Answer to Question #267134 in Operations Research for Bil

Question

At a one man barber shop, customers arrive according to poison distribution with amean arrival rate of 5 per hour and hair cutting time was exponentially distributedwith an average hair cutting time was exponentially distributed with an average haircut taking 10 minutes. It is assumed that because of excellent reputation, customerswere always willing to wait. Calculate the followinga. Average number of customers in the shop and average numbers waiting for a haircutb .Percentage of time arrival can walk in right without having to waitc. 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}$$=\frac{1}{12}$ per minute 

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

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

(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%.