Airline analytics 6. Let\'s look at overbooking. JetPurple Airlines\' first clas
ID: 2930232 • Letter: A
Question
Airline analytics 6. Let's look at overbooking. JetPurple Airlines' first class cabins have 10 seats per plane. JetPurple's overbooking policy is to sell up to 11 first class tickets, since cancellations and no-shows are always possible (and quite likely). For a given flight on JetPurple, there were 11 first class tickets sold. Suppose that each of the 11 persons who purchased tickets has a 20% chance of not showing up for the flight, and that the events that different persons show up for the flight are independent (e.g. the passengers are not traveling as part of a classy high school sports team) . What is the probability that at most 5 of the 11 persons who purchased first class tickets show up for . What is the probability that exactly 10 of the persons who purchased first class tickets show up for the . Suppose that there are 10 seats in first class available and that the cost of each first class ticket is the flight flight? 81,200. (This contributes entirely to profit since the variable cost associated with a passenger ona flight is close to zero.) Suppose further that any overbooked seat costs the airline $3,000, which is the cost of the free ticket plus some cost in damaged customer relations. (First class passengers do not expect to be bumped!) Thus, if 10 of the first class passengers show up for the flight, the airline's profit is $12,000. If 11 first class passengers show up, the profit is $9,000. What is the expected profit from first class passengers for this flight? Suppose that only 10 first class tickets were sold. What would be the expected profit from first class passengers for this flight? (Think) People often travel in groups of two or more. Does this affect the independence assumption about passenger behavior? Why or why not?Explanation / Answer
10 seats, 11 bookings, 20% chance each not showing up
1.SOLUTION
P( atmost 5 show up of 11) = P(0 showup) + P(1) +P(2)+P(3) +P(4)+P(5)
P(0 showup) = 0.2^11 , P( 1 showup) = 0.8 * (0.2)^10 , P(2) = (0.8^2) * (0.2^9) , P(3) = (0.8^3) * (0.2^8) ,
P(4) = (0.8^4) * (0.2^7) , P(5) =(0.8^5) * (0.2^6)
P( atmost 5 show up of 11) = (0.2^11) + ((0.8) * (0.2)^10 ) + ((0.8^2) * (0.2^9)) + ((0.8^3) * (0.2^8)) + ((0.8^4) * (0.2^7)) + ((0.8^5) * (0.2^6))
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2. SOLUTION
P(exactly 10 show up) = (0.8^10 ) * ( 0.2^1) = 0.107 * 0.2 = 0.0214
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3. SOLUTION
10 seats, 11 bookings, profit from ticket =$1200 , cost of overbooking = $3000
Expected profit =??
Expected Profit = P( 0 showup)* Profit + P( 1 showup)* Profit + P( 2 showup)* Profit + P( 3 showup)* Profit + P( 4 showup)* Profit + P( 5 showup)* Profit + P( 6 showup)* Profit + P( 7 showup)* Profit + P( 8 showup)* Profit + P( 9 showup)* Profit + P( 10 showup)* Profit + P( 11 showup) * Profit
= (0.2^11)*0 + ((0.8) * (0.2)^10 ) *1200 + ((0.8^2) * (0.2^9))*2400 + ((0.8^3) * (0.2^8))*3600 +
((0.8^4) * (0.2^7))*4800 + ((0.8^5) * (0.2^6))*6000 +((0.8^6) * (0.2^5))*7200 + ((0.8^7) * (0.2^4))*8400 +
((0.8^8) * (0.2^3))*9600 + ((0.8^9) * (0.2^2))*1,08,00 + ((0.8^10) * (0.2^1))*1,20,00 + (0.8^11)*(1,20,00-3,000)
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4. SOLUTION
If 10 tickets were sold only, expected profit =??
Profit from each ticket = $1200
Expected Profit = P( 0 showup)* Profit + P( 1 showup)* Profit + P( 2 showup)* Profit + P( 3 showup)* Profit + P( 4 showup)* Profit + P( 5 showup)* Profit + P( 6 showup)* Profit + P( 7 showup)* Profit + P( 8 showup)* Profit + P( 9 showup)* Profit + P( 10 showup)* Profit
= (0.2^10)*0 + ((0.8) * (0.2)^9 ) *1200 + ((0.8^2) * (0.2^8))*2400 + ((0.8^3) * (0.2^7))*3600 + ((0.8^4) * (0.2^6))*4800 + ((0.8^5) * (0.2^5))*6000 +
((0.8^6) * (0.2^4))*7200 + ((0.8^7) * (0.2^3))*8400 + ((0.8^8) * (0.2^2))*9600 + ((0.8^9) * (0.2^1))*1,08,00 + (0.8^10)*1,20,00
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