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Mathematical optimization studies extrema under constraints.

With the tail-positive incidence convention, a network edge's reduced cost is $r_{ij}=c_{ij}-{\pi }_i+{\pi }_j$. For uncapacitated minimum-cost flow, dual feasibility requires $r_{ij}\ge 0$ and complementary slackness requires $f_{ij}{r}_{ij}=0$. A negative non-tree reduced cost identifies an improving graph cycle pivot in the network simplex algorithm. This sign convention differs from the row-column potentials of a transportation problem.

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.

A transportation problem minimizes ${\sum }_{ij}{c}_{ij}{x}_{ij}$ subject to nonnegative shipments, prescribed row supplies, and prescribed column demands.

One optimal shipment table iswhich has the required row sums $(10,8,8)$ and column sums $(3,9,14)$ and costs $4(3)+6(1)+8(3)+3(3)+5(5)=\boxed{76}$.

For an optimality certificate, take row potentials $(1,3,3)$ and column potentials $(0,2,0)$. Their sums equal costs in occupied cells and do not exceed any cost:The dual value is $10+24+24+18=76$, so linear programming duality proves optimality.

A transportation problem chooses nonnegative shipments $x_{ij}$ from $n$ suppliers to $m$ consumers so that row sums equal supplies and column sums equal demands, while minimizing $\sum c_{ij}{x}_{ij}$.

The north-west corner rule givesIt has $3+4-1=6$ positive cells and its support contains no cycle, so it is a nondegenerate basic feasible solution.

A degenerate basic feasible solution isIts five positive cells form a forest with two balanced components; adding one zero cell that joins the components completes a basis of six cells. Thus at least one basic variable is zero.

The stated plan isIts six positive cells form a spanning tree of the supplier--consumer bipartite graph, so it is basic and feasible. Taking $u_1=0$, the basic-cell equations $u_i+v_j=c_{ij}$ giveThe complete matrix of reduced costs ${\bar{c}}_{ij}=c_{ij}-u_i-v_j$ isThe negative entry ${\bar{c}}_{12}=-2$ proves that the plan is not optimal.

Enter cell $(1,2)$. The alternating cycle isand the step is $\theta =\min (3,1)=1$. The new plan iswhose cost is $26$, down from $28$. New potentials areand the reduced-cost matrix isEvery reduced cost is nonnegative, so the transportation optimality criterion shows thatThe zero reduced cost in cell $(2,3)$ also indicates an alternative optimum.