Question

A university cafeteria line in the student union hall is a self-serve facility in which students select the food items they want and then form a single line to pay the cashier. Students arrive at a rate of about 4 per minute according to the Poisson distribution. The single cashier ringing up sales takes about 12 seconds per customer, following an exponential distribution. a) Assess the performance of the queue and write a report on your findings.

Students are threatening to boycott the student centre food service because the cashier is too slow. The Students’ Union President has therefore sought to improve the service by taking the following action: “Employ an Earn & Study Student, who will work at the same pace as the current cashier, but who will stand back to back in adjoining cubicles. This will eliminate students moving between lines (jockeying). Assume that student customers will divide themselves equally between each line, so that the arrival rate for each cashier will be half of the prior arrival rate for a single cashier. The service rate for each cashier will remain the same as the previous service rate. The Union Treasurer has estimated that this alternative will enable an additional 150 students to cash their lunches in a timely manner. These students would not otherwise have bought lunch in the student centre. The Treasurer has projected that the Union will earns $135 000 in additional income for 5 consecutive school days. However, employing the additional student cashier over that time would cost $6200 per day in wages plus a one-time cost of $12500 for the 5-day period. b) Would you advise the president to implement his new alternative plan? Your response must be derived from a cost benefit analysis which must be included as part of your response to the president.

Answer 1

a= µ=60/12=5 students/min=4 students / min

Step-by-step explanation:

Waiting time in Queue=
`\lambda`/µ(µ-

Number of students in the line L(q)= *W(q)= 4*.8= 3.2 students

Number of students in the system L(q)=
`\lambda` /(µ-

Probability of system to be empty= P0= 1-P= 1-0.8= 0.2

If the management decides to add one more cashier with the same efficiency then

µ= 6 sec/student= 10 students/min.

P= /µ =4/10=0.4

Now the probability that cafeteria is empty= P0= 1-0.4= 0.6

If we look at the above traits of the system, it is obvious that the line is no always empty and the students have to wait for 0.8 in waiting in the queue to place the order and get it, While there are 3.2 students in the queue on an average and there are 4 students in the entre cafeteria who are waiting to be served.

If the management decides to hire one more cashier with the same efficiency the probability of the cafeteria being free increases from 0.2 to 0.6 so it indicates that the management must hire one additional cashier.