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

if you use only 25 for your entertainment and only 200 for the groceries that you really need and take out the 50 per month for the computer then that should be enough