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.
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?
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.