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For my game theory class I have to determine the core of the following cooperative game:

$\mathrm{N}={1,2,3}$

$S=\{1\}$ gives $v(S)=2$

$S=\{2\}$ gives $v(S)=5$

$S=\{3\}$ gives $v(S)=4$

$S=\{1,2\}$ gives $v(S)=14$

$S=\{1,3\}$ gives $v(S)=18$

$S=\{2,3\}$ gives $v(S)=9$

$S=\{1,2,3\}$ gives $v(S)=24$

I don't know what I should do to find the core properly. I also have to draw it in a triangle.

What I found up until now is the following:

$$C=\{ (x_1,x_2,x_3) \in \mathbb{R}^3 | x_1 \geq 2,x_2 \geq 5, x_3 \geq 4, x_1+x_2 \geq 14,\\ x_1+x_3 \geq 18, x_2 + x_3 \geq 9, x_1+x_2+x_3=24 \}$$

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The core of the permutation game is given by the convex combination of the four extreme points:

9    5   10  
8    6   10
14    6    4
15    5    4

Since, it is a homework, check out by yourself that the result is correct! Moreover, you will find out that the Shapley value is located outside of the core. Why?

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  • $\begingroup$ Okay, so my question is how to find the extreme points. In this question did you find the first extreme point using the following: suppose $x_2=5$, then $x_1+x_2=x_1+5=14$ thus $x_1=9$ thus $x_3=24-5-9=10$? And then for the other points you did the same? And would this method work for all examples? $\endgroup$ – DeanTheMachine Jan 12 '15 at 11:44
  • $\begingroup$ The simplest approach is to find out which linear equations forms the boundary of the core (you will find four linear equations), and then look at the intersection points. $\endgroup$ – Holger I. Meinhardt Jan 12 '15 at 13:04

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