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:

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.

Given knowledge: One flight with a total capacity of 100 passengers.

Passengers paying full fare, the average ticket price of $150, mean of 56 passengers, SD of 23

Participants on a discount price, with a ticket cost of $100, a mean of 88 passengers, and a standard deviation of 44.

(a) Spite of the fact there is a hint to utilize the CDF because statistical power is not supplied, we will make some presumptions to minimize the complexity of the question whilst retaining statistical accuracy.

We can do so by hardly taking full-fare passengers, in which particular instance 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 that total full fare ticket passengers will be 56+23=79, leaving 21 tickets for discount passengers, in which case total revenues will be[tex]79\times150+21\times100=\$13,950.[/tex]

(b) This new limited program now assigns a specific number of seats to each passenger category: 44 for discount (total revenues [tex]44\times100[/tex]) and 56 for full-fare (total revenues [tex]56\times150[/tex]), both of which are within the probability (full fare mean=56, discount mean=88).

In this instance, total revenues will be [tex]44\times100+56\times150=\$12,800.[/tex]

(c) Gain is the differential between the excess earnings in both the ideal overall revenue and restricted seat policies or $13,950- $12,800=$1,150.

(d) Without a well-defined standard error, there is no real answer to this question. Another assumption we might make to make things easier is to treat the average passengers as projected bookings. In this instance, total revenues will be 44*100 from discount passengers and 56*150 from full rate passengers.

Due to our assumption/lack of limitations, our ideal booking will be 54 full-price tickets and 44 discount passenger tickets, which is comparable to the solution (c).

You may alternatively go for the worst-case scenario by subtracting the SD of each passenger type from the mean, or the best-case scenario by adding the SD of the full fare to the mean and using the pending seats (leftover from 100) to discount the fare for optimal revenue collection.

To know more about the discounted price of the airline's tickets, refer to the link below:

https://brainly.com/question/14506205

Grab Manufacturing Co. purchased a 10-ton draw press at a cost of $183,000 with terms of 4/15, n/45. Payment was made within the discount period. Shipping costs were $4,300, which included $270 for insurance in transit. Installation costs totaled $12,900, which included $3,700 for taking out a section of a wall and rebuilding it because the press was too large for the doorway. The capitalized cost of the 10-ton draw press is: Multiple Choice $192,880. $190,880. $196,380. $197,880.

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Please help ASAP giving BRAINLIEST , Did I get this correct?