# Expontential distribution

1. Jan 5, 2012

### skwey

1. The problem statement, all variables and given/known data
3 people John, Paul and Naomi enter simultaniously a postoffice. There are only 2 clerks there, john and paul go first. The service time is exponentially distributed with parameters lamda1 and lamda2. Naomi must wait uintil either John or Paul is finised.

The time that Naomi spends in the post office is less than for John or Paul provided that
max(lamda1,lamda2)>c*min(lamda1,lamda2) for some constant c. What is this constant?

2. Relevant equations
E(X1)=1/lamda1, E(X2)=1/lamda2, x1 and x2 are the service times

I also use P(clerk one finish first)=P(X1>X2)=X1/(x1+X2)

3. The attempt at a solution

I tried to do this by saying that the expected service time for Naomi has to be less than the expected service time for the "oponent", this is the person of John or Paul left standing. I compare expected values from when Naomi starts, because it is a meomryless distribution. So when Naomi is serviced I can say that the oponents distribution starts from scratch. The oponent can be either John or Paul, depending on who finishes first.

E(Naomi service time-oponent service time)=

E(Naomi service time-oponent service time|Naomi gets service with the fastest clerk)*P(Naomi gets service with the fastest clerk)+
E(Naomi service time-oponent service time|Naomi gets service with the slowest clerk)*P(Naomi gets service with the slowest clerk)
=
max(lamda1,lamda2)/(lamda1+lamda2)*[1/max(lamda1,lamda2)-1/min(lamda1,lamda2)]
+min(lamda1,lamda2)/(lamda1+lamda2)*[1/min(lamda1,lamda2)-1/max(lamda1,lamda2)]
I say that this has to be less than 0 (<0) because then it means that Naomi has the lowest extected time

this gives max(lamda1,lamda2)>min(lamda1,lamda2)

so c=1

2. Jan 6, 2012

### lanedance

The question is not posed that well...

Do you mean find c, such that the average time Naomi waits is less than the average of one of John or Paul?

Also are you counting the time she waits whilst they are being served, before her service begins? If so early one of them must be quicker than Naomi, each time and on average.

As a start you could simplify notation wlog by setting lamda2>lambda1

Note also if required you could easily find the joint pdf, as the diffferent tellers finishing are independent events.

And whilst you know the probability of each teller finishing first, the tricky part may be working out how much time she spends waiting on average prior to the teller finishing with Paul (or John)...

3. Jan 6, 2012

### skwey

Here is the question.

I think my calculations in the first post is wrong, there I look at the mean of the difference of each run(where we look at both naomis time vs john or pauls.) but I think we should look at the difference of the means of Naomi vs the slowest of paul or john. That is mean[difference on each run] is not the same as difference of means. So to your question, in the first postI didnt have to look at the service time before Naomi got serviced, because it would cancel in the difference, but when I look at difference of total means, I take it into account:
I tried it agan and looked at the mean waiting time for Naomi, and said like you that lamda2>lamda1. x1, and x2 are service times for the two clerks for John and Paul, x3 is service time for naomi at either clerk 1 or 2
-
E(Naomi, time)=E(Naomi|x1<x2)*P(x1<x2)+E(Naomi|x2<x1)*P(x2<x1)
=E(x1+x3|x1<x2)*P(x1<x2)+E(x2+x3|x2<x1)*P(x2<x1)
=[E(x1|x1<x2)+E(x3)]*P(x1<x2)+[E(x2|x2<x1)+E(x3)]*P(x2<x1)
Here I used that x3 is independent of x1 and x2, but the expected value of x1 and x2(given x1<x2 or x2<x1) depends on the fact that x1<x2 or x2<x1. Also note the fact that x3 either has paramter lamda1 or lamda 2, in the first part it has paramtere lamda1 for the first clerk, and in the second part, it has paramtere lamda 2 for the second clerk.

=1/lamda1+1/lamda2+2/(lamda1+lamda2)-lamda2/(lamda1+lamda2)^2-lamda2/lamda1/(lamda1+lamda2)-lamda1/(lamda1+lamda2)^2-lamda1/lamda2/(lamda1+lamda2).

There was some integration involved here. i wrote a matlab program which simulates this process, and the formula agrees, the avarage of many Naomi times, agrees with the formula. But what is correct to do now? Is it correct to set this smaller than 1/lamda1?(since it would be the mean of the longest of johns or pauls time) If I do that I get that lamda2>2*lamda1, that is the constant c=2. It is not then 2+sqrt(3) as the answer appendix says. The interesting part is that in the omputer program the mean time for Naomi is less than the maximum of (1/lamda1, 1/lamda2) when c=2, at c=2 it changes from bigger to smaller.

Last edited: Jan 6, 2012