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Computing Average revenue using Standard deviation

Can anyone help me in the illustrated problem? The airport branch of a car rental company maintains a fleet of 50 SUVs. The inter-arrival time between the requests for an SUV is 2.4 hrs, on an average, with a standard deviation of 2.4 hrs. There is no indication of a systematic arrival pattern over the course of a day. Suppose that, if all SUVs are rented, then customers are willing to wait until there is an SUV available. The SUV is rented, on an average, for 3 days, with a standard deviation of one day.

a. Determine the average number of SUVs parked in the company’s lot?

b. By using a marketing survey, the company has discovered that if it decreases its daily rental price of $80 by $25, the average demand would rise to 12 rental requests per day and the average rental duration will become 4 days. Supposing that the standard deviation values stay unchanged, should this company adopt this latest pricing policy? Give an analysis!

c. Determine the average time a customer has to wait to rent an SUV? Please employ the initial parameters instead of the information in (b).

d. How would the waiting time change when the company decides to limit all the SUV rentals to exactly 4 days? Suppose that if such a restriction is imposed, the average inter-arrival time will rise to 3 hours, with the standard deviation changing to 3 hrs.

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a) We know that a = 2.4 hours, p = 24 x 3 = 72 hours, CVa = (2.4/2.4) = 1, CVp = (24/72) = 0.33, and m = 50 cars

Percentage of cars used = (1/a)/(m/p) = (1/2.4)/(50/72) = 60%

Therefore, cars in use = 50 x 0.6 = 30

So, cars in parking lot = 20

b) a = 2, p = 96 hours,

Therefore, Percentage of cars used = (1/a)/ (m/p) = (1/2)/ (50/96) = 96%

So, average number of cars used = 50 x 0.96 = 48

Average revenue initially = 80 x 30 = $2400
Average revenue now = 48 x 55 = $2640

Therefore, the company should take the proposed step.

c) a = 2.4 hours, p = 24 x 3 = 72 hours, CVa = (2.4/2.4) = 1, CVp = (24/72) = 0.33, and m = 50 cars,

Waiting time = (p/m)(u^[{2(m+1)}1/2 – 1]/1-u)[CVa2 + CVp2/2]

= (72/50) (0.6^ {(102)1/2 – 1}/0.4) (1 + (0.33)2/2)

= 1.44 x (0.6) ^9 x 0.55/0.4 = 0.02 hours = 1.2 minutes

d) a = 3 hours, p = 24 x 4 =96 hours, CVa = (3/3) = 1, CVp = (24/96) = 0.25, and m = 50 cars

Percentage of cars used = (1/a)/(m/p) = (1/3)/(50/96) = 64%

Waiting time = (p/m) (u^ [{2(m+1)}1/2 – 1]/1-u) [CVa2 + CVp2/2]

= (96/50) (0.64^ {(102)1/2 – 1}/0.36) (1 + (0.25)2/2)

= 1.92 x (0.64) ^9 x 0.53/0.36 = 0.05 hours = 3 minutes

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