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For a transferable utility game, the characteristic function of a coalitional game assigns a coalition $S$ its largest total attainable revenue $v(S)$, with $v(\varnothing )=0$. The core of a cooperative game consists of payoff vectors $x$ satisfying efficiency, $x(N)=v(N)$, and coalition stability, $x(S)\ge v(S)$ for every $S$. A coalition receiving less could operate alone and distribute its extra value to make each member better off.

Use thousands of pounds as the monetary unit. The physical directed arcs are $A\to B$ and $C\to D$, owned by $i$, with capacities $3,2$; $A\to C$ and $B\to D$, owned by $j$, with capacities $1,1$; and $B\to C$, owned by $k$, with capacity $2$.

The physical network is acyclic. There are exactly three $A$–$D$ graph paths and one $B$–$C$ graph path. Let $u,v,w$ be the $A$–$D$ flows on $ABD$, $ACD$, $ABCD$, respectively, and let $z$ be the direct $B$–$C$ demand. Their capacities and revenue areEvery feasible traffic pattern decomposes into these graph paths, so these inequalities are exact. The first is redundant, since $u\le 1$ and $w\le 2$ imply it.

Formulate this as a minimum-cost flow through the two-resource path packing as minimum-cost circulation reduction. Use auxiliary vertices $s,L,R,t$ and the following arcs, whose entries are (capacity, cost):The first forward graph path carries $u$, $sLt$ carries $v$, $sLRt$ carries $w$, and $sRt$ carries $z$. Conservation at $L$ imposes $v+w\le 2$, and at $R$ imposes $w+z\le 2$; the individual finite-capacity arcs give the other bounds. The return arc closes the circulation. Its cost is exactly $-R$, so this is an ordinary minimum-cost circulation with precisely the required feasible traffic set. Unlimited capacities may all be replaced by $5$, the maximum possible total graph path flow.

An linear programming optimality certificate is particularly short:The feasible choice $u=v=w=z=1$ attains equality. Thus the actual traffic is three units from $A$ to $D$ and one from $B$ to $C$, andAll four positive terms in the bound must be tight at any optimum, so these graph path amounts are unique.

The two traffic demands must retain their identities. Merely adding fictitious return arcs $D\to A$ and $C\to B$ to the physical graph can create an invalid graph cycle combining both rewards: $A\to C\to B\to D\to A$ would credit the $j$-only network with revenue even though it carries neither requested demand. The resource-flow formulation above avoids that problem.

For the characteristic function, only complete physical graph paths owned by a coalition are available. Company $i$ or $j$ alone has no requested graph path; company $k$ carries two units of $B$–$C$ traffic. The pair $\{i,j\}$ carries one unit on each of $ABD,ACD$; the pair $\{i,k\}$ sends two units on $ABCD$ rather than selling them at the lower $B$–$C$ rate; and $\{j,k\}$ has only the two direct $B$–$C$ units. ThereforeWriting payments in thousands, all stable divisions of the revenue are exactlyEquivalently, $8\le x_k\le 10$, $0\le x_j\le 10$ and $x_i=22-x_j-x_k$. This is the entire core, not just one acceptable allocation.

The nucleolus minimizes in lexicographic order the sorted decreasing excesses of a coalition $e(S,x)=v(S)-x(S)$ over imputations. Empty and grand-coalition coalition excesses are fixed zeros and can be omitted. For the six other coalitions, efficiency givesThe two $k$-related coalition excesses sum to $-2$, so their maximum is at least $-1$. This bound is attainable, and attaining it forces $x_k=9$. All six coalition excesses are then at most $-1$ precisely when $1\le x_j\le 9$ and $x_i=13-x_j$. The fixed pair of coalition excesses equals $-1$ throughout these first-stage minimizers.

Minimize the largest of the remaining coalition excesses. They become $x_j-13,-x_j,x_j-10,-1-x_j$, whose maximum is $\max (-x_j,x_j-10)$. Its unique minimum is $-5$ at $x_j=5$. Hence $x_i=8$ and $x_k=9$. The sorted proper coalition excesses there are $(-1,-1,-5,-5,-6,-8)$, and