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svp
2 months ago
15

Granite State Airlines serves the route between New York and Portsmouth, NH, with a single-flight-daily 100-seat aircraft. The o

ne-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?
Business
2 answers:
stepan [3.5K]2 months ago
6 0

Answer:

The data indicates: One flight has a total of 100 seats.

Full fare passengers, ticket cost=$150, average=56 passengers, SD=23.

Discount fare passengers, ticket cost=$100, average=88 passengers, SD=44.

(a) The question suggests optimizing total revenue per flight (one way) by potentially only taking full fare passengers, which would yield $15,000. However, historical probabilities show an average of 56 with a standard deviation of 23, thus in an ideal scenario, total full fare passengers could reach 79. That would allocate 21 tickets for discount passengers, leading to total revenues of $13,950.

(b) Now with the new constrained policy, specific seat allocations for both categories are set—44 for discount (resulting in total revenues of 44*100) and 56 for full fare (resulting in total revenues of 56*150)—both within the previously mentioned probabilities. The total revenue in this case will be 44*100+56*150 = $12,800.

(c) The difference in excess revenues between both scenarios for optimal total revenues and limited seats policy is calculated as answer (a) - answer (b) = $13,950 - $12,800 = $1,150.

(d) Realistically, this question cannot be properly answered without a clear confidence interval. Another simplifying assumption is to take the mean number of passengers as expected bookings (which can later be adjusted with provided confidence intervals). The total revenues in this scenario will come from 44*100 from discount and 56*150 from full fare passengers. This remains similar to answer (c) due to the assumption of no constraints, so optimal bookings might total 54 full fare tickets and 44 discount tickets. Worst case scenarios could involve subtracting SD from each passenger type’s mean, or for better scenarios, add SD of full fare passengers to the mean and allocate remaining seats for discount fare in order to maximize revenue.

Free_Kalibri [3.7K]2 months ago
3 0

Answer:

Explanation:

The formula below is applied:

1 −

(

∗

) =

= 100, = $100, = $150

The distribution of full fare demand is normally with a mean of 56 passengers and a standard deviation of 23. The optimal y* is the maximum value that satisfies

1 −

(

∗

) ≥

100

150

= 0.667, thus

(

∗

) ≤ 0.333. Using the normal distribution table, we write that

∗−56

23

≤ −0.43, leading to

∗ = 46.

∗ = −

∗ →

∗ = 100 − 46 = 54.

b) Note that our formula shows allocation of seats relies only on the distribution of full fare demand, hence there will be no changes in this context and seat allocations remain the same.

c) In this case,

∗ remains unchanged since it is based on /.

However,

∗ will alter so that

∗ = −

∗.

Thus, for the new conditions,

∗ would total 74 seats.

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