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A basic feasible solution is degenerate when one or more of its basic variables is zero. A simplex algorithm pivot may then change the basis without moving the feasible point or changing the objective. Assignment networks force many zero-flow basic tree arcs.

If a linear program has an optimal solution, it has an optimal basic feasible solution. Choose an optimum with minimal positive support. A dependence among its active columns gives a feasible two-sided perturbation; optimality makes its objective slope zero, and moving until one coordinate vanishes contradicts minimality.

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.

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 $4/5$.

Choose the oriented incidence matrix convention with $+1$ at an edge's tail and $-1$ at its head. Let $b_i$ be net supply, so ${\sum }_i{b}_i=0$. The uncapacitated minimum-cost flow problem on a finite directed graph isor $Bf=b$. Negative $b_i$ denotes net demand. Costs may have either sign; the problem can be infeasible or unbounded.

For vertex network dual potentials $\pi$, the optimization Lagrangian isThe infimum over $f\ge 0$ is finite exactly when every network reduced cost $r_{ij}=c_{ij}-{\pi }_i+{\pi }_j$ is nonnegative. Thus the Lagrangian dual problem and complementary slackness areA feasible flow and feasible network dual potentials satisfying these equalities have equal costs and are optimal by weak duality.

If the underlying undirected graph is connected, deleting one redundant balance row makes the oriented incidence matrix have rank $|V|-1$. A set of $|V|-1$ independent edge columns is exactly a spanning tree. Set all non-tree flows to zero, solve the tree balances, and check nonnegativity to obtain a basic feasible solution. Basic tree edges may have zero flow, which is degeneracy in linear programming. Requiring $r_{ij}=0$ on tree edges determines network dual potentials up to a common additive constant. If every non-tree network reduced cost is nonnegative, the tree flow is optimal. This is the basis of the network simplex algorithm. If the graph is disconnected, solve the balances separately in each component; each component must have total net supply zero and uses its own spanning tree.

Choose a set $B$ of $m$ column indices for which the square matrix $A_B$ is invertible. The associated basic solution sets $x_j=0$ for $j\notin B$ and solves $A_B{x}_B=b$. It is a basic feasible solution when all its coordinates are nonnegative.

Let $x$ be a basic feasible solution with basis $B$. If $x=(y+z)/2$ for $y,z\in X(b)$, then $x_j=0$ and nonnegativity force $y_j=z_j=0$ for every $j\notin B$. Since $A_B{y}_B=A_B{z}_B=b$ and $A_B$ is invertible, $y=z=x$. Thus $x$ is an extreme point.

Conversely, let $x$ be extreme and let $S=\{j:x_j>0\}$. If $|S|>m$, the corresponding columns are linearly dependent, so there is a nonzero vector $d$ supported on $S$ with $Ad=0$. For sufficiently small $\epsilon >0$, both $x+\epsilon d$ and $x-\epsilon d$ remain nonnegative and feasible, contradicting extremality. Hence $|S|\le m$. Enlarge $S$ to a set $B$ of $m$ indices. By the hypotheses $A_B$ is invertible, and $x$ is the resulting basic feasible solution. Therefore

A vector $x$ is a basic feasible solution whenand the columns $A_i$ for which $x_i>0$ are linearly independent. Equivalently, after extending those columns to a basis of the column space of $A$, one sets all nonbasic variables to zero and solves for the basic variables, obtaining a nonnegative vector.

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.

One of the $m+n$ balance equations is redundant, and the remaining $m+n-1$ have full rank. A nondegenerate basic feasible solution therefore has exactly $m+n-1$ positive variables. Equivalently, its positive cells form a spanning tree of the complete bipartite row--column graph.

A simplex dictionary expresses the basic variables and the objective as affine functions of the nonbasic variables in a linear program. Setting the nonbasic variables to zero gives a basic feasible solution when all basic values are nonnegative. A simplex algorithm pivot exchanges one basic and one nonbasic variable. For maximization, a feasible dictionary with all nonbasic objective coefficients nonpositive certifies optimality.

The simplex method moves between adjacent basic feasible solutions along improving edges of a feasible polytope until no improving pivot remains.

A slack variable turns an inequality constraint $a^{T}x\le b$ into the equality $a^{T}x+s=b$ with $s\ge 0$. Its value measures the unused amount of the constraint. In the simplex algorithm, slack variables often supply the initial basic feasible solution.