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:
$(V(SUi)-V(S))$
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?
Thanks
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
$$cv=(100,160,200,160,200,200,200)$$
Computing the Shapley value, and we obtain
$$sh_{cv}=(100,190,310)/3. $$