44. Consider the priority queueing model of Section 8.6.2 but now suppose that if a type 2 customer is being served when a type 1 arrives then the type 2 customer is bumped out of service. This is called the preemptive case. Suppose that when a bumped type 2 customer goes back in service his service begins at the point where it left off when he was bumped.

(a) Argue that the work in the system at any time is the same as in the nonpreemptive case.

(b) Derive W1Q

.

577 Hint: How do type 2 customers affect type 1s?

(c) Why is it not true that V2Q = λ2E[S2]W2Q

(d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so W2Q = W2Q (nonpreemptive) + E[extra time]
where the extra time is due to the fact that he may be bumped.

(e) Let N denote the number of times a type 2 customer is bumped. Why is E[extra time|N] = NE[S1]
1 − λ1E[S1]
Hint: When a type 2 is bumped, relate the time until he gets back in service to a “busy period.”

(f) Let S2 denote the service time of a type 2. What is E[N|S2]?
(g) Combine the preceding to obtain W2Q = W2Q (nonpreemptive) + λ1E[S1]E[S2]
1 − λ1E[S1]