Return-Path: william@bourbon.usc.edu
Delivery-Date: Sun Oct 22 23:08:36 2006
X-Spam-Checker-Version: SpamAssassin 3.1.3 (2006-06-01) on merlot.usc.edu
X-Spam-Level:
X-Spam-Status: No, score=-2.9 required=5.0 tests=ALL_TRUSTED,AWL,BAYES_00,
NO_REAL_NAME autolearn=ham version=3.1.3
Received: from bourbon.usc.edu (bourbon.usc.edu [128.125.9.75])
by merlot.usc.edu (8.13.5/8.13.5) with ESMTP id k9N68aHx030045
for ; Sun, 22 Oct 2006 23:08:36 -0700
Received: from bourbon.usc.edu (localhost.localdomain [127.0.0.1])
by bourbon.usc.edu (8.13.5/8.13.5) with ESMTP id k9N65ZEE003071
for ; Sun, 22 Oct 2006 23:05:35 -0700
Message-Id: <200610230605.k9N65ZEE003071@bourbon.usc.edu>
To: cs551@merlot.usc.edu
Subject: Re: Fair queue question
Date: Sun, 22 Oct 2006 23:05:35 -0700
From: william@bourbon.usc.edu
Someone wrote:
> i have a question on the slide 28 of lecture 16.
> it says that FX1 = FZ1 = 2
> and AY2 = 1.5 but the FY2 = 2.5, why is that
> if we look at the rr equation it says F(i) = max( f(i-1), A(i) ) + P(i).
> over here the f(i-1) is 2 and (a(i) is 1.5, shudnt we consider 2, rather
> than 1.5 in f(i) for F(y2)
F(i-1) is the finish time of the previous packet *in the same queue*.
So, if F(i) is F_Y2, what's F(i-1)? Y2 is the first packet in queue
Y, therefore, there is no F(i-1) (or F(i-1) is negative infinity).
Therefore, max(F(i-1),A(i)) is simply A(i)=A_Y2 in this case.
--
Bill Cheng // bill.cheng@usc.edu
william@bourbon.usc.edu wrote:
> Someone wrote:
>
> > I've a question regarding the fairing queue example (pg 30).
> >
> > 1. At real time 3.5 and logical time 2, X1 and Z1 goes out (see
> > my calculation below). Then how did we get arrival time of Z4 to
> > be 2.24? Could you please explain. Also, could you please explain
> > how to handle the departure time (and its effect on the slope?
>
> It's not 2.24, it should be 2.25, I think. At real time 3.5
> and logical time 2, X1 and Z1 depart and Y2 and X3 are still
> in the system. So, the slope is 1/2. The next arrival event
> is at real time 4.0, so change in vertical axis is
> (4.0-3.5)/2=0.25. Therefore, real time 4.0 maps to logical
> time 2.25.
>
> I'm not sure what you meant by handling the departure time.
> --
> Bill Cheng // bill.cheng@usc.edu