Exact algorithms for single-machine scheduling with time windows and precedence constraints

Abstract

We study a single-machine scheduling problem that is a generalization of a number of problems for which computational procedures have already been published. Each job has a processing time, a release date, a due date, a deadline, and a weight representing the penalty per unit-time delay beyond the due date. The goal is to schedule all jobs such that the total weighted tardiness penalty is minimized and both the precedence constraints as well as the time windows (implied by the release dates and the deadlines) are respected. We develop a branch-and-bound algorithm that solves the problem to optimality. Computational results show that our approach is effective in solving medium-sized instances, and that it compares favorably with existing methods for special cases of the problem.

Appendix: Proofs

Proof of Lemma 1

Consider the set of constraints (10). For each $(i,j) \in A$, the following inequalities hold:

$$\begin{aligned}&x_{i1} + \dots + x_{in} \le 1 - x_{j1} = x_{j2}+ \dots + x_{jn} \\&x_{i2} + \dots + x_{in} \le 1 - x_{j1} -x_{j2} = x_{j3}+\dots + x_{jn} \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \vdots \\&x_{i(n-1)} + x_{in} \le 1 - x_{j1} - \dots - x_{j(n-1)} = x_{jn} \\&x_{in} \le 1 - x_{j1} - \dots - x_{jn} = x_{j1}+\dots + x_{jn} -1 \\ \end{aligned}$$

By adding the above inequalities, we obtain

$$\begin{aligned} x_{i1} + 2 x_{i2} + 3 x_{i3}&+ \dots + n x_{in} \le \\&x_{j1} + 2 x_{j2} + 3 x_{j3} + \dots + n x_{jn} - 1. \end{aligned}$$

This is exactly the associated constraint in the set of constraints (5). As a result, the solution space of the LP relaxation of ASF$^{\prime }$ is included in that of ASF. To show that the inclusion is strict, consider the following fractional values for the decision variables corresponding with a couple $(i,j)\in A$: $x_{i1}=x_{i5}=0.5$ and $x_{j4}=1$. These values can be seen to respect the weak but not the strong formulation. $\square$

Proof of Theorem 1

Since $1||\sum w_j T_j$ is a relaxation of $1|\beta |\sum w_j T_j$, the optimal value of $I^{\prime }$ and any valid lower bound for this optimal value is considered as a valid lower bound for $I$. $\square$

Proof of Theorem 2

The following equality holds:

$$\begin{aligned} \mathrm {TWT}^*(I) =&\sum _{j \in \mathcal {T}}{w_j (C_j(S^*) - d_j)} \\ =&\sum _{j \in N}{w_j (C_j(S^*) - d_j)} \\&- \sum _{j \in \mathcal {E}}{w_j (C_j(S^*) - d_j)} = \sum _{j \in N}{w_j C_j(S^*)} \\&- \sum _{j \in N}{w_j d_j} + \sum _{j \in \mathcal {E}}{w_j (d_j - C_j(S^*))}. \end{aligned}$$

Recall that $\bar{lb}_{I^{{\prime }{\prime }}}$ is a valid lower bound on the value $\sum _{j \in N}{w_j C_j(S^*)}$ and $lb^E$ is a valid lower bound on the value $\sum _{j \in \mathcal {E}}{w_j (d_j - C_j(S^*))}$. $\square$

Proof of Theorem 3

We argue that

$$\begin{aligned} \mathrm {LB}_0 = L_0(\lambda _\mathrm{PV}) = \hat{L}_1(\lambda _{\mathrm{PV}},\mu _0) \le \hat{L}_1(\lambda _{\mathrm{TPA}},\mu _{\mathrm{TPA}})\\ \le L_1(\lambda _{\mathrm{TPA}},\mu _{\mathrm{TPA}}) = \mathrm {LB}_1. \end{aligned}$$

The first inequality follows from Theorem 3 in van de Velde (1995), where it is shown that TPA generates a series of monotonically increasing lower bounds. The second inequality corresponds with Theorem 1. $\square$

Proof of Theorem 4

$\mathrm {LB}_1({\text {NO}})$ is obtained by solving $\mathrm {LR}_{\mathrm{P}_1}$ with $A^{\prime }=\emptyset$ and $A^{{\prime }{\prime }} = A$, so with the same multipliers the problem associated with $\mathrm {LB}_1({\text {NO}})$ is a relaxation of the problem associated with $\mathrm {LB}_1({\text {VSP}})$. The multipliers are updated with TPA in both cases, and will indeed be the same for a given $k_{\max }$, so the theorem holds. $\square$

Proof of Theorem 5

We introduce $g_{B_i}(\lambda ,\mu ,S)$ as follows:

$$\begin{aligned} g_{B_i}(\lambda ,\mu ,S) = \sum \limits _{j \in B_i}^{}{(w_j - \lambda _j) T_{j}} + \sum \limits _{j \in B_i}^{}{\lambda _j (C_j - d_j)}+ \\ \sum \limits _{j \in B_i}^{}{\sum _{k \in \mathcal {Q}_j} {\mu _{jk} (C_j + p_k - C_k)}}. \end{aligned}$$

Let $S_1^*$ be an optimal solution to $\mathrm {LB}^*_1$ and ${S_2^*} = ({S_{B_1}^*},\dots ,{S_{B_\kappa }^*})$ an optimal solution to $\mathrm {LB}^*_2$. The following result is derived.

$$\begin{aligned}&\mathrm {LB}^*_1 = g(\lambda ^*,\mu ^*,S_1^*) \le g(\lambda ^*,\mu ^*,{S_2^*}) \\&= \sum _{i=1}^{\kappa }{g_{B_i}(\lambda ^*,\mu ^*,S^*_{B_i})} \le \sum _{i=1}^{\kappa } {g_{B_i}(\lambda _{B_i}^*,\mu _{B_i}^*,S^*_{B_i})} = \mathrm {LB}^*_2. \end{aligned}$$

$\square$

Proof of Theorem 6

If $\min _{\pi \in \varPi _\mathcal {A}} \{C_{\max }(\sigma _B|\pi )\}$ is greater than $\max _{j\in \mathcal {A}}{\{\bar{\delta _j}\}}$ then at least one job in $\mathcal {A}$ cannot be scheduled before its deadline and the schedule $(\sigma _B,\sigma _E)$ is not feasible. $\square$

Proof of Theorem 7

If $\tau _2^{\prime } = 0$, then $\varGamma _{jk}(t,\tau _1,\tau _2) = \hat{\varGamma }_{jk} (t,\tau _1,\tau _2,\tau _2^{\prime })$ and the theorem holds based on Jouglet et al. (2004). If $\tau _2^{\prime } > 0$, all jobs in $\mathcal {B}$ are shifted to the left at least $\tau _2^{\prime }$ units. Also, $\tau _2$ equals zero because no job is shifted to the right. For all jobs $i \in \mathcal {B}^{\prime }$ we have $C_i(S_1) \ge C_i(S_1^{\prime }) \ge d_i$. Consequently, $\tau _2^{\prime } \sum _{i \in \mathcal {B}^{\prime }}{w_i} \ge 0$ is a lower bound for the gain of rescheduling jobs in $\mathcal {B}$. The value $w_j \max \{0,t+p_j-d_j\} - w_j \max \{0,\hat{r}_j+p_j-d_j\}$ equals the gain of rescheduling job $j$ and the value $w_k \max \{0,\hat{r}_k+p_k-d_k\}- w_k \max \{0,t+\tau _1+p_k-d_k\}$ equals the gain of rescheduling job $k$. By adding lower bounds for rescheduling gains of all jobs in $U = \mathcal {B}\cup \{j,k\}$, a lower bound for the gain of interchanging jobs $j$ and $k$ is obtained. $\square$

Proof of Theorem 8

We examine under which conditions the jobs belonging to the set $U = \mathcal {B} \cup \{j,k\}$ do not violate any of their deadlines and/or precedence constraints. Precedence constraints are not violated because jobs $j,k \in E_B$ are deliberately chosen such that $\mathcal {Q}_k \cap \mathcal {Q}_j = \mathcal {Q}_k$ and job $j$ does not violate its deadline simply because $C_j(S_1^{\prime }) \le C_j(S_1) \le \bar{\delta }_j$.

Condition 1 ensures that job $k$ does not violate its deadline. We argue that $t = st_j(S_1) \le \bar{\delta }_j - p_j$. If $\bar{\delta }_j - p_j \le \bar{\delta }_k - \tau _1 - p_k$ holds, then we infer $C_k(S_1^{\prime }) = t+\tau _1+p_k \le \bar{\delta }_k$. Also, if $\varPsi \le \hat{\delta }_k$, then all unscheduled jobs including $j$ and $k$ are completed at or before $\hat{\delta }_k$. Note that $\hat{\delta }_k$ is preferred over $\bar{\delta }_k$ because $\bar{\delta }_k \le \hat{\delta }_k$, thus condition 1 is less violated, and the inequality $\varPsi \le \hat{\delta }_k$ also implies $C_k(S_1^{\prime }) \le \bar{\delta }_k$.

Condition 2 verifies that no job in $\mathcal {B}$ violates its deadline. On the one hand, if $\tau _2 = 0$, then no job in $\mathcal {B}$ is shifted to the right, which means $C_i(S_1^{\prime }) \le C_i(S_1) \le \bar{\delta }_i$ for each job $i \in \mathcal {B}$. On the other hand, if $\tau _2 > 0$ and $\varPsi \le \min _{i \in \mathcal {B}}\{\hat{\delta }_i\}$, then for all jobs $i \in \mathcal {B}$ we conclude: $C_i(S_1^{\prime }) \le \varPsi \le \min _{i \in \mathcal {B}}\{\hat{\delta }_i\} \le \hat{\delta }_i$. Again, $\hat{\delta }_i$ is preferred over $\bar{\delta }_i$ because of the same reasoning as for the preference of $\hat{\delta }_k$ over $\bar{\delta }_k$. $\square$

Proof Lemma 4

Let $f$ have a global minimum at value $t^*$. Depending on the values of the parameters $\alpha ,\beta ,a$, and $b$, the function $f$ behaves differently. We discuss four possible cases for the parameter combinations to prove this lemma (see also Fig. 12). In the two first cases, we assume that $\alpha \ge \beta$. Case (a): in this case, $a \le b$, and then $f$ is constant on interval $[u,a]$ and is increasing on interval $[a,v]$, as shown in Fig. 12a. Case (b): $a > b, f$ is constant on interval $[u,b]$, decreasing on interval $[b,a]$ and increasing on interval $[a,v]$, in line with Fig. 12b. The following results are valid for these two cases: 1- If $u \le a \le v$ then $t^* = a$. 2- If $a < u, t^* = u$ because $f$ is always increasing on interval $[u,v]$. 3- If $a > v, t^* = v$ because $f$ is always decreasing on interval $[u,v]$. We conclude that $t^* = \min \{\max \{a,u\},v\}$ for the first two cases.

In the next two cases, we assume that $\alpha < \beta$. Case (c): $a < b, f$ is constant for $[u,b]$, increasing for $[b,a]$ and decreasing for $[a,v]$, as shown in Fig. 12c. In this case, $t^*$ equals either $u$ or $v$. On the one hand, if $\alpha (b - \max \{a,u\}) \ge (\beta - \alpha )(\max \{v,b\}-b) \Rightarrow \alpha (\bar{v} - \bar{u}) \ge \beta (\bar{v} - b)$ then $f(v) \ge f(u)$ is inferred and $t^* = u$ is concluded. On the other hand, if $\alpha (\bar{v} - \bar{u}) < \beta (\bar{v} - b)$ then $t^* = v$ is concluded. Case (d): $a \ge b, f$ is constant for $[u,b]$ and decreasing for $[b,v]$; see Fig. 12d. We find that $t^*$ equals $v$ for this case. $\square$

Fig. 12

Four possible cases for the parameter combinations in the proof of Lemma 4

Proof of Theorem 9 and Theorem 10

Similar to the proof of Theorem 8. $\square$