To M/M/2 system,n=2
?n?1(n?)m(n?)n?P0?????m!n!(1??)m?0??-1?1?2?2? ??1??2??1-???1-????，其中???1??n?2??(n?)mP,m?n??0m!Pn??nm?Pn?,m?n0?n!??1??(2?)m?,m?2?m!?1???? m?1???4?,m?2?2?1??其中???2?P0!(n?)n1Pq??n!1??1??(2?)21 ???1??21??2?2?1??P0(n?)n?1??(2?)2?Nq?????n!(1??)21??2(1??)2?2?1??23
2?3 Wq???2??(1??)?(1??)2?3 T??Wq??2???(1??)11Nq?Pq?2??3 N??T?T????(1??2)3.1
A customer that carries out the order(eats in the restaurant)stays for 5 mins(25mins).Therefore the
average customer time in the system is T=0.5*5+0.5*25=15.By Little’s Theorem the average number in the system is N=A*T=5*15=75.
3.5
The expected time in question equals
E{Time} = (5 + E{stay of 2nd student})*P{ 1st stays less or equal to 5 minutes}
+ (E(stayof 1st Istay of 1st ~ 5} + E{stay of2nd})* .P{1st stays more than 5 minutes}. We have E(stay of 2nd student} = 30, and, using the memoryless property of the exponential distribution,
E{stay of 1st I stay of lst ~ 5} = 5 + E(stay of 1st} = 35. Also
P{1st student stays less or equal to 5 minutes} = 1 - e-S/30 P{1st student stays more than 5 minutes}= e-S/3o. By substitution we obtain
E{Time} = (5 + 30)*(1 - e-S/3o) + (35 + 30)* e-SI3O =35 + 30*e-SI30 = 60.394. 3.6
(a) The probability that the person will be the last to leave is 1/4 because the exponential distribution is memoryless, and all customers have identical service time distribution. In particular, at the instant the customer enters service, the remaining service time ofeach of the other three customers served has the same distribution as the service time of the customer.
(b) The average time in the bank is 1 (the average customer service time) plus the expected
time for the first customer to finish service. The latter time is 1/4 since the departure process is statistically identical to that of a single server facility with 4 times larger service
rate. More precisely we have
P{no customer departs in the next t mins} P{I st does not depart in next t mins} * P{2nd does not depart in next t mins}* P{ 3rd does not depart in next t mins} * P{4th does not depart in next t mins}= (e-t)4 = e-4t. Therefore
P(the first departure occurs within the nextt mins} = 1-e-4t,
and the expected time to the next depature is 1/4. So the answer is 5/4 minutes.
(c) The answer will not change because the situation at the instant when the customer begins service will be the same under the conditions for (a) and the conditions for (c)
=
3.8
Packet Arrivals
r1 r2
Fix a packet. Let rl and r2 be the interarrival times between a packet and its immediate predecessor, and successor respectively as shown in the figure above. Let X 1and X2 be thelengths of the predecessor packet, and of the packet itself respectively. We have: P{No collision wi predecessor or successor) = P{r1> X 1, r2> X2} = P{ r1> X 1} P{ r2> X2}·
P{No collision with any other packet} = P1 P{ r2> X2} Where P1 =P{No collision with all preceding packets}. (a) For fixed packet lengths (= 20 msec) P{ r1> X 1}= P{ r2> X2}=e??*&20?e?0.2
P1 = P{ r1<=X 1} ·
Therefore the two probabilities of collision are both equal to e-0.4 = 0.67.
(b) ForX exponentially distributed packet length with mean
?1. we have ?P(r1?X1)?P(r2?X2)??P(r1?X|X1?X)p(X1?X)dX0??e??X?e??XdX?0?????
Substituting ?= 0.01 and ?= 0.05 we obtain P{ r1> X 1} = P{ r2> X2}= 5/6, and P{No collision w/ predecessor or successor} =(5/6)2 = 0.694.
Also PI is seen to be the steady-state probability of a customer finding an empty system inthe M/M/? system with arrival and service rate ?and ? respectively. Therefore P1?e????e?0.2.Therefare .
P{No collision with any other packet} = e-0.2 *(5/6)=0.682.
3.9