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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 $u+2v\ge 3/2$.

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]