At Letchworth Community College, one person, the registrar, registers students f
ID: 363713 • Letter: A
Question
At Letchworth Community College, one person, the registrar, registers students for classes. Students arrive at a rate of 10/h (Poisson arrivals), and the registration process takes 5 min on the average (exponential distribution). The registrar is paid $5 per hour, and the cost of keeping students waiting is estimated to be $2 for each student for each hour waited (not including service time). Develop a process-driven spreadsheet simulation to compare the estimated hourly cost of the following three systems. (See the hint in Exercise 7.3 and simulate 500 students in each case.)
a. The current system.
b. A computerized system that results in a service time of exactly 4 min. The computer leasing cost is $7 per hour.
c. Hiring a more efficient registrar. Service time could be reduced to an average of 3 min (exponentially distributed), and the new registrar would be paid $8 per hour.
Explanation / Answer
The student is arriving at the Rate of = 10/hours= 1/6 per minute
One student is arriving after Every 6 Minutes
Time to service on Student by Registrar = 5 Minutes
Hence,
Service Rate (Mu) = 12/hours
Arrival Rate (lamda) = 10/hours
Length of Queue= lamda^2/mu(mu-lamda) = (10*10)/12*(12-10) = 4.17
Expected waiting time in Queue = lamda/mu(mu-lamda) = 10/12*(12-10) = 1.666667
Hourly Cost = 5 +1.666667*4.17*2 = $ 18.90000278
In case of computer system
Hence,
Service Rate (Mu) = 15/hours
Arrival Rate (lamda) = 10/hours
Length of Queue= lamda^2/mu(mu-lamda) = (10*10)/15*(15-10) = 1.34
Expected waiting time in Queue = lamda/mu(mu-lamda) = 10/15*(15-10) = 3.333333
Hourly Cost = 7 +3.333333*1.34*2 = $ 15.93333
In case of new registrar
Hence,
Service Rate (Mu) = 20/hours
Arrival Rate (lamda) = 10/hours
Length of Queue= lamda^2/mu(mu-lamda) = (10*10)/20*(20-10) = 0.5
Expected waiting time in Queue = lamda/mu(mu-lamda) = 10/20*(20-10) = 5
Hourly Cost = 8 +5*0.5*2 = $ 13
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