Return-Path: william@bourbon.usc.edu
Delivery-Date: Mon Sep 11 07:05:40 2006
X-Spam-Checker-Version: SpamAssassin 3.1.0 (2005-09-13) on merlot.usc.edu
X-Spam-Level:
X-Spam-Status: No, score=-0.7 required=5.0 tests=ALL_TRUSTED,AWL,
MAILTO_TO_SPAM_ADDR,NO_REAL_NAME autolearn=ham version=3.1.0
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 k8BE5ere005938
for ; Mon, 11 Sep 2006 07:05:40 -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 k8BE2Dag001612
for ; Mon, 11 Sep 2006 07:02:13 -0700
Message-Id: <200609111402.k8BE2Dag001612@bourbon.usc.edu>
To: cs551@merlot.usc.edu
Subject: Re: Exponential distribution calculation
Date: Mon, 11 Sep 2006 07:02:13 -0700
From: william@bourbon.usc.edu
Someone wrote:
> I would like to get some clarification on the units part. The
> unit is always in seconds right?
>
> For example, if lambda = 0.5 and r = 0.5
>
> w (inter-arrival time)= (-1/0.5) * ln(0.5) = 1.386 seconds
>
> We need to convert them to milli seconds for display and sleep.
> Is this what you meant by getting units correct?
Yes. And if the lambda you kept around is in a different
unit, you need to go through a similar exercise to make
sure things are correct.
--
Bill Cheng // bill.cheng@usc.edu
----- Original Message -----
From: william@bourbon.usc.edu
Date: Sunday, September 10, 2006 11:00 pm
Subject: Re: Exponential distribution calculation
To: cs551@merlot.usc.edu
> Someone wrote:
>
> > I'm little bit confused about the exponential distribution
> > calculation. In the lecture notes it says,
> > y = 1 - e^(-mx).
> >
> > What is the parameter y?
>
> y is F(x). Usually, we plot F(x) on the vertical axis
> and x on the horizontal axis. The vertical axis is
> also known as the y-axis and the horizontal axis is
> also known as the x-axis.
>
> > I'm confused about calculating the w
> > given lamda or mu. Could you please illustrate the steps?
>
> r is an instance of y, to get w, you need F^{-1}(x), the
> inverse of F(x). To get F^{-1}(x) is easy since we have
> the equation for F(x):
>
> y = 1 - e^(-mx)
> e^(-mx) = 1 - y
> -mx = ln( 1 - y )
> x = (-1/m) ln( 1 - y )
>
> Therefore,
>
> w = F^{-1}(r) = (-1/m) ln( 1 - r )
>
> and m is either lambda or mu. I think I did the above
> correctly, but you should make sure what you get is
> correct (including the units).
> --
> Bill Cheng // bill.cheng@usc.edu
>