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A Supersonic Burger fast-food franchise is considering adding drive-through serv

ID: 462764 • Letter: A

Question

A Supersonic Burger fast-food franchise is considering adding drive-through service. Data collected at other Supersonic drive drive-throughs in the area indicate that customer arrivals follow a Poisson probability distribution, with a mean arrival rate of 24 cars per hour, and that service times follow an exponential probability distribution. In the proposed drive-through system, arriving customers will 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, which are in use at other Supersonic Burgers with drive-through service, are being considered: 1. 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. 2. 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. 3. A two-channel operation with two service windows and two employees. The employee at each service window fills the order and takes the money from the customer. The average service time for this alternative is 2 minutes. (a) For each alternative determine the probability that an arriving car will have to wait for service. (b) For each alternative determine the average number of cars waiting for service. (c) For each alternative determine the average time that a car spends in the drive-through system. (d) For each alternative determine the average time that a car spends waiting to place an order. (e) Because waiting time is costly in the fast-food business, Supersonic's management assigns a cost of $25 per hour to customer waiting time. From an expected-cost perspective, which of the three alternatives described above would you recommend if Supersonic drivethrough employees are paid $6.50 per hour and the operating cost per channel (equipment, space, etc) is $20 per hour? Explain your answer.

Explanation / Answer

Here Alternative One and three takes average time of 2 minutes to serve the customer. So both have same answer for the questions ((a),(b),(c)(d) but answer for (e) id different.

First Let me consider Alternantive 1 and 3

Average arrival rate A= 24 Cars

Service rate ( 2 Minutes per customer means , 60/2 = 30 customers can be served) S = 30

a) Probability that a cusotmer has to wait = Probability that system is not idle = A/S = 24/30 = 0.8

b) Average number of cars waiting for the service = A/(S-A) = 24/(30-24) = 24/6 = 4 Cars

c) Average time a car spent on the system = 1/(S-A) = 1/(30-24) = 1/6 hour = 10 ,minutes

d) Average time in Queue = A/S(S-A) = 24/30(30-24) = 24/180 hour = 8 Minutes

e) The cost of the first alternative is $6.50*1+$20*1= $26.50 (One Employee and One sytem)

The cost second alternative is $6.5*2+$20*2 = $53 (Two employee and two sytem)

Answers for Alternative 2

Arrival rate A = 24

Service rate S= 48 (1.25 minutes per car a total 60/1.25 =48 cars in hour

a) Probability that a cusotmer has to wait = Probability that system is not idle = A/S =24/48 =0.5

b) Average number of cars waiting for the service = A/(S-A) = 24/(48-24) =24/24=1

c) Average time a car spent on the system = 1/(S-A) = 1/(48-24) = 1/24 hour = 2.5 minutes

d) Average time in Queue = A/S(S-A) = 24/(48(48-24) = 24/(48*24) = 1/48 hour = 1.25 minutes = 1 Minutes and 15 seconds

e) Cost of operation = 2 employess + One operation channel = 2*$6.5+1*$20 = $33

Analysis

In alternative one and three total time a customer waits (10+8 ) =18 minutes

The cost of waiting in alternative one and three = 18/60 * $25 = $ 7.5

But in alternative 2 waiting time is just (2.5 +1.25) = 3.75 minutes

The cost of waiting = 3.75/60 *$25 = 1.5625

These cost should be added to the total cost

Cost at alternative = $34

Cost at alternative two =$33+$1.5625 = $34.5625

Cost at alternative three = $53+$7.5 = $60.5

Here Least cost is alternative 2 and it should be selected

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