can anyone help me on this?

I am trying to understand how can I find the Shapley values for the following airport problem.

There are 3 types of airplanes, type 1, type 2 and type 3 and they require the following length of the runway to land:

Type 1: 100

Type 2: 160

Type 3: 200

I would like to know how much each airplane should pay to land in the airport.

I know that the Shapley formula is:

$\phi= \sum{\frac{|S|! (|N|-|S|-1)!}{|N|!}}* (v(SU{i})-v(S)) $

where in this case:

N={1,2,3}=3, so for the first part of the formula I have that:

$\phi= \frac{|S|! (|3|-|S|-1)!}{|N|!}$=

$\phi= \frac{|S|! (|2|-|S|)!}{|3|!}$

To find the Shapley value for type 1:

I did:

for the first part of the formula and if S= $\emptyset$

$ \frac{|S|! (|2|-|S|)!}{|3|!}$ = $\frac{| \emptyset| ! (|2|-|0|)!}{|3|!}$= $\frac{0! * 2!}{3!}= \frac{1}{3}$

And the same for S={2}

$ \frac{|S|! (|2|-|S|)!}{|3|!}$ = $\frac{|1| ! (|2|-|1|)!}{|3|!}$= $\frac{1! * 1!}{3!}= \frac{1}{6}$

The same if S={3}

The result is the same as S=2

For S={2,3}

$ \frac{|S|! (|2|-|S|)!}{|3|!}$ = $\frac{|2| ! (|2|-|2|)!}{|3|!}$= $\frac{2! * 0!}{3!}= \frac{1}{3}$

Now I need to apply the second part of the formula which is:


I believe that for S=$\emptyset$

the formula will be: $v(\emptyset U 1) $- $V(\emptyset)$= v(1)-v($\emptyset$)=100-0=100

But then for

S={2}, S={3} and S={2,3} I cannot understand how can I apply the formula. Should it be:

For S={2}

$v(1 U 1) $- $V(1)$= v(1)-v($\emptyset$)=(100+100)-100=200-100=100 ?

And how can I know what is the value of S={2,3} ?

Can anyone give a hint how to calculate these values?



Your airport TU-game is not completely defined. For instance, if we define the airport game from the cost vector $c=(100,160,200)$ through

$$C(S)=\max_{i \in S}\, c_{i} \qquad \forall S \subseteq N,$$

then the associated cost game is given by


Computing the Shapley value, and we obtain

$$sh_{cv}=(100,190,310)/3. $$

  • $\begingroup$ is this for type 1 airplanes? Can you clarify what you did please? thanks $\endgroup$ – user290335 Nov 28 '17 at 21:18
  • $\begingroup$ I don't get your point. For each composition of airplanes I compute the cost charged by the above formula. For instance, for S=$\{1,2\}$, the cost incurred to S is $max(100,160) = 160$. Having this accomplished, apply the formula of the Shapley value. Then the fee imposed by the Shapley value for airplane 1 is $100/3$, for airplane 2 it is $190/3$, and finally for the third, a fee of $310/3$ is charged. $\endgroup$ – Holger I. Meinhardt Nov 29 '17 at 9:47
  • $\begingroup$ I got it. I was asking for the cost of airplane 1 because the others will follow the same logic, So for S={2,3} the cost will be 200 right? Thank you for your help. $\endgroup$ – user290335 Nov 29 '17 at 10:14
  • $\begingroup$ is this the cost or the runaway in meters they need to land? because additionally, I know that the price to repair the damage done by one landing to 1m of the runaway is 10pences $\endgroup$ – user290335 Nov 30 '17 at 13:08
  • $\begingroup$ Yes, this is simplest way one can approach the construction of an airport game. In the literature, one can find different forms to get the associated airport game. $\endgroup$ – Holger I. Meinhardt Nov 30 '17 at 13:16

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