Question

SCM 366

Revenue Management Assignment II

II. San Francisco Express Airlines, SaFE for short, flies from PHL to SFO. On a Thursday evening flight, the number of last-minute no-shows and cancellations is Poisson distributed with mean 7.5. SaFE has an unlimited number of low fare travelers who pay $300. The cost of bumping such a passenger is estimated to be $350 (due lost goodwill as well as the cost of routing their itinerary through other airlines).   SaFE offers this low fare because it also comes with a cancellation/rebooking fee of $150 – if a customer doesn’t show up for the flight or cancels her reservation, she must pay $150 to use the ticket on another flight.

To maximize revenue from this flight, how many seats should the airline overbook?

Customers are more reliable on the Friday evening flight. On that flight, the average number of no-shows and cancellations is Poisson with mean 4.5. Suppose SaFE overbooks that flight by 6 seats. What is the probability that at least 1 passenger will be bumped from this flight?

PLEASE ANSWER IT IN EXCEL SHEET WITH EXPLANATION

Answer 1

1.

Cost of overbooking, Co = $350 - $300 = $50

Cost of underbooking, Cu = $300 (revenue lost)

Service Level, SL = Cu/(Cu + Co) = 300/(300+50) = 0.857143 or 85.7143%

Z = NORMSINV(SL) = NORMSINV(0.857143) = 1.067571

Mean = 7.5

Standard Deviation = sqrt(7.5) = 2.7386

Thus, we should overbook by = Mean + Z*Std_Dev = 7.5 + 1.067571*2.7386 = 10.42366 ~ 10.42

So, the airline should overbook by 10.42 seats on Thursday for maximising revenues.

Attached is the table from excel -

	Cu =	300
	Co =	50	350-300
	SL =	0.8571	300/(300+50)
	Z =	1.0676	NORMSINV(0.8571)
	Mean =	7.5
	Std Dev. =	2.7386	SQRT(7.5)
	Overbookings =	10.4237	7.5 + Z*2.7386

2. Here, the probability that at least 1 passeger will be bumped with 6 overbookings = Probability that cancellations will be < 5. So, we will calculate the service level (probability of serving customers) with max 5 cancellations. Calculations are shown below, copied and pasted from excel with explanations:

	Mean =	4.5
	Std Dev. =	2.1213	SQRT(4.5)
	Overbookings =	6
=>	4.5 + Z*2.1213 =	6-1
=>	Z =	(5-4.5)/2.1213
=>	Z =	0.2357
So,	SL =	NORMSDIST(Z)
	SL =	0.593168

So, the required probability that at least 1 passenger will be bumped from this flight is 59.32%.