Question

There are 200 seats on the Boeing 747 flight from Dallas to Houston. Assume that the number of passengers who buy tickets but do not show up at the airport is normally distributed with mean 30 and standard deviation 15. The average price of a ticket is $475. The manager decides to overbook the flight. He estimates that it costs $800 on average to compensate for a customer if there are not enough seats.
a. Using the technique learned in OPRE 3310, how many reservations should the manager accept? (Number of reservations = capacity + overbook). Round Down the number of overlook.
b. Assume that the manager approves 220 reservations. What is the probability that the manager will have to compensate for customers when there are not enough seats?

Answer 1

Answer:

a. The number of reservations should the manager accept is 225 bookings

b. The probability that the manager will have to compensate for customers when there are not enough seats is 0.2317

Step-by-step explanation:

According to the given data we have the following:

Cost of Overage Co = 800

Cost of Underage Cu = 475

Hence, Critical Ratio = Cu/(Cu+Co)

Critical Ratio = 475/1275

Critical Ratio = 0.3725

To calculate the number of reservations should the manager accept we would have to calculate first the Optimal Number of Overbooking as follows:

Z = NORM.S.INV(0.3725) = -0.3251

Optimal Number of Overbooking = Mean of No show + Z*Sd of No show

Optimal Number of Overbooking = 30 - 0.3251*15

Optimal Number of Overbooking = 25.12

Optimal Number of Overbooking = 25

a. Therefore, Number of reservation = Capacity + Overbooking = 200+25 = 225 bookings

b. Number of Overbooking = 220 - 200 = 20

Manager will have to compensate if there will be less than or equal to 19 No show.

Therefore, to calculate the probability that the manager will have to compensate for customers when there are not enough seats we have to calculate the following formula:

P(Compensation) = P(No Show <= 19) = P(Z <= (19-30/15))

P(Compensation) = P(Z<= - 0.7333)

P(Z< - 0.7333) = NORM.S.DIST(-0.7333,1)

P(Z< - 0.7333) = 0.2317

P(compensation) = 0.2317

Probability of Compensation = 0.2317