Question

Potential customers arrive at a full-service, one-pump gas station at a Poisson rate of20 cars per hour. However, customers will only enter the station for gas if there are no more than two cars (including the one currently being attended to) at the pump. Suppose that the amount of time required to service a car is exponentially distributed with a mean of five minutes.

Required:
a. What fraction of the attendant’s time will be spent servicing cars?
b. What fraction of potential customers are lost?

Answer 1

a) 19/20

b) 1/20

Explanation:

a) calculate the fraction of the attendants time will be spent servicing cars

l / L > 0 = Ц / [ Ц - λ ] = 20 --- ( 1 )

Ц (service ) = 15 hour

∴ λ ( arrival rate ) = 14.25

P = λ / Ц = 14.25 / 15 = 0.95 = 19/20

b) Fraction of potential customers lost

Ls = P / ( 1 - P )

where ; P = 19/20

Ls ( fraction of potential customers lost ) = 19/20 / ( 1 - 19/20 ) = 19

hence proportion of customers lost = 1 - 19/20 = 0.05 = 1/20