Question

Dee F. is considering building a drive-up/drive thru coffee stall at a location she researched as a viable location. The location can accommodate a maximum of 10 cars. Based on her research, customer arrivals follow a Poisson probability distribution, with a mean arrival rate of 25 cars per hour, and that service times follow an exponential probability distribution. Arriving customers will place their orders at an intercom station as soon as they enter the lot where the stall will be located, and then drive to the service window to pay for and receive their orders. Three service alternatives are being considered:

• A single channel operation in which one employee, whose hourly wage rate is $15, 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. Each employee has an hourly wage of $15. The average service time for this alternative is 1.50 minutes
• A two-channel operation with two service windows and two employees in each service window. Each of the employees has an hourly wage of $15. The employees 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 1.5 minutes for each channel.

1. Complete the table below comparing the three alternatives on the following service characteristics. Show your outputs used in completing the table.

Waiting Line Effectiveness Measures	Option 1: One window, one employee	Option 2: One window, two employees	Option 3: Two windows, two employees
Probability that there will be no car in the system.
Average number of cars waiting for service
Average time (in minutes) a car waits for service
Average time for a car to be in the system
Average number of cars in the system
The probability that an arriving car will have to wait for service

2. Which of the three options will you recommend Dee F. to use? Support your answer.

Answer 1

1.

Scenario 1

lambda ( arrival rate) = 24

mu ( service rate ) =60/2 =30 per hour

utilisation rho = 24/30 =0.8

a) Prob. that there is no customer is the system = 1-rho = 1-0.8 =0.2

b) No. of cars waiting Lq = (lambda)² / mu ( mu-lambda)

= 24x24 / 30x6 = 3.2

c) Waiting time for a car in line = Lq / lambda = 3.2 / 24 =0.1333 hour = 8 min.

d) avg time in system = waiting time + service time =8min + 2 min = 10 min

e) avg number of cars in system = lambda / mu-lambda = 24 /6 =4

f) Prob that a car need to wait = rho =0.8

Scenario 2

lambda = 24

mu = 60/1.2 = 50

utilisation rho = 24/50 = 0.48

a) prob of no car = 1-0.48 =0.52

b) No. of cars waiting = 24x24 /50x26 =0.443

c) avg wtg time in line = 0.443 / 24 =0.0184 hours = 1.107 min

d) avg time in system = wtg time + service time = 1.2+1.107 =2.307 min

e) avg cars in system = 24 /26 =0.923

f) Prob of waiting = rho =0.48

Scenario 3

Prob that there is no car in system P0 = 1/ [ 1+rho+(rho)² x mu / (2xmu-lambda)]

a) = 1/ [ 1+0.8+0.64x30 / 36]

= 0.4285

b) No. of cars waiting Lq = P0 x [lambda x mu x (rho)² / ( 2xmu-lambda)²]

= 0.4285x [ 24x30x0.64 / 36x36] =0.1523

c) Waiting time in line = Lq /lambda = 0.1523 /24 =0.00634 hours = 0.3808 min.

d) avg time in system =(Lq + rho) /lambda = 1.1808 /24 =0.0492 hour = 2.95 min

e) avg cars in system =Lq + rho = 0.3808 +0.8 =1.1808

f) prob that car will have to wait = (1-P0) = 1-0.4285 =0.5715

________________________________________________________

2.

The best option will depend on the cost of waiting and the cost of employees engaged in both cases. In the second case, it will be employee wages only, while in the third case, the other necessary equipment will also be needed. Since the waiting time is less with 2 separate windows, it is the better option, but owing to lesser output it will be less economical. For determination of the right option between options 2 and 3, cost data also needs to be taken into consideration.