The ticket price for an OKC-Houston flight is $150. Each plane can hold up to 10

Question

The ticket price for an OKC-Houston flight is $150. Each plane can hold up to 100passengers. Usually, some of the passengers who have purchased tickets for a flight failto show up (no shows). Assume that anybody who does not use a ticket gets $150 refund.To protect against no-shows, the airline will try to sell more than 100 tickets for eachflight. Any ticketed customer who is unable to board the plane due to its capacity beingreached is entitled to compensation of $100. Past data indicate that the number of no-shows for OKC-Houston flight, D, is normally distributed with a mean of 15 and astandard deviation of 5 customers. To maximize expected revenue less compensationcosts, how many tickets should the airline sell for each flight? Hint: Let q denote thenumber of tickets sold for the flight and d the actual number of no shows. Hence q-d isthe number of customers actually showing up to board the plane which can be less than orequal to or greater than the capacity of the plane, which is 100.

Explanation / Answer

If the only concern is to maximize expected revenue less compensation costs, how many tickets should the airline sell for each flight in this question does not give a finite solution, See how,

Let TR be the total revenue, we a/q only the no. of no show ups is a random variable and the no. of tickets (N) to be sold is to be decided. So,

TR=150N-(150D+(N-D-100)100) »TR=50N-50D+10000

»E(TR)=50N-50E(D)+10000

»E(TR)=50N-50*15+10000

»E(TR)=50N-750+10000

»E(TR)=50N+9250.

»E(TR) is a increasing function of N and will be greter for greater values of N.

SO, E(TR) is maximum for N= INFINITY

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