Strong duality and sensitivity analysis in semi-infinite linear programming

Abstract

Finite-dimensional linear programs satisfy strong duality (SD) and have the “dual pricing” (DP) property. The DP property ensures that, given a sufficiently small perturbation of the right-hand-side vector, there exists a dual solution that correctly “prices” the perturbation by computing the exact change in the optimal objective function value. These properties may fail in semi-infinite linear programming where the constraint vector space is infinite dimensional. Unlike the finite-dimensional case, in semi-infinite linear programs the constraint vector space is a modeling choice. We show that, for a sufficiently restricted vector space, both SD and DP always hold, at the cost of restricting the perturbations to that space. The main goal of the paper is to extend this restricted space to the largest possible constraint space where SD and DP hold. Once SD or DP fail for a given constraint space, then these conditions fail for all larger constraint spaces. We give sufficient conditions for when SD and DP hold in an extended constraint space. Our results require the use of linear functionals that are singular or purely finitely additive and thus not representable as finite support vectors. We use the extension of the Fourier–Motzkin elimination procedure to semi-infinite linear systems to understand these linear functionals.

Appendix

This appendix contains three technical lemmas used in the proof of Theorem 5.13.

Lemma 7.1

Let $b^1, b^2 \in \mathbb R^I$ such that $OV(b^1)<\infty ,$ $OV(b^2) < \infty$ and denote ${\tilde{b}}^1 = \overline{FM}(b^1)$ and ${\tilde{b}}^2 = \overline{FM}(b^2)$. Suppose $\{h_m\}_{m\in \mathbb N}$ is a sequence in $I_4$ such that $\lim _{m\rightarrow \infty } {\tilde{b}}^j(h_m) = L(b^j)$ for $j=1,2$ and $\lim _{m\rightarrow \infty } \sum _{k=\ell }^n |{\tilde{a}}^k(h_m)| \rightarrow 0$. Then for every $\lambda \in [0,1], b_\lambda := \lambda b^1 + (1-\lambda )b^2$ has the property that

$$\begin{aligned} \lim _{m\rightarrow \infty } {\tilde{b}}_\lambda (h_m) = L(b_\lambda ) = \lambda L(b_1) + (1-\lambda )L(b_2), \end{aligned}$$

where ${\tilde{b}}_\lambda = \overline{FM}(b_\lambda )$.

Moreover, suppose $\{h_m\}_{m\in \mathbb N}$ is a sequence in $I_3$ such that $\lim _{m\rightarrow \infty } {\tilde{b}}^j(h_m) = S(b^j)$ for $j=1,2$. Then for every $\lambda \in [0,1], b_\lambda := \lambda b^1 + (1-\lambda )b^2$ has the property that

$$\begin{aligned} \lim _{m\rightarrow \infty } {\tilde{b}}_\lambda (h_m) = S(b_\lambda ) = \lambda S(b_1) + (1-\lambda )S(b_2), \end{aligned}$$

where ${\tilde{b}}_\lambda = \overline{FM}(b_\lambda )$.

Proof

By Lemma 3.7 L is sublinear and therefore convex which implies

$$\begin{aligned} \begin{array}{rcl} L(b_\lambda ) &{}\le &{}\lambda L(b^1) + (1-\lambda )L(b^2) \\ &{} = &{} \lambda \lim _{m\rightarrow \infty } {\tilde{b}}^1(h_m) + (1-\lambda )\lim _{m\rightarrow \infty } {\tilde{b}}^2(h_m) \\ &{} = &{} \lim _{m\rightarrow \infty } (\lambda {\tilde{b}}^1(h_m) + (1-\lambda ){\tilde{b}}^2(h_m))\\ &{} \le &{} L(\lambda b^1 + (1-\lambda )b^2)\\ &{} = &{} L(b_\lambda ), \end{array} \end{aligned}$$

where the second inequality follows from Lemma 3.9.

Thus, all the inequalities in the above are actually equalities. In particular, $\lim _{m\rightarrow \infty } (\lambda {\tilde{b}}^1(h_m) + (1-\lambda )b^2(h_m)) = L(b_\lambda ) = \lambda L(b_1) + (1-\lambda )L(b_2)$. Since $\overline{FM}$ is a linear operator, $\overline{FM}(b_\lambda ) = \lambda \overline{FM}(b^1) + (1-\lambda ) \overline{FM}(b^2)$ and so ${\tilde{b}}_\lambda (h_m) = \lambda {\tilde{b}}^1(h_m) + (1-\lambda )b^2(h_m)$ for all $m \in \mathbb N$. Hence, $\lim _{m\rightarrow \infty } {\tilde{b}}_\lambda (h_m) = \lim _{m\rightarrow \infty } (\lambda {\tilde{b}}^1(h_m) + (1-\lambda )b^2(h_m)) = L(b_\lambda )$.

For the second part of the result concerning S, completely analogous reasoning (except now $\left\{ h_m\right\}$ is a sequence in $I_3$ instead of $I_4$ and we use Lemma 3.10 instead of Lemma 3.9) shows $\lim _{m\rightarrow \infty } {\tilde{b}}_\lambda (h_m) = S(b_\lambda )$. $\square$

Lemma 7.2

Let $b, d \in \ell _\infty (I)$ such that $OV(b) < \infty , OV(d) < \infty$ and $-\infty < L(b), L(b+d) < \infty .$ Assume DP.2 and that $\overline{FM}(\ell _\infty (I)) \subseteq \ell _\infty (H)$. Then there exists ${\hat{\epsilon }} > 0$ and a sequence $\{h_m\}_{m\in \mathbb N} \subseteq I_4$ such that for all $\epsilon \in [0,{\hat{\epsilon }}]$:

$$\begin{aligned} {\tilde{d}}_{\epsilon } (h_m) \rightarrow L(b + \epsilon d) \text { and } \sum _{k=\ell }^{n} |\tilde{a}^k(h_m)| \rightarrow 0, \end{aligned}$$

where ${\tilde{d}}_{\epsilon } := \overline{FM}(b + \epsilon d)$.

Proof

Define

$$\begin{aligned} \alpha := L(b) - \sup \{ \limsup \{\tilde{b} (h_{m})\}_{m \in \mathbb N} : \{h_{m}\}_{m \in \mathbb N} \in {\mathcal {H}}_L \}. \end{aligned}$$

By hypothesis, $-\infty < L(b) < \infty$ so $I_{4}$ is not empty and then by assumption DP.2 $\alpha$ is a positive real number.

• 1. Since ${\tilde{d}} = \overline{FM}(d) \in \ell _\infty (H)$ there exists ${\hat{\epsilon }} > 0$ such that

  $$\begin{aligned} {\hat{\epsilon }} \sup _{h \in I_4} | {\tilde{d}}(h)| < \frac{\alpha }{3}. \end{aligned}$$

• 2. Claim: $L(b) - \frac{\alpha }{3} \le L(b + {\hat{\epsilon }} d) \le L(b) + \frac{\alpha }{3}$. Proof:

  $$\begin{aligned} \begin{array}{rcl} L(b + {\hat{\epsilon }} d) &{} = &{}\lim _{\delta \rightarrow \infty }\sup \left\{ \tilde{b}(h) + {\hat{\epsilon }}{\tilde{d}}(h)- \delta \sum _{k=\ell }^{n} |\tilde{a}^k(h)| \, : \, h \in I_4 \right\} \\ &{}\ge &{}\lim _{\delta \rightarrow \infty }\sup \left\{ \tilde{b}(h) - \frac{\alpha }{3} - \delta \sum _{k=\ell }^{n} |\tilde{a}^k(h)| \, : \, h \in I_4 \right\} \\ &{} = &{} \lim _{\delta \rightarrow \infty }\sup \left\{ \tilde{b}(h) - \delta \sum _{k=\ell }^{n} |\tilde{a}^k(h)| \, : \, h \in I_4 \right\} - \frac{\alpha }{3} \\ &{} = &{} L(b) -\frac{\alpha }{3}. \end{array} \end{aligned}$$

  (7.1)

  Similarly, one can show $L(b + {\hat{\epsilon }} d) \le L(b) + \frac{\alpha }{3}$.

• 3. Consider $\overline{FM}(b + {\hat{\epsilon }} d) = \overline{FM}(b) + {\hat{\epsilon }} \overline{FM}(d) = {\tilde{b}} + {\hat{\epsilon }}{\tilde{d}}.$ By Claim 2, $L(b + {\hat{\epsilon }} d)$ is finite. By Lemma 3.9, there exists a sequence $\{h'_m\}$ such that ${\tilde{b}}(h'_m) + {\hat{\epsilon }}{\tilde{d}}(h'_m) \rightarrow L(b + {\hat{\epsilon }} d)$ and $\sum _{k=\ell }^{n} |\tilde{a}^k(h'_m)| \rightarrow 0$.

• 4. Claim: $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} = L(b).$ Proof: first show $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} \le L(b).$ If

  $$\begin{aligned} \limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} > L(b) \end{aligned}$$

  then there is subsequence of indices $\{ h^{\prime \prime } \}_{m \in \mathbb N}$ from $\{ h^{\prime } \}_{m \in \mathbb N}$ such that $\lim _{m \rightarrow \infty }{\tilde{b}}(h''_m) > L(b).$ But $\sum _{k=\ell }^{n} |\tilde{a}^k(h'_m)| \rightarrow 0$ so $\sum _{k=\ell }^{n} |\tilde{a}^k(h''_m)| \rightarrow 0.$ This directly contradicts Lemma 3.9 so we conclude $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} \le L(b).$

Since $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} \le L(b)$ it suffices to show $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} = L(b)$ by showing $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N}$ cannot be strictly less than L(b). From Step 3. above, we know $\{ {\tilde{b}}(h'_m) + {\hat{\epsilon }}{\tilde{d}}(h'_m) \}_{m \in \mathbb N}$ is a sequence that converges to $L(b + {\hat{\epsilon }} d).$ This implies

$$\begin{aligned} L(b + {\hat{\epsilon }} d)= & {} \lim _{m \rightarrow \infty } \left( {\tilde{b}}(h'_m) + {\hat{\epsilon }}{\tilde{d}}(h'_m) \right) \end{aligned}$$

(7.2)

$$\begin{aligned}= & {} \limsup \left\{ {\tilde{b}}(h'_m) + {\hat{\epsilon }}{\tilde{d}}(h'_m) \right\} _{m \in \mathbb N} \end{aligned}$$

(7.3)

$$\begin{aligned}\le & {} \limsup \left\{ {\tilde{b}}(h'_m)\right\} _{m \in M} + \limsup \left\{ {\hat{\epsilon }}{\tilde{d}}(h'_m) \right\} _{m \in \mathbb N} \end{aligned}$$

(7.4)

$$\begin{aligned}< & {} \limsup \left\{ {\tilde{b}}(h'_m) \right\} _{m \in M} + \frac{\alpha }{3}. \end{aligned}$$

(7.5)

If $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} < L(b),$ then by definition of $\alpha$,

$$\begin{aligned} \limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} \le L(b) - \alpha . \end{aligned}$$

Then from (7.2) to (7.5)

$$\begin{aligned} L(b + {\hat{\epsilon }} d) < \limsup \{ {\tilde{b}}(h'_m) \}_{m \in M} + \frac{\alpha }{3} \le L(b) - \alpha + \frac{\alpha }{3} = L(b) - \frac{2}{3}\alpha , \end{aligned}$$

which cannot happen since from Step 2, $L(b + {\hat{\epsilon }} d) \ge L(b) - \frac{\alpha }{3} > L(b) - \frac{2}{3} \alpha .$ Therefore $\limsup \{ {\tilde{b}}(h'_m) \}_{m \in \mathbb N} = L(b).$ Then by Lemma 3.9 there is subsequence of indices $\{ h^{\prime \prime } \}_{m \in \mathbb N}$ from $\{ h^{\prime } \}_{m \in \mathbb N}$ such that

$$\begin{aligned} \begin{array}{l} {\tilde{b}}(h''_m) \rightarrow L(b) \text { and }\\ \sum _{k=\ell }^{n} |\tilde{a}^k(h''_m)| \rightarrow 0 \end{array} \end{aligned}$$

and from Claim 3 since $\{ h^{\prime \prime } \}_{m \in \mathbb N}$ is a subsequence from $\{ h^{\prime } \}_{m \in \mathbb N}$

$$\begin{aligned} \begin{array}{l} {\tilde{b}}(h''_m) + {\hat{\epsilon }} {\tilde{d}}(h''_m) \rightarrow L(b + {\hat{\epsilon }} d) \text { and } \\ \sum _{k=\ell }^{n} |\tilde{a}^k(h''_m)| \rightarrow 0. \end{array} \end{aligned}$$

5. Claim:

$$\begin{aligned} \begin{array}{l} {\tilde{b}}(h''_m) + \epsilon {\tilde{d}}(h''_m) \rightarrow L(b + \epsilon d) \text { and } \\ \sum _{k=\ell }^{n} |\tilde{a}^k(h''_m)| \rightarrow 0. \end{array} \end{aligned}$$

holds for all $\epsilon \in [0,{\hat{\epsilon }}]$. Proof: this is because for every $\epsilon \in [0, {\hat{\epsilon }}], b + \epsilon d$ is a convex combination of the sequences b and $b + \hat{\epsilon } d$. The claim follows by applying Lemma 7.1 with $b^1 = b$ and $b^2 = b + {\hat{\epsilon }} d$. $\square$

Lemma 7.3 is an analogous result for sequences in $I_{3}$ converging to S(b).

Lemma 7.3

Let $b, d \in \ell _\infty (I)$ such that $OV(b) < \infty , OV(d) < \infty$ and $-\infty < S(b), S(b+d) < \infty .$ Assume DP.1 and $\overline{FM}(\ell _\infty (I)) \subseteq \ell _\infty (H)$. Then there exists ${\hat{\epsilon }} > 0$ and a sequence $\{h_m\}_{m\in \mathbb N} \subseteq I_3$ such that for all $\epsilon \in [0,{\hat{\epsilon }}]$:

$$\begin{aligned} {\tilde{d}}_{\epsilon } (h_m) \rightarrow S(b + \epsilon d), \end{aligned}$$

where ${\tilde{d}}_{\epsilon } := \overline{FM}(b + \epsilon d)$.

Proof

The proof is analogous to Lemma 7.2. Replace L with $S, I_4$ with $I_3$, and redefine $\alpha$ as

$$\begin{aligned} \alpha := S(b) - \sup \{ \limsup \{\tilde{b} (h_{m})\}_{m \in \mathbb N} : \{h_{m}\}_{m \in \mathbb N} \in {\mathcal {H}}_S \}. \end{aligned}$$

By hypothesis, $-\infty < S(b) < \infty$ so $I_{3}$ is not empty and then by assumption DP.1, $\alpha$ is a positive real number. The result follows from DP.1 and noting $\sum _{k=\ell }^{n} |\tilde{a}^k(h_m)| = 0$ for all sequences $\left\{ h_m\right\}$ in $I_3.$ $\square$