等候理論的問題（queueing theory)

1. Suppose that a queueing system has 2 servers. an exponential interarrival time distribution with a mean of 2 hours

and an exponential service-time distribution with a mean of 2 hour. Futhermore

a customer has just arrived at 12:00 noon.a) What is the probability that the next arrival will come before 1:00 p.m. ?Between 1:00 and 2:00 pm? After 2:oo pm?b)Suppose that no additional customers arrive before 1:00 p.m.. Now what is the probability that the next arrival will come between 1:00 and 2:00 pm?c)What is the probability that the numner of arrivals between 1:00 and 2:00 p.m. will be zero? One? Two or more?2. Consider a variation of the M/M/1 model where customers renege (leavethe queueing system without being served) if their waiting time in the queue grows too large. In particular

assume that the time each customer is willing to wait in the queue before reneging has an exponential distribution with a mean of 1/x. Construct the rate diagram for this queueing system.
a) The inter-arrival times are distributed exponentially with λ =1/2 the probability that the next arrival will come before 1:00 p.m. is 1-e^(-1/2)=0.3935 the probability that the next arrival will come Between 1:00 and 2:00 pm will be ( (1-e^(-2/2))-(1-e^(-1/2)) = 0.6321-0.3935 =0.2387 the probability that the next arrival will come After 2:oo pm is 1- ( (1-e^(-2/2)) =1-0.6321 =0.3679b) by the memoryless property

it does not matter that no additional customer has arrivedbefore 1:00. That is to say that the probability that the next customer arrives in the nexthour will be the same at 1:00 as it was a noon. so it will be 1-e^(-1/2)=0.3935c) m = λt =1/2 (2:00-1:00) =1/2 [Poisson distributed random variable] so P(N=0) = ((1/2)^0 e^(-1/2)) / 0! = e ^(1/2) = 0.6065 P(N=1)= ((1/2)^1 e^(-1/2)) / 1! = 1/2 e ^(1/2) =0.3033 P(N greater than 2) = sum of ((1/2)^n e^(-1/2)) / n!) = 1- [P(N=0) P(N=1)] = 1-1/2 e ^(1/2)- e ^(1/2) =0.09022. can