At Letchworth Community College, one person, the registrar, registers students f

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 com- puter 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

Answer

The student is inward at the pace of = 10/hours= 1/6 per minute

One student is incoming after each 6 Minutes

Time to check on Student by Registrar = 5 Minutes

therefore,

repair Rate (Mu) = 12/hours

advent Rate (lamda) = 10/hours

time-span of Queue= lamda^2/mu(mu-lamda) = (10*10)/12*(12-10) = 4.17

probable waiting time in line = lamda/mu(mu-lamda) = 10/12*(12-10) = 1.666667

Hourly asking price = 5 +1.666667*4.17*2 = $ 18.90000278

In case of mainframe system

therefore,

examine Rate (Mu) = 15/hours

Arrival speed (lamda) = 10/hours

extent of Queue= lamda^2/mu(mu-lamda) = (10*10)/15*(15-10) = 1.34

Predictable for the future moment in Queue = lamda/mu(mu-lamda) = 10/15*(15-10) = 3.333333

Hourly price = 7 +3.333333*1.34*2 = $ 15.93333

In crate of new registrar

and so,

check Rate (Mu) = 20/hours

advent Rate (lamda) = 10/hours

extent of Queue= lamda^2/mu(mu-lamda) = (10*10)/20*(20-10) = 0.5

Expected coming up time in Queue = lamda/mu(mu-lamda) = 10/20*(20-10) = 5

Hourly price = 8 +5*0.5*2 = $ 13

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