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A cutting-plane method repeatedly adds valid inequalities that exclude a current relaxed solution while preserving all feasible integer solutions. Gomory fractional cuts provide such inequalities from a fractional simplex dictionary.

Introduce nonnegative surplus variablesAt the proposed starting basic feasible solution, the basic variables are $x_1,r_1,r_2$ and the nonbasic variables are $x_2,x_3,r_3$. Its simplex dictionary, with objective $z=x_1+x_2+x_3$, isIncrease $x_3$ to decrease $z$. The simplex ratio test gives limits $1/5$ from $x_1$, $1/21$ from $r_1$, and $3/31$ from $r_2$. The smallest is $1/21$, so $x_3$ enters and $r_1$ leaves. After this single pivot the dictionary isAll objective coefficients of nonbasic variables are strictly positive. Thus the simplex method has reached its unique optimum:The feasible dual vector $y=(1/21,0,4/21)$ has the same objective, providing an independent weak duality certificate. Normalization givesThese are all the equilibria: the dual's strict slack in row two forces $p_2=0$, and the primal's strict slack in column two forces $q_2=0$. Equality of the two active row and column payoffs then fixes the displayed probabilities. The value is the first player's expected net loss; the second gains $4/5$.

Introduce nonnegative slack variables $s_1,s_2,s_3$ and start the simplex algorithm at $x=y=z=0$, with slacks $9,8,2$. The objective's largest positive entering coefficient belongs to $y$. Its ratio test is $9/3=3$, $8/2=4$, $2/1=2$, so $s_3$ leaves and $y$ enters.

Solve that pivot row for $y$ and substitute into the other rows and the objective $v$:This is a feasible simplex dictionary at nonbasic variables $x=z=s_3=0$. All their objective coefficients are negative, so no feasible increase can improve the objective. ThusThe negative reduced coefficients also prove uniqueness of this optimum.

The dual linear program, with nonnegative variables $u_1,u_2,u_3$, isThe feasible choice $\boxed{(u_1,u_2,u_3)=(0,0,5)}$ has objective $10$. Weak duality proves that it and the primal point are optimal. Complementary slackness also forces $u_1=u_2=0$ because the first two primal slacks are positive, and then the positive $y$ forces $u_3=5$.

Introduce integer-valued slack variables $s_1=11-2x_1-x_2$ and $s_2=6+x_1-2x_2$. Write $Z=4x_1+3x_2$. The initial simplex dictionary isIn the simplex algorithm, let $x_1$ enter. The first constraint is the only limiting one, so $s_1$ leaves at $x_1=11/2$. Substitution givesNow $x_2$ enters and $s_2$ leaves: its ratio $23/5$ is smaller than the first row's ratio $11$. The resulting simplex dictionary isAll nonbasic objective coefficients are nonpositive, and the basic values are nonnegative. Thus the continuous linear program has optimum

For integer programming, a row $x_B+{\sum }_j{a}_j{x}_j=b$ with integer variables yields the Gomory fractional cut ${\sum }_j\{a_j\}{x}_j\ge \{b\}$, where $\{u\}=u-\lfloor u\rfloor$ is the fractional part. In the first row, the coefficient of $s_2$ on the left is $-1/5$, whose fractional part is $4/5$, not $-1/5$. The two possible initial Gomory fractional cuts are thereforeIndeed $s_1+2s_2=23-5x_2$. Choose the second row: $x_2\le 4$ implies $x_2\le 9/2$, so its cut removes a strictly larger portion of the feasible polygon. This compares both printed alternatives before any further integer rounding.

Use the normalized integer slack variable $s_3=4-x_2$, rather than an unnecessarily scaled cut slack. At the current basis,The simplex dictionary is dual feasible but primal infeasible. In the dual simplex algorithm, $s_3$ leaves; the ratios of objective loss to improvement of this negative basic value are $11$ for $s_1$ and $1$ for $s_2$, so $s_2$ enters. The pivot givesThe relaxed optimum is now $(7/2,4)$, still nonintegral. The fractional $x_1$ row gives the next Gomory fractional cutThe negative left-hand coefficient $-1/2$ of $s_3$ also has fractional part $1/2$. Normalize the new integer slack variable as $s_4=7-x_1-x_2$. Its dictionary row is $s_4=-1/2+s_1/2+s_3/2$. The dual simplex algorithm chooses $s_3$ to enter: its ratio is $1/(1/2)=2$, compared with $2/(1/2)=4$ for $s_1$. Substitution givesSetting the nonbasic $s_1,s_4$ to zero gives a feasible integer solution. The final objective row bounds every point in the cut relaxation by $25$, so the integer optimum isBoth cuts are valid for every original integer feasible point, making this also a certificate for the original integer program.