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A flow balance prescribes net outgoing minus incoming flow at a vertex. Positive $b_i$ is supply and negative $b_i$ is demand; zero is ordinary flow conservation. Summing all balances requires total supply to equal total demand. In a directed graph, these equations are represented by the oriented incidence matrix.

Assign a real value $x_i$ to each vertex. The Graph Laplacian quadratic form measures disagreement across edges:This makes the Graph Laplacian a positive semidefinite matrix. The energy is small when neighbours have similar values, even if distant vertices have very different values; it depends on connections rather than where the graph is drawn. In the three-vertex path graph, take $x=(2,0,-1{)}^T$. Thenand the two edges contribute $(2-0{)}^2=4$ and $(0-(-1){)}^2=1$.

For a short proof, expand $x^T(D-A)x$. Each edge contributes $x_i^2+x_j^2$ through the degree matrix and $-2x_i{x}_j$ through the adjacency matrix, yielding $(x_i-x_j{)}^2$. The sum over unordered edges has no factor $1/2$; a sum over all ordered pairs does:Equivalently, let $B$ be an oriented incidence matrix, with one row per edge. Then $Bx$ records endpoint differences and $L=B^{T}B$, independently of the chosen orientations. This translates a local comparison into a global quadratic form and makes its nonnegativity visible.

The graph total variation measures absolute variation of a vertex signal across edges. It is independent of the orientation used for the oriented incidence matrix. It vanishes exactly on signals constant on each connected component of a graph. A graph total variation denoising estimator combines this penalty with squared error, favouring signals that are piecewise constant on connected regions.

A vertex-edge incidence matrix records which vertices are endpoints of each edge of a graph. The unsigned convention uses entries one at the endpoints, with a convention for loops. Choosing an orientation gives the oriented incidence matrix, with minus one at the initial endpoint and plus one at the terminal endpoint. These are different conventions and should be specified when using the matrix in a computation.

For a connected graph, the Moore-Penrose inverse of its oriented incidence matrix satisfies $D^{†}D=I-{\Pi }_1$, because this product is the orthogonal projection onto $(\ker D{)}^{\perp }$. Thus every signal decomposes into its constant mean and $D^{†}D\theta$. Writing $L={\max }_e\Vert (D^{†}{)}_{\cdot ,e}{\Vert }_2$, a sub-Gaussian random vector noise gives a simultaneous bound on $\Vert (D^{†}{)}^{\top }z{\Vert }_{\infty }$ with scale $L\sqrt{\log (m/\delta )}$.

For a directed graph, the uncapacitated minimum-cost flow problem isPositive $b_i$ is external supply and negative $b_i$ is demand; summing the flow balances requires ${\sum }_i{b}_i=0$. There are no finite upper arc capacities. Let $B$ be the oriented incidence matrix with tail $+1$ and head $-1$. For free dual potentials ${\pi }_i$, the Lagrangian isIts infimum over nonnegative flows is ${\pi }^{T}b$ exactly when every network reduced cost $r_{ij}=c_{ij}-{\pi }_i+{\pi }_j$ is nonnegative; otherwise the infimum is minus infinity. Thus the dual maximizes ${\pi }^{T}b$ subject to these inequalities. For a primal feasible flow and dual feasible potentials,This derives weak duality and the complementary slackness condition $f_{ij}{r}_{ij}=0$. A feasible pair satisfying it has equal objectives and certifies optimality.

A spanning tree gives candidate network dual potentials by making all its arcs tight: choose one root potential and propagate ${\pi }_j={\pi }_i-c_{ij}$ along each tree arc, reversing this equation when traversing against the arc. An arbitrary tree does not automatically make the non-tree reduced costs nonnegative; those inequalities must be checked. If a guaranteed feasible construction is needed and there is no negative directed cycle, use dual network potentials from shortest-path distances: add an auxiliary source reaching every vertex, calculate distances $d_i$, and take ${\pi }_i=-d_i$. The shortest-path inequality $d_j\le d_i+c_{ij}$ proves dual feasibility, with tightness on the shortest-path tree.

For the printed dashed tree, set all non-tree flows to zero and solve the balances from the leaves. This givesThe remaining arcs $46,67,52,28,89$ have zero flow. The total initial cost is $20$. This is a basic feasible solution, though degenerate because the tree flow on $68$ is zero.

Normalize ${\pi }_1=0$. Tree tightness givesThe non-tree reduced costs areThe negative reduced cost on $89$ identifies an improving entering arc for the network simplex algorithm. Adding it to the tree creates a unique undirected cycle. Sending $\theta$ along $8\to 9$ increases the flows on $15,56,68,89$ by $\theta$ and decreases those on $14,43,37,79$ by $\theta$; all other flows are unchanged. This balanced displacement preserves every node equation. Nonnegativity limits $\theta$ to the minimum of the four decreasing flows, namely $2$. Choose arc $14$ to leave among the tied limiting arcs.

After this one pivot the nonzero flows areand every other arc flow is zero. The objective decreases by $\theta r_{89}=-2$, giving cost $18$.

For the new tree, tight potentials normalized at vertex 1 areAll tree reduced costs are zero. For the non-tree arcs $14,46,67,52,28$ they are $1,1,1,2,1$, respectively, so the potentials are dual feasible. Their dual objective ismatching the primal cost. Complementary slackness therefore provesThis is the stopping test of the network simplex method: maintain a feasible tree flow, compute tight potentials, enter a negative-reduced-cost non-tree arc, make a maximal feasible cycle adjustment and exchange a limiting tree arc, until all reduced costs are nonnegative. Here exactly one improving step suffices.

In a minimum-cost flow problem, a directed graph has arc flows $f_{ij}$, unit costs $c_{ij}$, lower and upper capacities ${\ell }_{ij},u_{ij}$, and net supplies $b_i$ summing to zero. The linear program minimizes ${\sum }_{ij}{c}_{ij}{f}_{ij}$ subject to the flow balances ${\sum }_j{f}_{ij}-{\sum }_j{f}_{ji}=b_i$ and ${\ell }_{ij}\le f_{ij}\le u_{ij}$.

Give each graph vertex a network dual potential ${\pi }_i$. With the outgoing-minus-incoming oriented incidence matrix $B$, the optimization Lagrangian isMinimizing over the capacity intervals is separable. A candidate feasible flow minimizes this optimization Lagrangian if it takes its lower bound whenever the network reduced cost is positive, its upper bound whenever that network reduced cost is negative, and any allowed value when that network reduced cost is zero. Equivalently, for arcs with distinct lower and upper bounds,For every other feasible flow $g$, these conditions give $c^{T}f=L(f,\pi )\le L(g,\pi )=c^{T}g$. This proves the Lagrangian sufficiency theorem certificate, rather than merely stating the capacitated flow optimality conditions.

Put $t=f_{23}$. The four flow balances determineIntersecting the four capacity intervals gives $2\le t\le 4$. In particular,is a feasible flow. Write $C=c_{12}+c_{23}+c_{34}+c_{41}$. Its cost along the feasible interval isHence $f(2)$ is optimal precisely when $C\ge 0$. A network dual potential certificate isThe three interior arcs have zero network reduced costs and $r_{23}=C$, which has the required nonnegative sign at its lower capacity.

For the network simplex algorithm, start with the network simplex tree basis consisting of arcs $12,34,41$, and fix nonbasic arc $23$ at its lower bound 2. These are exactly the potentials just found. If $C\ge 0$, no improving pivot is needed. If $C<0$, arc $23$ enters and its fundamental graph cycle increases all four flows by the same amount. At $f(2)$ the remaining upper capacities are $(3,3,2,4)$, so the simplex ratio test allows an increase of 2. Arc $34$ reaches its upper bound and leaves the tree. The new network simplex tree basis has arcs $12,23,41$, with potentialsNow only the nonbasic arc $34$ has a possibly nonzero network reduced cost, namely $r_{34}=C<0$. It is at its upper bound, so the sign is optimal. Thus for all real costs the complete answer isWhen $C=0$, both endpoint basic feasible solutions and every intermediate feasible flow have the same cost.

To count basic solutions, a network simplex tree basis omits one arc of the four-cycle and fixes that arc at either bound. The possibilities areThere are eight tree/bound basic representations, seven distinct basic flow vectors, and two basic feasible solutions. The distinction in the first count matters: the upper bounds on $12$ and $23$ both give the identical flow $(8,5,6,7)$ at $t=5$. Thus if “basic solutions” counts vectors rather than basis representations, the answer to (i) is seven; the frequently used count of basis/bound choices is eight. The answer to (ii) is unambiguously two, since only $t=2,4$ satisfy all capacities.

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.

Each row of the oriented incidence matrix records one signed endpoint difference, henceReversing an edge changes the sign of its row, and permuting the edges permutes rows; neither operation changes the sum of absolute differences. Thus the graph total variation is independent of the chosen orientation and ordering. If $D\theta =0$, values agree at the endpoints of every edge. Any two graph vertices of a connected graph are joined by a graph path, so their values agree. Therefore $\ker D=\mathrm{span}\{1\}$.

An uncapacitated minimum-cost flow has nonnegative edge flows without finite upper capacities. The flow balances prescribe net supply at each vertex. For the oriented incidence matrix convention with tail $+1$ and head $-1$, vertex network dual potentials have network reduced costs $c_{ij}-{\pi }_i+{\pi }_j$. Nonnegative network reduced costs and complementary slackness certify a feasible optimum. A feasible negative-cost directed cycle makes the problem unbounded below.