Overbooking a 180-seat flight
A small airline operates a 180-seat aircraft on a route where, historically, 6% of passengers who have booked do not turn up. The airline is considering selling 190 tickets for the flight. The no-show behaviour of different passengers can be treated as independent, and the probability of a no-show is the same for each passenger. If more passengers turn up than there are seats, the airline must deny boarding to the excess passengers and pay compensation. The airline’s revenue per ticket is £95, and the compensation cost is £220 per denied passenger, plus reputational damage that is difficult to quantify. The operations manager wants a simple estimate of the risk of overbooking and a recommendation on how many tickets to sell. The finance director argues that selling more tickets is always better because no-shows are lost revenue. The safety regulator requires that every passenger who turns up is either seated or compensated. The airline can change the number of tickets sold but not the aircraft.
1. Using a normal approximation to the binomial distribution, estimate the probability that more than 180 passengers turn up when 190 tickets are sold. State the mean and standard deviation used.
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Let X be the number who turn up. Then X ~ Binomial(n = 190, p = 0.94).
Mean = 190 × 0.94 = 178.6.
Variance = 190 × 0.94 × 0.06 = 10.716, so sd = 3.27.
Using a continuity correction:
P(X > 180) = P(X ≥ 181) ≈ P(Z ≥ (180.5 − 178.6)/3.27) = P(Z ≥ 0.58) = 0.281.
The estimated risk of overbooking is about 28%.
2. Find the largest number of tickets the airline can sell so that this probability is at most 5%. Use the normal approximation.
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Try n = 186. Mean = 174.84, variance = 186 × 0.94 × 0.06 = 10.49, sd = 3.24.
P(X ≥ 181) ≈ P(Z ≥ (180.5 − 174.84)/3.24) = P(Z ≥ 1.75) = 0.040, which is below 5%.
Try n = 187. Mean = 175.78, variance = 10.55, sd = 3.25.
P(X ≥ 181) ≈ P(Z ≥ (180.5 − 175.78)/3.25) = P(Z ≥ 1.45) = 0.073, which is above 5%.
The largest number is therefore 186 tickets.
3. Advise the airline, taking account of expected compensation and the limitations of the model.
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The normal approximation suggests about 0.57 expected denied boardings when 190 tickets are sold, compared with about 0.05 when 186 are sold. The extra four tickets bring 4 × £95 = £380 in revenue, while the extra expected compensation is about (0.57 − 0.05) × £220 = £114. On pure expected money, selling 190 is better by about £266 per flight.
However, 190 has about a 28% chance of any overbooking, which may damage goodwill, while 186 keeps the risk near 4%. A risk-averse airline should sell 186; an airline that values the £266 margin and can handle compensation smoothly could sell 190. The binomial model assumes independent passengers and a constant 6% no-show rate; group bookings, seasonality, or changed conditions would make these risk estimates unreliable.