A small town with one hospital has two ambulances to supply ambulance service. Requests for ambulances during nonholiday weekends average .45 per hour and tend to be Poisson-distributed. Travel and assistance time averages two hours per call and follows an exponential distribution. Find: a. System utilization.

b. The average number of customers waiting.

c. The average time customers wait for an ambulance.

d. The probability that both ambulances will be busy when a call comes in.

Explanation:

Given: -

The number of ambulances is(m) =  2.

Arrival rate = 0.45 per hour

Service time = 2 per hour

Service rate =?

service time = 2 (Travel and assistance time averages two hours per call)

Therefore, Service rate will be 1/2 = 0.5 per hour.

a). System utilization(p) = arrival rate/mean of ambulances*service time

p = 0.45/2×0.5 = 0.45.

(b) :- The average number of customer waiting or waiting time for an ambulance is equal to:-

Arrival rate divided by the service rate ( on its corresponding service time as per table value)

I.e. Arrival time = 0.45 ÷ service rate =0.5

= 0.9 ( see table value for 0.9 with service rate as 2)

It comes to 0.229.

Therefore, the average number of customers waiting. 0.229.

(c) :- The average time customers wait for an ambulance is equal to :

No. of customers waiting for ÷ arrival rate

0.229 ÷ 0.45 = 0.508

Or 0.509 (approx. )

D) probability of Both ambulances is idle is Po = 0.378 (from the table for the value of and M=2)

So Probability of both ambulance is busy = 1-Po

= 1 - 0.378

= 0.622

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