Question

Donna Shader, manager at the Winter Park Hotel, is considering how to restructure the front desk to reach an optmum level of staff efficiency and guest service. Presently, the hotel has five clerks on duty, each with a separate waiting line, during the peak check in time of 3:00 P.M to 5:00 P.M.

Observation of arrivals during this time show that an average of 90 guests arrive each hour (although there is no upward limit on the number that could arrive at any given time). It takes an average of 3 minutes for the front-desk clerk to register each guest.

Donna is considering three plans for improving guest service by reducing the length of time guests spend waiting in line.

The first proposal would designate on employee as a quick service clerk for guests registering under corporate accounts, a market segment that fills about 30% of all occupied rooms. Because corporate guests are preregistered, their registration takes just 2 minutes. With these guests separated from the rest of the clientele, the average time for registering a typical guest would climb to 3.4 minutes. Under plan 1, noncorporate guests would choose any of the remaining four lines.

The second plan is to implement a single line system. All guests could form a single waiting line to be served by whichever of the five clerks became available. This option would require sufficient lobby space for what could be a substantial queue.

The third proposal using an automatic teller machine (ATM) for check-ins. This ATM would provide approximately the same service rate as a clerk would. Given that initial use of this technology might be minimal, Shader estimated that 20% of customers, primarily frequent guests, would be willing to use the machines.

(This might be a conservative estimate if the guests perceive direct benefits from using the ATM, as bank customers do. Citibank reports that 95% of its Manhattan customers use its ATMs.) Donna would set up a single queue for customers who prefer human check-in clerks. This would be served by the five clerks although Donna is hopeful that the machine will allow a reduction to four.

Required:

1. Determine the average amount of time that a guest spends checking in.
2. What is the check in time under each plan?
3. What plan would you recommend.

Answer 1

Answer :

1.   

Which of the two plans appears to be better? The current system has five clerks each with his or her own waiting line. This can be treated as five independent queues each with an arrival time of λ = 90/5 = 18 per hour. The service rate is one every 3 minutes, or μ = 20 per hour. Assuming Poisson arrivals and exponential service times, the average amount of time that a guest spends waiting and checking in is given by

hour, or 30 minutes

If 30% of the arrivals [that is, λ = 0.3(90) = 27 per hour] are diverted to a quick-serve clerk who can register them in an average of 2 minutes (μ = 30 per hour) their average time in the system will be 20 minutes. The remaining 63 arrivals per hour would distribute themselves equally among the four remaining clerks (λ  = 63/4 = 15.75 per hour), each of whose mean service time is 3.4 minutes (or 0.05667 hour), so that μ = 1/0.05667 = 17.65 per hour. The average time in the system for these guests will be 0.53 hour or 31.8 minutes. The average time for all arrivals would be 0.3(20) + 0.7(31.8) = 28.3 minutes.

A single waiting line for the five clerks yields an M/M/5 queue with λ = 90 per hour, μ = 20 per hour. The calculation of average time in the system gives W = 7.6 minutes. This plan is clearly faster.

2.

Use of an ATM with the same service rate as the clerks (20 per hour) by 20 percent of the arrivals (18 per hour) gives the same average time for these guests as the current systems-30 minutes. The remaining λ = 72 per hour form an M/M/4 or M/M/5 queuing system. With four servers, the average time in the system is 8.9 minutes, resulting in an overall average of:

0.2 × 30 + 0.8 × 8.9 = 13.1 minutes

With five servers, the average time is 3.9 minutes resulting in an overall average of: 0.2 × 30 + 0.8 × 3.9 = 9.1 minutes