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Decompose the required runway into successive layers. The first layer has cost $s$ and is needed by all four planes. The next costs $2s$ and is needed by the three medium-or-long planes. The final layer costs $4s$ and is needed only by the longest plane.

For each layer, the first requiring plane in a uniformly random ordering contributes its entire cost. Each of its requiring planes has equal probability of being first. By linearity of the Shapley value, divide each layer cost equally among those requiring it. The payments are thereforeThey sum to $7s=c(N)$. The equality of the two medium payments follows from their identical runway requirements; the random-order argument proves the amounts, not just their equality.

Suppose the distinct required runway lengths are $0=s_0<s_1<\cdots <s_k$, with $n_i>0$ users of type $i$, constant cost $\lambda \ge 0$ per unit length, and suffix counts $N_i={\sum }_{\ell =i}^k{n}_{\ell }$. Then every type-$j$ user has the displayed Shapley value. For layer $i$, exactly $N_i$ users require it. The first of them in a uniformly random permutation contributes its entire cost $\lambda (s_i-s_{i-1})$. Each requiring user has probability $1/N_i$ of being first. Linearity of the Shapley value sums these expected marginal contributions across the layers that user needs.

The allocation lies in the cost core. Its payments sum to the full runway cost. For any coalition $S$, its total payment toward layer $i$ is $\lambda (s_i-s_{i-1})|S\cap R_i|/N_i$, where $R_i$ is the set of users requiring that layer. This is zero if $S$ needs no such layer, and at most the full layer cost otherwise. Summing gives ${\sum }_{j\in S}{\varphi }_j\le c(S)$. Suffix counts followed by prefix sums compute all distinct type payments in $O(k)$ arithmetic operations.