Winter Park

Winter park hotel case ANALYSIS

Prepared by:

Submitted to:Prof. Mahesh K. C

2WINTER PARK HOTEL CASE ANALYSIS

CASE OVERVIEW

Donna shader is manager of the Winter Park Hotel, is considering how to restructure the front desk to reach an optimum level of staff efficiency and guest service. At present the hotel has 5 clerks on duty each with a separate waiting line, during peak check-in time of 3:00 to 5:00 p.m. On average 90 guests arrive each hour. And it takes average 3 minutes for the front-desk clerk to register the guest. Donna is considering 3 plans for improving guest service by reducing the length of time guest spend waiting in line.

1. First proposal

Designate 1 employee as a quick service clerk for guests registering under corporate accounts, a market segment that fills about 30% of all occupied rooms. As corporate guests are preregistered, their registration will take only 2 minutes. With these guests separated from the rest, the average time for registering a typical guest would be 3.4 minutes. Non-corporate guest can choose any line from the remaining lines.

2. Second proposal

To implement single line system. All guests could form a single waiting line to be served by whichever of five clerks became available. This option requires sufficient lobby space for what could be a substantial queue.

3. Third proposal

It involves using ATM for check-ins. It will provide same service rate as a clerk would. Initial use of technology might be minimal, Shader estimated that 20% of customers would be willing to use machines (this might be a conservative estimate if the guest perceive direct benefits from using the ATM, as bank customers do. Citibank reports that some 95% of its Manhattan customers use its ATMs.) Donna would set up a single queue for customers who prefer human check-in-clerks, although Donna is hopeful that the machine will allow a reduction to four.

DISCUSSION QUESTIONS:

1. Determine the average amount of time that a guest spends checking in. How would this change under each of the stated option?

2. Which option do you recommend?


SOLUTION

The Current System:

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 90/5 = 18 per hour.

The service rate is one every 3 minutes, or 20 per hour.

Number of servers = 1

Operating Characteristics

Average server utilization = 0.9000

Average number of customers in the queue (Lq) = 8.1000

Average number of customers in the system (L) = 9.0000

Average waiting time in the queue (Wq) = 0.4500

Average time in the system (W) = 0.5000

Probability (% of time) system is empty (P0) = 0.1000

First proposal

1. 30% arrivals

Arrival rate 0.3(90)= 27 per hour

Service rate (2 minutes)= 30 per hour

2. For remaining 63.

Arrival rate 63/4=15.75

Service time (3.4 minutes)=17.65 per hour.

Number of server= 1

Operating Characteristics (1)


Average server utilization (r) 0.9000



Average waiting time in the queue (Wq) = 0.3000

Average time in the system (W) = 0.3333(20 mins)

Probability (% of time) system is empty (P0) = 0.1000

Operating Characteristics (2)

Average server utilization (r) = 0.8925



Average waiting time in the queue (Wq) =0.4705

Average time in the system (W) =0.5271(31.8 mins)

Probability (% of time) system is empty (P0)=0.1075

The average time for all arrivals would be 0.3(20) + 0.7(31.8) = 28.3 minutes.

SECOND PROPOSAL

A single waiting line for the five clerks


queue with arrival time90 per hour

Service rate=20 per hour.

Number of server= 5



Average number of customers in the queue (Lq) =6.8624

Average number of customers in the system (L)= 11.3624

Average waiting time in the queue (Wq)= 0.0762

Average time in the system (W)= 0.1262(7.6 mins)

Probability (% of time) system is empty (P0)= 0.0050

THIRD PROPOSAL

1) 20% arrivals


Service rate 20%

Arrival rate 0.2(90)= 18 per hour

It is same as current system

Average time in the system (W)= 0.5000

The remaining arrival rate=72 hours

Service rate=20

Number of servers= 4



Average number of customers in the queue (Lq)= 7.0898


Average waiting time in the queue (Wq) =0.0985

Average time in the system (W)= 0.1485 (8.9 mins)


With four servers, the average time in the system is 8.9 minutes, resulting in an

overall average 0.2*30 + 0.8*8.9 = 13.1 minutes

2) The remaining

arrival rate=72 hours

Service rate=20


Number of servers= 5

OPERATING CHARACTERISTICS


Average number of customers in the queue (Lq) =1.0553


Average waiting time in the queue (Wq)= 0.0147

Average time in the system (W) =0.0647(3.9 mins)


With five servers, the average time is 3.9 minutes resulting in an overall average

0.2*30 + 0.8*3.9 = 9.1 minutes

RESULT & CONCLUSION:

Waiting time is lower in second proposal it is just 7.6 minutes. So the option that is to be recommended is 2nd.

Documents

Winter Park