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The practice of overbooking in the airline industry is a strategic approach aimed at maximizing revenue despite the uncertainty of passenger no-shows. Airlines, understanding that a certain percentage of ticketed passengers will not appear for their flights, sell more tickets than the actual number of available seats. However, overbooking carries the risk of overselling and the potential for denied boarding if more passengers show up than there are seats. Therefore, the airlines must carefully determine the optimal number of tickets to sell to balance revenue and customer satisfaction, while adhering to safety and regulatory constraints.
United Airlines (UAL), being notorious for cost-cutting and substandard customer service, follows a typical industry practice of overbooking. Given that only 85% of ticket holders tend to show up, the airline operates under the assumption that the number of passengers arriving follows a binomial distribution. If we let n be the number of tickets sold, then the number of passengers showing up is a random variable X ~ Binomial(n, p), where p = 0.85.
In this scenario, UAL needs to determine the maximum value of n such that the probability of all passengers who show up being accommodated (i.e., X ≤ 234) is at least 95%. Mathematically, this can be expressed as:
P(X ≤ 234) ≥ 0.95
Since X follows a binomial distribution with parameters n and p, calculating this probability directly for large n can be cumbersome. Therefore, we typically use a normal approximation to the binomial distribution for computational efficiency, especially when n is large. The normal approximation states that X ≈ N(µ, σ²), where the mean µ = np, and the standard deviation σ = √(np(1 - p)).
To find the maximum n satisfying the probability constraint, we proceed with the approximation:
P(X ≤ 234) ≈ Φ((234 + 0.5 - µ) / σ) ≥ 0.95
Here, Φ denotes the cumulative distribution function (CDF) of the standard normal distribution, and the +0.5 is a continuity correction.
Now, express µ and σ explicitly:
µ = n * 0.85
σ = √(n * 0.85 * 0.15)
The inequality becomes:
Φ((234 + 0.5 - 0.85n) / √(0.1275n)) ≥ 0.95
Next, we find the z-score corresponding to the 95th percentile of the standard normal distribution, which is approximately 1.645. Rearranging the inequality:
(234.5 - 0.85n) / √(0.1275n) ≥ 1.645
Multiplying both sides by √(0.1275n):
234.5 - 0.85n ≥ 1.645 * √(0.1275n)
Define s = √(0.1275n), then n = s² / 0.1275, and rewrite the inequality in terms of s:
234.5 - 0.85 * (s² / 0.1275) ≥ 1.645 * s
Calculate 0.85 / 0.1275 ≈ 6.6667. Therefore:
234.5 - 6.6667 * s² ≥ 1.645 * s
Bring all terms to one side:
-6.6667 * s² - 1.645 * s + 234.5 ≥ 0
Multiply through by -1 to get a standard quadratic form:
6.6667 * s² + 1.645 * s - 234.5 ≤ 0
This is a quadratic inequality in s, which we can solve to find the permissible range of s. The quadratic equation is:
6.6667s² + 1.645s - 234.5 = 0
Apply the quadratic formula:
s = [-1.645 ± √(1.645² - 4 * 6.6667 * (-234.5))] / (2 * 6.6667)
Calculate discriminant D:
1.645² ≈ 2.707
4 * 6.6667 * 234.5 ≈ 4 * 6.6667 * 234.5 ≈ 4 * 1,564.1667 ≈ 6,256.6668
Since the last term is negative times positive, the discriminant is:
D = 2.707 + 6,256.6668 ≈ 6,259.3738
Then, s becomes:
s = [-1.645 ± √6,259.3738] / 13.3334
√6,259.3738 ≈ 79.13, so the two solutions are:
s = [-1.645 + 79.13] / 13.3334 ≈ 77.485 / 13.3334 ≈ 5.813
s = [-1.645 - 79.13] / 13.3334 ≈ -80.775 / 13.3334 ≈ -6.066
< p>Since s represents a standard deviation component (and must be positive), we take s ≈ 5.813. Recall that s = √(0.1275n), which implies:
√(0.1275n) ≈ 5.813
Squaring both sides:
0.1275n ≈ 5.813² ≈ 33.785
Finally, solve for n:
n ≈ 33.785 / 0.1275 ≈ 265.2
Given that n must be an integer, the maximum number of tickets that can be sold while maintaining at least a 95% probability of accommodating all passengers who show up is approximately 265 seats.
Since the aircraft seating capacity is 234 seats, overbooking up to 265 tickets allows a margin that meets the specified probability threshold, though practically, the airline might choose a conservative number
slightly below this estimate to account for operational uncertainties.
In conclusion, UAL can confidently sell around 265 tickets for the Boring 878 flight while ensuring, with 95% confidence, that they can handle all arriving passengers without violating safety margins or FAA regulations.
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