Question

5. A flight from New York to Atlanta has 146 seats. Advance tickets purchased cost $74. Last-minute tickets cost $114. Demand for full-fare tickets is normally distributed with a mean of 92 and standard deviation of 30. What booking limit maximizes expected revenues? Assume there are no no-shows and always enough advanced purchasers to fill the flight.

Answer 1

Here total seats = 146

Advance ticket cost = $ 74

Last minute cost = $ 114

mean demand for tickets = 92

standard deviation = 30

Here if Q would be the last minute tickets, 146 - Q would be advanced tickets

Total profit = (146 - Q) * 74 + 114 Q

Here we define F(Q), the probability number of last minute reservations as less than equal to Q.As we assumed all rooms can be filled as advanced tickets, we would atleast get $ 74 for each ticket.

With probability 1 - F(Q), airlines would sell the Q + 1 seats at higher price. Therefore, the airlines should reduce the protcetion level from Q + 1 to Q if and only if

74 > = (1 - F(Q) * 114

F(Q) > =(114 - 74)/114

F(Q) > = 40/114 = 0.351

Now we will use normsinb function for the given critical fractile

z = NORMSINV(0.351) = -0.383

so here

Q₀ = 92 - 0.383 * 30 = 80.52 or 81 tickets

so 81 tickets should be advanced tickets that would be booking limit and rest are last time tickets.