= Solution
For a <coalition> $S$, its <coalition excess> at an allocation is $e(S,x)=v(S)-x(S)$, measuring its complaint. Sort all these <excesses of a coalition> in decreasing order. The <nucleolus> is the <imputation> whose sorted <vector> is smallest in <lexicographic order>: minimize the greatest complaint first, then the next among ties, and continue. Including the empty and grand <coalitions> merely inserts two fixed zero entries and does not change the minimizing allocation.
For every efficient allocation, the <excesses of a coalition> of $\{2\}$ and $\{1,3\}$ sum to zero:
$$
e(\{2\},x)+e(\{1,3\},x)
=2-x_2+10-(12-x_2)=0.
$$
The largest proper-coalition <coalition excess> is therefore at least zero. Since the <core of a cooperative game> from part (b) is nonempty, this first-stage minimum is exactly zero and its minimizers are precisely that <core of a cooperative game>. Restrict to $x=(t,2,10-t)$, $1\le t\le6$. The six proper-coalition <excesses of a coalition> are
$$
\begin{array}{c|rrrrrr}
S&1&2&3&12&13&23\\ \hline
e(S,x)&1-t&0&t-7&1-t&0&t-6
\end{array}
$$
The two zero <excesses of a coalition> are fixed. Among the remaining <excesses of a coalition>, $t-7<t-6$, so the next largest complaint is $\max\{1-t,t-6\}$. It is minimized by equalizing those two affine expressions:
$$
1-t=t-6,\qquad t=\frac72,\qquad
\max\{1-t,t-6\}=-\frac52.
$$
This minimizer is unique, so later lexicographic stages cannot change it. Thus
$$
\boxed{\nu(v)=\left(\frac72,2,\frac{13}{2}\right).}
$$
Its sorted proper-coalition <coalition excess> <vector> is $(0,0,-5/2,-5/2,-5/2,-7/2)$. The calculation both identifies the <nucleolus> and proves uniqueness for this game without replacing the requested computation by the general uniqueness theorem.
Back to article page
