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These feasible potentials give the lower bound ${\sum }_i{\lambda }_i-{\sum }_j{\mu }_j$ on every assignment cost. An assignment using only equality edges attains this bound and is optimal by weak duality and complementary slackness.

For finite convex total variation and a positive squared-error budget, the Slater condition supplies a multiplier $\lambda \ge 0$ such that every constrained minimizer also minimizes $\mathrm{TV}(u)+\lambda \Vert u-g{\Vert }^2$. Conversely, a minimizer of that penalized objective with $\lambda >0$ minimizes total variation under the budget equal to its own squared error. Complementary slackness proves the first implication; a comparison of objective values proves the second. This does not assert uniqueness of the multiplier or equality of all minimizer sets when $\lambda =0$.

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.

For a minimization problem with $g_i(u)\le 0$ and $h_j(u)=0$, the optimization Lagrangian is $L=f+{\sum }_i{\lambda }_i{g}_i+{\sum }_j{\mu }_j{h}_j$ with ${\lambda }_i\ge 0$ and unrestricted equality Lagrange multipliers. Its infimum over the original variable domain gives a dual lower bound. For a maximization problem, use $L=f-{\sum }_i{\lambda }_i{g}_i-{\sum }_j{\mu }_j{h}_j$ to obtain upper bounds. The Lagrangian sufficiency theorem combines a global extremum of this function with feasibility and complementary slackness. This is distinct from a Lagrangian density in variational physics.

For a minimization problem on a domain $D$, write the equality constraints as $h(x)=0$ and the inequality constraints as $g(x)\le 0$. Define the optimization LagrangianThe Lagrangian sufficiency theorem states that a feasible $x^{\ast }$ is globally optimal if it globally minimizes $L(\cdot ,{\lambda }^{\ast },{\mu }^{\ast })$ on $D$ for some multipliers with ${\mu }^{\ast }\ge 0$ and satisfies complementary slackness, ${\mu }_j^{\ast }{g}_j(x^{\ast })=0$ for every $j$. Equality multipliers have unrestricted signs.

For any feasible $x$, the multiplier sign gives $L(x,{\lambda }^{\ast },{\mu }^{\ast })\le f(x)$. Feasibility and complementary slackness at $x^{\ast }$ give $L(x^{\ast },{\lambda }^{\ast },{\mu }^{\ast })=f(x^{\ast })$. HenceThis proves sufficiency. Global Lagrangian minimization, rather than stationarity alone, is the certificate. No convexity or constraint qualification is needed for this implication; convexity is one way to verify the required global minimum.

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.