6.8 Suppose customers arrive in a one-server queue according to a Poisson distri

Question

6.8 Suppose customers arrive in a one-server queue according to a Poisson distribution with rate A- (in hours). Suppose that the service times equal 1/4 hour, 1/2 hour, or one hour each with probability 1/3. expected amount of time until that customer leaves? expected amount of time until the queue is empty again? the queue? (a) Assume that the queue is empty and a customer arrives. What is the (b) Assume that the queue is empty and a customer arrives. What is the (c) At a large time t what is the probability that there are no customers in

Explanation / Answer

a) Since the queue is empty, the customer will not have to wait. The expected amount of time until customer leaves = Mean Service time = (0.25 + 0.5+1) / 3 = 1.75/3 = 0.583 hours

b) Expected length of busy time = 1 / p - arrival rate, where p = 1/ service rate

Hence, Expected time till queue is empty againg = [1 / (1/0.0583) - 1] = 1 / 0.71 = 1.4 hours

c) Probability that there are no customers = 1 - Service rate / arrival rate = 1 - 0.583 = 0.417 or 41.7%

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.