Question

Problem 6. Janet is planning to open a small two-bay car-wash operation, and she must decide how much space to provide for waiting cars. Janet estimates that customers would arrive ran- domly (i.e., a Poisson input process) with a mean rate of 1 every 4 minutes, unless the waiting area is full, in which case the arriving customers would take their cars elsewhere. The time that can be attributed to washing one car has an exponential distribution with a mean of 3 minutes. Compare the expected fraction of potential customers that will be lost because of inadequate waiting space if (a) 2 spaces, and (b) 4 spaces were provided.

Answer 1

lambda = arrival rate = 1 every 4 minutes = 1/4 per minute mu = service rate = 1 every 3 minutes = 1/3 per minute S = number of servers = 1 We use M/M/s queuing model First we find P_0 = probability that there are no customers in the system = ratio traffic intensity = lambda/mu = 1/4/1/3 = 3/4 = 0.75 P_0 = [sigma^s-1_i = 0 ^i/i! + ^s/s!(smu/smu - lambda)]^-1 = [sigma^0_i = 0 (0.75) ^0/0! + (0.75) ^1/1! (1 times 1/3/1 times 1/3 - 1/4)]^-1 = [1 + 0.75(12/3(1))]^-1 = [1 + 3]^-1 = (4) ^-1 = 1/4 P_0 = 0.25 a) Number of spaces excluding the car being washed = 0 therefore K = 0 + 1 = 1 (included the car being washed) To find expected fraction of customers last = P (there is 1 car in the system including one being washed) = P_1 = ^k/k! times P_0 = (0.75) ^1/1! times 0.25 = 0.1875 P_n = ^n/n! P_0 n lessthanorequalto s = ^n/s!s^n-s P_0 n > s

here a) 0 space

b) 2 spaces

c) 4 spaces

hence corretc answers

a) 2 spaces = 0.1055

b) 4 spaces = 0.0593