Scheduling parallel-machine batch operations to maximize on-time delivery performance

Abstract

In this paper we study the problem of minimizing total weighted tardiness, a proxy for maximizing on-time delivery performance, on parallel nonidentical batch processing machines. We first formulate the (primal) problem as a nonlinear integer programming model. We then show that the primal problem can be solved exactly by solving a corresponding dual problem with a nonlinear relaxation. Since both the primal and the dual problems are NP-hard, we use genetic algorithms, based on random keys and multiple choice encodings, to heuristically solve them. We find that the genetic algorithms consistently outperform a standard mathematical programming package in terms of solution quality and computation time. We also compare the smaller problem instances to a breadth-first tree search algorithm that gives evidence of the quality of the solutions.

Appendix: Nonlinear penalty function method

In this section we discuss the nonlinear penalty function method. We study the more general discrete optimization problem and show that the (primal) optimization problem can be solved exactly by solving a dual problem with a nonlinear relaxation, which establishes the theoretical foundation of using the Multiple Choice Genetic Algorithm to solve the original scheduling problem (OP) defined in Sect. 3.

Certain notations will be used. Let ${\mathbb {N}} = \{1,2,3,\ldots \}$ be the set of natural numbers and ${\mathbb {R}}$ be the set of real numbers. Let ${\mathbb {R}}_{+}$ and ${\mathbb {R}}_{++}$ denote the set of nonnegative and positive real numbers, respectively. Denote by ${\mathbb {R}}^n$ the set of n-dimensional real vectors. If $u \in {\mathbb {R}}^n$, then $u \le 0$ and $u=0$ mean that all components of u are nonpositive and zeros, respectively. We use the symbol min to refer to either “minimize (an optimization problem)” or “minimum (value).” Similar simplification applies to the symbol max. The exact meaning will be clear from context. If $(\cdot )$ stands for an optimization problem, then $v(\cdot )$ is its optimal value.

Consider the following discrete optimization problem, which we call the primal problem (P):

$$\begin{aligned} \begin{array}{lll} (P)~~~~ &{} \min \limits _{x}~ &{} f(x) \\ &{} \mathrm{s.t.} &{} g(x) \le 0 \\ &{} &{} h(x) = 0 \\ &{} &{} x \in {\mathscr {X}} \end{array} \end{aligned}$$

where $x=(x_{1},\dots ,x_{n}) \in {\mathbb {R}}^n$ is a vector of decision variables, with each component taking some discrete values. Both $g(x)=\left( g_1(x),\dots ,g_q(x)\right)$ and $h(x)=\left( h_1(x),\dots ,h_l(x)\right)$ are vector functions with q and l components, where $q, l \in {\mathbb {N}}$. Functions f, $g_{1}$, ..., $g_{q}$, and $h_{1}$, ..., $h_{l}$ are bounded and real-valued functions that can be arbitrary nonlinear or nonconvex. The feasible set of the problem is denoted by ${\mathscr {S}} = \{\,x \in {\mathbb {R}}^n \mid g(x) \le 0, h(x) = 0\,\} \cap {\mathscr {X}}$. The set ${\mathscr {S}}$ consists of explicit constraints $g(x) \le 0$ and $h(x) = 0$ and other constraints represented by the set ${\mathscr {X}}$, which is a finite and nonempty subset of ${\mathbb {R}}^n$. The set ${\mathscr {X}}$ might represent some simple constraints that can be easily handled, such as lower and upper bounds on the variables. Assume that the constraints $g(x) \le 0$ and $h(x) = 0$ are “complicating” in terms of solving the problem, while ${\mathscr {X}}$ is “easy.” Any vector (point) in ${\mathscr {S}}$ is called feasible, while any point not in ${\mathscr {S}}$ is called infeasible. A point $x^*$ is called optimal or a solution if it solves the problem, and $f(x^*)=v(P)$ is the optimal value.

The penalty function method drops the complicating constraints $g(x) \le 0$ and $h(x) = 0$ by introducing a weighted penalty for constraint violations. In general, for a minimization problem, a penalty function shall incur a positive penalty for infeasible solutions and no penalty for feasible solutions (Bazaraa et al. 2006).

Definition 1

A function $\alpha (x) :{\mathbb {R}}^n \mapsto {\mathbb {R}}$ is called a penalty function for problem (P) if it satisfies: (i) $\alpha (x) > 0$ if $g(x) > 0$ or $h(x) \ne 0$; (ii) $\alpha (x) = 0$ if $g(x) \le 0$ and $h(x) = 0$.

Various forms of penalty functions satisfying the above definition exist. A suitable nonlinear penalty function is defined by $\alpha (x) {:=}\sum _{i=1}^{q}\left[ \max \left( 0,g_i(x)\right) \right] ^2 + \sum _{i=1}^{l}\left[ h_i(x)\right] ^2. \nonumber$

The transformed objective function $F(x,\lambda ) :{\mathbb {R}}^n \times {\mathbb {R}}_{+} \mapsto {\mathbb {R}}$ is defined by $F(x,\lambda ) {:=}f(x) +\lambda \alpha (x),$ where $\lambda \ge 0$ is called the penalty parameter.

Rather than solve the problem (P), we consider the following penalty problem:

$$\begin{aligned} \begin{array}{lll} (PP_{\lambda })~~~~ &{} \min \limits _{x} ~ &{} F(x,\lambda ) \\ &{} \mathrm{s.t.} &{} x \in {\mathscr {X}} \end{array} \end{aligned}$$

Hadj-Alouane and Bean (1997) extended the results of penalty function methods from continuous nonlinear program to multiple choice integer program with linear objective function and linear constraints. General linear constraints are relaxed by a nonlinear penalty function and the corresponding dual problem has weak and strong duality. We now generalize their results to finite discrete optimization with nonlinear objective function and nonlinear constraints. This generalization is meaningful, since many practical problems, including the batch scheduling problems, can be formulated as finite discrete optimization models.

Proposition 1

(Weak Duality) $v(PP_{\lambda }) \le v(P)$ for all $\lambda \ge 0$; that is, the optimal value of the penalty problem provides a lower bound on the optimal value of the primal problem.

Proof

Let $\bar{x}$ be a feasible solution to the primal problem (P); that is, $\bar{x} \in {\mathscr {S}}$. So $\bar{x} \in {\mathscr {X}}$. Since $\bar{x} \in {\mathscr {S}}$, $g(\bar{x}) \le 0$ and $h(\bar{x})=0$. It follows that $\alpha (\bar{x}) = 0$. The following relations then hold for $\lambda \ge 0$:

$$\begin{aligned} v(PP_{\lambda }) = \min _{x \in {\mathscr {X}}} \{ f(x) + \lambda \alpha (x) \} \le f(\bar{x}) + \lambda \alpha (\bar{x}) = f(\bar{x}). \end{aligned}$$

Since the above relations hold for all the feasible solutions to (P), they must also hold for $x^*$, which is the optimal solution to (P). That is, $v(PP_{\lambda }) \le f(x^*) = v(P)$. $\square$

Definition 2

Consider a general optimization problem

$$\begin{aligned} \begin{array}{lll} (GP)~~~~ &{} \min \limits _x ~ &{} f(x) \\ &{} \mathrm{s.t.} &{} x \in {\mathscr {X}} \end{array} \end{aligned}$$

where f is a real-valued function defined on ${\mathbb {R}}^n$ and ${\mathscr {X}}$ is a nonempty subset of ${\mathbb {R}}^n$. For $\epsilon \in {\mathbb {R}}_{++}$, we call $\bar{x}$ an $\epsilon$-optimal solution to (GP) if $\bar{x} \in {\mathscr {X}}$ and $f(\bar{x}) \le \inf _{x\in {\mathscr {X}}} f(x) + \epsilon$.

Lemma 1 establishes the relationship between $v(PP_{\lambda })$ and v(P) for a given $\lambda \ge 0$.

Lemma 1

• (a) For a given $\lambda \ge 0$, if $\bar{x}$ is $\epsilon$-optimal to $(PP_{\lambda })$, $g(\bar{x}) \le 0$, and $h(\bar{x}) = 0$, then $\bar{x}$ is also $\epsilon$-optimal to (P).

• (b) For a given $\lambda \ge 0$, if $x^*$ is optimal to $(PP_{\lambda })$, $g(x^*) \le 0$, and $h(x^*) = 0$, then $x^*$ is also optimal to (P).

Proof

• (a) Since $\bar{x}$ is feasible for $(PP_{\lambda })$, $\bar{x} \in {\mathscr {X}}$. Moreover, $g(\bar{x}) \le 0$ and $h(\bar{x}) = 0$. Thus $\bar{x}$ is also feasible for (P) and $\alpha (\bar{x}) = 0$. If $\bar{x}$ is $\epsilon$-optimal to $(PP_{\lambda })$, then $f(\bar{x}) + \lambda \alpha (\bar{x}) \le v(PP_{\lambda }) + \epsilon$ by Definition 2. Therefore, $f(\bar{x}) \le v(PP_{\lambda }) + \epsilon$. It follows that $f(\bar{x}) \le v(P) + \epsilon$ since $v(PP_{\lambda }) \le v(P)$ by Proposition 1.

• (b) If $x^*$ is optimal to $(PP_{\lambda })$, then $v(PP_{\lambda }) = f(x^*)+\lambda \alpha (x^*)$. Since $g(x^*) \le 0$, $h(x^*) = 0$, and $x^* \in {\mathscr {X}}$, $x^*$ is also feasible for (P). Then $\alpha (x^*) = 0$ and $f(x^*) \ge v(P)$. Hence $v(PP_{\lambda }) = f(x^*)$. Then $f(x^*) \le v(P)$ since $v(PP_{\lambda }) \le v(P)$ by Proposition 1. It follows that $f(x^*) = v(P)$. $\square$

The penalty problem provides lower bounds for the primal problem. Under certain conditions, strong duality holds; that is, we can obtain the optimal solution to the primal problem by optimally solving the corresponding penalty problem. Proposition 2 establishes conditions for the existence of $\lambda$ for which strong duality holds.

Proposition 2

(Strong Duality) Let ${\mathscr {S}}_\lambda$ be the finite set of optimal solutions to $(PP_{\lambda })$. If (P) is feasible, then there exists $\bar{\lambda } \ge 0$, such that for all $\lambda \ge \bar{\lambda }$, there exists $x^* \in {\mathscr {S}}_\lambda$ such that $\alpha (x^*) = 0$ and $x^*$ is optimal to (P).

Proof

The proof is similar to that of Theorem 2 in Hadj-Alouane and Bean (1997). Since by assumption f(x), g(x), and h(x) are bounded and real-valued functions, (P) is also bounded. And since ${{\mathscr {X}}}$ is finite, there exists an $x^* \in {\mathscr {X}}$, such that $\alpha (x^*)=0$, and $x^*$ is optimal to (P). It remains to show that there exists $\bar{\lambda } \ge 0$ such that $x^* \in {\mathscr {S}}_\lambda$ for all $\lambda \ge \bar{\lambda }$.

Define ${\mathscr {Y}} {:=}\left\{ \, x \in {\mathbb {R}}^n \mid \alpha (x) = 0 \,\right\}$. The complement of ${\mathscr {Y}}$ is ${\mathscr {Y}}^c {:=}\left\{ \,x \in {\mathbb {R}}^n \mid \alpha (x) > 0 \,\right\}$. We have ${\mathscr {Y}} \cup {\mathscr {Y}}^c = {\mathbb {R}}^n \supset {\mathscr {X}} = ({\mathscr {X}} \cap {\mathscr {Y}} ) \cup ({\mathscr {X}} \cap {\mathscr {Y}}^c )$, and $x^* \in {\mathscr {X}} \cap {\mathscr {Y}}$.

$$\begin{aligned} v(PP_{\lambda })&= \min _{x \in {{\mathscr {X}}}}\left\{ f(x)+\lambda \alpha (x)\right\} \\&= \min \Bigl \{\min _{x\in \left( {\mathscr {X}} \cap {\mathscr {Y}} \right) } \left\{ f(x)+\lambda \alpha (x)\right\} ,\\&\quad \min _{x\in \left( {\mathscr {X}} \cap {\mathscr {Y}}^c \right) } \left\{ f(x)+\lambda \alpha (x)\right\} \Bigr \}\\&= \min \Bigl \{\min _{x\in \left( {\mathscr {X}} \cap {\mathscr {Y}} \right) } f(x), \min _{x\in \left( {\mathscr {X}} \cap {\mathscr {Y}}^c \right) } \left\{ f(x)+\lambda \alpha (x)\right\} \Bigr \}\\&= \min \Bigl \{f(x^*), \min _{x\in \left( {\mathscr {X}} \cap {\mathscr {Y}}^c \right) } \left\{ f(x)+\lambda \alpha (x)\right\} \Bigr \} \\&= \min \Bigl \{f(x^*), \min _{x\in {{\mathscr {X}}},\,\alpha (x) > 0} \left\{ f(x)+\lambda \alpha (x)\right\} \Bigr \}. \end{aligned}$$

Let $\lambda _0 {:=}\max _{x\in {{\mathscr {X}}},\,\alpha (x) > 0}[f(x^*)-f(x)]/\alpha (x) \in {\mathbb {R}}$. Since ${\mathscr {X}}$ is finite, f, g, and h are bounded, and $\alpha (x) > 0$, $\lambda _0$ exists and is bounded. And $\alpha (x) > 0$ implies that x is infeasible for (P). Thus, the sign of $f(x^*)-f(x)$ is indefinite, as is the sign of $\lambda _0$. Choose $\bar{\lambda }=\max \left\{ \lambda _0,0 \right\}$. Thus, $\bar{\lambda }$ is also bounded, $\bar{\lambda } \ge \lambda _0$, and $\bar{\lambda } \ge 0$.

$\forall \lambda \ge \bar{\lambda }$, let $\bar{x}_\lambda = \hbox {arg} \,\hbox {min}_{x\in {{\mathscr {X}}},\,\alpha (x) > 0}\left\{ f(x)+\lambda \alpha (x) \right\}$. We see that $\bar{x}_\lambda$ exists, since ${\mathscr {X}}$ is finite and nonempty, f, g, and h are bounded. Also notice that $\alpha (\bar{x}_\lambda ) >0$. Since $\lambda _0 {:=} \max _{x\in {\mathscr {X}},\,\alpha (x) > 0}\frac{f(x^*)-f(x)}{\alpha (x)}$ and $\bar{x}_\lambda \in \left\{ x \in {\mathscr {X}} \mid \alpha (x) >0 \right\}$, it follows that $\lambda _0 \ge \frac{f(x^*)-f(\bar{x}_\lambda )}{\alpha (\bar{x}_\lambda )}$.

Combining the above results, we have $\lambda \ge \bar{\lambda } \ge \lambda _0$. For any $x \in {\mathscr {X}}$ such that $\alpha (x) > 0$, we have $f(x)+\lambda \alpha (x) \ge f(\bar{x}_\lambda )+\lambda \alpha (\bar{x}_\lambda ) \ge f(\bar{x}_\lambda )+\bar{\lambda }\alpha (\bar{x}_\lambda ) \ge f(\bar{x}_\lambda )+\lambda _0\alpha (\bar{x}_\lambda ) \ge f(\bar{x}_\lambda )+\frac{f(x^*)-f(\bar{x}_\lambda )}{\alpha (\bar{x}_\lambda )}\cdot \alpha (\bar{x}_\lambda ) = f(x^*).$

That is, $\min _{x\in {\mathscr {X}},\,\alpha (x) > 0} \left\{ f(x)+\lambda \alpha (x)\right\} \ge f(x^*)$.² Thus,

$$\begin{aligned} v(PP_{\lambda }) = \min \Bigl \{f(x^*), \min _{x\in {\mathscr {X}},\,\alpha (x) > 0} \left\{ f(x)+\lambda \alpha (x)\right\} \Bigr \} = f(x^*). \end{aligned}$$

Since $x^* \in {\mathscr {X}}$, $x^*$ is feasible for $(PP_\lambda )$. The above result implies that $x^*$ is optimal to $(PP_\lambda )$; that is, $x^* \in {\mathscr {S}}_\lambda$. $\square$

We do not assume that the optimization problems are convex in the above results. The results are valid for both convex and nonconvex optimization problems. The only assumptions are that the objective and constraints functions are discrete and the general set ${\mathscr {X}}$ is finite and nonempty.