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The objective function $\theta$ is nonnegative. We now construct a feasible zero-objective point, distinguishing the input problem's possible states.

If the primal is feasible and bounded below, strong duality and attainment for a linear program give $x_{\ast }\ge 0$, ${\lambda }_{\ast }$ and $s_{\ast }=c-A^T{\lambda }_{\ast }\ge 0$ withSet $\theta =\kappa =0$, $\tau =1$, $x=x_{\ast }$, $s=s_{\ast }$, $\lambda ={\lambda }_{\ast }$, andThe first three embedding equations hold. Substituting their equalities into the fourth gives precisely $-e^T{x}_{\ast }-e^T{s}_{\ast }-1+N\rho =0$.

If the primal is infeasible, the Farkas alternative supplies $\lambda$ with $A^T\lambda \le 0$ and $b^T\lambda >0$. Take $x=\tau =\theta =0$, $s=-A^T\lambda \ge 0$, $\kappa =b^T\lambda >0$, and $\rho =(e^{T}s+\kappa )/N$. These satisfy the same equations. If the primal is feasible and unbounded below, it has a direction in the recession cone $x\ge 0$ with $Ax=0$ and $c^{T}x<0$. Set $\lambda =s=\tau =\theta =0$, $\kappa =-c^{T}x>0$, and $\rho =(e^{T}x+\kappa )/N$. This is again a feasible nonzero embedding vector. Normalize the constructed vector by the sum of its nonnegative coordinates. In every case,The facts about optimal pairs and recession directions follow from the polyhedral alternatives of linear programming duality; they do not require knowing their values or witnesses to write the transformed coefficients.

To recover information from the transformed optimum, multiply the first three original embedding equations by ${\lambda }^T$, $x^T$, and $\tau$, respectively, and add. Their cross terms cancel; the fourth equation givesAt $\theta =0$, nonnegativity forces $x^{T}s=0$ and $\tau \kappa =0$. The equations also implyThus any normalized zero-objective point has $\rho >0$: if $\rho =0$, all these coordinates vanish, and then $q=\lambda =\theta =0$, contradicting normalization.

If $\tau >0$, division by $\tau$ gives a primal-feasible point $x/\tau$ and dual-feasible point $\lambda /\tau$ with equal objective function values; hence they are optimal. If $\kappa >0$, then $\tau =0$, $Ax=0$, $A^T\lambda =-s\le 0$, andTherefore either $b^T\lambda >0$ certifies primal infeasibility, or $c^{T}x<0$ is an improving recession direction certifying dual infeasibility and unboundedness whenever the primal is feasible. If only the latter certificate is obtained and primal feasibility is unknown, apply the same construction to the feasibility problem with objective function $c=0$; it gives either a feasible point or a direct primal infeasibility certificate.

A degenerate optimal point can have $\tau =\kappa =0$ and convey neither conclusion. Choose an optimum in the relative interior of the optimal face of a linear program. The explicit existence argument above gives a point of this face with $\tau +\kappa >0$. A nonnegative coordinate which is positive somewhere on a face of a convex set is positive throughout its relative interior, so the chosen point has $\tau +\kappa >0$ and can be decoded. The assertion follows because a small relative-interior displacement in the direction opposite a positive-coordinate point would otherwise make that coordinate negative. Equivalently one may obtain a maximum-support optimum; choosing an arbitrary degenerate boundary optimum is insufficient.

This reduction is useful because an interior-point method now has a known strictly feasible start, a compact standard simplex slice, and a known target objective function. Projective rescalings can recenter positive iterates, and the transformed problem supplies optimality or infeasibility information without a separate guessed optimum. Polynomial bit-complexity claims need rational data with a finite encoding length. The printed bound $U$ on arbitrary real coefficients by itself does not bound denominators or the size of nonzero determinants; it is not a substitute for that encoding assumption.

A reduction must encode primal solutions and infeasibility or unboundedness information; merely declaring a known optimum or subtracting an unknown optimum from the objective function is not a construction. The following homogeneous self-dual embedding of a linear program works without assuming the input problem has an optimum.

First remove linearly dependent equality rows by Gaussian elimination. If an eliminated row has an inconsistent right-hand side, the original problem is already certified infeasible. Return the one-variable special-form program $\min 0$ with $A^{\ast }=0$, $y=1$, and retain that certificate; no original mathematical optimization remains. Otherwise the remaining matrix has full row rank; write it as $A$ again. Put $e=1\in {\mathbb{R}}^n$ andUse nonnegative variables $x,s\in {\mathbb{R}}^n$ and $\tau ,\kappa ,\theta ,\rho \in \mathbb{R}$, with a temporary unrestricted multiplier $\lambda \in {\mathbb{R}}^m$. Impose the homogeneous linear systemAll coefficients are computable from the input. The temporary free multiplier can be eliminated without introducing cancelling positive/negative variables. DefineReplace the second equation by $(I-A^{T}H)q=0$, and substitute $Hq$ in the other equations. Because $A$ has full row rank, $A^T\lambda =q$ is equivalent to these conditions. For no remaining equality rows, simply omit $\lambda$ and its empty equations.

Let $y=(x,s,\tau ,\kappa ,\theta ,\rho )$ and $n^{\ast }=2n+4$. Collect these substituted equations into $A^{\ast }y=0$, impose $1^{T}y=1$, and let $c^{\ast }$ select the coordinate $\theta$. The resulting Karmarkar standard form isThe next two parts prove its known positive starting point and zero optimal value. Notice that normalization is legitimate for any nonzero nonnegative embedding vector because all the equations are homogeneous; it does not require a prior bound on the original optimizer.

Project scheduling asks how quickly a collection of dependent activities can finish. Adding their durations usually overestimates the answer, because independent activities can run together; taking only the longest single activity usually underestimates it, because some activities must wait. The critical path method identifies the chain of dependencies that determines the finish time. Its simplest model assumes known, fixed durations, unlimited parallel processing, and precedence requirements saying that each predecessor must finish before its successor starts. These are the assumptions of the basic model in Hendrickson, Project Management for Construction, §10.2.

Represent each activity by a vertex of a finite Directed acyclic graph; an edge $i\to j$ means that activity $j$ waits for activity $i$. Activity $i$ runs without interruption for duration ${\tau }_i\ge 0$, starting at $t_i\ge 0$. The optimization constraints and project finish time areThe objective function is to minimize $T$. All durations and start times use the same time unit, and the project starts at time zero. The Directed acyclic graph assumption allows a topological order in which every predecessor is processed first.

Consider this small, abstract project:

Activities A and E may start together. After A finishes, B and C may run together, but D must wait for both. F must similarly wait for both D and E: a merge means all incoming requirements must hold. This activity-as-vertex convention keeps durations on the activities and uses arrows only to express precedence.

Unlimited parallel processing is essential to the equality between the longest graph path duration and the optimal finish time. For example, two independent chains, each containing two activities of duration 3, have longest graph path duration 6. Two processors can finish both chains in 6, but one processor needs 12 because all four activities must run sequentially. Thus with limited resources the longest graph path still gives a lower bound, but it need not be achievable. Such resource requirements belong to the broader subject of scheduling.

// Target: mathematical-optimization.bigb

Consider the minimization problemThe objective function is a strictly convex function with its unconstrained optimization minimum at $x=2$. If $b<2$, it decreases throughout the feasible set until the bound $x=b$ is reached. If $b\ge 2$, the feasible set already contains $2$. HenceWith $g(x)=x-b$, the optimization Lagrangian and KKT conditions areSubstituting $x_{\ast }(b)$ gives the unique inequality Lagrange multiplierFor $b<2$, the active constraint has ${\lambda }_{\ast }>0$. For $b>2$, there is strict slack $b-x_{\ast }=b-2$ and ${\lambda }_{\ast }=0$. At $b=2$, the active constraint has zero Lagrange multiplier: the boundary happens to pass through the unconstrained optimum. Feasibility and complementary slackness must both be checked; solving only $2(x-2)+\lambda =0$ would allow infeasible candidates.

An independent certificate follows by completing the square:The Lagrange dual function is a concave function of $\lambda$. Its maximum over $\lambda \ge 0$ occurs at ${\lambda }_{\ast }(b)$ and equals $V(b)$. Thus every case has an explicit lower bound attained by a feasible point; the KKT conditions provide a certificate that can be checked without searching all real numbers.

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.

A value function assigns each dynamic programming state and time the best objective function value attainable over the remaining decisions. It is a score, whereas a control policy specifies the actions producing that score. In the 0-1 knapsack problem above, $V_3(5)=7$ is the score, and choosing the items of weights $2$ and $3$ attains it. In sequential mathematical optimization, the Bellman equation computes a value function by comparing immediate reward plus continuation value. Use a supremum or infimum when the best score need not be attained by any admissible control policy.