Question

There are 3 servers in the checkout area. The average interarrival time of customers is 2 minutes. The processing time is 5 minutes. The coefficients of variation for the arrival process and the service process are 1 and 0.85 respectively. What is the average number of people waiting for service? A.) 14.42; B) 15.42; C) 16.42; D) 13.42

Answer 1

Number of Servers m = 3

Processing Time P = 5 minutes

Interarrival Time a = 2 minutes

CVa = 1

CVp = 0.85

Utilization U = P/(am) = 5/(2*3) = 0.83

Average waiting time = (P/m)*((U^((2*(m+1))^(1/2)-1)/(1-U))*(CVa^2 + CVp^2)/2

Average waiting Time = (5/3)*((0.83^((2*(4))^(1/2)-1)/(1-0.83))*(1 + 0.85^2)/2 = 6.1693

Customer Being Served = m*U = 3*0.83 = 2.49

Customers Waiting = 6.006/2 = 3.003

Average Number of people = 2.49 + 3.003 = 5.493

Hence none of the options are correct.