Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 42 / 5 / c

For three players, the predecessor-set weights are $1/3$ for sizes zero and two, and $1/6$ for size one. ThusTherefore the Shapley value is $\boxed{(19/3,17/6,17/6)}$, whose coordinates sum to $12$.

For the nucleolus, use imputations $x_1+x_2+x_3=12$, $x_1\ge 4$, $x_2\ge 3$, $x_3\ge 2$. The excess of $\{1,3\}$ is $11-(12-x_2)=x_2-1\ge 2$. Thus the smallest possible largest excess is at least $2$. It is attained when $x_2=3$, $x_3=9-x_1$, and $5\le x_1\le 7$: the remaining proper-coalition excesses then areall at most $2$, while $e_{13}=2$. Conversely, a largest excess of $2$ forces $x_2=3$ and precisely this interval for $x_1$.

On this first-stage face the top excess $2$ is fixed. The next largest excess is $7-x_1\ge 0$, because the other varying excesses are nonpositive and $e_2=0$ is fixed. Its unique minimum is zero at $x_1=7$. No further lexicographic tie remains. Hence the nucleolus is $\boxed{(7,3,2)}$. Its proper-coalition excesses, sorted decreasingly, are $(2,0,0,0,0,-3)$.

Finally, a core allocation would require $x_2\ge 3$ and $x_1+x_3\ge 11$, whose sum contradicts the efficient total $12$. Thus the core of a cooperative game is empty. The game is not superadditive, so neither core nonemptiness nor individual rationality of the Shapley value should be presumed. Under the distinct prenucleolus convention the answer would be $(15/2,2,5/2)$. Balancing $e_2=3-x_2$ and $e_{13}=x_2-1$ gives a first-stage maximum of $1$ and forces $x_2=2$. The remaining first-stage constraints restrict $x_1$ to $[7,8]$. The next largest complaints are $e_{12}=8-x_1$ and $e_{23}=x_1-7$, balanced at $x_1=15/2$, with $x_3=5/2$. This is not an imputation and is not the nucleolus under the definition in (a).