A fast-food franchise is considering operating a drive-up window food-service op
ID: 446911 • Letter: A
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 21 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:
System A: A single-server 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.
System B: A single-server 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 1.5 minutes.
System C: A two-server operation with two service windows and two employees. The employee stationed at each window fills the order and takes the money for customers arriving at the window. The average service time for this alternative is 2 minutes for each server.
The following cost information is available.
Customer waiting time is valued at $34 per hour to reflect the fact that waiting time is costly to the fast-food business.
The cost of each employee is $6.5 per hour.
To account for equipment and space, an additional cost of $22 per hour is attributable to each server.
What is the lowest-cost design for the fast-food business? **ROUND TO THE NEAREST CENT!!.
Note: Use P0 values from Table 11.4 to answer the questions below.
Total Cost System A $ System B $ System C $Explanation / Answer
Formulae common to both System A and System B (or for any single server system, these formulae hold good) But there may be a slight change for System B as there is a second employee involved:
Arrival rate = Lambda
Service Rate = Miu
Utilization Facttor = Chi = Lambda / Miu
Chi must be < 1 for Single server (System A and B in our case)
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Ls = Length of System = Number of customers being serviced and waiting
Lq = Just the number of customers waiting only
Ws = Total time spent by customers in our system both being served and waiting
Wq = Total time spent by customers in our system just waiting only
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Ls = Chi / (1-Chi)
Lq = Ls – Chi
Ws = Ls / Lambda
Wq = Lq / Lamda
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System A:
Lambda = 21 cars per hour
Miu = 2 minutes per car – must be converted to cars per hour
In 2 minutes, they serve 1 car;
Hence in 60 minutes, they can serve 1 * 60 / 2 = 30 cars
Hence Miu = 30 cars per hour
Chi = Lambda / Miu = 21/30 = 0.7
Ls = Chi / (1-Chi) = 0.7 / (1-0.7) = 2.3333 customers per hour
Lq = Ls – Chi = 2.3333 – 0.7 = 1.6333 customers waiting
Ws = Ls / Lambda = 2.3333 / 21 = 0.1111
Wq = Lq / Lamda = 1.6333 / 21 = 0.0777
Ws of 0.1111 means a customer spends 0.1111 hours in our system = 0.1111 * 60 minutes = 6.666 minutes (both being served and waiting)
Wq of 0.0777 means a customer waits for 0.07777 hours = 0.0777 * 60 = 4.662 minutes
6.666 – 4.662 = 2.004 minutes of service time
This double confirms that we are on the right track as it matches with the average service time of 2 minutes per car in System A.
Customer waiting time = $34 per hour; When a customer waits for 0.07777 hours, it costs us 0.0777 * $34 = $2.642
Salary = $6.5 per hour
Server cost = $22 per hour
Assume that Employees work for 40 hours per week – but that is not relevant, we can just calculate the hourly cost.
Cost per hour (Pay + Server) = 22 + 6.5 = $28.5
(Remember we already said there 1.63333 customers waiting in our Queue)
Customer wait cost = $2.642 * 1.6333 = $4.31517
28.5 + 4.31517 = $32.82
System A Total Cost = 28.5 + 4.31517 = $32.8
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System B:
Miu = 1.5 minutes per car = 1.5 * 60/2 = 45 cars per hour
Chi = 21 / 45 = 0.4666
Ls = Chi / (1-Chi) = 0.4666 / (1-0.4666) = 0.87476 customers per hour
Lq = Ls – Chi = 0.87476 – 0.4666 = 0.41 customers waiting
Ws = Ls / Lambda = 0.87476 / 21 = 0.041655
Wq = Lq / Lamda = 0.41 / 21 = 0.0195
Cost per hour (Pay + Server):
Customer wait cost = Lq * $34 * Wq
= 0.41 * 34 * 0.0195 = $0.272
Server cost = the same $22
As there are two employees, Employee cost = 2 * 6.5 = $13
Total cost = 0.272 + 22 + 13 = $35.272
System B Total Cost = $35.272
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System C:
Formula is different because multi server
Let C = Server count = 2
Chi = 21/30 = 0.7
From table, P0 = 0.456
Lq = P0 * ( Chi ^ (C+1) ) / ( (C-1)! * (C-chi)^2 )
= 0.456 * ( ( 0.7 ^ 3 ) / ( ( 1 * ( 2 – 0.7 ) ^ 2 )
= 0.09255
Ls = Lq + Chi = 0.09255 + 0.7 = 0.79255
Wq = Lq / Lambda = 0.09255 / 21 = 0.0044
Ws = 0.79255 / 21 = 0.0377
Pay = two employees = 2*6.5 = $13
2 servers cost 2 * 22 = $44
wait cost = Lq * $34 * Wq
= 0.09255 * 34 * 0.0044 = $0.014
Total cost = 13 + 44 + 0.014 = $57.014 = $57
Server C total cost = $57
Though the System C is highly customer friendly, System A is the most economical for us costing the minimum amount of $32.8 and it is the lowest cost design
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