Consider the concierge service at a large resort. The concierge helps guests with various tasks, including reservations at local
restaurants or offering advice on various local activities. On average, 15 guests arrive at the concierge desk per hour with a coefficient of variability of 0.5. The hotel has only one concierge on service. It takes the concierge, on average, 3 minutes to help the guest with their questions. The standard deviation of this time is 3 minutes.
a) What is the average time a guest has to wait before talking to the concierge?
b) How many guests will, on average, be at the concierge’s desk (either talking to the concierge or waiting in line)? (Your answer does NOT have to be an integer.)
c) If the hotel decides to add a 2nd concierge, what will be the impact ?
a) What is the average time a guest has to wait before talking to the concierge?
b) How many guests will, on average, be at the concierge’s desk (either talking to the concierge or waiting in line)? (Your answer does NOT have to be an integer.)
c) If the hotel decides to add a 2nd concierge, what will be the impact ?
1 answer:
Answer:
Step-by-step explanation:
Arrival rate of guests: R = 15 per hour.
Number of concierge services: c =1, mean = 3 min, Standard deviation = 3 min , so
Rp = c/Tp = 1/3 per min or 20 per hour.
U= R/Rp = 15/20= 0.75
Average service time: Tp = 3 min
Standard deviation of service time: Sp = 3 min
Coefficient of variation for service time: Cp = Sp /Tp = 3/3 =1
Average inter-arrival time: Ta = 1/R = 1/15 hr = 4 min
Ca=0.5
Considering the formula, for M/G/1 queue which is attached to this solution.
B ) Putting the values in the formula , we get I = 2.875 – Number of customers in the waiting line
How long a customer wait in the concierge line?
RT=I, rate of arrival x time = Number of customers in the waiting line
15xT=2.875, solving for T
T=2.875*60/15= 11.5 minutes
c) the number of service = C= 2
Number of concierge services: c =1, mean = 3 min, Standard deviation = 3 min, so
Rp = c/Tp = 2/3 per min or 40 per hour.
U= R/Rp = 15/40= 0.375
Average service time: Tp = 3 min
Standard deviation of service time: Sp = 3 min
Coefficient of variation for service time: Cp = Sp /Tp = 3/3 =1
Average inter-arrival time: Ta = 1/R = 1/15 hr = 4 min
Ca=0.5
Considering the formula, for M/G/2 queue ,
Substituting values in above equation,
I = 0.769 , Number of customers in the waiting line. Also,
RT=I, rate of arrival x time = Number of customers in the waiting line
15xT=0.769, solving for T
T=0.769*60/15= 3.07 minutes , so the wait time reduces , as evident
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