Return-Path: william@bourbon.usc.edu Delivery-Date: Thu Dec 11 08:17:47 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 mBBGHl5l026756 for ; Thu, 11 Dec 2008 08:17:47 -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 mBBGKGXv028242 for ; Thu, 11 Dec 2008 08:20:16 -0800 Message-Id: <200812111620.mBBGKGXv028242@bourbon.usc.edu> To: cs551@merlot.usc.edu Subject: Re: Lecture 17, Slide 15 Date: Thu, 11 Dec 2008 08:20:16 -0800 From: Bill Cheng 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