Question

Problem 10-16 Customers enter the camera department of a store at the average rate of nine per hour. The department is staffed by one employee, who takes an average of 3.0 minutes to serve each arrival. Assume this is a simple Poisson arrival, exponentially distributed service time situation. Use Exhibit 10.9. a-1. As a casual observer, how many people would you expect to see in the camera department (excluding the clerk)? (Round your answer to 2 decimal places.) Number of people T .82 a-2. How long would a customer expect to spend in the camera department (total time)? (Round your answer to 1 decimal place.) Average total time 5.5 minutes b. What is the utilization of the clerk? (Round your answer to 1 decimal place.) Utilization 45 %

The first three answers are correct. Need the final two.

Answer 1

Average rate of arrivals per hour 4 = 1/A= 9 Customers per hour Time between arrival of Customers(A) = 0.1111 hrs = 6.6667 mins 1/B = 20 Customer per hour Average rate of service per hour t = Time taken to serve one Customer (B) = = 3 mins 0.05 hrs 9.00 Utilization rate,p, =— = — 0.45 45.00% 20 2 Average number of Customers in the system per hour = Ls = F = 0.8182 Customers Average number of Customers in the queue per hour = L = = uu-4). 0.3682 Customers 1 Average waiting time of Customers in the system = Ws=- 0.0909 hrs - = U-2 5.4545 mins

Probability that the system is empty, P(O) 1 - Utilization rate 0.55 Probability that the Customers has to wait in the system = P(wait) = 1 - P(0) 0.45 Probability that the system has n Customers,P(n) = Pn = (1 - p)p" Probability that the system has 1 Customers,P(1) = P(1)= ( 1 -0.45)*(0.4541 ) = 0.2475 Probability that the system has 2 Customers,P(2) = P(2)= (1 -0.45)*(0.45^2) 0.1114 If we need to find the probability that more than two are waiting, we need to find the probability that less than or equal 2 customers wait in the system and subtract that from total proability = 1 - (P(O) + P(1) +P(2)) = 1 - (0.55 +0.2475 + 0.114) = 0.0885 = 8.85% = 8.9%(rounded)

Fortwo clerks, we have Time between arrival of Customers = 0.11 hrs = 6.667 mins = 9 Customers per hour Average rate of arrivals per hour 2 Time taken to serve one Customers Average rate of service per hour ytl 0.15 hrs = 9 mins = = 6.67 Customers per hour For 2 servers, P ,= 0.67 67.00% 2*11 13.34 - 2p = 1+p Utilization rate, 0.8 80.00% Average number of Customers in the system per hour 2p 1 - p2 2.43 Customers Average number of Customers in the queue per hour = L = XW 2 03 1- p2 1.0915 Customers Average waiting time of Customers in the system = W = 2p ; 1(1 - p2) = 0.2702 hrs 0.2702 hrs 16.2 mins (1 - p2) """