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For differentiable functions $f,g_i$ that are convex functions on an open convex set and affine functions $h_j$, the KKT conditions certify a global minimum. Once Lagrange multipliers satisfy them, no extra Slater condition is needed.^[ref]

The certificate is a short chain of inequalities. Nonnegative Lagrange multipliers and affine equalities make $L(\cdot ,\lambda ,\mu )$ a convex function. Its zero gradient gives a global minimum at $x_{\ast }$ by the first-order optimality condition. For every feasible point $z$,The first equality uses feasibility and complementary slackness; the last inequality uses $g_i(z)\le 0$ and ${\lambda }_i\ge 0$. Equivalently, the Lagrange dual function attains the same value as the primal objective function, proving strong duality. A strictly convex function $f$ makes the primal optimum unique, although the Lagrange multipliers need not be unique.

Affine equality optimization constraints matter: an arbitrary signed multiple of a nonlinear equality can destroy convexity of the optimization Lagrangian. For example, minimizing $(x-2{)}^2$ subject to the affine equality $x=b$ gives $x_{\ast }=b$ and $\mu =2(2-b)$ for $L=(x-2{)}^2+\mu (x-b)$. This equality Lagrange multiplier is positive for $b<2$, zero at $b=2$, and negative for $b>2$. Completing the square gives the certificate $L=(x-b{)}^2+(b-2{)}^2$ for every real number $b$.

Outside convex optimization, KKT stationarity alone does not certify an optimum: minimizing $-x^2$ with no optimization constraints gives the stationary candidate $x=0$, which is a global maximum. The Lagrangian sufficiency theorem remains applicable whenever a global minimum of the optimization Lagrangian can be proved directly.

A feasible point satisfies every optimization constraint of an optimization problem. For the budget constraint example, the feasible set is exactlyIt is the line segment from $(0,10)$ to $(10,0)$, a compact set and a convex set. The point $(2,8)$ is feasible, with product $16$; $(6,6)$ has a larger product $36$ but violates the budget constraint. Feasibility answers whether a decision is allowed; optimality asks whether any allowed decision is better. The extreme value theorem guarantees that this continuous objective function attains its maximum on this nonempty compact set.

Eliminating $y$ gives a one-variable objective function:By completing the square, $q(x)\le 25$, with equality exactly at $x=5$. Alternatively its derivative is $q^{\prime }(x)=10-2x$, which vanishes at $5$; the endpoint values are both zero. Completing the square proves global optimality directly, whereas the first-order optimality condition only identifies a candidate. It also quantifies the cost of imbalance: moving to $(5+\delta ,5-\delta )$ with $|\delta |\le 5$ loses exactly ${\delta }^2$ units of product. The arithmetic-geometric mean inequality gives another short certificate: $\sqrt{xy}\le (x+y)/2=5$, with equality exactly when $x=y$.

For a transportation problem with capacity inequalities, feasibility requires nonnegative shipments, row totals no greater than factory capacities and column totals equal to shop demands. Feasibility alone does not prove minimal cost: a matching transportation dual certificate with capacity inequalities provides an optimality proof. The feasible set can be empty, and a finite infimum need not be attained even when the set is nonempty.

For the link-route incidence matrix $A$, the reduced-load approximation and Erlang loss formula giveIntroduce $y_j=-\log (1-B_j)$ and accepted route flows $x_r={\nu }_r{e}^{-(A^{T}y{)}_r}$. Multiplying the reduced load by the link's own acceptance probability givesThese are precisely the first-order optimality conditions for the convex potential for the Erlang fixed point, sinceEach exponential term is a convex function, and the integral of the strictly increasing function $U(\cdot ,C_j)$ is a strictly convex function. Hence $F$ is a strictly convex function. Moreover $U(z,C_j)\to C_j>0$ as $z\to \infty$, by the saturation of the carried load of an Erlang loss resource. Its integrals tend to infinity at least linearly, so $F$ is a coercive function on the nonnegative orthant and has a unique minimizer.

At $y_j=0$, $U(0,C_j)=0$. If link $j$ is used by any positive-rate route, then ${\partial }_{j}F<0$, excluding a boundary minimizer there. An unused link instead minimizes at $y_j=0$ and has zero derivative. Thus the minimizer satisfies the displayed flow equations for every link. Conversely, those equations imply $\nabla F=0$ and therefore minimize the convex potential for the Erlang fixed point. Finally $U(y,C)=a[1-E(a,C)]$ with $y=-\log (1-E(a,C))$ converts the flow equations back to $B_j=E(a_j,C_j)$. The Erlang fixed point approximation has exactly one solution under fixed routing.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

Optimization with no additional optimization constraints on the chosen variable domain. At an interior local extremum of a differentiable function, the first-order optimality condition is $\nabla f=0$.