Answer to Question #218969 in Operations Research for ANU

Question

At a one man barber shop, customers arrive according to poison distribution with a

mean arrival rate of 5 per hour and hair cutting time was exponentially distributed

with an average hair cutting time was exponentially distributed with an average hair

cut taking 19 minutes. It is assumed that because of excellent reputation, customers

were always willing to wait. Calculate the following

a. Average number of customers in the shop and average numbers waiting for a hair

cut

b .Percentage of time arrival can walk in right without having to wait

c. The percentage of customers who have to wait before getting into the barber’s chair

Answer 1

Given information:

Mean arrival rate=5 per hour

λ=$\frac{5}{60}per minute$

$=\frac{1}{12}per minute$

Mean service rate$=\frac{1}{\mu}=19minutes$

$\mu=\frac{1}{19}per mimutes$

(a)

$L_s=\frac{\lambda}{\mu-\lambda}$

$=\frac{1}{12}\div(\frac{1}{19}-\frac{1}{12})$

=2.7143

Average number of customers in the shop is 2.71.

Expected number of customers in the queue =$\frac{\lambda^2}{\mu(\mu-\lambda)}$

$=\frac{(0.083)^2}{0.053(0.053-0.083)}$

=4.2976

=4.3

(b) P(Customer directly goes to the barber's chair)$=1-\frac{\lambda}{\mu}$

$=1-(\frac{1}{12}\div\frac{1}{19})$

=0.58333

=0.58333×100%

The percentage of time arrival can walk in right without having to wait is 58.33%.

(c) P(Customer directly goes to the barber's chair)=1-P(W=0)

1-0.58333

=0.41667

=0.41667×100%

=41.67%

The percentage of customers who have to wait before getting into the barber’s chair is 41.67%.