A fast-food franchise is considering operating a drive-up window food-service op

Question

A fast-food franchise is considering operating a drive-up window food-service operation. Assume that customer arrivals follow a Poisson probability distribution, with an arrival rate of 24 cars per hour, and that service times follow an exponential probability distribution. Arriving customers place orders at an intercom station at the back of the parking lot and then drive to the service window to pay for and receive their orders. The following three service alternatives are being considered A single-channel operation in which one employee fills the order and takes the money from the customer. The average service time for this alternative is 2 minutes A single-channel operation in which one employee fills the order while a second employee takes the money from the customer. The average service time for this alternative is 1.25 minutes A two-channel operation with two service windows and two employees. The employee stationed at each window fills the order and takes the money from customers arriving at the window. The average service time for this alternative is 2 minutes for each channel Answer the following questions and recommend an alternative design for the fast-food franchise (a) What is the utilization of the system (p)? (b) What is the probability that no cars are in the system? (c) What is the average number of cars waiting for service? (d) What is the average number of cars in the system? (e) What is the average time a car waits for service? (f) What is the average time in the system?

Explanation / Answer

Option A

Option B

Option C

n =2

Hence Option C is the best in terms of all parameter.

Comment Arrival rate A 24 Service rate S 30 2 mins = 60/2 =30 U Utilization ratio U=A/S 0.80000 <1 Length of queue diminishing Ans A Ls Expected number of people in system Ls A/(S-A) 4 Ans D Ws Avg waiting time in the system Ws 1/(S-A) 0.16667 hours 10.0000 mins Ans F Wq Avg waiting time in the Queue Wq A/S(S-A) 0.13333 hous 8.0000 mins Ans E Lq Avg no of customer waiting to be served Lq A^2/S(S-A) 3.2000 Ans C Po Prob of 0 units in system Po Ans B 1-(A/S) 0.20000 Pw Prob of people have to wait 1-Po 0.80000 Pn Prob of people n are in system (1-U)U^n P0 Prob of people 0 are in system 0.20000 p1 Prob of people 1 are in system 0.16 p2 Prob of people 2 are in system 0.128 p3 Prob of people 3 are in system 0.1024

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