Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 38 / 1

The dual linear program is the minimization problemIndeed, for $y\ge 0$ the optimization Lagrangian $c^{T}x+y^T(b-Ax)$ bounds the primal objective above on its feasible points. Its supremum over $x\ge 0$ is finite precisely when $A^{T}y\ge c$, and then equals $b^{T}y$. The two labelled parts below establish the relevant weak duality and strong duality conclusions.

For the subsequent infeasibility argument, introduce nonnegative slack variables $s$ and the convex coneThis is generated by the columns of $A$ and the coordinate vectors. Here is why this finitely generated cone is closed. Any representation with a minimal number of positive coefficients uses linearly independent generators: otherwise subtract a suitable multiple of a dependence until one positive coefficient vanishes, preserving nonnegativity. For each fixed independent selection of generators, its nonnegative combinations form a closed set, because the inverse coordinate map on its span is continuous. There are finitely many selections, so their union $C$ is closed. This is the closedness of finitely generated cones used in Farkas' lemma.

When the primal linear program is infeasible, $b\notin C$. Let $p$ be the nearest point of $C$ to $b$, and set $h=p-b$. The minimizing property gives $h^T(z-p)\ge 0$ for every $z\in C$. Taking $z=0,2p$ gives $h^{T}p=0$. ConsequentlyIn particular $h\ge 0$ and $A^{T}h\ge 0$. Rescaling $h$ by the positive number $-h^{T}b$ produces the requested Farkas' lemma certificate:Now a feasible point of the relaxed inequalities would satisfySince ${\lambda }^{T}b=-1$, the sum of the nonnegative ${\lambda }_i$ is positive. Thus robust infeasibility under uniform constraint relaxation givesThe ellipsoid method makes progress by reducing the volume of a containing ellipsoid. A nonempty linear polyhedron may have empty interior, so small volume alone cannot certify that the original system is infeasible. Relaxing the inequalities remedies this difficulty, but must not make an infeasible system appear feasible. The displayed positive gap proves exactly that sufficiently small relaxations preserve infeasibility. For rational input, determinant bounds give a Farkas' lemma certificate of polynomial encoding length, so a sufficiently small rational relaxation can be chosen from the input encoding length. Together with a bounding box and bounds on the volume of a relaxed feasible system, this lets the ellipsoid method distinguish infeasibility from a feasible system of very small or zero original volume in polynomial time. The certificate argument alone does not assert a polynomial precision bound for arbitrary unencoded real coefficients.

Fix a primal feasible point $x_0$. For every dual feasible point $y$, nonnegativity and the primal and dual inequalities giveHence the dual objective is bounded below by $c^T{x}_0$ wherever it is feasible, which is weak duality. This is the meaning of “bounded” in this assertion; it does not say that the dual feasible set is bounded, or that it is nonempty. For example, with $A=0$, $b=0$, $c=1$ and a single variable, the primal is feasible and unbounded above, while its dual would require $0\ge 1$ and is infeasible.

Let the finite primal supremum be $v$. First it is attained. The setis a finitely generated cone, hence closed by the argument in the root Solution. For a sequence of primal feasible points whose values approach $v$, the pairs $(b,c^{T}x)$ belong to $K$. Their limit $(b,v)$ therefore belongs to $K$. It gives a feasible point $x_{\ast }$ with $c^T{x}_{\ast }\ge v$, so necessarily $c^T{x}_{\ast }=v$.

Let $I$ index the tight primal inequalities at $x_{\ast }$ and let $Z$ index its zero coordinates. Consider the closed convex cone generated by $A_i^T$ for $i\in I$ and $-e_j$ for $j\in Z$. If $c$ were outside this cone, separation from a closed convex cone would give a direction $d$ withFor sufficiently small positive $\delta$, $x_{\ast }+\delta d$ remains a feasible point: the active constraints have the required signs, and every inactive constraint retains its positive slack. Its larger objective contradicts optimality. Therefore there are $y_i\ge 0$, supported on $I$, and ${\mu }_j\ge 0$, supported on $Z$, such thatThis makes $y$ a dual feasible point. Moreover, the support restrictions giveTogether with weak duality, this proves strong duality and dual attainment: