Finding the Nucleolus Given the following table of values and excesses of coalitions S and imputation $\vec{x} = (9,6,9)$:

How do I find the Nucleolus? My book wasnt clear on the method of calculating it, so Id like to find a general approach. The vector $\vec{x}$ was chosen by me from the imputation set 
$\begin{align*}
I = \{ (x_{1}, x_{2}, x_{3}) \in \mathbb{R}^{3} \; |  & \;  x_{1} \geq 2, x_{2} \geq 5, x_{3} \geq 4 \\ 
& \; x_{1} + x_{2} + x_{3} = 24 \}
\end{align*}$
as originally only the values of the coalitions were given. 
 A: I created game theory package "gt.mac" on free CAS Maxima
(%i1)   load(gt);
(%o1)   "C:/Users/aleksas/maxima/gt.mac"
Example
(%i2)   G:[{1,2,3},
    [v({1})=2,
    v({2})=5,
    v({3})=4,
    v({1,2})=14,
    v({1,3})=18,
    v({2,3})=9,
    v({1,2,3})=24
    ]]$
(%i3)   core(G);
(%o3)   convexhull([[8,6,10],[9,5,10],[14,6,4],[15,5,4]])
(%i4)   least_core(G);
(%o4)   [-1/2,convexhull([[9,11/2,19/2],[14,11/2,9/2]])]
(%i5)   next_core(G,%);
(%o5)   [-3,[23/2,11/2,7]]
(%i6)   next_core(G,%);
(%o6)   [-3,[23/2,11/2,7]]
(%i7)   nucleolus(G);
(%o7)   [23/2,11/2,7]
(%i8)   ShapleyValue(G);
(%o8)   [19/2,13/2,8]
For used method see:
 E.N.Barron, Game theory. An Introduction, second ed., John Wiley & Sons, Inc., 2013
Nucleolus computation details:


*

*We compute least core. Solve problem:
minimize $\epsilon$ subject to 


$2-{x_1}\leqslant\epsilon,\;5-{x_2}\leqslant\epsilon,\;4-{x_3}\leqslant\epsilon,-{x_2}-{x_1}+14\leqslant\epsilon,\\-{x_3}-{x_1}+18\leqslant\epsilon,\;-{x_3}-{x_2}+9\leqslant\epsilon,\;{x_3}+{x_2}+{x_1}=24$.
Solution is: $\epsilon=-\frac12,\quad [x_1,x_2,x_3]\in \operatorname{convexhull}\left( [[9,\frac{11}{2},\frac{19}{2}],[14,\frac{11}{2},\frac{9}{2}]]\right) $.


*

*$x_2=\frac{11}{2}$

*Delete coalitions {2} and {1,3}.

*We compute next core. Solve problem:
minimize $\epsilon$ subject to


$2-{x_1}\leqslant\epsilon,\;4-{x_3}\leqslant\epsilon,\;\frac{17}{2}-{x_1}\leqslant\epsilon,\;\frac{7}{2}-{x_3}\leqslant\epsilon,\;{x_3}+{x_1}+\frac{11}{2}=24$.
Solution is: $\epsilon=-3$, ${x_1}=\frac{23}{2}$, ${x_3}=7$.


*

*If next core consists of exactly one point we stop computation.


$$Nucleolus=\left[ \frac{23}{2}, \frac{11}{2}, 7\right].$$
