Parametric Algorithm for Finding a Guaranteed Solution to a Quantile Optimization Problem

Abstract

The problem of stochastic programming with a quantile criterion for a normal distribution is studied in the case of a loss function that is piecewise linear in random parameters and convex in strategy. Using the confidence method, the original problem is approximated by a deterministic minimax problem parameterized by the radius of a ball inscribed in a confidence polyhedral set. The approximating problem is reduced to a convex programming problem. The properties of the measure of the confidence set are investigated when the radius of the ball changes. An algorithm is proposed for finding the radius of a ball that provides a guaranteeing solution to the problem. A method for obtaining a lower estimate of the optimal value of the criterion function is described. The theorems are proved on the convergence of the algorithm with any predetermined probability and on the accuracy of the resulting solution.

APPENDIX

Proof of Theorem 1. Conditions 2 and 3 ensure that all constraints in the problem (5) are active. This means that all faces of the set C_r touch the ball B_r. As r increases on the segment [0, R] the faces of the set C_r are transferred in parallel, touching the ball B_r. This means that the set C_r expands as r increases. Therefore, the function h, defined as the measure C_r, is non-decreasing. Theorem 1 is proved.

Proof of Theorem 2. Let γ ∈ (0, 1). The set ${{C}_{{{{\rho }_{\gamma }}}}}$ is defined as the intersection of k half-planes of measure no less than γ. Denote these half-planes by L_i, i = $\overline {1,k}$. Then

$$h({{\rho }_{\gamma }}) = {\mathbf{P}}\left\{ {X \in \bigcap\limits_{i = 1}^k {{{L}_{i}}} } \right\} = 1 - {\mathbf{P}}\left\{ {X \in \bigcup\limits_{i = 1}^k {({{\mathbb{R}}^{m}}{{\backslash }}{{L}_{i}})} } \right\}\; \geqslant \;1 - \sum\limits_{i = 1}^k {{\mathbf{P}}\{ X \notin {{L}_{i}}\} } = 1 - (1 - \gamma )k.$$

Thus, h(ρ_γ) $\geqslant$ α for α $\leqslant$ 1 – (1 – γ)k, which is equivalent to γ $\geqslant$ β = 1 – $\frac{{1 - \alpha }}{k}$. Theorem 2 is proved.

Proof of Theorem 3. Since at each iteration the segment of the search for a solution narrows two times, the number of iterations K of the algorithm can be found as the minimum natural number K, that satisfies the inequality

$$\frac{{\left| {{{{\bar {R}}}_{\alpha }} - {{\rho }_{\alpha }}} \right|}}{{{{2}^{K}}}}\;\leqslant \;\delta .$$

It follows from this inequality that K = $\left\lceil {{{{\log }}_{2}}\frac{{\left| {{{{\bar {R}}}_{\alpha }} - {{\rho }_{\alpha }}} \right|}}{\delta }} \right\rceil$. The algorithm can make an error in its work only if at some iteration it turns out that $\hat {h}(r)$ $\geqslant$ α + ε, although in fact h(r) < α. It is easy to see that the random variable s(r) is distributed according to the binomial law with the success probability h(r) – μ(r). The inequality is known ([15], Chapter 1, Section 6):

$${\mathbf{P}}\{ \hat {h}(r) - h(r)\; \geqslant \;\varepsilon \} = {\mathbf{P}}\left\{ {\frac{{s(r)}}{N} - (h(r) - \mu (r))\; \geqslant \;\varepsilon } \right\}\;\leqslant \;{{e}^{{ - 2N{{\varepsilon }^{2}}}}}.$$

Therefore, if we assume that h(r) < α, then P{$\hat {h}$(r) $\geqslant$ α + ε} $\leqslant$ ${{e}^{{ - 2N{{\varepsilon }^{2}}}}}$. Since the samples used to evaluate the measure are independent, the probability that the algorithm will work correctly is at least ${{\left( {1 - {{e}^{{ - 2N{{\varepsilon }^{2}}}}}} \right)}^{K}}$. Hence it follows that, in order to ensure the probability p of successful operation of Algorithm 1, the inequality

$$p\;\leqslant \;{{\left( {1 - {{e}^{{ - 2N{{\varepsilon }^{2}}}}}} \right)}^{K}} \Leftrightarrow N\; \geqslant \;\frac{{\ln \left( {{\text{1/}}\left( {1 - \sqrt[K]{p}} \right)} \right)}}{{2{{\varepsilon }^{2}}}}.$$

must be satisfied.

Theorem 3 is proved.

Proof of Theorem 4. Let Ψ(u, r) $\triangleq$ ${{\max }_{{x \in {{B}_{r}}}}}\Phi (u,x)$ = Φ(u, x⁰(r)), where x⁰ is the point on the boundary of the ball B_r, where the specified maximum is reached. Since B_ρ ⊂ B_R, Ψ(u, ρ) $\leqslant$ Ψ(u, R) holds. Since the point y = $\frac{\rho }{R}{{x}^{0}}(R)$ lies on the boundary of the ball B_ρ, Φ(u, y) $\leqslant$ Ψ{u, ρ). That’s why

$$0\;\leqslant \;\Psi (u,\,\,R) - \Psi (u,\,\,\rho )\;\leqslant \;\Phi (u,\,\,{{x}^{0}}(R)) - \Phi (u,\,\,y)\;\leqslant \;L\left\| {{{x}^{0}}(R) - y} \right\| = (R - \rho )L.$$

Thus, the inequalities

$$\Psi (u,\,\,\rho )\;\leqslant \;\Psi (u,\,\,R)\;\leqslant \;\Psi (u,\,\,\rho ) + (R - \rho )L$$

(A.1)

are true. Minimizing the left and right parts of the first inequality in (A.1) with respect to u ∈ U so that ${{\max }_{{j = \overline {1,{{k}_{2}}} }}}${b_2j(u) + ||B_2j(u)||R} $\leqslant$ 0 (constraints of the problem (4) for r = R), we obtain the first inequality to be proved ψ(ρ) $\leqslant$ ψ(R) (here we take into account that ψ(ρ) is defined at least on a wider set). From (11) and the second inequality in (A.1) it follows that

$$\psi (R)\;\leqslant \;\Psi (u(\rho ),\,\,R)\;\leqslant \;\Psi (u(\rho ),\,\,\rho ) + (R - \rho )L = \psi (\rho ) + (R - \rho )L.$$

This estimate implies the second inequality to be proved. Theorem 4 is proved.