so to start,
i need to find the probability before getting the expectations
but im not sure what the probability should be?
A post office is run by 2 clerks.
When person C arrives, A is served by one clerk and B served by the other. C gets served as soon as A or B leave (no one else waiting).
The time it takes server i to serve customers is exponentially distributed with mean 1/lambda(i) , i = 1, 2. lambda(i) = lambda subscript i
What is the expected time until all 3 customers have left the post office?
I need hint on how to start this problem.
I thought about it... and this is what I have.
4 ways:
ABC, BAC, BCA, ACB
expected time for ABC = 2/lambda(1)
E(T for BAC) = 2/lambda(2)
E(T for BCA) = 1/lambda(1)
E(T for ACB) = 1/lambda(2)
add them up... and total expected time is
E(T) = 3/lambda(1) + 3/ lambda(2)
is that right?
That does not look good. Suppose clerk 1 works very fast so that we can set lambda(1) to infinity. Then that clerk will serve A taking no time, and then serve C also taking no time. The total time is then determined solely by how long it takes clerk 2 to serve B. This will be 1/lambda(2). Your result however gives 3/lambda(2).
You can calculate the time-dependent pdf's for the different situations that you can have (all three customers, only A and C, only B and C, etc.). This involves setting up and solving a master equation. From the pdf that all customers have been served you can then derive the time that all customers have left. That gives you all information you can possibly want, but, as you only want the expected time, the following may be simpler.
There are three stages. The first consists of C waiting to be served. The second consists of C being served and both clerks serving a customer. The third consists of only one clerk serving a customer. The expected times of each stage can be calculated rather easily, and the results can be added.
When both clerks are serving a customer, then the expected time until one of them finishes is 1/(lambda(1)+lambda(2)). Hence the expected times for stages one and two are both equal to this number.
The third stage is a bit more complicated, because you have to take into account which of the clerks is still serving a customer. Clerk 1 takes 1/lambda(1) and clerk 2 takes 1/lambda(2). The probability that clerk 1 is still serving a customer is equal to the probability the clerk 2 has finished and is equal to lambda(2)/(lambda(1)+lambda(2)). We get a similar probability that clerk 2 is still serving a customer. The expected time for stage three is then (lambda(1)/lambda(2)+lambda(2)/lambda(1))/(lambda(1)+lambda(2)).
Adding all times gives then finally the simple result 1/lambda(1)+1/lambda(2). This result simplifies to what one would expect in the limiting cases where one clerk is infinitely fast, or both clerks work equal fast.