Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 40 / 1 / Solution

Introduce nonnegative slack variables $s_1=2+x_1-x_2$, $s_2=8-x_1-2x_2$ and $s_3=8-2x_1-x_2$, and write the objective as $Z$. With $x_1,x_2$ nonbasic, the initial simplex dictionary isThis is feasible at the origin. Enter $x_1$. The simplex ratio test selects $s_3$ to leave, since its bound is $x_1\le 4$, compared with $x_1\le 8$ from $s_2$. Pivoting givesNow enter $x_2$. The ratio test chooses $s_2$ to leave: its bound $8/3$ is smaller than $4$ from $s_1$ and $8$ from $x_1$. The resulting dictionary isAll basic values are nonnegative and all nonbasic objective coefficients are nonpositive. The simplex algorithm therefore certifiesThe negative objective coefficients also make this continuous optimum unique.

For integer $x_1,x_2$, all three original slacks are integers. Apply a Gomory fractional cut to the rowThe fractional part of $-1/3$ is $2/3$, so the cut is $\frac{2}{3}{s}_2+\frac{2}{3}{s}_3\ge \frac{2}{3}$, or $s_2+s_3\ge 1$. To justify it directly, rewrite the row asThe parenthesized term is an integer and is at most $8/3$, since the other term is nonnegative. It is therefore at most $2$, proving the cut. This retains every feasible integer solution and excludes the current fractional optimum. Since $s_2+s_3=16-3(x_1+x_2)$, the cut is equivalently $x_1+x_2\le 5$.

Introduce the integer-valued nonnegative slack $t=s_2+s_3-1$. Its dictionary row is $t=-1+s_2+s_3$, so primal feasibility fails while the objective row remains dual feasible. In the dual simplex algorithm, $t$ leaves; $s_2,s_3$ tie in the dual ratio test, each with ratio $1/3$. Choose $s_2$ to enter. Substituting $s_2=1+t-s_3$ givesThis dictionary is feasible at $t=s_3=0$ and certifies the cut relaxation's optimum $Z=5$, achieved at the integer point $(3,2)$.

To find every optimal integer point, the objective row forces $t=0$. Nonnegativity of all the remaining rows then reduces to $0\le s_3\le 1$. On this optimal edge, $(x_1,x_2)=(3-s_3,2+s_3)$. Since $s_3$ must be integral, only its two endpoints are possible: