Question

A small barbershop, operated by a single barber, has room for at most two customers. Potential customers arrive according to a Poisson process with rate three customers per hour. The successive service times are independent exponential random variables with mean 15 minutes.

(a) Model this system as a CTMC.

(b) What is the proportion of time the barber is busy?

(c) What is the expected long-run average number of customers in the shop?

(d) If the barber could work twice as fast, how would your answers to parts (b) and (c) would change?

Answer 1

SOLUTION:
If we say that the system is in state n whenever n customers are in the shop, then the preceding is a birth and death process with parameters 0, n=0 3, n-0 4. n-2 0, n-2 For the birth and death process we have the limiting probabilities First fin<d 丛丛나 3,(33) 4 (4)(4 21 16

Therefore the limiting probabilities are 1 1+21/16 16 37 4(1+21/16) (303) 12 37 37 The average number of customers in the shop is 37 37 37 30 37

The proportion of potential customers that enter the shop is equal to the sum of the probability when there are no customers in the shop and the probability when there is only 1 customer in the shop, because if there are two customers in the shop then the potential customer doesn't enter the shop. Therefore, the proportion of potential customers that enter the shop is 16 12 o37 37 28 37 If the barber could work twice as fast, the parameters of the process would be 0, n 0 3, n-0 8, n 2 0, n-2 For the new process, first find 3(3(3) 8 (8)(8) 64 Therefore the limiting probabilities of the new process are

1+33/64 64 97 2022 n- 8 (1+33/64) (8) (8) (1+21/16) 24 97 9 97 If the barber could worktwice as fast, the proportion ofpotential customers that enter the shop would be 64 24 97 97 97 The rate of added customers is 88 28 88 28 97 37 97 37 0.45 Thus, if the barber could work twice as fast, he would do by 0.45 customers more per hour