In a queuing system with three servers each customers is ser

In a queuing system with three servers each customers is served by one of the servers and them leaves. The service times at server I,II and III are independent exponential random variables with rates Lambda1,Lambda2,and Lambda3, respectively. There are four customers A,B,C being servers I,II,III respectively and D waiting. Find the expected time until one of the customers has first finished the service. Find the probability that B is the first to finish the service Find the probability that D finishes earlier than k. Find the expected time when D finishes the service. Find the expected time when all the Com customs have finish their services.

Solution

1)rates = a1 ,a2 , a3 respectively...then......expectation of exponential distrbtn = 1/ai...

expected time until 1 of the customer has 1st finished the service = E( time to finish the quickest service)
= 1 / min ( a i)....i = 1,2,3...

2) they are all equally likely to finish..So, p( B finished the service 1st] = 1/3...

3) p [ D finished before A ]
D can\'t start until 1 of B,C has finished their service.

Prob. of B or C finishing 1st = (1/3+1/3)=2/3.. and then due to lack of memory property of exponential distribution, A and D has the same prob. of finishing first...WHICH IS 1/2..
So prob[ D finished before A] = (2/3)*(1/2) =1/3........

D) Expected time when D finished = 1/ (2*min(ai ) )....

e) expected time when all of then has finished = 1 / [ a1+a2+a3+ min(ai) ]....i =1,2,3