The arrival times of vehicles at the ticket gate of a sports stadium may be assumed to be Poisson with a mean of 30 veh/hr. It takes an average of 1.5 min for the necessary tickets to be bought for occupants of each car. (a) What is the expected length of queue at the ticket gate, not including the vehicle being served? (b) What is the probability that there are no more than 5 cars at the gate, including the vehicle being served? (c) What will be the average waiting time of a vehicle?

Question

What will be an ideal response?

Answer 1

q = 30 veh/h (arrival rate)
Q = 40 veh/h (service rate)
a) Expected queue length
Using Equation 6.71,
E(m) = q2/[Q(Q – q)] = (30)2/[40(40 – 30)] = 2.25 vehicles
b) Probability of no more than 5 cars
Using Equation 6.76,
P(n>N) = (q/Q)N+1 (Probability of more than N)
P(n>5) = (30/40)6 = 0.178
for P(n<=5) = 1 – 0.178 = 0.822
c) Average waiting time per vehicle
Using Equations 6.73 and 6.72,
E(v) = 1/(Q – q) = 0.1 hr = 6 minutes wait time including queue time and
service time
E(w) = q/(Q(Q – q)) = 0.075 hr = 4.5 minutes wait time in the queue