An improved test set approach to nonlinear integer problems with applications to engineering design

Abstract

Many problems in engineering design involve the use of nonlinearities and some integer variables. Methods based on test sets have been proposed to solve some particular problems with integer variables, but they have not been frequently applied because of computation costs. The walk-back procedure based on a test set gives an exact method to obtain an optimal point of an integer programming problem with linear and nonlinear constraints, but the calculation of this test set and the identification of an optimal solution using the test set directions are usually computationally intensive. In problems for which obtaining the test set is reasonably fast, we show how the effectiveness can still be substantially improved. This methodology is presented in its full generality and illustrated on two specific problems: (1) minimizing cost in the problem of scheduling jobs on parallel machines given restrictions on demands and capacity, and (2) minimizing cost in the series parallel redundancy allocation problem, given a target reliability. Our computational results are promising and suggest the applicability of this approach to deal with other problems with similar characteristics or to combine it with mainstream solvers to certify optimality.

Appendix: A Proof of Theorem 3

Let $A$ be the matrix associated with the restrictions for Problem (LSP-II). Then $A$ can be divided in five blocks according to the five sets of variables $y_{ij}, a_{ij}, z_{ij}, b_{ij}$ and $c_{ij}$. Let $I$ denote the identity matrix of size $mn$ and $\varvec{O}$ denote a matrix with all entries zero. Let $D$ be the block diagonal $(0,1)$-coefficient matrix associated with constraints (7). Let $-MI$ denote the coefficient matrix of the variables $z_{ij}, i=1,\ldots ,n, j=1,\ldots ,m$ in constraints (8). We can then write $A$ as the $(n+3mn) \times 5mn$ integer matrix

$$\begin{aligned} A = \left( \begin{array}{c|c|c|c|c} D &{} \varvec{O} &{} \varvec{O} &{} \varvec{O} &{}\varvec{O} \\ I &{} I &{} -MI &{} \varvec{O} &{} \varvec{O} \\ \varvec{O} &{} \varvec{O} &{} I &{} I &{} \varvec{O} \\ -I &{} \varvec{O} &{} I &{} \varvec{O} &{} I \end{array} \right) \end{aligned}$$

We note $t = (t_y, t_a, t_z, t_b, t_c), S = \{ t ~|~ A t = 0 \}$ and the block $t_y$ represents the variables $(Y_{11}, \ldots , Y_{mn})$, noted in capital letters. Similar for the other blocks. Consider the following cases:

• 1. $K_1 = \{ t \in S ~|~ t_y = t_b = 0 \}$. Let $S' = \ker A'$, where

  $$\begin{aligned} A' = \left( \begin{array}{ccc} I &{} -MI &{} \varvec{O} \\ \varvec{O} &{} I &{} \varvec{O} \\ \varvec{O} &{} I &{} I \end{array} \right) \rightarrow \left( \begin{array}{ccc} I &{} -MI &{} \varvec{O} \\ \varvec{O} &{} I &{} \varvec{O} \\ \varvec{O} &{} \varvec{O} &{} I \end{array} \right) \!\!, \end{aligned}$$

  which is not singular. Then $K_1 = 0$ and there is no binomial $x^\alpha - x^\beta$ such that contains only the variables $a_{ij}, z_{ij}, c_{ij}$.

• 2. $K_2 = \{ t \in S ~|~ t_b = 0 \}$. Let $S''= \ker A''$, where

  $$\begin{aligned} A'' = \left( \begin{array}{cccc} D &{} \varvec{O} &{} \varvec{O} &{} \varvec{O} \\ I &{} I &{} -MI &{} \varvec{O} \\ \varvec{O} &{} \varvec{O} &{} I &{} \varvec{O} \\ -I &{} \varvec{O} &{} I &{} I \end{array} \right) . \end{aligned}$$

  By the previous case, we can assume $t_y \ne 0$. The rows in the third block imply that $t_z = 0$. Then $t \in K_2$ if and only if $(t_y,t_a,t_c) \in S''' = \ker A'''$, for

  $$\begin{aligned} A''' = \left( \begin{array}{ccc} D &{} \varvec{O} &{} \varvec{O} \\ I &{} I &{} \varvec{O} \\ -I &{} \varvec{O} &{} I \end{array} \right) . \end{aligned}$$

  Let $Y_{ip}$ be the leftmost nonzero component in $t_y$. We can assume $Y_{ip} > 0$, because $S'''$ is a vector space. The row block corresponding to $D$ implies that there exists $q > p$ such that $Y_{iq} < 0$. Then

  $$\begin{aligned} y_{ip} \text { divides } x^{t^+}, y_{iq} \text { divides } x^{t^-}. \end{aligned}$$

  The second block of rows in $A'''$ implies $A_{ip} = - Y_{ip} < 0, A_{iq} = -Y_{iq} > 0$. From the third block of rows, $C_{ip} = Y_{ip} > 0, C_{iq} = A_{iq} < 0$. Then

  $$\begin{aligned} y_{ip} a_{iq} c_{ip} \text { divides } x^{t^+}, y_{iq} a_{ip} c_{iq} \text { divides } x^{t^-}. \end{aligned}$$

  The initial term in $x^{t^+} - x^{t^-}$ with respect to $>$ is $x^{t^+}$ because $y_{ip}$ divides $x^{t^+}$ and $y_{ip}$ it the greatest monomial in the binomial. Then the initial term of

  $$\begin{aligned} y_{ip} a_{iq} c_{ip} - y_{iq} a_{ip} c_{iq} \end{aligned}$$

  (12)

  divides the initial term of $x^{t^+} - x^{t^-}$.

• 3. Let $S = \ker A$. By (a) and (b) we suppose $t_b \ne 0$. Let $B_{rs}$ the leftmost nonzero component in $t_b$. As in the previous case, take $B_{rs} > 0$. Because $b$ is the set of greatest variables for our term order, $x^{t^+}$ is the leading term. We also have $Z_{rs} = -B_{rs} < 0$.

  • (a) If $Y_{rs} \ge 0$ then

    $$\begin{aligned} -Y_{rs} + Z_{rs} + C_{rs} = 0 \end{aligned}$$

    then

    $$\begin{aligned} b_{rs}c_{rs} \text { divides } x^{t^+} \end{aligned}$$

    and the initial term of

    $$\begin{aligned} b_{rs} c_{rs} - z_{rs} a_{rs}^{M_r}. \end{aligned}$$

    (13)

    divides $x^{t^+}$.

  • (b) $Y_{rs} < 0$. Then there exists $Y_{rq}, \,s\ne q$ with $Y_{rq}=-Y_{rs}>0$ .

    • i. If $C_{rq} > 0$, then $b_{rs}y_{rq}c_{rq}$ divides $x^{t^+}$. The initial term of

      $$\begin{aligned} b_{rs} y_{rq} c_{rq} - z_{rs} y_{rs} a_{rs}^{M_r-1} a_{rq}, s \ne q. \end{aligned}$$

      (14)

      divides the initial term of $x^{t^+}-x^{t^-}$.

    • ii. If $C_{rq}\le 0$ then $0<Y_{rq}\le Z_{rq}$, which implies $B_{rq}<0$ so $s < q$, because $B_{rs}$ was the first nonzero element. The rows in the second block imply that

      $$\begin{aligned}&Y_{rq} + A_{rq} - M_r Z_{rq} = 0, \text { so } A_{rq} = M_r Z_{rq} - Y_{rq} \ge M_r Y_{rq} - Y_{rq} \\&\quad = (M_r -1) Y_{rq}. \end{aligned}$$

      Then the initial term of

      $$\begin{aligned} b_{rs} y_{rq} z_{rq} a_{rq}^{M_r-1} - z_{rs} y_{rs} b_{rq} a_{rs}^{M_r-1}, s < q \end{aligned}$$

      (15)

      divides the initial term of $x^{t^+}-x^{t^-}$.

We have reviewed all the cases, so the sets (12), (13), (14) and (15) form the reduced Gröbner basis.