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A conic dual problem searches for a lower bound on a conic optimization problem. A feasible point exhibits an attainable cost; a feasible dual variable explains why every allowed decision costs at least a specified amount. If the two costs agree, the pair certifies a global minimum without relying on how the points were found.

Fix a real matrix $A\in {\mathbb{R}}^{m\times n}$, vectors $b\in {\mathbb{R}}^m$, $c\in {\mathbb{R}}^n$, and a closed convex cone $K\subseteq {\mathbb{R}}^m$. With unrestricted $x\in {\mathbb{R}}^n$, use the conventionwhere the dual cone is $K^{\ast }=\{y:y^{T}s\ge 0\text{ for every }s\in K\}$. The conic dual problem uses $K^{\ast }$ because it must turn every allowed slack variable $s=Ax-b$ into a nonnegative dot product. Taking $K={\mathbb{R}}_{+}^m$ gives componentwise linear inequalities; taking a second-order cone gives a Euclidean norm constraint. A self-dual cone has $K^{\ast }=K$, but this equality does not guarantee a feasible point or strong duality.

The Lagrangian supplies a short derivation. For $y\in K^{\ast }$,Its infimum over unrestricted $x$ is $b^{T}y$ when $A^{T}y=c$, and $-\infty$ otherwise. Maximizing the resulting Lagrange dual function gives the stated conic dual problem. The minus sign matches the convention $Ax-b\in K$.

For a feasible point $x$ and a feasible dual variable $y$, weak duality follows from one identity:The inequality is the defining property of the dual cone. Every feasible pair therefore encloses the optimal objective function value:Both optima are finite when such a pair exists. Consequently,These are absolute cost bounds, rather than bounds on distance to an optimizing vector. Zero gap certifies both optima immediately. A small positive gap bounds the remaining improvement even if no strong duality theorem has been invoked.

For the nonnegative-orthant optimality certificate example, $x=(2.9,0.2{)}^T$ is feasible and costs $3.3$. Pairing it with $y=(0,1,1{)}^T$ gives $g=0.3$, so its cost exceeds the optimum by at most $0.3$. Here the optimum is known to be $3$, and the bound is exact. Pairing the same $x$ with the weaker $r=1/2$ certificate gives $g=1.8$: a valid but less informative guarantee.

The identity assumes exact primal and dual feasibility. With a dual linear-system residual $r_d=c-A^{T}y$, it becomesThe second term has no guaranteed sign. Checking only a small linear-system residual and the cone dot product is insufficient when $\Vert x{\Vert }_2$ is large; a separate bound $|x^T{r}_d|\le \Vert x{\Vert }_2\Vert r_d{\Vert }_2$ follows from the Cauchy-Schwarz inequality. Approximate cone membership likewise needs a quantified correction before claiming a certificate.

For an exact arithmetic example, minimize $x$ subject to $x\ge 10^9$. The point $x=10^9$ is optimal. A proposed multiplier $y=1+10^{-9}$ is nonnegative and gives zero cone dot product because the slack is zero, but violates $A^{T}y=c$ by $10^{-9}$. Its purported lower bound is $10^9+1$, exceeding the optimum. The missing term $x^T{r}_d=-1$ explains the failure.

A dual variable is a multiplier associated with a primal constraint in a dual linear program. In a minimization linear program with equality constraints $Ax=b$ and nonnegative primal variables, the Lagrangian $c^{T}x+y^T(b-Ax)$ yields the dual objective $b^{T}y$ and inequalities $A^{T}y\le c$; the equality-constraint multipliers $y$ are unrestricted in sign. For a transportation problem, row and column dual variables give potentials $u_i,v_j$ satisfying $u_i+v_j\le c_{ij}$. Complementary slackness makes occupied routes attain equality.

Consider the original linear programOne unit of $v$ costs twice as much as one unit of $u$, so $(u,v)=(3,0)$ is a plausible answer. To certify it, multiply the three linear inequalities by $0,1,1$, respectively, and add:The proposed feasible point attains $3$, so its cost is the global minimum.

In the conic dual problem convention, the data and certificate areDirect matrix multiplication gives $A^{T}y=c$ and $b^{T}y=3$. At $x=(3,0{)}^T$, the slack variable is $s=(3,0,0{)}^T$, so $s^{T}y=0$. The first linear inequality has positive slack and zero multiplier; the two positively weighted linear inequalities are tight. This is complementary slackness.

The entire dual feasible set can also be read off: write $y=(1-r,2-r,r{)}^T$ with $0\le r\le 1$. Then $A^{T}y=c$ and $b^{T}y=3r$. Maximizing this lower bound gives $r=1$. A point such as $r=1/2$ is still a valid certificate of the weaker bound $u+2v\ge 3/2$.

Remove the integer requirements from the previous formulation to obtain its linear programming relaxation. Give each edge constraint a nonnegative dual variable $y_e$, and each bound for $v\in Z$ a nonnegative dual variable $z_v$. Since all right-hand sides are one, the dual linear program isAn integer feasible point selects a multiset of edges touching every nonisolated vertex, together with at least one singleton for each isolated vertex. Repeated edges or singleton choices can be discarded without losing coverage, so an optimum has a genuine edge cover of the nonisolated part and one singleton per isolated vertex. If $G$ has no isolated vertices, this is exactly an edge cover, with the objective counting selected edges including any unnecessary multiplicities. The isolated-singleton extension is necessary because a graph with an isolated vertex admits no ordinary edge cover.

This interpretation concerns integer feasible dual solutions. The linear programming relaxation can have fractional optima, so its value need not equal the integer independent set optimum. For example, the triangle has relaxed independent-set value $3/2$, whereas its maximum independent set has size $1$ and its minimum edge cover has size $2$.

Regard the two kinds of agents as the two sides of a complete bipartite graph, with edge weight $c_{ij}$. Pairing everyone is a perfect matching, and its maximum total weight is $P$. Consider the capacitated linear programIt is feasible and bounded. Every optimum has all row and column sums equal to one: otherwise the equal total deficits of the two sides give a row $i$ and column $j$ both below one. Increasing $x_{ij}$ by the smaller of their positive deficits remains feasible and increases the objective because $c_{ij}>0$, a contradiction.

Thus an optimum is a doubly stochastic matrix. By the Birkhoff-von Neumann theorem it is a convex combination of permutation matrices, so at least one corresponding perfect matching has weight at least its objective. Conversely every such perfect matching gives a feasible matrix. Hence $P_{\mathrm{LP}}=P$. One way to justify the decomposition is to apply Hall's marriage theorem to the positive-entry support, subtract the smallest entry on a supported perfect matching, and repeat after rescaling the nonzero remainder; each subtraction removes a positive entry, so the process terminates.

The dual linear program assigns nonnegative dual variables $p_i,q_j$ to the row and column capacity constraints:These are exactly the permitted payments and all the pairwise acceptance conditions, so its optimal value is $Q$. It is feasible, for example by taking $p_i={\max }_j{c}_{ij}$ and $q_j=0$. Strong duality now gives $P=P_{\mathrm{LP}}=Q$. In particular, the nonnegative payment restriction has been retained throughout; using capacity inequalities rather than equality constraints avoids introducing unrestricted prices.

For a direct weak duality check, sum $p_i+q_{\pi (i)}\ge c_{i,\pi (i)}$ over any pairing $\pi$. Its revenue never exceeds the company's total offered payment. The linear programming duality and integrality argument prove that this lower bound on payments is attained.

Use the standard maximum form of a linear program convention $\max c^{T}u$ subject to $Mu\le b$ and $u\ge 0$. Split the unrestricted variables as $x=x^{+}-x^{-}$ and $y=y^{+}-y^{-}$, with all four new variables nonnegative. The resulting linear program isIf equality-constraint standard form is preferred, add a nonnegative slack variable to each inequality.

Give the two inequalities nonnegative dual variables $\alpha ,\beta$. The dual linear program isThe paired inequalities encode equalities for the two original unrestricted variables. Equivalently, the dual linear program has $2\alpha =2$, $-\alpha +\beta =3$, $\beta \ge 4$, $5\alpha +\beta \ge 7$ and $\alpha ,\beta \ge 0$. Thus its only feasible point is $(\alpha ,\beta )=(1,4)$, of objective function value $13$.

Associate free real dual variables $u_i$ and $v_j$ with the row and column equality constraints. The Lagrangian for the transportation problem isIts infimum over $x_{ij}\ge 0$ is finite exactly when $u_i+v_j\le c_{ij}$ for every pair. The linear programming duality therefore givesThe dual variables have no sign restrictions because the primal constraints are equalities.

A feasible shipment $x$ is optimal if and only if there exist feasible transportation dual potentials satisfying complementary slackness:Thus every occupied route must have $u_i+v_j=c_{ij}$. To see sufficiency explicitly, the primal minus dual objective isand complementary slackness makes it zero. This is the weak duality certificate. Necessity follows from strong duality for the feasible, bounded linear program. If the total supply $M$ is positive, $x_{ij}=a_i{b}_j/M$ gives feasibility; if $M=0$, $x=0$ does. The feasible shipments form a closed bounded set, so the primal optimum is attained. This also covers zero supplies and demands. The one redundant equality merely leaves the dual gauge freedom $(u,v)\mapsto (u+t,v-t)$.

If $A$ has $m$ rows and $n$ columns, attach a nonnegative dual variable $y_i$ to each of the $m$ upper-bound constraints. The dual linear program isHere $A^T$ is the transpose of $A$, and all vector inequalities hold componentwise. Nonnegative dual variables preserve the direction of the original inequalities when taking weighted sums.

There are two primal inequalities, so there are two nonnegative dual variables. The five primal variables yield five dual inequalities:Each coefficient column of the primal linear program becomes one row of this dual linear program.

At ${\mathbf{y}}^{\ast }=(4/3,2)$, the first three dual left sides are $10/3$, $2/3$ and $14/3$, strictly above their bounds $1$, $-1$ and $-5$. Thus complementary slackness forces $x_1=x_2=x_3=0$. Both dual variables are positive, so both primal inequalities are equalities:Solving this linear system givesAll components are nonnegative, the two original upper bounds are met exactly, and the objective function equals the dual value. The weak duality bound therefore certifies optimality. The forced zero components and unique solution of the two remaining equalities also prove that the primal optimum is unique.

A curved feasible set can have the same certificate structure. ConsiderWrite $x=(t,z{)}^T$ and use the second-order cone $K=\{(s_0,s_1,s_2):s_0\ge \sqrt{s_1^2+s_2^2}\}$. In this coordinate ordering,Self-duality of a second-order cone makes $K^{\ast }=K$. The equality $A^{T}y=c$ forces $y_0=1$ and $y_2=0$, leaving $-1\le y_1\le 1$. Its dual objective function is $b^{T}y=-y_1$, maximized at $y=(1,-1,0{)}^T$. Thus every feasible point satisfies $t\ge 1$, and $(t,z)=(1,0)$ attains that bound: $\boxed{p_{\ast }=d_{\ast }=1}$.

The zero dot product is $(1,1,0)\cdot (1,-1,0)=0$. Individual coordinate products need not be nonnegative here: the cancellation $1-1=0$ occurs across coordinates. The valid complementary slackness condition is cone-level orthogonality, rather than the coordinatewise interpretation of the nonnegative orthant.

For a proper cone $K$, suppose a point satisfies $Ax-b\in \mathrm{int}K$ and $p_{\ast }$ is finite. This generalized Slater condition gives strong duality and attainment of the conic dual problem's optimum. It does not by itself assert primal attainment.^[ref]

The two worked examples have strict primal points: $(u,v)=(2,2)$ has slack $(2,2,1)$ in the interior of the nonnegative orthant, and $(t,z)=(2,0)$ has slack $(2,1,0)$ in the interior of the second-order cone. Both optima were already certified by explicit feasible pairs. Slater's condition explains why a matching optimal dual variable can be expected under these hypotheses; checking a supplied zero-gap pair needs only weak duality.

Changing cone geometry preserves the bound mechanism. For a positive semidefinite cone, the dual cone pairing is the Frobenius inner product of two matrices. This connects the same certificate idea to semidefinite programming, while the central path uses strictly interior primal and dual points to approach vanishing gap.

In 1995, Lieven Vandenberghe and Stephen Boyd presented a primal-dual interior-point method for convex optimization with matrix inequalities. Their algorithmic study included exploiting problem structure and allowing incomplete conjugate gradient method solves within an iteration. This illustrates how the primal-dual viewpoint guides computation as well as checking a final answer.^[ref]