Question

Alexandra Marcus, manager of the Sky Club Hotel, has requested your assistance on a queuing issue to improve the guest service at the hotel. Alexandra Marcus is considering how to restructure the front desk to reach an optimal level of staff efficiency and guest service. Observation of arrivals during the peak check-in time of 3:00PM to 5:00PM shows that an average of 80 guests arrive each hour. It takes an average of 3 minutes for the front-desk clerk to register each guest. At present, the hotel has five clerks on duty, each with a separate waiting line.

PLEASE SHOW CALCULATIONS!!

a. Average utilization rate of a server (p)

b. The probability of no. customers in the system (Po)

c. Average number of customers in the system (L)

d. Average time in the system (W)

e. Average waiting in line (Wq)

f. Average number of customers in line waiting Lq

Answer 1

Given

Observation of arrivals during the peak check-in time of 3:00PM to 5:00PM shows that an average of 80 guests arrive each hour.

Thus = 80   Per hour

It takes an average of 3 minutes for the front-desk clerk to register each guest.

Thus 0 = 60 / 3 = 20 Per hour

At present, the hotel has five clerks on duty, each with a separate waiting line.

Hence = 20 * 5 = 100 Per Hour

So we have average arrive each hour i.e Arrival Rate =   = 80  

For five clerks on duty to register each guest i.e Service Rate = = 100

a. Average utilization rate of a server (p)

p = / = 80 / 100 = 0.8

Thus Average utilization rate of a server is p = 0.8

b. The probability of no customers in the system (Po)

Now if there are n customers in the system , its probability is given by = ( 1 - p ) pⁿ

Thus probability of no customers in the system (Po) is = ( 1 - p ) * p ⁰

                                                                                       = ( 1 - p ) * 1

                                                                                       = 1 - 0.8 = 0.2

Hence probability of no customers in the system (Po) is 0.2

c. Average number of customers in the system (L)

Average number of customers in the system is given by L = = = 4

Hence Average number of customers in the system (L) = 4

d. Average time in the system (W)

Average time in the system is given by W = = = 0.05

Thus Average time in the system (W) is 0.05 hours or 3 minutes      { 0.05*60 =3 }

e. Average waiting in line (Wq)

Average waiting in line is given by Wq = = = 0.04

Thus Average waiting in line (Wq) is 0.04 Hours or 2.4 minutes      { 0.04 * 60 = 2.4 minutes }

f. Average number of customers in line waiting Lq

It is given by Lq = . = . = 0.8 * 4 = 3.2

Thus Average number of customers in line waiting is Lq = 3.2