Question

Granite State Airlines serves the route between New York and Portsmouth, NH, with a single-flight-daily 100-seat aircraft. The one-way fare for discount tickets is $100, and the one-way fare for full-fare tickets is $150. Discount tickets can be booked up until one week in advance, and all discount passengers book before all full-fare passengers. Over a long history of observation, the airline estimates that full-fare demand is normally distributed, with a mean of 56 passengers and a standard deviation of 23, while discount-fare demand is normally distributed, with a mean of 88 passengers and a standard deviation of 44.
a) A consultant tells the airline they can maximize expected revenue by optimizing the booking limit. What is the optimal booking limit? (Hint: Use the standard normal cumulative distribution table)
b) The airline has been setting a booking limit of 44 on discount demand, to preserve 56 seats for full-fare demand. What is their expected revenue per flight under this policy? (Hint: First find the expected revenue when b= 0. Here you can assume Probability{df = k} = Ff(k+0.5) – Ff(k-0.5) and use a spreadsheet. Then using the recursive formula, find the expected revenue if b is increased by 1 until it reaches b=44 using a spreadsheet)
c) What is the expected gain from the optimal booking limit over the original booking limit?
d) A low-fare competitor enters the market and Granite State Airlines sees its discount demand drop to 44 passengers per flight, with a standard-deviation of 30. Full-fare demand is unchanged. What is the new optimal booking limit?

Answer 1

Answer:

Given data: One flight with total seats = 100

Full fare passengers, cost per ticket=$150, mean=56 passengers, SD=23

Discount fare passengers, cost per ticket=$100, mean=88 passengers, SD=44

(a) Here, though there is a hint to use the CDF, since the confidence interval is not given we will make some simplying assumptions that will reduce the complexity of the question, of course keeping the question statistically correct.

this question wants us to maximize total revenue per flight (one way), we can do that by taking only full fare passengers or total revenue will be 150*100=$15,000, but since historical probability shows a mean of 56 with a standard deviation of 23, we can assume in best case scenario total full fare ticket passengers will be 56+23=79, leaving 21 tickets for discount passenger, in this case the total revenues will be 79*150+21*100=$13,950

(b) Now, the new constrained policy is giving a clear cut number of seats to each category of pasengers, 44 for discount (total revenues 44*100) and 56 for full fare (total revenues 56*150) both of which are within the probabilities given earlier (full fare mean=56, discount mean=88). Total revenues in case will be 44*100+56*150=$12,800.

(c) Gain is the difference of the excess revenues in both cases of optimal total revenues and limited seats policy or answer (a) - answer (b) = $13,950- $12,800=$1,150

(d) Realistically speaking, there is no answer for this question without a clear cut confidence interval. Another simplifying assumption we can make here is taking the mean passengers as expected bookings (can be tweaked once confidence interval or degree of significance is given). so total revenues in this case will be 44*100 from discount and 56*150 from full fare passengers. That is still similar to answer (c) due to our assumption/lack of constraints, so our optimal booking will be 54 full fare tickets and 44 discount passenger tickets. You can also take worst case scenario by subtracting SD of each passenger type from the mean or go the best case scenario in which SD of full fare will be added to the mean while the pending seats (left over from 100) will be the total to discount fare for optimal revenue collection.

Answer 2

Answer:

Explanation:

We use the following formula:

1 −

(

∗

) =

= 100, = $100, = $150

Full fare demand is normally distributed with a mean of 56 passengers and a standard deviation

of 23. The optimal y* is the largest value that satisfies

1 −

(

∗

) ≥

100

150

= 0.667 and thus

(

∗

) ≤ 0.333 By using the normal distribution table we

can write that

∗−56

23

≤ −0.43 and as a result

∗ = 46

∗ = −

∗ →

∗ = 100 − 46 = 54 .

b) Observe that according to our formula, allocation of seats is dependent to full fare demand

distribution only. Therefore, nothing would change in such a situation and the seats allocations

will remain as the same.

c) Here, our

∗ will again remain as the same since it is determined according to /.

However, our

∗ will change such that

∗ = −

∗

.

As a result, in the new case

∗ will be 74 seats.