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Write the top row as $(u,v,5-u-v)$. The bottom row is then $(3-u,3-v,u+v-1)$, and nonnegativity becomesThe feasible set is a hexagon in this plane. Its six vertices, hence all basic feasible solutions of this transportation problem, areThere are four independent constraints; each displayed support consists of four cells forming a transportation spanning tree, so each is a nondegenerate basic solution. Conversely, a basic feasible solution is a vertex, so the hexagon enumeration is exhaustive.

For any feasible $x^{\prime }$, the proposed dual inequalities giveIf equality holds on every cell with $x_{ij}>0$, this last quantity equals $f(x)$. Hence $f(x)\le f(x^{\prime })$ for every feasible $x^{\prime }$. This is weak duality with complementary slackness.

The Northwest corner method first allocates three units to $(1,1)$, then two to $(1,2)$, one to $(2,2)$ and four to $(2,3)$:Choose ${\lambda }_1=0$, ${\lambda }_2=2$, $({\mu }_1,{\mu }_2,{\mu }_3)=(a,2,-1)$. Equality holds on its four occupied cells. The other inequalities are $-1\le 3$ and $a+2\le b$. Thus $a+2\le b$ implies that $x^{\ast }$ minimizes $f$.

The last printed clause changes from minimization to maximization. To answer that wording, use the transportation simplex algorithm for maximization, equivalently minimize the negative costs. There are only two possible entering cells at $x^{\ast }$:

Put $d=b-a$. On the feasible hexagon,If $d\le -2$, this expression is maximized by taking $u=3,v=0$, so $E$ is a global maximizer. If $d>-2$, the feasible vertex $(u,v)=(1,0)$ has objective $2(d+2)$ larger than $E$, so $E$ is not a maximizer. Nor can $C$ ever be one: comparison with $x^{\ast }$ requires $d\ge 2$, while comparison with the vertex $(0,3)$ requires $d\le -2$. Therefore one pivot to a maximizer is possible exactly when $a\ge b+2$, by entering $(1,3)$. Its reduced cost for maximization is $4>0$; the other reduced cost is $d-2\le -4$, so the improving pivot is unambiguous.

If “transportation algorithm” is required to retain the preceding minimization objective, no such step can reach a maximizer: every minimizing step is nonincreasing, whereas the feasible point $E$ always has $f(E)=f(x^{\ast })+8$. If the final word was intended to be “minimizes”, the corresponding genuinely improving one-pivot condition would instead be $a-2\le b<a+2$; at $b=a+2$, $x^{\ast }$ is already optimal and a zero-cost pivot is possible. These distinctions preserve the printed maximization request without silently altering it.