar with counting sort. > Consider now the third case (uniform[0,1]): > > I do not see how you can sort real numbers with complexity less than NlogN. > I think on average you could have something better than NlogN, but this will not > be the worst case, so that this will not be the complexity. I believe the complexity > to sort N real numbers is always NlogN, no matter the prob. distribution. > You may have something better on average (according to the prob dist), but is > this the complexity ?? I have not looked at the third part in detail, but maybe you could have the complexity to sort a list, given that the list has a certain distribtuion. This would not be the complexity to sort *any* list, but the complexity to sort a list, with specified property (ie, certain distribution). Hope some of this helped you out! Christine > > Hope someone can reply with some comments. > Thanks! > > Regards, > Ana. > > From jsucec@ece.rutgers.edu Wed Nov 19 02:44:40 1997 X-UIDL: 4ea83e53844c6bca657ef6371ac44029 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25255; Wed, 19 Nov 97 02:43:38 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA07780; Wed, 19 Nov 1997 02:36:37 -0500 Date: Wed, 19 Nov 1997 02:36:36 -0500 (EST) From: John Sucec Reply-To: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Probabilistic Sorting In-Reply-To: <347272DC.21E2@winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Hi, Ana. I see you've already received a response on the first 2 cases so I'll try to answer to the 3rd case. (Note, I have not yet seen the web page solution posting, so I hope I am not repeating what's posted on the web page.) Right now, I'm looking at my old algorithms book (Introduction to Algorithms by Cormen, Leiserson and Rivest), and I see that the "Bucket Sort runs in linear time on the average." Bucket Sort is very similar to the Counting Sort (which is how a list of integers 1 to N is best sorted). The idea of Bucket Sort is to subdivide the interval [0,1] into N "buckets". We then "toss" each of our N independently chosen and uniformly distributed numbers over [0,1], into one of the N buckets, where each of these buckets represents an interval in the range {[0,1/N], [1/N,2/N], ..., [(N-1)/N,N]}. Tossing these N random numbers into its proper bucket requires O(N) time, just as when a list of integers 1 to N is sorted. Next, sort the numbers within each bucket. Define n_i to be the number of numbers in bucket i. Using Insertion Sort, these n_i elements can be sorted in O(n_i^2) time. However, it can be shown (I refer to my old text again), that E[n_i^2] = O(1). Therefore, the average time to sort the elements in bucket i will be O(1). We then have, at most, O(n) time spent bucketizing the original N elements and then O(1) time, on average, spent sorting the n_i elements in each of the N buckets. --> Bucket Sort requires O(n) inspections/comparisons, on average. I guess in the worst case, Bucket Sort might require O(N^2) comparisons. I guess this would occur if all N numbers happened to fall in a single bucket, i --> n_i=N, n_j=0 (j not equal to i). But this is the worst case, and supposedly, will rarely happen. Besides, you can always have a good old merge sort algorithm working in parallel to sort the case 3 sequence in O(N*logN) comparisons, just in case your program was racing a computer sorting a sequence of N arbitrary numbers. Anyway, I hope I explained this algorithm well enough. If not, perhaps relook at case 3 by considering the fact that since we have some a priori information on the sequence (we know it is confined to the range [0,1] with uniform distribution), its binary entropy may be less (and it is) than that associated with a sequence of N arbitrary unsorted numbers. And if the binary entropy (i.e., unpredictability) of case 3 is less than that of case 1, then we may be able to (and we can) sort case 3 in less time than case 1, at least ON AVERAGE. I hope this helps. ...John On Tue, 18 Nov 1997, Ana Lucia Pinheiro wrote: > Hello all, > > I would like to make some comments/questions about Problem 2 of exam (probabilistic > sorting): > > We had 3 cases: N real numbers, N integer numbers, N real numbers between 0 and 1. > > .... > > Consider now the third case (uniform[0,1]): > > I do not see how you can sort real numbers with complexity less than NlogN. > I think on average you could have something better than NlogN, but this will not > be the worst case, so that this will not be the complexity. I believe the complexity > to sort N real numbers is always NlogN, no matter the prob. distribution. > You may have something better on average (according to the prob dist), but is > this the complexity ?? > > Hope someone can reply with some comments. > Thanks! > > Regards, > Ana. > > From cwrice@att.com Wed Nov 19 09:00:31 1997 X-UIDL: 89dd729a661012f7943200635cd89173 Return-Path: Received: from zen.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25464; Wed, 19 Nov 97 09:00:30 EST Received: from att.com (kcgw1.att.com [192.128.133.151]) by zen.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id IAA11330 for ; Wed, 19 Nov 1997 08:53:48 -0500 Received: by kcgw1.att.com; Wed Nov 19 07:55 CST 1997 Received: from hoccson.ho.att.com (hoccson.ho.att.com [135.16.2.30]) by kcig1.att.att.com (AT&T/GW-1.0) with SMTP id HAA02403 for ; Wed, 19 Nov 1997 07:47:29 -0600 (CST) Received: from nj-mailnet.ho.att.com (mailnet.ho.att.com) by hoccson.ho.att.com (4.1/EMS-1.1.1 SunOS) id AA14268; Wed, 19 Nov 97 09:02:43 EST Received: by mailnet.ho.att.com with Internet Mail Service (5.0.1458.49) id ; Wed, 19 Nov 1997 08:58:15 -0500 Message-Id: From: "Rice, Chris" To: crose@ece.rutgers.edu Cc: "Dodley, JP" Subject: ECE542 in the Spring Date: Wed, 19 Nov 1997 08:57:51 -0500 X-Priority: 3 Mime-Version: 1.0 X-Mailer: Internet Mail Service (5.0.1458.49) Content-Type: text/plain Status: RO Dr. Rose, Two questions: 1) Will you be teaching the ECE542 Informaton Theory and Coding Class in the Spring?, and 2) If so, what is the book that you will be using? I sent this to you directly because I felt it was not of general interest to the class. Please feel free to post this message with your answer if you think the class would be interested. Sincerely, Chris Rice From alap@winlab.rutgers.edu Wed Nov 19 14:46:58 1997 X-UIDL: 8b2cb1e8580f58f7d019200583f01b9b Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA25766; Wed, 19 Nov 97 14:44:35 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id OAA12069; Wed, 19 Nov 1997 14:41:59 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id OAA24404; Wed, 19 Nov 1997 14:39:58 -0500 Date: Wed, 19 Nov 1997 14:39:58 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711191939.OAA24404@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Prob. Sorting Status: RO Thank you for all you that replied my message. About the N integers, I think I got confused in what is actually ordering a list , which is not necessary write down the list ordered, it may be to create an array that "position" your numbers in order, which then would take N-1 inspections. About the Uniform distribution, I still have a question in what is the definition of complexity. Do you have something like "average complexity" in opposite to "worst case complexity" ? I understand John's explanation, but that would be an "average complexity". Is that right? Thanks again. Ana. From crose@ece.rutgers.edu Wed Nov 19 15:37:25 1997 X-UIDL: d80563bf09859d9f9f234df3621945c1 Return-Path: Received: from localhost.localdomain (ppp-10.ts-5.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA25840; Wed, 19 Nov 97 15:37:06 EST Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id QAA00890; Wed, 19 Nov 1997 16:23:36 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <347358D7.C10E014C@ece.rutgers.edu> Date: Wed, 19 Nov 1997 16:23:35 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: Ana Lucia Pinheiro , 330_543@mogli.rutgers.edu Subject: Re: Probabilistic Sorting References: <347272DC.21E2@winlab.rutgers.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Ana Lucia Pinheiro wrote: > > Hello all, > > I would like to make some comments/questions about Problem 2 of exam (probabilistic > sorting): > > We had 3 cases: N real numbers, N integer numbers, N real numbers between 0 and 1. > > Consider the second case (N integers numbers from 1 to N): > > I can see 2 situations here: > 1) N numbers without repetition, > 2) N numbers with repetition (in the range 1 to N) > > Professor Rose's solution assume case (1), which I believe is because it's a "set" > of integers (no repetitions). But I think that if you know N and know that > the numbers are from 1 to N (and no repetitions), then you actually need no > inspections to write down your answer (just write the numbers in order). Hi Ana, you can't assume the numbers are in order. > Actually he is using this same argument when he says that you need > only (N-1) inspections. He assumes that if you inspect (N-1) numbers, you know the last. > But I think if you inspect (N-2) then you also know the 2 last, and so on. > So you need no inspections. (?) Nope. Once you've sorted the first N-1, you know the last (assumption 1) > > For case (2) you could use counting sort, which would take N inspections (which I > think it's a more real situation). Yes, repetition screws things up in that case. > > Consider now the third case (uniform[0,1]): > > I do not see how you can sort real numbers with complexity less than NlogN. > I think on average you could have something better than NlogN, but this will not > be the worst case, so that this will not be the complexity. I believe the complexity > to sort N real numbers is always NlogN, no matter the prob. distribution. > You may have something better on average (according to the prob dist), but is > this the complexity ?? > > Hope someone can reply with some comments. > Thanks! > > Regards, > Ana. The basic idea is that when the numbers are chosen at random from an interval, the closer a number is to the boundary, the more you can infer about it's position ON AVERAGE. The way to approach this is to inspect the numbers one by one and figure out the conditional probability on placement in the list. If the number we pick happens to be 1, then the probability it goes to the head of the list is 1 (same idea with 0). Continue this argument as you move away from 1. Another way to think about it is that the probability mass function on placement of numbers into bins decreases differently for the uniform on (0,1) case than for the free-for-all all reals case. -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** From crose@ece.rutgers.edu Wed Nov 19 15:39:26 1997 X-UIDL: 22225ac44ceba4f52bb1fda45067d30f Return-Path: Received: from localhost.localdomain (ppp-10.ts-5.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA25851; Wed, 19 Nov 97 15:39:02 EST Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id QAA00901; Wed, 19 Nov 1997 16:25:25 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <34735945.82F6DF4B@ece.rutgers.edu> Date: Wed, 19 Nov 1997 16:25:25 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: Christine Kleiwerda , 330_543@mogli.rutgers.edu Subject: Re: Probabilistic Sorting References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Christine Kleiwerda wrote: > > Ana - > > In the exam, I also had some problems with #2. I interpretted the > problem to be the second situation you describe. > > > Consider the second case (N integers numbers from 1 to N): > > > > I can see 2 situations here: > > 1) N numbers without repetition, > > 2) N numbers with repetition (in the range 1 to N) > > > > Professor Rose's solution assume case (1), which I believe is because it's a "set" > > of integers (no repetitions). But I think that if you know N and know that > > the numbers are from 1 to N (and no repetitions), then you actually need no > > inspections to write down your answer (just write the numbers in order). > > Actually he is using this same argument when he says that you need > > only (N-1) inspections. He assumes that if you inspect (N-1) numbers, you know the last. > > But I think if you inspect (N-2) then you also know the 2 last, and so on. > > So you need no inspections. (?) > > > > I think your problem lies in the statement above "just write the > numbers in order" You are given an unordered list and must put it in > order. You look at one number and then know which position it must go > in. (number k goes in position k). When you have filled in N-1 > positions in the ordered list, the final number must go in the last > position, so you only need N-1 inspections. After N-2 inspections, > you have 2 numbers left to put in the ordered list and you will not > know which of the 2 open spaces to put them in, unles you look at one > of them. "Writing the numbers in order" means that you have to look > at the numbers, thereby "inspecting" them. > > > For case (2) you could use counting sort, which would take N inspections (which I > > think it's a more real situation). > > > > I'm not familiar with counting sort. > > > Consider now the third case (uniform[0,1]): > > > > I do not see how you can sort real numbers with complexity less than NlogN. > > I think on average you could have something better than NlogN, but this will not > > be the worst case, so that this will not be the complexity. I believe the complexity > > to sort N real numbers is always NlogN, no matter the prob. distribution. > > You may have something better on average (according to the prob dist), but is > > this the complexity ?? > > I have not looked at the third part in detail, but maybe you could > have the complexity to sort a list, given that the list has a certain > distribtuion. This would not be the complexity to sort *any* list, > but the complexity to sort a list, with specified property (ie, > certain distribution). > > Hope some of this helped you out! > > Christine > > > > > > Hope someone can reply with some comments. > > Thanks! > > > > Regards, > > Ana. > > > > Nicely put! -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** From crose@ece.rutgers.edu Wed Nov 19 15:41:03 1997 X-UIDL: 790a76dbbc53ff02044dc62133ebd160 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25866; Wed, 19 Nov 97 15:41:02 EST Received: from localhost.localdomain (root@ppp-10.ts-5.nyc.idt.net [169.132.97.154]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id PAA16290 for ; Wed, 19 Nov 1997 15:33:56 -0500 Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id QAA00913; Wed, 19 Nov 1997 16:27:29 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <347359C1.95EB8B71@ece.rutgers.edu> Date: Wed, 19 Nov 1997 16:27:29 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: 330_543@mogli.rutgers.edu Subject: KEEP COMMENTS COMING! Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Keep comments coming on sorting problem. It's really a very interesting one. Ana's question helped me to see the glimerings of a formulation of the problem. More colloquy might help us all see farther (given my unfortunately limited time right now). -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** From crose@ece.rutgers.edu Wed Nov 19 15:52:30 1997 X-UIDL: 0c5e3a94ca784b92384d1b0d9bcf0e58 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25911; Wed, 19 Nov 97 15:52:29 EST Received: from localhost.localdomain (root@ppp-10.ts-5.nyc.idt.net [169.132.97.154]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id PAA16887 for ; Wed, 19 Nov 1997 15:45:23 -0500 Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id QAA00933; Wed, 19 Nov 1997 16:32:43 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <34735AFB.22C469CF@ece.rutgers.edu> Date: Wed, 19 Nov 1997 16:32:43 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: "Rice, Chris" , 330_501@mogli.rutgers.edu, 330_543@mogli.rutgers.edu Subject: Re: ECE542 in the Spring References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Rice, Chris wrote: > > Dr. Rose, > > Two questions: > > 1) Will you be teaching the ECE542 Informaton Theory and Coding Class > in the Spring?, and > 2) If so, what is the book that you will be using? > > I sent this to you directly because I felt it was not of general > interest to the class. Please feel free to post this message with your > answer if you think the class would be interested. > > Sincerely, > > Chris Rice Hi Chris, (Great Name!) Unfortunately, I'm on sabbatical next term and won't be teaching anything. However, I plan to offer IT in spring 1999. It think you'd LOVE the course since it's analytic but tells you a lot about the way the world operations. For example, in the networks course there was a problem on sorting which is intimately related to IT concepts. The book (WHICH I ABSOLUTELY LOVE) is Tom Cover's (Stanford) book (and Joy Thomas, one of Tom's old students) entitled Elements of Information Theory (WILEY) it's only "flaw" is that it has NO error correction coding. But for me, that's a benefit as opposed to a liability since it allows the freedom to explore the many ramifications of IT (statistics, stock market, thermodynamics among others). Cheers, Chris Rose -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** From chriskly@er4.rutgers.edu Wed Nov 19 16:16:55 1997 X-UIDL: 0231249fcda958dd3cdda5f55ea5e383 Return-Path: Received: from er4.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25962; Wed, 19 Nov 97 16:16:41 EST Received: (from chriskly@localhost) by er4.rutgers.edu (8.8.5/8.8.5) id QAA26153; Wed, 19 Nov 1997 16:13:00 -0500 (EST) Date: Wed, 19 Nov 97 16:13:00 EST From: Christine Kleiwerda To: alap@liman.Rutgers.EDU (Ana Lucia Pinheiro) Cc: 330_543@mogli.rutgers.edu Subject: Re: Prob. Sorting In-Reply-To: Your message of Wed, 19 Nov 1997 14:39:58 -0500 Message-Id: Status: RO > > Thank you for all you that replied my message. > > About the N integers, I think I got confused in what is actually > ordering a list , which is not necessary write down the list ordered, > it may be to create an array that "position" your numbers in order, > which then would take N-1 inspections. > > About the Uniform distribution, I still have a question in what is the > definition of complexity. Do you have something like "average complexity" > in opposite to "worst case complexity" ? I understand John's explanation, > but that would be an "average complexity". Is that right? > > Thanks again. > Ana. > The way I understand is that you can have an average case number of comparisons and a worst case number of comparisons. When refering to *complexity* I think it is usually assumed to be the worst case. This might be one of those issues where a word can have different shades of meaning, so that when you are talking iwth someone, you should be sure that you are both on the same page. If it was not specified, I would assume that complexity was refering to the worst case scenario. Anyone disagree? Christine From crose@ece.rutgers.edu Wed Nov 19 15:37:14 1997 X-UIDL: 455df61fec030dfe5026814ffe9fd6a3 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA25841; Wed, 19 Nov 97 15:37:13 EST Received: from localhost.localdomain (root@ppp-10.ts-5.nyc.idt.net [169.132.97.154]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id PAA16074 for ; Wed, 19 Nov 1997 15:30:07 -0500 Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id QAA00890; Wed, 19 Nov 1997 16:23:36 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <347358D7.C10E014C@ece.rutgers.edu> Date: Wed, 19 Nov 1997 16:23:35 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: Ana Lucia Pinheiro , 330_543@mogli.rutgers.edu Subject: Re: Probabilistic Sorting References: <347272DC.21E2@winlab.rutgers.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Ana Lucia Pinheiro wrote: > > Hello all, > > I would like to make some comments/questions about Problem 2 of exam (probabilistic > sorting): > > We had 3 cases: N real numbers, N integer numbers, N real numbers between 0 and 1. > > Consider the second case (N integers numbers from 1 to N): > > I can see 2 situations here: > 1) N numbers without repetition, > 2) N numbers with repetition (in the range 1 to N) > > Professor Rose's solution assume case (1), which I believe is because it's a "set" > of integers (no repetitions). But I think that if you know N and know that > the numbers are from 1 to N (and no repetitions), then you actually need no > inspections to write down your answer (just write the numbers in order). Hi Ana, you can't assume the numbers are in order. > Actually he is using this same argument when he says that you need > only (N-1) inspections. He assumes that if you inspect (N-1) numbers, you know the last. > But I think if you inspect (N-2) then you also know the 2 last, and so on. > So you need no inspections. (?) Nope. Once you've sorted the first N-1, you know the last (assumption 1) > > For case (2) you could use counting sort, which would take N inspections (which I > think it's a more real situation). Yes, repetition screws things up in that case. > > Consider now the third case (uniform[0,1]): > > I do not see how you can sort real numbers with complexity less than NlogN. > I think on average you could have something better than NlogN, but this will not > be the worst case, so that this will not be the complexity. I believe the complexity > to sort N real numbers is always NlogN, no matter the prob. distribution. > You may have something better on average (according to the prob dist), but is > this the complexity ?? > > Hope someone can reply with some comments. > Thanks! > > Regards, > Ana. The basic idea is that when the numbers are chosen at random from an interval, the closer a number is to the boundary, the more you can infer about it's position ON AVERAGE. The way to approach this is to inspect the numbers one by one and figure out the conditional probability on placement in the list. If the number we pick happens to be 1, then the probability it goes to the head of the list is 1 (same idea with 0). Continue this argument as you move away from 1. Another way to think about it is that the probability mass function on placement of numbers into bins decreases differently for the uniform on (0,1) case than for the free-for-all all reals case. -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** ************************************************* From crose Wed Nov 19 16:42:09 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA26027; Wed, 19 Nov 97 16:41:57 EST Date: Wed, 19 Nov 97 16:41:57 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711192141.AA26027@MOGLI.rutgers.edu> To: alap@liman.Rutgers.EDU, chriskly@eden.rutgers.edu Subject: Re: Prob. Sorting Cc: 330_543@mogli.rutgers.edu Status: RO INTERSTING! You know, there's a cultural divide here. THe CS folks look at worst case whereas the EE folks look at average! I was thinking average in the worst case, you ALWAYS HAVE N LOG N From alap@winlab.rutgers.edu Wed Nov 19 16:59:22 1997 X-UIDL: 62b200ccde85cc39591a5ac06693efb6 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA26054; Wed, 19 Nov 97 16:59:07 EST Received: from raleigh.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA14717; Wed, 19 Nov 1997 16:56:55 -0500 Received: by raleigh.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id QAA03968; Wed, 19 Nov 1997 16:56:55 -0500 Date: Wed, 19 Nov 1997 16:56:55 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711192156.QAA03968@raleigh.winlab.rutgers.edu> To: 330_543@mogli Subject: Re: Prob. Sorting Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Content-Md5: dR0/9fouaKphu8HV9bQJdg== Status: RO > From crose@MOGLI Wed Nov 19 16:39:48 1997 > Date: Wed, 19 Nov 97 16:41:57 EST > From: Christopher Rose > To: alap@winlab.rutgers.edu, chriskly@eden > Subject: Re: Prob. Sorting > Cc: 330_543@mogli > > INTERSTING! > > You know, there's a cultural divide here. THe CS folks look at worst > case whereas the EE folks look at average! I was thinking average > > in the worst case, you ALWAYS HAVE N LOG N > Yes, I believe the sorting complexity is always NlogN but if the numbers are integers then you can do better (complexity N). If you have the prob.distrib. then you can do better ON AVERAGE, and then the kind of sorting you should use will depend on the prob.dist. (e.g. for the uniform case see John's e-mail). But the worst case would still be NlogN. Thanks, Ana. From crose@ece.rutgers.edu Wed Nov 19 18:33:23 1997 X-UIDL: 967a6a61a20f92678765a0cd8a21e131 Return-Path: Received: from localhost.localdomain (ppp-10.ts-5.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA26110; Wed, 19 Nov 97 18:33:13 EST Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id TAA01522; Wed, 19 Nov 1997 19:19:40 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <3473821B.A568D7AD@ece.rutgers.edu> Date: Wed, 19 Nov 1997 19:19:39 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Prob. Sorting References: <199711192156.QAA03968@raleigh.winlab.rutgers.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Ana Lucia Pinheiro wrote: > Yes, I believe the sorting complexity is always NlogN but if the numbers > are integers then you can do better (complexity N). > If you have the prob.distrib. then you can do better ON AVERAGE, > and then the kind of sorting you should use will depend on the prob.dist. > (e.g. for the uniform case see John's e-mail). But the worst case would > still be NlogN. > Thanks, > Ana. Whoops! I meant the worst case for an unknown set of numbers is N log N. If you know all the numbers (like the set from 1 to N) then you take only (N-1) inspections (which is still of O(N) :) Cheers, Chris Rose PS: There is a tight entropic argument for all this. But you should stop searching for it. At least for the continous distribution it depends on the differential entropy (for continous distributions) and that's not worth your looking into. From jsucec@ece.rutgers.edu Wed Nov 19 23:07:02 1997 X-UIDL: bd6c5c2befdbf145ffa0a502ec777230 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA26314; Wed, 19 Nov 97 23:06:36 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id WAA04314; Wed, 19 Nov 1997 22:59:26 -0500 Date: Wed, 19 Nov 1997 22:59:26 -0500 (EST) From: John Sucec To: Christine Kleiwerda Cc: Ana Lucia Pinheiro , 330_543@MOGLI.rutgers.edu Subject: Re: Prob. Sorting In-Reply-To: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Christine, yes I agree that "complexity" may be a confusing term, here. I prefer to use "run-time" when referring to how computationally intensive an algorithm might be. Anyway, "worst-case" and "averarge" run-time for a particular algorithm may be the same order of magnitude (as is the case for the Counting Sort algorithm) or may be of different order of magnitude (as is the case for the Bucket Sort algorithm). For example, employing the Counting Sort, one can always sort a list of integers, 1 to N, in not more than O(N) run-time. Further, O(N) also happens to be the average run-time of the Counting Sort algorithm because there is no particular arrangement of integers, 1 to N, that sorts any faster than any other arrangement via Counting Sort. Now, in the case of the Bucket Sort, it is statistically possible that a disproportionate fraction of the randomly generated numbers on [0,1] lie in a single sub-interval (bucket) i. The Bucket Sort algorithm specifies next that Insertion Sort be used to sort the n_i numbers in bucket i. The Insertion Sort requires O(n_i^2) run-time to sort the n_i elements. Therefore, if we're "unlucky" and have all N numbers fall in bucket i, then the Bucket Sort algorithm runs in O(N^2) time... This is the worst-case run time for the Bucket sort. Fortunately, however, given N occurrences of a uniform[0,1] random number generator, the possibility of being presented such a pathological arrangement of numbers is statistically remote. The arrangement of the N numbers will, on average, result in n_i=1. This will allow us to complete the Bucket Sort in O(N) time. In contrast with the Counting Sort algorithm, therefore, the run-time of the Bucket Sort algorithm IS dependent on the arrangement of the N randomly generated numbers from Uniform[0,1]. Thus, worst case run-time does not equal average case run-time for the Bucket Sort. Note, I believe that as N-->Infininty, the probability that a worst case arrangement of numbers occurs, goes to zero. (Otherwise, it wouldn't make much sense to use Bucket Sort and we would be better off using the Merge Sort.) Well, that's how I think of run-time (i.e. complexity). That's also how I think of worst case and average run-times for a particular algorithm. Also, it is good to remember that we have to carefully choose the algorithm we use to match the discipline (e.g., we can't use Counting Sort to order N arbitrarily generated numbers). ...John On Wed, 19 Nov 1997, Christine Kleiwerda wrote > > > The way I understand is that you can have an average case number of > comparisons and a worst case number of comparisons. When refering to > *complexity* I think it is usually assumed to be the worst case. > > This might be one of those issues where a word can have different shades of > meaning, so that when you are talking iwth someone, you should be sure that > you are both on the same page. If it was not specified, I would assume that > complexity was refering to the worst case scenario. > > Anyone disagree? > > Christine > From crose Thu Nov 20 00:39:03 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA26480; Thu, 20 Nov 97 00:38:48 EST Date: Thu, 20 Nov 97 00:38:48 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711200538.AA26480@MOGLI.rutgers.edu> To: chriskly@eden.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Prob. Sorting Cc: 330_543@MOGLI.rutgers.edu, alap@liman.Rutgers.EDU Status: RO John, That was a VERY nice treatment of sorting. There is still an information theoretic answer lurking out there (for the average case), but your careful comparison of the various possibilities was nice. It would be nice if we could just use combination, bounds etc on the known sorting algorithms on the (0,1) uniform problem. However, then the answer is dependent upon the particular algorithm used. With an IT approach the answer is INDEPENDENT of the method used and forms a true lower bound on average "complexity" suitably defined. Cheers, Chris Rose From alap@winlab.rutgers.edu Tue Nov 25 12:54:05 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA02574; Tue, 25 Nov 97 12:53:02 EST Received: from exponent.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id MAA09970; Tue, 25 Nov 1997 12:49:50 -0500 Received: by exponent.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id MAA05017; Tue, 25 Nov 1997 12:41:48 -0500 Date: Tue, 25 Nov 1997 12:41:48 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199711251741.MAA05017@exponent.winlab.rutgers.edu> To: 330_543@mogli Subject: Questions in Markov Chain Example Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Content-Md5: nxrjDgJIdfl2PDp0KjoG7w== Status: RO I have questions related to Markov chain: Example: Weather Forecast (From INTRODUCTION TO PROBABILTY MODELS, Sheldon Ross) Let state 0 be "rainning today" and state 1 "not rainning today" and the weather tomorrow depends only on the weather today. Let the transition from 0 to 0 (rains today and rains tomorrow) equal to 0.7 and from 1 to 0 (does not rain today and rains tomorrow) equal to 0.4. The transition matrix is: P= | 0.7 0.6 | | 0.3 0.4 | The steady state is: P_0 = 2/3 P_1 = 1/3 What is the probability that it will rain for 4 days from today given that it is rainning today ? I thought that would be (0.7)^4 = 0.24 But I saw in a book that I should use Chapman-Kolmogorov equation and that will be to multiply the matrix P by itself 4 times and pick P_00, which is equal to 0.5749. I believe that's because the book is considering that it will rain for 4 days and I am considering that it will rain for 4 consecutives days. In the book solution we could have transitions from 0 to 1 and then back to 0, where in my solution I want to be in state 0 for 4 consecutive days. Is that the difference? If the information that "it is rainning today" was not given we would have to multiply our result by P_0 in both cases. Is that correct? From jsucec@ece.rutgers.edu Mon Nov 24 20:47:41 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA01835; Mon, 24 Nov 97 20:47:40 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id UAA24146 for ; Mon, 24 Nov 1997 20:39:39 -0500 Date: Mon, 24 Nov 1997 20:39:39 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu Subject: h(t) + p(t) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, concerning the question I asked about the particular solution of the bowl example done in 501 on Thursday, I have since checked my differential equations notes and see by particular solution, you refer to the contribution of the drive term, in this case, the force of gravity... It has simply been a while since my undergraduate differential equations class days. Thanks for the clarification. (A reply to this E-Mail is not anticipated.) ...John From crose Tue Nov 25 13:55:06 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA02684; Tue, 25 Nov 97 13:54:33 EST Date: Tue, 25 Nov 97 13:54:33 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9711251854.AA02684@MOGLI.rutgers.edu> To: 330_543, alap@winlab.rutgers.edu Subject: Re: Questions in Markov Chain Example Status: RO COudl you compute the matrix multiply. Remember, there are a number of sequences from the current day which lead to rain 4 days from now. You computed the probability that it will rain 4 days in a row (you already said this I know). Cheers, Chris Rose From alap@winlab.rutgers.edu Tue Nov 25 15:30:25 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA02841; Tue, 25 Nov 97 15:30:01 EST Received: from fading.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id PAA12958; Tue, 25 Nov 1997 15:26:49 -0500 Received: from fading by fading.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id PAA08389; Tue, 25 Nov 1997 15:18:51 -0500 Message-Id: <199711252018.PAA08389@fading.winlab.rutgers.edu> Date: Tue, 25 Nov 1997 15:18:51 -0500 (EST) From: Ana Lucia Pinheiro Reply-To: Ana Lucia Pinheiro Subject: Re: Questions in Markov Chain Example To: 330_543@mogli Mime-Version: 1.0 Content-Type: TEXT/plain; charset=us-ascii Content-Md5: TGQbp8KKjPItGCLv9BbEnw== X-Mailer: dtmail 1.2.0 CDE Version 1.2 SunOS 5.6 sun4u sparc Status: RO I made a mistake in last e-mail, the transition matrix is: P= | 0.7 0.4 | | 0.3 0.6 | Then P_0 = 4/7 and P_1 = 3/7 The others numbers are correct. Just correcting because someone may try to do the problem and check with my answers and it would not match! Thanks Ana. > Date: Tue, 25 Nov 1997 12:41:48 -0500 > From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) > To: 330_543@mogli > Subject: Questions in Markov Chain Example > Mime-Version: 1.0 > Content-Transfer-Encoding: 7bit > Content-Md5: nxrjDgJIdfl2PDp0KjoG7w== > > I have questions related to Markov chain: > > Example: Weather Forecast (From INTRODUCTION TO PROBABILTY MODELS, Sheldon Ross) > > Let state 0 be "rainning today" and state 1 "not rainning today" and the > weather tomorrow depends only on the weather today. > Let the transition from 0 to 0 (rains today and rains tomorrow) > equal to 0.7 and from 1 to 0 (does not rain today and rains tomorrow) > equal to 0.4. > The transition matrix is: > > P= | 0.7 0.6 | > | 0.3 0.4 | > > The steady state is: > P_0 = 2/3 > P_1 = 1/3 > > What is the probability that it will rain for 4 days from today given that > it is rainning today ? > I thought that would be (0.7)^4 = 0.24 > But I saw in a book that I should use Chapman-Kolmogorov equation and that will > be to multiply the matrix P by itself 4 times and pick P_00, which is equal > to 0.5749. > I believe that's because the book is considering that it will rain for > 4 days and I am considering that it will rain for 4 consecutives days. > In the book solution we could have transitions from 0 to 1 and then back to 0, > where in my solution I want to be in state 0 for 4 consecutive days. > Is that the difference? > If the information that "it is rainning today" was not given we would have to > multiply our result by P_0 in both cases. Is that correct? From jsucec@ece.rutgers.edu Sun Nov 30 13:31:15 1997 X-UIDL: fb1752f17fa4fa4a5535b5eb6a2fe085 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA07621; Sun, 30 Nov 97 13:29:32 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id NAA04131 for <330_501@MOGLI.rutgers.edu>; Sun, 30 Nov 1997 13:20:29 -0500 Date: Sun, 30 Nov 1997 13:20:28 -0500 (EST) From: John Sucec To: 330_501@MOGLI.rutgers.edu Subject: Question on Ch5P8 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO I have a question on the text problem Ch5P8 that was assigned for Homework set #8. I'd appreciate comments by others who have worked this problem, too... It seems to me that this problem can be very easily solved by simply multiplying both of sides of Eq. 5.19 by (s*I - A), then doing the necessary algebra to isolate U(s) and finally taking the inverse Laplace transform to find u(t). Doing this, I get u(t) = -(1 + t)*unit_step(0) where unit_step(0) is 1 for t>=0, and 0 otherwise. My question is, is this a valid approach for finding u(t)? The reasons why I doubt this approach are because when I back substitute my solution for u(t) into Eq. 3.19, I get something very messy that doesn't appear to work, and also because the solution statement for this problem does not consider this approach. Anyway, the approach I outlined above makes sense to me and seems more direct than the solution statement because it does not require INV(s*I - A) to be calculated. However, I have not been able to verify the results successfully and it is quite possible that my approach is somehow inherently flawed. If this approach is incorrect, I'd appreciate it if anyone could explain to me where it is flawed. Thanks. ...John From anagha@er7.rutgers.edu Mon Dec 1 11:19:08 1997 X-UIDL: 3d153fb917edd77471a2777ce224d18d Return-Path: Received: from er7.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA08549; Mon, 1 Dec 97 11:14:03 EST Received: (from anagha@localhost) by er7.rutgers.edu (8.8.5/8.8.5) id LAA25902 for 330_501@mogli.rutgers.edu; Mon, 1 Dec 1997 11:09:51 -0500 (EST) Date: Mon, 1 Dec 97 11:09:51 EST From: Anagha Kelkar To: 330_501@mogli.rutgers.edu Subject: Fixed points Message-Id: Status: RO Prof. Rose, What is the difference between fixed point and limit points? When are they the same?? Regards Anagha From crose@ece.rutgers.edu Mon Dec 1 15:52:59 1997 X-UIDL: caecf0d1fe8c60a600a151b8976b5454 Return-Path: Received: from localhost.localdomain (ppp-26.ts-1.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA08797; Mon, 1 Dec 97 15:49:37 EST Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id QAA00943; Mon, 1 Dec 1997 16:33:41 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <34832D34.5ACE6CCE@ece.rutgers.edu> Date: Mon, 01 Dec 1997 16:33:40 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: Anagha Kelkar Cc: 330_501@mogli.rutgers.edu Subject: Re: Fixed points References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Anagha Kelkar wrote: > > Prof. Rose, > > What is the difference between fixed point and limit points? > When are they the same?? > > Regards > Anagha HI, Fixed point is any point where the system is stationary (mapping maps to the same point ---> this means xdot = 0 for diff eqs and x(n+1) = x (n) for discrete maps). A Limit point is a fixed point which is approached through state evolution. So for example, in the inverted pendulum problem you have two fixed points but only one limit point ( when there is damping that is). Cheers -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** From jsucec@ece.rutgers.edu Mon Dec 1 22:08:32 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA09138; Mon, 1 Dec 97 22:08:15 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id VAA29444; Mon, 1 Dec 1997 21:58:57 -0500 Date: Mon, 1 Dec 1997 21:58:57 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu, 330_543@MOGLI.rutgers.edu Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: R Prof. Rose (and class), I am trying to get a grip on the material covered in Ch7. There was a lot of intense mathematics done in class and in the text. I would like to capture the key concepts, factoids, results, etc. that were generated in this analysis. I'd appreciate it if you or anyone else would comment or add to the list below, of Ch7 results that I consider to be the tools we should assimilate from this class. * Finite state models, state transitions and steady state probabilities - Semi-Markov --> not necessarily exponential holding time - Markov --> exponential holding time - Discrete time Markov --> self-transitions allowed - Alternating state renewal --> one state has 0 holding time - Steady state probabilities proportional to mean holding times * W(t) = aggregate traffic = sum of independent random variables (RVs) * Superposition of i.i.d. processes --> get Poisson process * "Transform of a sum of RVs is given by the product of the transforms of the RVs." (section 7.5, page 192) --> moment generating function for W(t). * Poisson point process (filtered Poisson process) has moment generating function that "is the exponential of a sum of exponentials". (section 7.6, page 197). * Markov Inequality, Law of Large Numbers, Central Limit Theorem. * Chernoff Bound for estimating traffic tail density. Again, I'd appreciate any comments, corrections or additions offered by you or the class. I also have one specific question in mind... In class, back on 11-Nov, I thought I heard you say something to the effect that Large Deviation Theory --> VERY IMPORTANT. Now I have read section 7.9 (Appendix - Improved Large Deviation Approximation), but I find the content to be very esoteric. Were you referring to the general result of this section or any of the detail of this section as being the Large Deviation material that is important, or did you have the Chernoff Bound, Markov Inequality, Law of Large Numbers stuff more in mind? Thanks. ...John From crose Tue Dec 2 04:13:55 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA09415; Tue, 2 Dec 97 04:12:39 EST Date: Tue, 2 Dec 97 04:12:39 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712020912.AA09415@MOGLI.rutgers.edu> To: 330_543@MOGLI.rutgers.edu, crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Greato! Status: R Hi John, Nice summary. Now as for the large deviations, the improved RESULT is more important. Forget about the derivation of it since as you point out, it's a pain in the bottom. The Chernoff bound is the workhorse and Joe's (and a guy named Weiss at Bell Labs) large deviations stuff are useful as well, but the details... One correction on your list however taht I caught in passing. The steady stae probability is is not proportioanl to the mean holding time. It's the average RESIDENCE time over the life of the process. Think of a continous time markov chain (birth death process) with C states, each with transition rate lambda to the right and mu to the left. If holding time were teh only determinant of steady state probability then the central states would all have the same steady state probs (do the math to see what I mean). You have no choice but to solve teh probability flow balance equations to determine the steady state. Cheers From crose@localhost.localdomain Sat Dec 6 19:26:09 1997 X-UIDL: 4c2c5a8c4ca39a80a8b5f08487cce60b Return-Path: Received: from localhost.localdomain (ppp-26.ts-11.nyc.idt.net) by boom.rutgers.edu (4.1/SMI-4.1) id AA04219; Sat, 6 Dec 97 19:26:03 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id UAA12348; Sat, 6 Dec 1997 20:11:32 -0500 Date: Sat, 6 Dec 1997 20:11:32 -0500 From: Christopher Rose Message-Id: <199712070111.UAA12348@localhost.localdomain> To: zwf@liman.Rutgers.EDU Subject: problem sets Cc: crose@boom.rutgers.edu Status: RO hi Wenfeng, The next problem sets are CHapter 7: problems 1 2 and 5 Chapter 8: problems 2 3 4 6 10 and 11 We'll need solutions posted quickly. I'll provide problems from chapter 10 soon as well. Cheers, Chris Rose PS: Class, GET CRACKING!!!!!!!!!!!!!!!!! From crose@localhost.localdomain Sat Dec 6 19:29:49 1997 Return-Path: Received: from localhost.localdomain (ppp-26.ts-11.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA15461; Sat, 6 Dec 97 19:29:27 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id UAA12355 for 330_543@mogli.rutgers.edu; Sat, 6 Dec 1997 20:12:25 -0500 Date: Sat, 6 Dec 1997 20:12:25 -0500 From: Christopher Rose Message-Id: <199712070112.UAA12355@localhost.localdomain> To: 330_543@mogli.rutgers.edu Subject: problems sets Status: R The next problem sets are CHapter 7: problems 1 2 and 5 Chapter 8: problems 2 3 4 6 10 and 11 I'll provide problems from chapter 10 soon as well. Cheers, Chris Rose PS: Class, GET CRACKING!!!!!!!!!!!!!!!!! PPS: Wenfeng is working on solutions. From zwf@winlab.rutgers.edu Sun Dec 7 21:53:59 1997 X-UIDL: b0d5dc6006a6bab70f2feba34c76bde5 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA16605; Sun, 7 Dec 97 21:53:58 EST Received: from dilbert.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id VAA07173; Sun, 7 Dec 1997 21:48:40 -0500 Received: from winlab.rutgers.edu by dilbert.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id VAA29391; Sun, 7 Dec 1997 21:48:38 -0500 Sender: zwf@winlab.rutgers.edu Message-Id: <348B6003.66E878CE@winlab.rutgers.edu> Date: Sun, 07 Dec 1997 21:48:35 -0500 From: Wenfeng Zhang X-Mailer: Mozilla 4.03 [en] (X11; I; SunOS 5.6 sun4u) Mime-Version: 1.0 To: Christopher Rose Subject: Re: problem sets References: <199712070111.UAA12348@localhost.localdomain> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: R Hi, Christopher, Please get the solution files: set5.tex, set5.ps set6.tex, set6.ps from /common/ECE543. Please tell me the problems from Chapter 10 ASAP. Regards, Wenfeng Christopher Rose wrote: > > hi Wenfeng, > > The next problem sets are > > CHapter 7: problems 1 2 and 5 > > Chapter 8: problems 2 3 4 6 10 and 11 > > We'll need solutions posted quickly. > > I'll provide problems from chapter 10 soon as well. > > Cheers, > > Chris Rose > > PS: Class, GET CRACKING!!!!!!!!!!!!!!!!! -- Wenfeng Zhang Post: BPO 23888, PO BOX 1119, Piscataway, NJ 08855 Phone: (732)463-1964 (H) http://www.eden.rutgers.edu/~wfzhang From jsucec@ece.rutgers.edu Sun Dec 7 03:12:22 1997 X-UIDL: 4b868d5a4eba86576511221488c091e6 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA15842; Sun, 7 Dec 97 03:12:21 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id DAA22167 for ; Sun, 7 Dec 1997 03:02:07 -0500 Date: Sun, 7 Dec 1997 03:02:06 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu Subject: Comments on Proof Notation Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, I am quite impressed with the proof presented in class (20-Nov-97) of the theorem that p_W(w) depends only on f(a). However, I have some doubts about how the notation of the aggregate measure function, f(a), that was seemingly modified a few times during the course of the proof... DEF: sum(i=1,n){*} = "sum over i = 1 to n of the function inside the { }" sum(v<=x){*} = "sum over all v <= x of the function inside the { }" The aggregate measure function, f(a), is equivalently defined in the text (Eq. 7.6.2) and in class (18-Nov-97) as f(a) = sum(i=1,n){lambda_i*b_i(a)) where n is the number of superposed Poisson processes, b_i(a) is the measure function of the ith process and lambda_i is the traffic arrival rate for the ith process. Now again, I am have no issues with the logic of the proof presented in class, but I have noticed in my notes, the following notational discrepencies which occurred over the course of the proof, presented on 20-Nov-97, concerning the aggregate measure function, f(a): I. "p_W(w) depends only on f(a) = sum(x<=a){b(x)}" II. f(a) = sum(x<=a){lambda_i*b_i(a)} III. f(x) = sum(v<=x){sum(i=1,K){lambda_i*b_i(v)}} IV. "So, phi_W(s) depends only on f(x) = sum(v<=x){lamda_i*b_i(x)}" It is possible that one or two of the above 4 discrepencies are due to transcription errors on my part. However, I suspect that at least 2 or 3 of these apparent discrepencies were due to being written on the board incorrectly. Anyway, here is how I have reconciled the discrepencies in my mind so that the proof remains consistent with aggregate measure function, defined in equation 7.6.2: I. This may have been either board typo or transcription error, because I have, in my notes, just 4 lines below this discrepency the theorem statement, "p_W(w) depends only on f(a)=sum(i=i,K){lambda_i*b_a(a)}". The other possibility is that perhaps what you really wrote or meant to write was "P_W(w) depends only on sum(x<=a){f(a)}". That is, the c.d.f. for W(t) depends only on the sum of the aggregate measure functions. Is this possible? II. I think this may have been a case where you were writing quickly and perhaps merged the definition of f(a) with the expression, T_i = sum(all a){b_i(a)}, which you also happened to have been writing on the board, a few lines previously. III. I believe the correct description of the term you were isolating at time, is sum(v<=x){f(v)}. IV. I think the wording of this QED statement should have been simply "phi_W(s) depends only on f(v)", or alternatively, "phi_W(s) depends only on sum(v<=x){f(v)}". Again, I feel comfortable with the logic of the proof, so I am not overly worried about these discrepencies, which I believe can be attributed to either transcription errors or board gremlins. Therefore, if my rationalizations appear to be reasonable, it is not necessary for you to respond to this E-Mail. However, if you think any of the rationalizations for the above discrepencies is incorrect, please feel free to comment on them so that I may be able update my reasoning. I have not addressed this E-Mail to the discussion group because it is somewhat lengthy and I don't want to clutter the discussion group if it turns out somehow that my rationalizations are bogus or all of the discrepencies happened to have been due to transcription errors on my part. However, if you think it is useful to post this E-Mail, please feel welcome to do so. Thanks. ...John From crose@ece.rutgers.edu Sun Dec 7 05:58:55 1997 X-UIDL: 70dda82fb4c61b5cfbdaaf156c7e059b Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA15948; Sun, 7 Dec 97 05:58:54 EST Received: from localhost.localdomain (root@ppp-26.ts-11.nyc.idt.net [169.132.99.170]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id FAA23107; Sun, 7 Dec 1997 05:48:32 -0500 Received: from ece.rutgers.edu (localhost [127.0.0.1]) by localhost.localdomain (8.8.5/8.8.5) with ESMTP id GAA14728; Sun, 7 Dec 1997 06:41:38 -0500 Sender: crose@mogli.rutgers.edu Message-Id: <348A8B71.B3E9AE09@ece.rutgers.edu> Date: Sun, 07 Dec 1997 06:41:37 -0500 From: Christopher Rose Organization: Rutgers.University X-Mailer: Mozilla 4.03 [en] (X11; I; Linux 2.0.30 i586) Mime-Version: 1.0 To: John Sucec Subject: Re: Comments on Proof Notation References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO John Sucec wrote: > > Prof. Rose, I am quite impressed with the proof presented in class > (20-Nov-97) of the theorem that p_W(w) depends only on f(a). However, I > have some doubts about how the notation of the aggregate measure function, > f(a), that was seemingly modified a few times during the course of the > proof... > > DEF: sum(i=1,n){*} = "sum over i = 1 to n of the function inside the { }" > sum(v<=x){*} = "sum over all v <= x of the function inside the { }" > > The aggregate measure function, f(a), is equivalently defined in the text > (Eq. 7.6.2) and in class (18-Nov-97) as > > f(a) = sum(i=1,n){lambda_i*b_i(a)) > > where n is the number of superposed Poisson processes, b_i(a) is the > measure function of the ith process and lambda_i is the traffic arrival > rate for the ith process. > > Now again, I am have no issues with the logic of the proof presented in > class, but I have noticed in my notes, the following notational > discrepencies which occurred over the course of the proof, > presented on 20-Nov-97, concerning the aggregate measure function, f(a): > > I. "p_W(w) depends only on f(a) = sum(x<=a){b(x)}" > II. f(a) = sum(x<=a){lambda_i*b_i(a)} > III. f(x) = sum(v<=x){sum(i=1,K){lambda_i*b_i(v)}} > IV. "So, phi_W(s) depends only on f(x) = sum(v<=x){lamda_i*b_i(x)}" > > It is possible that one or two of the above 4 discrepencies are due to > transcription errors on my part. However, I suspect that at least 2 or 3 > of these apparent discrepencies were due to being written on the board > incorrectly. Anyway, here is how I have reconciled the discrepencies in > my mind so that the proof remains consistent with aggregate measure > function, defined in equation 7.6.2: > > I. This may have been either board typo or transcription error, because > I have, in my notes, just 4 lines below this discrepency the theorem > statement, "p_W(w) depends only on f(a)=sum(i=i,K){lambda_i*b_a(a)}". > The other possibility is that perhaps what you really wrote or > meant to write was "P_W(w) depends only on sum(x<=a){f(a)}". That is, > the c.d.f. for W(t) depends only on the sum of the aggregate measure > functions. Is this possible? > > II. I think this may have been a case where you were writing quickly and > perhaps merged the definition of f(a) with the expression, > T_i = sum(all a){b_i(a)}, which you also happened to have been writing > on the board, a few lines previously. > > III. I believe the correct description of the term you were isolating at > time, is sum(v<=x){f(v)}. > > IV. I think the wording of this QED statement should have been simply > "phi_W(s) depends only on f(v)", or alternatively, "phi_W(s) depends > only on sum(v<=x){f(v)}". > > Again, I feel comfortable with the logic of the proof, so I am not overly > worried about these discrepencies, which I believe can be attributed to > either transcription errors or board gremlins. Therefore, if my > rationalizations appear to be reasonable, it is not necessary for you to > respond to this E-Mail. However, if you think any of the rationalizations > for the above discrepencies is incorrect, please feel free to comment on > them so that I may be able update my reasoning. > > I have not addressed this E-Mail to the discussion group because it is > somewhat lengthy and I don't want to clutter the discussion group if > it turns out somehow that my rationalizations are bogus or all of the > discrepencies happened to have been due to transcription errors on my > part. However, if you think it is useful to post this E-Mail, please feel > welcome to do so. > > Thanks. > > ...John HI John, Yup there are errors. The only one which is correct is III! The problem with I is that the lambda_i is missing. The problem with II and IV is that the arg of b() is wrong --- the argument should be the free variable. I think I need to provide a latex version of the proof since it's not in the book. Cheers, Chris ROse PS: I've taken the liberty of posting this to the class. -- ********************************************************************** * Dr. Christopher Rose * * * Associate Professor of * \ / -----------> \ / * * Electrical & Computer Engineering * | | * * Rutgers University -- WINLAB * | | * * (732) 445-5250 * * * crose@ece.rutgers.edu ******************************** * http://winwww.rutgers.edu/pub/about/people/staff/crose/crose.html * ********************************************************************** From shahid@er7.rutgers.edu Mon Dec 8 22:04:37 1997 X-UIDL: ead171183562d96c965c563f39fe6d0f Return-Path: Received: from er7.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA18068; Mon, 8 Dec 97 22:04:36 EST Received: from localhost (shahid@localhost) by er7.rutgers.edu (8.8.5/8.8.5) with SMTP id VAA16919 for ; Mon, 8 Dec 1997 21:59:08 -0500 (EST) Date: Mon, 8 Dec 1997 21:59:08 -0500 (EST) From: Shahid Kagal To: crose@MOGLI.rutgers.edu Subject: Chap 7 : Equation 7.4.1 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Respected Prof. Rose, I was going through the text book section 7.4 Superposition of Traffic. Here I am confused about two points. One is that while considering the superposition of traffic at an interface of a network are we supposed to consider all the states of a particular terminal connected to that interface or only one particular state as mentioned in the text Rk(t)=a if Sk(t)=s. If it is the latter case than I cannot understand how in Equation 7.4.1 we get the product part i.e. Prod(k=1 to K) Pk(a). What I tried to work out was that the product part involved all the states of each terminal but that argument fails when I consider the fact that the product goes from k=1 to K which would mean my assuming out of the blue that each terminal has K finite states. So I argued in another way that each of the K terminal has one particular state and since each has a distribution independent of the others hence we get the product form. But here again the confusing part becomes the summation. If my second argument has to hold than I cannot justify the summation part. One more question is whether equation 7.3.2 is correct. Should it not be fs's instead of fss' given in the text. In this context Sir, could you also elaborate on how to calculate frequency of occurrence of states in an arbitrary process. I will appreciate your reply, Thanks, Shahid. From jsucec@ece.rutgers.edu Tue Dec 9 21:55:29 1997 X-UIDL: ec8c73814991a0535d1bedd848da44f2 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA19351; Tue, 9 Dec 97 21:55:20 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id VAA04238; Tue, 9 Dec 1997 21:44:35 -0500 Date: Tue, 9 Dec 1997 21:44:35 -0500 (EST) From: John Sucec To: Christopher Rose Cc: 330_543@MOGLI.rutgers.edu Subject: Re: question In-Reply-To: <9712091739.AA18780@MOGLI.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: R I agree with the questioning of the f_ss' notation used in equation 7.3.2. Obviously, the author means f_ss' to be the discrete time transition probailities of going from state s' to state s. However, on the top of page of 183, the author defines, "Given a process is in the state s, it would visit the next state s' with probability f_ss'." I think the author is being inconsistent in the use of this notation. The only justification I can see for this is perhaps because the author is assuming an equilibrium conditions in the formulation of 7.3.2, then the discrete time transition probabilities f_ss' and f_s's will be the same. Therefore, equation 7.3.2 is valid. Either way, I do not like the seemingly loose notation usage. Concerning section 7.4, I believe that the author has simplified the notation for the random traffic process, R_k(t), by representing it is a distribution, p_k(a_k), which is also equivalent to saying that it is dependent on the state of the traffic process, s_k. So, instead of representing the distribution of the random traffic process, R_k(t), as a probability mass function of the state, s_k, the author represents it as probability mass funtion of the rate, a_k. This is allowable because there is a one to one mapping between the state, s_k, and the rate, a_k. I think the motivation for this "simplification" of notation done by the author, comes from the fact that the aggregrate traffic process, W(t), is proportional to the weighted sums of the rates, a_k. If, on the other hand, the random process being analyzed had nothing to due with the rates, a_k, the author may defined his left probability mass distributions defined in terms of the state, s_k, or some other characteristic that also has a one to one mapping with the state. ...John On Tue, 9 Dec 1997, Christopher Rose wrote: > > > > Prof. Rose, > > I was going through the text book section 7.4 Superposition of Traffic. > Here I am confused about two points. One is that while considering the > superposition of traffic at an interface of a network are we supposed to > consider all the states of a particular terminal connected to that > interface or only one particular state as mentioned in the text Rk(t)=a > if Sk(t)=s. If it is the latter case than I cannot understand how in > Equation 7.4.1 we get the product part i.e. Prod(k=1 to K) Pk(a). What I > tried to work out was that the product part involved all the states of > each terminal but that argument fails when I consider the fact that the > product goes from k=1 to K which would mean my assuming out of the blue > that each terminal has K finite states. So I argued in another way that > each of the K terminal has one particular state and since each has a > distribution independent of the others hence we get the product form. But > here again the confusing part becomes the summation. If my second argument > has to hold than I cannot justify the summation part. > > One more question is whether equation 7.3.2 is correct. Should it not be > fs's instead of fss' given in the text. In this context Sir, could you > also elaborate on how to calculate frequency of occurrence of states in an > arbitrary process. > > From crose Fri Dec 12 14:48:55 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA22484; Fri, 12 Dec 97 14:45:42 EST Date: Fri, 12 Dec 97 14:45:42 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712121945.AA22484@MOGLI.rutgers.edu> To: abei@us.ibm.com, crose@MOGLI.rutgers.edu Subject: Re: final Cc: 330_501, 330_543 Status: R Hi Abe, I've never had an open book final. And yest, you can have thre sheets both sides of notes! Cheers, Chris ROse From alap@winlab.rutgers.edu Fri Dec 12 18:24:10 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA22806; Fri, 12 Dec 97 18:23:47 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id SAA07276; Fri, 12 Dec 1997 18:17:40 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id SAA23238; Fri, 12 Dec 1997 18:17:39 -0500 Date: Fri, 12 Dec 1997 18:17:39 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712122317.SAA23238@gsm.winlab.rutgers.edu> To: 330_543@mogli Subject: Chp 8 Status: R Hello, Chernoff bound in Chp 7 gives a lower bound for log(1/P_b) log(1/P_b) >= sc - sum(...) Call admission police in Chp 8 tell us to admit a call if log(1/P_b) <= sc - sum(...) How can this happen if chernoff bound gives us this term as a lower bound ? Thank you, Ana. From jsucec@ece.rutgers.edu Sat Dec 13 01:36:04 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA23139; Sat, 13 Dec 97 01:35:41 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA12279; Sat, 13 Dec 1997 01:24:19 -0500 Date: Sat, 13 Dec 1997 01:24:18 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Chp 8 In-Reply-To: <199712122317.SAA23238@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: R Hello, it looks like you are comparing the inverse of Equation 7.8.1 with Equation 8.6.3, in our text. I am not sure if I can clarify anything adequately, here, but I am fairly certain the author is being consistent, in this case. I believe equation 8.6.3 implies that we admit a call if the offered traffic does NOT exceed the bound specified on the right hand side of Equation 7.8.1. That is, taking the inverse of 8.6.3, we get: ln(P_b) >= -s*C + u_W(s*) <--> P_b >= e^(-(s*C - u_W(s*)) So I think what the author is saying is that in developing our call admission policy, we first agree upon a tolerable P_b. Therefore, our traffic policing policy (specified/implied by 8.6.3) requires the RHS of 7.8.1 to be <= to the LHS of 7.8.1. Then when a new call admission request is submitted to the network, we admit the new call if the new call does NOT cause the RHS of Equation 7.8.1 to become larger than the LHS Equation 7.8.1. In other words, if the inequality expressed in 8.6.3 is ever violated, then we will have a case where the allowable P_b we agreed upon will be exceeded by accepting the most recent call admission request. I don't know if my explaination is sufficiently clear, but your question got me thinking on the right track about how the Chernoff bound is being applied to estimating/bounding blocking probabilities. So, thanks very much for raising this issue. ...John On Fri, 12 Dec 1997, Ana Lucia Pinheiro wrote: > > Hello, > > Chernoff bound in Chp 7 gives a lower bound for log(1/P_b) > > log(1/P_b) >= sc - sum(...) > > Call admission police in Chp 8 tell us to admit a call if > > log(1/P_b) <= sc - sum(...) > > How can this happen if chernoff bound gives us this term as a lower bound ? > > Thank you, > Ana. > From alap@winlab.rutgers.edu Sat Dec 13 15:14:37 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA23666; Sat, 13 Dec 97 15:13:07 EST Received: from monk-a-asy-8.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id PAA11322; Sat, 13 Dec 1997 15:06:50 -0500 Message-Id: <3493149D.1228@winlab.rutgers.edu> Date: Sat, 13 Dec 1997 15:05:01 -0800 From: Ana Lucia Pinheiro Organization: Rutgers University X-Mailer: Mozilla 2.02 (Win16; I) Mime-Version: 1.0 To: 330_543@mogli Subject: Product form, truncated chain, detailed balance Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: R Hello, In deriving the expression for the joint probability ( P_N(n) ) for L truncated independent chains it was shown that it is a product form with normalization because of the truncation. Professor Rose said this is true because of the detailed balance. I don't see which part is true because of detailed balance: the product or the normalization. I also do not understand when we can use the detailed balance (or better, when we cannot). I appreciate any help. Regards, Ana. From crose Sat Dec 13 16:08:35 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA23761; Sat, 13 Dec 97 16:08:16 EST Date: Sat, 13 Dec 97 16:08:16 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712132108.AA23761@MOGLI.rutgers.edu> To: alap@liman.Rutgers.EDU, jsucec@ece.rutgers.edu Subject: Re: Chp 8 Cc: 330_543@mogli.rutgers.edu Status: R Right on JOHN From jsucec@ece.rutgers.edu Sun Dec 14 01:46:55 1997 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA24200; Sun, 14 Dec 97 01:46:42 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA12677; Sun, 14 Dec 1997 01:35:10 -0500 Date: Sun, 14 Dec 1997 01:35:09 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@mogli.rutgers.edu Subject: Re: Product form, truncated chain, detailed balance In-Reply-To: <3493149D.1228@winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: R Ana, here's my take on the matter... Detailed balance in the steady state is an analytical assumption we make in order to achieve the concise state probability results. I think the justification of this assumption is based on the general dynamic equilibrium principle that if detailed flow balance did not hold, then the system would not be in steady state. That is, equilibrium would not yet have been achieved. So, I think the underlying assumption here is that the probability expressions we derive are for a system in steady state. The normalization factor is employed, as you probably know, becuase the sum (or integral) of a probability mass function (or probability density function) over all possible values, must be 1. Although I also see something in my notes to the effect of "(Will turn out, however, that we still have product form due to a normalization factor)", I think the real reason why product form is maintained for the truncated chain, is due to the fact that we can systematically delete states from the infinite chain without disrupting the ratios between the probabilities of the remaining states. Therefore, the state probability function of the truncated chain can be obtain by simply scaling the state probability function of the infinite chain by a normalization factor. ...John On Sat, 13 Dec 1997, Ana Lucia Pinheiro wrote: > Hello, > > In deriving the expression for the joint probability ( P_N(n) ) for > L truncated independent chains it was shown that it is a product form > with normalization because of the truncation. > > Professor Rose said this is true because of the detailed balance. > > I don't see which part is true because of detailed balance: > the product or the normalization. I also do not understand when we > can use the detailed balance (or better, when we cannot). > > I appreciate any help. > > Regards, > Ana. > > From crose Mon Dec 15 00:25:35 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA25055; Mon, 15 Dec 97 00:25:13 EST Date: Mon, 15 Dec 97 00:25:13 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712150525.AA25055@MOGLI.rutgers.edu> To: alap@liman.Rutgers.EDU, jsucec@ece.rutgers.edu Subject: Re: Product form, truncated chain, detailed balance Cc: 330_543@mogli.rutgers.edu Status: R IT's the systematic deletion which matters AND the fact that detailed balance happens to hold. that is, if the steady state of the unperturbed chain obeys detailed balance, then we're all set for deletions. If not, then you've got problems. Might be a good quiz problem to provide a pair of chains one of which obeys detailed balance and the other which does not. From alap@winlab.rutgers.edu Tue Dec 16 02:00:41 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA26190; Tue, 16 Dec 97 01:59:08 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA26633; Tue, 16 Dec 1997 01:52:27 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA29282; Tue, 16 Dec 1997 01:52:27 -0500 Date: Tue, 16 Dec 1997 01:52:27 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712160652.BAA29282@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: 330_543@mogli Status: R Does someone understand problem 10 (chp 8) , why P_B,i = exp(s a_i) P_w(W>C) ?? Thanks Ana. From crose Tue Dec 16 02:45:23 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA26230; Tue, 16 Dec 97 02:44:56 EST Date: Tue, 16 Dec 97 02:44:56 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712160744.AA26230@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Want your 50 cents back? Cc: 330_501@MOGLI.rutgers.edu Status: R HI John, Consider a ball at height h=0. You get the usual mgh as it's potential energy. So what happens when the ball is lower than h=0? PE is usually referenced to the zero force point, but can be referenced anywhere as long as you keep track of where. SO, I think taht soda is long gone and you earned it! Cheers, Chris Rose From cwrice@att.com Tue Dec 16 09:03:48 1997 Return-Path: Received: from att.com (kcgw2.att.com) by MOGLI.rutgers.edu (4.1/25-eef) id AA26449; Tue, 16 Dec 97 09:01:54 EST Received: by kcgw2.att.com; Tue Dec 16 07:39 CST 1997 Received: from hoccson.ho.att.com (hoccson.ho.att.com [135.16.2.30]) by kcig2.att.att.com (AT&T/GW-1.0) with SMTP id HAA01593 for <330_501@mogli.rutgers.edu>; Tue, 16 Dec 1997 07:44:09 -0600 (CST) Received: from nj-mailnet.ho.att.com (mailnet.ho.att.com) by hoccson.ho.att.com (4.1/EMS-1.1.1 SunOS) id AA01185; Tue, 16 Dec 97 08:59:49 EST Received: by mailnet.ho.att.com with Internet Mail Service (5.0.1458.49) id ; Tue, 16 Dec 1997 08:55:15 -0500 Message-Id: From: "Rice, Chris" To: 330_501@mogli.rutgers.edu Subject: Exam Questions Date: Tue, 16 Dec 1997 08:55:25 -0500 X-Priority: 3 Mime-Version: 1.0 X-Mailer: Internet Mail Service (5.0.1458.49) Content-Type: text/plain Status: R Dr. Rose, Two questions: 1) In class we said, if Re(Lambdai)<0, then asymptotically stable. Should this not be GLOBALLY ASYMPTOTICALLY STABLE? If not, why not? 2) Again in class, THM: Using the characteristic eqn (SUM(pk*sk), k=0 to n) If matrix A has Re(Lambdai)<0, then Pk>0. SHOULD this not be Pk>=0. If not, why not? That's all for now. Thanks, > Christopher W. Rice > AT&T, Advanced Communications Laboratory > 67 Whippany Road, WH 15F-215 > Whippany, NJ 07981 > P: (973) 386-4488 > F: (973) 386-7831 > E: cwrice@att.com > > From alap@winlab.rutgers.edu Tue Dec 16 13:05:13 1997 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA26718; Tue, 16 Dec 97 13:01:17 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id MAA00921; Tue, 16 Dec 1997 12:53:47 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id MAA29876; Tue, 16 Dec 1997 12:53:41 -0500 Date: Tue, 16 Dec 1997 12:53:41 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712161753.MAA29876@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: 330_543@mogli Status: R On Tue, 16 Dec 1997, Ana Lucia Pinheiro wrote: > Does someone understand problem 10 (chp 8) , why > P_B,i = exp(s a_i) P_w(W>C) > ?? P_B,i = P_U(U>C-ai) = P_w(W>C-ai) = exp(-(s(C-ai)-u_w(s))) => exp(s*ai)*exp(-(sC-u_w(s))) = exp(s*ai)P_w(W>C) Jason > > Thanks > Ana. > > Jason, Thank you. My problem is to figure out the equality P_w(W>C-ai) = exp(-(s(C-ai)-u_w(s))) where I believe you are using chernoff bound but writing as an equality (instead of <= ). Actuaaly the book does the same in some parts (equation 8.5.1, for example). Why can you use the bound with "=" sign, if in class Dr. Rose even showed that this bound sometimes is not very thight ?? Thanks again. Ana. From alap@winlab.rutgers.edu Wed Dec 17 16:42:57 1997 X-UIDL: 49e7f569ac2be964a73593918d5c2313 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA27886; Wed, 17 Dec 97 16:41:44 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id QAA14499; Wed, 17 Dec 1997 16:34:47 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id QAA03047; Wed, 17 Dec 1997 16:34:41 -0500 Date: Wed, 17 Dec 1997 16:34:41 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712172134.QAA03047@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: 330_543@mogli Status: RO Queuing Theory: When deriving Erlang blocking formula in class Prof. Rose did: sum(rho^i/i!) = e^rho I believe this is true only when the sum is from zero to infinity (in our example we had the sum from 0 to C). Is that right? Also, I cannot think about one example with birth-death process with infinit number of states. Any ideas ? I think that's the only case you can write the sum as an exponential, isn't it ? Thank you, Ana. ps. By the way, I am waiting if someone can reply my last e-mail (about the chernoff bound). I would appreciate. Thanks again. Ana. From crose Wed Dec 17 18:44:35 1997 X-UIDL: 0ff602d32dc29315c5291e7e166ec4f0 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA27927; Wed, 17 Dec 97 18:44:34 EST Date: Wed, 17 Dec 97 18:44:34 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712172344.AA27927@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: Yup Cc: crose@MOGLI.rutgers.edu Status: RO Yup e^x = sum from 0 to inf of x^k/k! Board error? I think the e^x came in (and canceled) for the truncated version. There's no such thing as infinity but if your waiting room is terribly large, it effectively looks infinite. From crose Wed Dec 17 18:44:55 1997 X-UIDL: ff710af519024946d3c8ff402c41b7d3 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA27931; Wed, 17 Dec 97 18:44:54 EST Date: Wed, 17 Dec 97 18:44:54 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712172344.AA27931@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: chernoff Cc: crose@MOGLI.rutgers.edu Status: RO I thought John answered the chernoff question From alap@liman.Rutgers.EDU Wed Dec 17 19:56:55 1997 X-UIDL: 7a19f8056cf9a42f443cda6d006a2110 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA27958; Wed, 17 Dec 97 19:56:54 EST Received: from liman (liman.rutgers.edu [128.6.110.6]) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with ESMTP id TAA15094 for ; Wed, 17 Dec 1997 19:44:41 -0500 Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id TAA16860; Wed, 17 Dec 1997 19:49:55 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id TAA03593; Wed, 17 Dec 1997 19:49:55 -0500 Date: Wed, 17 Dec 1997 19:49:55 -0500 From: alap@liman.Rutgers.EDU (Ana Lucia Pinheiro) Message-Id: <199712180049.TAA03593@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: crose@ece.rutgers.edu Status: RO Yes, Jason answered that quastion, but I still had a problem in understanding. Here follows the e-mail. Thank you very much. --- On Tue, 16 Dec 1997, Ana Lucia Pinheiro wrote: > Does someone understand problem 10 (chp 8) , why > P_B,i = exp(s a_i) P_w(W>C) > ?? P_B,i = P_U(U>C-ai) = P_w(W>C-ai) = exp(-(s(C-ai)-u_w(s))) => exp(s*ai)*exp(-(sC-u_w(s))) = exp(s*ai)P_w(W>C) Jason > > Thanks > Ana. > > Jason, Thank you. My problem is to figure out the equality P_w(W>C-ai) = exp(-(s(C-ai)-u_w(s))) where I believe you are using chernoff bound but writing as an equality (instead of <= ). Actuaaly the book does the same in some parts (equation 8.5.1, for example). Why can you use the bound with "=" sign, if in class Dr. Rose even showed that this bound sometimes is not very thight ?? Thanks again. Ana. From alap@winlab.rutgers.edu Wed Dec 17 20:28:46 1997 X-UIDL: 0c9507608fd4425b2e50ec5f98ec6812 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA27971; Wed, 17 Dec 97 20:28:20 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id UAA17041; Wed, 17 Dec 1997 20:21:21 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id UAA03787; Wed, 17 Dec 1997 20:21:21 -0500 Date: Wed, 17 Dec 1997 20:21:21 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712180121.UAA03787@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: 330_543@mogli Status: RO Product form: If we have C channels to share among L independent processes, we proved that we still have product form: P_n(N) = product{ (1-rho_i)^n_i / (1-rho_i)^(C_i+1) } --------------------------------------------- sum{all n_j st sum n_j <= C} (P-n(N)) That means, it's a truncated version (normalized) My problem is to figure out what is C_i in this case. This case is as if you were truncating twice: first you truncate from infinity to C_i and then you truncate over the "diagonal" of the chain. So, is in this case the c_i's given ?? How do you, or could you work with the problem if the c-i's are not given?? I just try to see the physical meaning of having different c_i's. In my opinion they only make sense if they are less than C, if not the first truncation is ntot necessary. Any coments ? REgards, Ana. From chriskly@er6.rutgers.edu Thu Dec 18 00:28:54 1997 X-UIDL: 617325a7abd4a48edcdfeef3120e0976 Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28026; Thu, 18 Dec 97 00:28:36 EST Received: (from chriskly@localhost) by er6.rutgers.edu (8.8.5/8.8.5) id AAA23246 for 330_543@mogli.rutgers.edu; Thu, 18 Dec 1997 00:21:35 -0500 (EST) Date: Thu, 18 Dec 97 0:21:35 EST From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Message-Id: Status: RO Ana - > When deriving Erlang blocking formula in class Prof. Rose did: > sum(rho^i/i!) = e^rho > I believe this is true only when the sum is from zero to infinity > (in our example we had the sum from 0 to C). Is that right? Yes, sum (x^i / i!) = e^x only when summed from 0 to infinity. I cannot find in my notes where Prof Rose does the inequality for 0 to C (maybe you can send the date? or maybe you made a mistake copying down?) The Erlang blocking formula comes from the probablity that there are C trunks being used. Looking at the M/M queue wiht infinite servers and infinte capacity, we get the probablity of C users as (rho^k * e^-rho)/k! When we normalize for the case that there is finite capacity of C, we get the Erlang blocking formula. > Also, I cannot think about one example with birth-death process > with infinit number of states. Any ideas ? I think that's the only > case you can write the sum as an exponential, isn't it ? I think that for our general purpose in this class as related to networks, the queues we are dealing with all have finite capacity, since in real life our resources will probably always be limited. IF you look into other topics, there are cases where infinite queues could be appropriate (as a quick example, say people waiting in line at the movies... theoreticallyk, the line could stretch for miles around the block...) For us, we needed to first look at the infinite queues so we could then understand the finite queues and obtain their probabilities through normalization of the infinite case. Hope this helps.... Christine From chriskly@er6.rutgers.edu Thu Dec 18 00:32:32 1997 X-UIDL: bc4a0017bd6a2fdc5597f8ce34072238 Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28030; Thu, 18 Dec 97 00:32:16 EST Received: (from chriskly@localhost) by er6.rutgers.edu (8.8.5/8.8.5) id AAA23576 for 330_543@mogli.rutgers.edu; Thu, 18 Dec 1997 00:25:15 -0500 (EST) Date: Thu, 18 Dec 97 0:25:15 EST From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Subject: exam 1, extra problem Message-Id: Status: RO Professor Rose - Have you gotten around to giving the answer and grading that extra credit problem from the first exam? christine From chriskly@er6.rutgers.edu Thu Dec 18 00:48:34 1997 X-UIDL: 6beb6fa616aa1a7f247ddc2cd0cb3c3e Return-Path: Received: from er6.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28038; Thu, 18 Dec 97 00:48:19 EST Received: (from chriskly@localhost) by er6.rutgers.edu (8.8.5/8.8.5) id AAA25324 for 330_543@mogli.rutgers.edu; Thu, 18 Dec 1997 00:41:19 -0500 (EST) Date: Thu, 18 Dec 97 0:41:18 EST From: Christine Kleiwerda To: 330_543@mogli.rutgers.edu Subject: truncation formula Message-Id: Status: RO Ana - > Product form: > If we have C channels to share among L independent processes, we proved that we > still > have product form: > P_n(N) = product{ (1-rho_i)^n_i / (1-rho_i)^(C_i+1) } > --------------------------------------------- > sum{all n_j st sum n_j <= C} (P-n(N)) > That means, it's a truncated version (normalized) Hmm... very interesting, looking back at my notes (Dec 2), in one spot I have written the formula as you write it and in a second place I have it written with C_i replaced by C (so there is no C_i in the second formula I have). I'm not sure which formula is correct and I cannot find it in our book. > My problem is to figure out what is C_i in this case. This case is as if you wer > e > truncating twice: first you truncate from infinity to C_i and then you truncate > over the "diagonal" of the chain. So, is in this case the c_i's given ?? How do > you, or could you work with the problem if the c-i's are not given?? I just try > to see the physical meaning of having different c_i's. In my opinion they only > make sense if they are less than C, if not the first truncation is ntot > necessary. Any coments ? I also cannot get a good handle on what the C_i would represent. My guess is that the equation with the C_i is incorrect and it should be C, which we usually know (or else are trying to find out). Anyone else? -- christine From jsucec@ece.rutgers.edu Thu Dec 18 01:56:31 1997 X-UIDL: 6f59b1c47d53eb50db91e8d77bf2247c Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28061; Thu, 18 Dec 97 01:56:00 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id BAA19106; Thu, 18 Dec 1997 01:43:43 -0500 Date: Thu, 18 Dec 1997 01:43:43 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@MOGLI.rutgers.edu Subject: Product Form and C_i's In-Reply-To: <199712180121.UAA03787@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Ana, I believe the C_i's are a design criteria based on some sort of cost function of the various call types. Recall in class (9-Dec-97), Dr. Rose presented a weighting function to use to calculate the optimal P_b,i terms. The C_i terms are then calculated from the these corresponding P_b,i terms via some sort of call admission policy, e.g., equation 8.6.3. I think you have the right basic idea of how C_i terms fit into the product form of the truncated traffic distribution. However, I do disagree somewhat with the form you presented. Here is how I look at it: COMB(C_i,n_i)*(1 - pho_i)*pho_i^n_i P_Ni(n_i) = ----------------------------------- 1 - pho_i^(C_i + 1) where COMB(C_i,n_i) = C_i!/((C_i - n_i)!*n_i!) Assuming L indempendent processes, the truncated multiprocess traffic distribution is the product of the P_Ni(n_i) distributions given above. This matches equation 8.1.5 in our text. By the way, I notice that in our class notes, the COMB(C_i,n_i) term does not appear in any of the product form traffic distributions expressions for the Bernoulli case, presented on 2-Dec-97. This contradicts my basic understanding of the Bernoulli distribution and also the expression 8.1.5 presented in the text... I know that this term vanishes in the limit as we assume infinite number of users to yield a Poisson expression (8.1.6), but when we are still in the Bernoulli world (finite number of users/terminals) I think we should include the COMB(C_i,n_i) term in our aggregate traffic distribution expressions. Can anyone provide independent confirmation of this fact? ...John On Wed, 17 Dec 1997, Ana Lucia Pinheiro wrote: > Product form: > > If we have C channels to share among L independent processes, we proved that we still > have product form: > P_n(N) = product{ (1-rho_i)^n_i / (1-rho_i)^(C_i+1) } > --------------------------------------------- > sum{all n_j st sum n_j <= C} (P-n(N)) > > That means, it's a truncated version (normalized) > > My problem is to figure out what is C_i in this case. This case is as if you were > truncating twice: first you truncate from infinity to C_i and then you truncate > over the "diagonal" of the chain. So, is in this case the c_i's given ?? How do > you, or could you work with the problem if the c-i's are not given?? I just try > to see the physical meaning of having different c_i's. In my opinion they only > make sense if they are less than C, if not the first truncation is ntot > necessary. Any coments ? > > REgards, > Ana. > > From alap@winlab.rutgers.edu Thu Dec 18 01:59:59 1997 X-UIDL: aea85fd2b75b7eeb300a4fbe7124ff65 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA28066; Thu, 18 Dec 97 01:59:47 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA18239; Thu, 18 Dec 1997 01:52:46 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA04178; Thu, 18 Dec 1997 01:52:46 -0500 Date: Thu, 18 Dec 1997 01:52:46 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712180652.BAA04178@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: 330_543@mogli Status: RO Prof Rose, Section 10.6 it's very complicated ! Are we responsabile for that ? Thanks, Ana. From alap@winlab.rutgers.edu Thu Dec 18 02:06:29 1997 X-UIDL: 3c4f37b81d8e564c3be1bdcaf0c3a7f5 Return-Path: Received: from liman (liman.rutgers.edu) by MOGLI.rutgers.edu (4.1/25-eef) id AA28072; Thu, 18 Dec 97 02:06:17 EST Received: from gsm.winlab.rutgers.edu by liman (SMI-8.6/SMI-SVR4) id BAA18260; Thu, 18 Dec 1997 01:59:16 -0500 Received: by gsm.winlab.rutgers.edu (SMI-8.6/SMI-SVR4) id BAA04195; Thu, 18 Dec 1997 01:59:15 -0500 Date: Thu, 18 Dec 1997 01:59:15 -0500 From: alap@winlab.rutgers.edu (Ana Lucia Pinheiro) Message-Id: <199712180659.BAA04195@gsm.winlab.rutgers.edu> Content-Type: text Apparently-To: 330_543@mogli Status: RO John, The combination term does not exist. Recall that P_n was calculated using the markov flow balance. REgards, Ana. From jsucec@ece.rutgers.edu Thu Dec 18 02:23:02 1997 X-UIDL: fa0b47459c3291ecd5a9d1151639b22f Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28078; Thu, 18 Dec 97 02:22:44 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA19280; Thu, 18 Dec 1997 02:10:27 -0500 Date: Thu, 18 Dec 1997 02:10:27 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu Cc: 330_543@MOGLI.rutgers.edu Subject: Product Form and C_i's (fwd) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, as per below, I believe there may have been a term missing in the product form of the truncated traffic probability distributions you presented in class, 2-Dec-97. According to the text (8.2.4), the truncated distribution expression for each type i traffic of binomial nature should be as given below. For the aggregate of L such i.i.d traffic types, the truncated distribution is simply the product of the individual truncated distributions, given below. Anyway, on 2-Dec-97 you presented on two occasions a binomial product form expression without the COMB(C_i,n_i) term that appears in equation 8.2.4. Can you please explain why you omitted this? I may be overlooking something very obvious here... I apologize if I have misunderstood something and have caused some confusion for others. Please advise. Thanks. ...John ---------- Forwarded message ---------- Date: Thu, 18 Dec 1997 01:43:43 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@MOGLI.rutgers.edu Subject: Product Form and C_i's Ana, I believe the C_i's are a design criteria based on some sort of cost function of the various call types. Recall in class (9-Dec-97), Dr. Rose presented a weighting function to use to calculate the optimal P_b,i terms. The C_i terms are then calculated from the these corresponding P_b,i terms via some sort of call admission policy, e.g., equation 8.6.3. I think you have the right basic idea of how C_i terms fit into the product form of the truncated traffic distribution. However, I do disagree somewhat with the form you presented. Here is how I look at it: COMB(C_i,n_i)*(1 - pho_i)*pho_i^n_i P_Ni(n_i) = ----------------------------------- 1 - pho_i^(C_i + 1) where COMB(C_i,n_i) = C_i!/((C_i - n_i)!*n_i!) Assuming L indempendent processes, the truncated multiprocess traffic distribution is the product of the P_Ni(n_i) distributions given above. This matches equation 8.1.5 in our text. By the way, I notice that in our class notes, the COMB(C_i,n_i) term does not appear in any of the product form traffic distributions expressions for the Bernoulli case, presented on 2-Dec-97. This contradicts my basic understanding of the Bernoulli distribution and also the expression 8.1.5 presented in the text... I know that this term vanishes in the limit as we assume infinite number of users to yield a Poisson expression (8.1.6), but when we are still in the Bernoulli world (finite number of users/terminals) I think we should include the COMB(C_i,n_i) term in our aggregate traffic distribution expressions. Can anyone provide independent confirmation of this fact? ...John On Wed, 17 Dec 1997, Ana Lucia Pinheiro wrote: > Product form: > > If we have C channels to share among L independent processes, we proved that we still > have product form: > P_n(N) = product{ (1-rho_i)^n_i / (1-rho_i)^(C_i+1) } > --------------------------------------------- > sum{all n_j st sum n_j <= C} (P-n(N)) > > That means, it's a truncated version (normalized) > > My problem is to figure out what is C_i in this case. This case is as if you were > truncating twice: first you truncate from infinity to C_i and then you truncate > over the "diagonal" of the chain. So, is in this case the c_i's given ?? How do > you, or could you work with the problem if the c-i's are not given?? I just try > to see the physical meaning of having different c_i's. In my opinion they only > make sense if they are less than C, if not the first truncation is ntot > necessary. Any coments ? > > REgards, > Ana. > > From jsucec@ece.rutgers.edu Thu Dec 18 02:25:16 1997 X-UIDL: f6a31b030fb2af30ac53a4781bd8580e Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28083; Thu, 18 Dec 97 02:25:05 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA19289; Thu, 18 Dec 1997 02:12:48 -0500 Date: Thu, 18 Dec 1997 02:12:48 -0500 (EST) From: John Sucec To: Ana Lucia Pinheiro Cc: 330_543@MOGLI.rutgers.edu Subject: Re: your mail In-Reply-To: <199712180659.BAA04195@gsm.winlab.rutgers.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Thank you, Ana. I guess that answers my latest E-Mail to Prof. Rose... I guess I'll have to review the balanced flow equations once more. Again, thanks. ...John On Thu, 18 Dec 1997, Ana Lucia Pinheiro wrote: > John, > > The combination term does not exist. Recall that P_n was calculated using > the markov flow balance. > REgards, > Ana. > > From jsucec@ece.rutgers.edu Thu Dec 18 02:35:19 1997 X-UIDL: 2e75a7e430f321074cccc35a0cc1a3d8 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28089; Thu, 18 Dec 97 02:35:08 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA19357; Thu, 18 Dec 1997 02:22:51 -0500 Date: Thu, 18 Dec 1997 02:22:51 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu Cc: 330_543@MOGLI.rutgers.edu Subject: Ch8, Figure 8 (convexity) Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, remember how in class on 9-Dec-97 I asked about what seemed to be a contradiction between the way you and the author of our text were drawing convexity sketches? Well, upon further review, I believe there is no contradiction after all. Figure 8 of Ch8 is the drawing by the author that seemed to contradict your convex hull sketches. Actually, what Figure 8 shows is the region R(P_b) and the convex curve drawn shows the boundary for the region where the probability of blocking, P_b, exceeds a particular threshold. What I think you were saying with your convex functions was that we design our call admission policy so that the region of allowable call states is convex. Therefore, both your drawings and Figure 8 should have convex curves sketched... No contradiction as you and the author appear to have been addressing different issues of the traffic engineering problem. ...John From jsucec@ece.rutgers.edu Thu Dec 18 02:50:29 1997 X-UIDL: ef80276a8f41b0dc70e3a255c7adbe90 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28097; Thu, 18 Dec 97 02:50:16 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA19397; Thu, 18 Dec 1997 02:37:59 -0500 Date: Thu, 18 Dec 1997 02:37:58 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu Cc: 330_543@MOGLI.rutgers.edu Subject: General Question about non SS conditions Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, I have a general question about traffic modelling for the conditions where we do not have balanced flow equations. If you have time to read this note and give a brief (sentence or two), general answer, that would be great... Most of our formulations (i.e., probabilities of blocking, expected delays, expected queue sizes, average throughput, etc.) are based on the assumption of steady state (i.e., balanced flow equilibrium equations). However, what becomes of our formulations under non steady state conditions? That is, what kind of model is require when, for example: a) Our network is starting from initial conditions (i.e. no network traffic) b) Equilibrium is disrupted by massive user bursting. Do these non steady state conditions ever matter? That is, do we care about what happens as network traffic grows from zero loading to steady state? And, is massive user bursting ever an issue in a well designed broadband network (i.e., if the network design already accounts for delay and blocking events under peak load conditions, is user bursting a moot issue at this point)? Anyway, I am taking our steady state model equations as gospel for now. I am just curious to know under what conditions (if any) would different formulations be required. Thanks. ...John From jsucec@ece.rutgers.edu Thu Dec 18 02:50:29 1997 X-UIDL: ef80276a8f41b0dc70e3a255c7adbe90 Return-Path: Received: from ece.rutgers.edu by MOGLI.rutgers.edu (4.1/25-eef) id AA28097; Thu, 18 Dec 97 02:50:16 EST Received: from localhost (jsucec@localhost) by ece.rutgers.edu (8.6.12+bestmx+oldruq+newsunq/8.6.6) with SMTP id CAA19397; Thu, 18 Dec 1997 02:37:59 -0500 Date: Thu, 18 Dec 1997 02:37:58 -0500 (EST) From: John Sucec To: crose@MOGLI.rutgers.edu Cc: 330_543@MOGLI.rutgers.edu Subject: General Question about non SS conditions Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Status: RO Prof. Rose, I have a general question about traffic modelling for the conditions where we do not have balanced flow equations. If you have time to read this note and give a brief (sentence or two), general answer, that would be great... Most of our formulations (i.e., probabilities of blocking, expected delays, expected queue sizes, average throughput, etc.) are based on the assumption of steady state (i.e., balanced flow equilibrium equations). However, what becomes of our formulations under non steady state conditions? That is, what kind of model is require when, for example: a) Our network is starting from initial conditions (i.e. no network traffic) b) Equilibrium is disrupted by massive user bursting. Do these non steady state conditions ever matter? That is, do we care about what happens as network traffic grows from zero loading to steady state? And, is massive user bursting ever an issue in a well designed broadband network (i.e., if the network design already accounts for delay and blocking events under peak load conditions, is user bursting a moot issue at this point)? Anyway, I am taking our steady state model equations as gospel for now. I am just curious to know under what conditions (if any) would different formulations be required. Thanks. ...John From crose Thu Dec 18 12:44:53 1997 X-UIDL: c0ba6c84d3f3ef51cd19486bc56e58e6 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28274; Thu, 18 Dec 97 12:44:25 EST Date: Thu, 18 Dec 97 12:44:25 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712181744.AA28274@MOGLI.rutgers.edu> To: alap@winlab.rutgers.edu Subject: NO Cc: 330_543 Status: R NO not responsible for all of 10.6. Just basic idea behind HOL Blocking and most basic analysis of it. From crose Thu Dec 18 12:46:21 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28279; Thu, 18 Dec 97 12:45:59 EST Date: Thu, 18 Dec 97 12:45:59 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712181745.AA28279@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: Ch8, Figure 8 (convexity) Cc: 330_543@MOGLI.rutgers.edu Status: R Yup! (I think ... need to look at Joe's picture more carefully). Cheers, Chris ROse From crose Thu Dec 18 12:49:11 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28283; Thu, 18 Dec 97 12:48:58 EST Date: Thu, 18 Dec 97 12:48:58 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712181748.AA28283@MOGLI.rutgers.edu> To: crose@MOGLI.rutgers.edu, jsucec@ece.rutgers.edu Subject: Re: General Question about non SS conditions Cc: 330_543@MOGLI.rutgers.edu Status: R Think a little abot the traffic model (the measure theory stuff). You can model a bursty source as one with high peak rate but low duration. So you can still get steady state (averaged over long times and many callers) results (look at Campbell's theorem for example). IN general however, burstiness is a pain in the BUTT. From crose Thu Dec 18 12:57:50 1997 X-UIDL: 36544091a247fe94acaad09c2bdfeefe Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA28292; Thu, 18 Dec 97 12:57:33 EST Date: Thu, 18 Dec 97 12:57:33 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712181757.AA28292@MOGLI.rutgers.edu> To: 330_543 Subject: unable to come Status: R HI FOlks, The exam is still on tonite (12/18/97 in TCB-109 from 4-7) but I'll not be able to come owing to a medical emergency here at home. I wish you the best of luck. I'll post grades by next week. CHeers, CHris ROse From crose Tue Dec 23 12:53:27 1997 Return-Path: Received: by MOGLI.rutgers.edu (4.1/25-eef) id AA00965; Tue, 23 Dec 97 12:52:29 EST Date: Tue, 23 Dec 97 12:52:29 EST From: Christopher Rose Full-Name: Christopher Rose Message-Id: <9712231752.AA00965@MOGLI.rutgers.edu> To: cwrice@att.com Subject: Re: Exam Solutions and Grades Cc: 330_501, 330_543 Status: R Hi Chris, Lots happened since tuesday last. I'll have to disappoint and get grades in by official Jan 5 deadline (SORRY!). If sooner I'll post of course. So try to have a happy and a merry and all that without the grades. Cheers (ALL) Chris Rose From crose@localhost.localdomain Mon Jan 5 03:10:58 1998 Return-Path: Received: from localhost.localdomain (ppp-17.ts-6.nyc.idt.net) by MOGLI.rutgers.edu (4.1/25-eef) id AA03431; Mon, 5 Jan 98 03:10:24 EST Received: (from crose@localhost) by localhost.localdomain (8.8.5/8.8.5) id DAA03277 for 330_543@mogli.rutgers.edu; Mon, 5 Jan 1998 03:47:29 -0500 Date: Mon, 5 Jan 1998 03:47:29 -0500 From: Christopher Rose Message-Id: <199801050847.DAA03277@localhost.localdomain> To: 330_543@mogli.rutgers.edu Status: R Grades are generous and FINAL. No borderline appeals since I've already bent over backwards to assign benefit-of-doubt grades. You can pick up your exam from dept sec'y later in the term. Sorry, but gotta go... Cheers Chris Rose SSN Q1 Q2 F WEIGHTED GRADE 1850 145 115 240 546.294 A 3431 145 110 225 523.787 A 2692 160 110 230 544.328 A 1171 134 46 235 448.576 B 7837 95 70 237 440.028 B 4926 145 87 230 502.27 B+ 1403 135 61 235 467.223 B 1154 135 82 200 453.137 B 6974 100 79 240 470.706 B 7058 95 105 185 423.571 B 8296 160 110 230 544.328 A 0831 125 51 240 451.000 B 7793 145 95 240 522.765 A 8674 145 87 195 463.478 B 8441 135 97 210 481.868 B+ 2674 125 117 240 528.647 A 1930 120 47 240 441.294 B 2108 50 45 125 241.483 C 2111 115 60 240 451.588 B 2552 120 74 225 456.434 B 8931 135 145 240 571.588 A 1653 105 64 240 446.294 B 0505 115 85 220 458.833 B 0486 140 117 240 543.647 A 1368 130 80 240 490.118 B+