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Assume that there are $m$ different types of aircrafts. For each type $l$, we know the number of movements, $n_{l}$, and the overall runway costs, $c_{l}$, associated with type $l$ aircrafts. Since we want to find out how much each individual movement should be charged, we interpret each of them as a single player. $N_{l}$ be the set of "players" associated with movements of type $l$ aircrafts.

We assume that types are arranged in an increasing order with respect to the costs, i.e $0<c_{1}<c_{2}<...<c_{m}$. Let $N$ be the (finite) set of all movements, namely $N:= \displaystyle\bigcup_{l=1}^{m}{N_l }$.

The airport game associated with this is the TU-game with player $N$ and characteristic function $c$ defined, for each $S\subset N$,$S\ne \varnothing$ by $v(S):=- max\{c_{l}:S\cap N_{l}\ne \varnothing \}$.

Questions

1) Could you explain why the negative sign?.

2) Could you give me some help to verify that airport games is superadditive and convex?

Superadditive. If $A,B\subset N$, $A\cap B=\varnothing$, $v(A\cup B)\ge v(A)+v(B)$.

Convex.If $A,B\subset N$, $v(A\cup B)\ge v(A)+v(B)-v(A\cap B)$

Thank you.

Edit: The negative sign is because the cost function is not superadditive but its negative if it is. But why?

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    $\begingroup$ Hi Franciscolli, this looks like a homework problem. The policy of this forum is to not just provide an answer key to homework problems. If you want you can add more details as to what have you tried doing and what is confusing. People in the forum will be more akin to help you that way. Also, it looks like something is off with your notation. The characteristic function is defined for any subset of movements, $S$, but seems to depend on a particular aircraft, since it depends on $N_l$. If that is the case, what is he value of $v(S)$ if $S\cap N_l=\emptyset$? $\endgroup$ – Regio Jul 3 at 14:31
  • $\begingroup$ Notation is of book "An introductory course on mathematical game theory".I also find it confusing. $\endgroup$ – Franciscolli Jul 6 at 3:42
  • $\begingroup$ If $S∩N_{l}=∅?$ I think that $v(S)=0$ $\endgroup$ – Franciscolli Jul 7 at 1:56

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