The airport branch of a car rental company maintains a fleet of 50 SUVs. The int

Question

The airport branch of a car rental company maintains a fleet of 50 SUVs. The inter-arrival time

between requests for an SUV is 2.4 hours, on average, with a standard deviation of 2.4 hours.

Assume that, if all SUVs are rented, customers are willing to wait until there is an SUV

available. An SUV is rented, on average, for 3 days, with a standard deviation of 1 day. The

company opens for 24 hours a day.

4. What is the average customer waiting time (to get a SUV)?

5. What is the new waiting time if the company decides to limit all SUV rentals to exactly 4

days? Assume that if such a restriction is imposed, the average inter-arrival time will

increase to 3 hours, with the standard deviation changing to 3 hours. [Hint: “exactly”

means there is no variations in rental time.

Explanation / Answer

4)Using the wait time formula,

Waiting time for multiple servers, m

Tq = (Activity time/m)x (Utilization ^(sqrt(2(m+1)) -1)/(1-utilization)) x ((CVa^2+CVp^2)/2)

Activity time= average time car taken for rent = 3 days = 72 hours

m = 50 cars in the fleet

CVp= Coefficient of variation of service times

CVa = st. dev of inter-arrival times/a

Utilization = Flow Rate/ Capacity = p/ma

p=Average activity time

a= average inter-arrival time

Tq = (72/50) (0.6^(sqrt(2(50+1))-1) / (1-0.6) )*((1^2 + 0.33^2)/2) = 0.019 hours or 1.15 minutes.

5) Stdev_p = 0
p = 4 days = 96 hours
a = 3 hours
stdev_a = 3 hours
U = p/(ma)=96/(50x3)=0.64
Using the wait time formula the average wait time is computed as
Tq = (96/50)(0.64^(sqrt(2(50+1))-1) / (1-0.64) )*((1^2+0^2)/2) = 0.046 hours.

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