Question

A hair salon is run by two stylists, Darrel and Jill, each capable of serving four customers per hour, on average. Six customers, on average, arrive at the salon each hour. (Assume Poisson arrivals and exponential service times)

a) If all arriving customers wait in a common line for the next available stylist, how long would a customer wait in line, on average, before being served?

b) Suppose that 50 percent of the customers want to be served only by Darrel and that the other 50% want to be served only by Jill. How long would a customer wait in line on average, before being served by Darrel? By Sloan? What is the average time a customer spends waiting in line?

c) Is there a difference in the answers to part a) and b)? If so, explain what causes this difference.

Answer 1

Arrival rate lambda= 6

service rate mu= 4 per hour

rho ( utilisation) = 6/4 =1.5

Probability that there is no customer in the line

P(0) = 1/ [ 1+rho+ (rho)²*mu / (2*mu-lambda)]

= 1/ [ 1+1.5+2.25 x4 / 2] = 0.1428

Number of customers waiting in line Lq =[ lambda *mu x (rho)² / (2*mu-lambda)²] *P(0)

   = P(0) *4x6x2.25/(8-6)²

= 0,1428x13.5 =1.9278

waiting time in line = Lq / lambda = 1.9278 /6 = 0.3213 hour = 19.27 minutes

(ii)If there are two separate lines, each line will get half customers right from the beginning

The new lambda for each time = 3

service time for each line =4

Waiting time in line = lambda / mu ( mu-lambda)

   = 3/4(4-3) = 3/4 hours = 45 minutes

(iii) The time in second case ( two separate lines) is higher because the lines are separated from the beginning itself, and they act as two single lines, where customers are divided in half. However, in first case, the customers wait in a pool where the next available channel is joined. This is more efficient process where even slightest delay in service time affects everyone waiting in that line, affecting the overall average waiting time.

Lq =