i am having trouble figuring out how to answer this question.
A big Boeing jumbo jet has 400 seats for a trans–Atlantic flight from JFK Airport in New York. The probability that any particular passenger will not show up for this flight is 0.03, independent of other passengers. The average fare paid by a passenger who succeeds in boarding is \$400. A passenger who shows up but cannot board is given \$800 and a free flight later on.
Given that the airline can sell 400 + m tickets, calculate the m that maximizes expected revenue.
the answer is between 10 and 14 inclusive. i can't figure out how to set this up.
so far i have made one function for revenue:
$$\text{revenue}(m) = (400 + m) \cdot 400$$
because i assume that a minimum of 400 tickets will be sold. $m$ represents the number of extra tickets sold.
the cost function is where i'm really struggling.
i think each make-up ticket will cost the airline 400 because the value is 800, but the passenger has already paid 400, which leaves an additional cost of 400 for the airline to pick up.
but i'm not sure how to use the binomial distribution here. i know that i need to multiply it by the cost to the airline of each individual reimbursement, which i think is 400. but i'm not sure what the number of trials or successes should be, or even if this is how i should be thinking about this question.
any advice? thanks in advance.
If $400+m$ people are sold tickets, and each of these people fail to show up independent of each other with probability $0.03$, then the number of people who won't show up (which we'll denote as $N$) follows a $\text{Binomial}(400+m,0.03)$ distribution.
If $N \ge m$, then the number of people who showed up is $400+m-N \le 400$, so everyone has a seat. Thus, the airline's revenue is $400(400+m)$.
If $N < m$, then the number of people who showed up is $400+m-N > 400$. The airline gets $400(400+m)$ dollars from the ticket sales, but loses $1200(m-N)$ dollars from having to give $m-N$ passengers $800$ dollars in compensation and a free flight later on.
So the total revenue is $$\text{revenue}(m,N) = \begin{cases}400(400+m) & \text{if} \ N \ge m \\ 400(400+m) - 1200(m-N) & \text{if} \ N < m\end{cases}. $$
Since $N$ follows a $\text{Binomial}(400+m,0.03)$ distribution, $$\text{Pr}\{N = n\} = \dbinom{400+m}{n}\cdot 0.03^{n} \cdot 0.97^{400+m-n},$$ and so, the expected revenue by selling $m$ tickets is $$\sum_{n = 0}^{400+m}\text{Pr}\{N = n\} \cdot \text{revenue}(m,n).$$ Now, you just need to figure out what value of $m$ maximizes this sum.