Approximation schemes for r-weighted Minimization Knapsack problems

Abstract

Stimulated by salient applications arising from power systems, this paper studies a class of non-linear Knapsack problems with non-separable quadratic constrains, formulated in either binary or integer form. These problems resemble the duals of the corresponding variants of 2-weighted Knapsack problem (a.k.a., complex-demand Knapsack problem) which has been studied in the extant literature under the paradigm of smart grids. Nevertheless, the employed techniques resulting in a polynomial-time approximation scheme (PTAS) for the 2-weighted Knapsack problem are not amenable to its minimization version. We instead propose a greedy geometry-based approach that arrives at a quasi PTAS (QPTAS) for the minimization variant with boolean variables. As for the integer formulation, a linear programming-based method is developed that obtains a PTAS. In view of the curse of dimensionality, fast greedy heuristic algorithms are presented, additionally to QPTAS. Their performance is corroborated extensively by empirical simulations under diverse settings and scenarios.

Appendix

Proof of Fact 1

(i) Indeed,

$$\begin{aligned} \dfrac{(a+b)^T\xi }{\Vert a+b\Vert _2\Vert \xi \Vert _2}&= \dfrac{\Vert a\Vert _2}{\Vert a+b\Vert _2}\cdot \dfrac{a^T\xi }{\Vert a\Vert _2\Vert \xi \Vert _2}+\dfrac{\Vert b\Vert _2}{\Vert a+b\Vert _2}\cdot \dfrac{b^T\xi }{\Vert b\Vert _2\Vert \xi (s)\Vert _2}\\&\ge \dfrac{\Vert a\Vert _2+\Vert b\Vert _2}{\Vert a+b\Vert _2}\cdot (1-\varepsilon )\\&\ge 1-\varepsilon , \end{aligned}$$

by the triangular inequality.

(ii) Let ${\bar{a}}:=\frac{a}{\Vert a\Vert _2}$, ${\bar{b}}:=\frac{b}{\Vert b\Vert _2}$, and ${{\bar{\xi }}}:=\frac{\xi }{\Vert \xi \Vert _2}$. If $a,b,\xi$ all lie in the same two-dimensional subspace, then the claim follows since the angle between a and b is no more than the sum of the angles between a and $\xi$, and b and $\xi$, which is at most is $\cos ^{-1}((1-\varepsilon )^2-\varepsilon (2-\varepsilon ))\le \cos ^{-1}(1-4\varepsilon )$. Otherwise, let ${\bar{b}}={\widehat{b}}+{\tilde{b}}$ be the orthogonal decomposition of ${\bar{b}}$ with respect to the 2-dimensional subspace formed by the two vectors ${\bar{a}}$ and $\bar{\xi }$, where ${\widehat{b}}$ is the projection of ${\bar{b}}$ into this space, and ${\tilde{b}}$ is the orthogonal component. Then

$$\begin{aligned} \frac{{\widehat{b}}^T\bar{\xi }}{\Vert {\widehat{b}}\Vert _2}=\frac{\bar{b}^T\bar{\xi }}{\Vert {\widehat{b}}\Vert _2}\ge \frac{1-\varepsilon }{\Vert \widehat{b}\Vert _2}\ge 1-\varepsilon , \end{aligned}$$

(since $\Vert {\widehat{b}}\Vert _2\le \Vert \bar{b}\Vert _2=1$), which also implies that $\Vert {\widehat{b}}\Vert _2\ge 1-\varepsilon$. Since a, ${\widehat{b}}$, and $\xi$ lie in the same subspace, it follows by the above argument that $\frac{{\widehat{b}}^T\bar{a}}{\Vert {\widehat{b}}\Vert _2}\ge (1-4\varepsilon )$, and hence,

$$\begin{aligned} {\bar{b}}^T{\bar{a}}={\widehat{b}}^T{\bar{a}}\ge (1-4\varepsilon )\Vert \widehat{b}\Vert _2\ge (1-4\varepsilon )(1-\varepsilon ), \end{aligned}$$

implying the claim. $\square$

Proof of Fact 2

Since the statement is invariant under rotation, we may assume, w.l.o.g., that $\eta =\mathbf{1}_j$, the jth-dimensional unit vector in ${\mathbb {R}}^r$. Write $b:={\widehat{b}}+{\tilde{b}}$, where ${\tilde{b}}$ is the vector orthogonal to a in the subspace spanned by a and b. Then $\Vert \mathtt{Pj}_{\eta }(b)\Vert _2=b^j$ is the jth component of b, and $\Vert \mathtt{Pj}_{\eta }({\widehat{b}})\Vert _2={\widehat{b}}^j$. Since

$$\begin{aligned} \Vert {\widehat{b}}\Vert _2\ge \frac{a^T\widehat{b}}{\Vert a\Vert _2}=\frac{a^Tb}{\Vert a\Vert _2}\ge (1-5\varepsilon )\Vert b\Vert _2, \end{aligned}$$

it follows that

$$\begin{aligned} \Vert {\tilde{b}}\Vert _2= \sqrt{\Vert b\Vert _2^2-\Vert {\widehat{b}}\Vert _2^2}&\le \sqrt{5\varepsilon (2-5\varepsilon )}\Vert b\Vert _2\le \frac{\sqrt{5\varepsilon (2-5\varepsilon )}}{1-5\varepsilon }\Vert \widehat{b}\Vert _2\\&=\frac{\lambda \sqrt{5\varepsilon (2-5\varepsilon )}}{1-5\varepsilon }\Vert a\Vert _2. \end{aligned}$$

Since $|b^j-{\widehat{b}}^j|=|{\tilde{b}}^j|\le \Vert {\tilde{b}}\Vert _2$, and ${\widehat{b}}=\lambda \Vert a\Vert _2 \cdot a$, the claim follows. $\square$