Return-Path: william@bourbon.usc.edu Delivery-Date: Thu Dec 11 10:09:26 2008 X-Spam-Checker-Version: SpamAssassin 3.2.3 (2007-08-08) on merlot.usc.edu X-Spam-Level: X-Spam-Status: No, score=-2.4 required=5.0 tests=AWL,BAYES_00 autolearn=ham version=3.2.3 Received: from bourbon.usc.edu (bourbon.usc.edu [128.125.9.75]) by merlot.usc.edu (8.14.1/8.14.1) with ESMTP id mBBI9Po6027931 for ; Thu, 11 Dec 2008 10:09:25 -0800 Received: from bourbon.usc.edu (localhost.localdomain [127.0.0.1]) by bourbon.usc.edu (8.14.2/8.14.1) with ESMTP id mBBIBu2C029714 for ; Thu, 11 Dec 2008 10:11:56 -0800 Message-Id: <200812111811.mBBIBu2C029714@bourbon.usc.edu> To: cs551@merlot.usc.edu Subject: Re: Lecture 17, Slide 15 Date: Thu, 11 Dec 2008 10:11:56 -0800 From: Bill Cheng Someone wrote: > A follow up to the last question: > > > (X1, Z1, Y2, X3 where in Queue so Logical time would advance by 0.25) > > Isn't this incorrect, as the slope is determined by the number of active > flows/queues, and not the number of packets? So even though there are four > packets, there are only three flows/queues, so the logical time would > advance by 0.333? Sorry that I didn't answer that part in my previous e-mail since I answered it in my message with timestamp "Sat 06 Dec 14:01". I was focusing on the rest of the previous e-mail. You are correct and so were the slides. > Also, is there ever an interest for mapping the logical time back to real > time? By that I mean (at least as I understand it), the logical clock only > advances after each round (a 'pass through' of all the active flows/queues), > and as such, X1 and Z1 both depart at logical time 2. But, as there is only > one actual channel to exit on, either X1 must go first, or Z1 must go first, > so they must have different physical departure times, even though they have > the same logical departure times. Am I thinking of this correctly? If so, > how do you determine which packet goes first, and is there ever an interest > for mapping the logical time back to real time? Your thinking was correct. When I talked about slide 16 of lecture 17, I mentioned that a simple tie breaking rule was used since the finish time for packets X1 and Z1 are the same. We simply used the rule that, in case there is a tie, X goes before Y and Y goes before Z. The physical schedule comes from the sort order deteremined by the finish times of the packets. -- Bill Cheng // bill.cheng@usc.edu On Thu, Dec 11, 2008 at 8:20 AM, Bill Cheng wrote: > Someone wrote: > > > At Physical time = 4, Z4 arrives.Before that at Physical Time = 3, Logical > > time was 1.833 (X1, Z1, Y2, X3 where in Queue so Logical time would advance > > by 0.25). > > so at Physical time = 4, Logical time would be 1.833 + 0.25 = 2.083. > > > > But in the slide it is shown Arrival time of Z4 as 2.25 instead of 2.083 ??? > > > > 'm not sure if it means the same or not but was just confused :( > > It's probably more clear if you look at slide 14. > > When logical clock equals 2, packets X1 and Z1 left the system > and only packets X3 and Y2 are in the system. So, the slope > starting at logical clock = 2 (and physical clock = 3.5) is 1/2. > > The blue text in the picture shows what packets are currently > being served (and that determines the slope). > -- > Bill Cheng // bill.cheng@usc.edu