Single-machine scheduling with periodic due dates to minimize the total earliness and tardy penalty

Abstract

We consider a single-machine scheduling problem such that the due dates are assigned to each job depending on its order, and the lengths of the intervals between consecutive due dates are identical. The objective is to minimize the total penalty for the earliness and tardiness of each job. The early penalty proportionally increases according to the earliness amount, while the tardy penalty increases according to the step function. We show that the problem is strongly NP-hard, and furthermore, polynomially solvable if the two types of processing times exist.

Appendix A. The proof of Claim 3 in Theorem 1

(i) Suppose that two jobs $J_k$ and $J_l$ in ${\mathcal {J}}_{y}$ are processed in ${\hat{\pi }}_i$ for some $i\in \{1,2, \ldots ,m\}$. We assume that $J_l$ is processed later than $J_k$. Then, by inequalities (9) and (10), we have

$$\begin{aligned} E_{l}({\hat{\sigma }}) \ge E_{k}({\hat{\sigma }}) + (\Delta - p_l) \ge (\Delta - p_k) + (\Delta - p_l) > 2T^{3}. \end{aligned}$$

(12)

By inequality (9) and $p_{j}<M-T^{3}$ for $J_{j}\in {\mathcal {J}}_{y}$, we have

$$\begin{aligned} E_{j}({\hat{\sigma }})> T^{3} \quad \text {for}~~J_{j} \in {\mathcal {J}}_{y}. \end{aligned}$$

(13)

By inequalities (12) and (13),

$$\begin{aligned} z({\hat{\sigma }})\ge \sum _{J_{j} \in {\mathcal {J}}_{y}}E_{j}({\hat{\sigma }})> (m+1)T^{3}>K. \end{aligned}$$

This is a contradiction. Thus, we have the following result.

Result 1:

Exactly one job in ${\mathcal {J}}_{y}$ is processed in ${\hat{\pi }}_i$.

For simplicity, let $J_{2m+{\hat{a}}_{i}(3)}$ be that job. By equation (10), furthermore, the later the small job is processed, the better. Thus, since $J_{2m+{\hat{a}}_{i}(3)}$ has the smallest processing time in ${\hat{\pi }}_i$, it is processed last.

(ii) Suppose that two jobs $J_k$ and $J_l$ in ${\mathcal {J}}_{x}$ are processed in ${\hat{\pi }}_i$ for some $i\in \{1,2, \ldots ,m\}$. We assume that $J_l$ is processed later than $J_k$. Then, by inequalities (9) and (10), we have

$$\begin{aligned} E_{l}({\hat{\sigma }}) \ge E_{k}({\hat{\sigma }}) + (\Delta - p_l) \ge (\Delta - p_k) + (\Delta - p_l) > 2T^{2}. \end{aligned}$$

(14)

By inequality (9) and $p_{j}<M-T^{2}$ for $J_{j}\in {\mathcal {J}}_{x}$, we have

$$\begin{aligned} E_{j}({\hat{\sigma }})> T^{2}~~\text {for}~~J_{j} \in {\mathcal {J}}_{x}. \end{aligned}$$

(15)

By inequalities (14) and (15),

$$\begin{aligned} \sum _{J_{j} \in {\mathcal {J}}_{x}}E_{j}({\hat{\sigma }}) > (m+1)T^{2}. \end{aligned}$$

(16)

Also, by inequalities (9) and (10) and Result 1, we have

$$\begin{aligned} E_{2m+{\hat{a}}_{i}(3)}({\hat{\sigma }}) > n_i T^{2}+T^{3}~~\text {for}~~J_{2m+{\hat{a}}_{i}(3)} \in {\mathcal {J}}_{y}, \end{aligned}$$

(17)

where $n_i$ is the number of jobs of ${\mathcal {J}}_{x}$ in ${\hat{\pi }}_i$. By Result 1, and inequalities (16) and (17),

$$\begin{aligned} z({\hat{\sigma }}) \ge \sum _{J_{j} \in {\mathcal {J}}_{x}}E_{j}({\hat{\sigma }})+\sum _{i=1}^{m} E_{2m+{\hat{a}}_{i}(3)}({\hat{\sigma }})> (m+1)T^{2} + \sum _{i=1}^{m} (n_i T^{2} + T^{3})>K. \end{aligned}$$

This is a contradiction. Thus, we have the following result.

Result 2:

Exactly one job in ${\mathcal {J}}_{x}$ is processed in ${\hat{\pi }}_i$.

For simplicity, let $J_{m+{\hat{a}}_{i}(2)}$ be that job. Since $J_{2m+{\hat{a}}_{i}(3)}$ has the smallest processing time in ${\hat{\pi }}_i$, it is processed immediately before job $J_{2m+{\hat{a}}_{i}(3)}$.

(iii) Suppose that two jobs $J_k$ and $J_l$ in ${\mathcal {J}}_{w}$ are processed in ${\hat{\pi }}_i$ for some $i\in \{1,2, \ldots ,m\}$. We assume that $J_l$ is processed later than $J_k$. Then, by inequality (9) and (10), we have

$$\begin{aligned} E_{l}({\hat{\sigma }}) \ge E_{k}({\hat{\sigma }}) + (\Delta - p_l) \ge (\Delta - p_k) + (\Delta - p_l) > 2T. \end{aligned}$$

(18)

By inequality (9) and $p_{j}<M-T$ for $J_{j}\in {\mathcal {J}}_{w}$, we have

$$\begin{aligned} E_{j}({\hat{\sigma }})> T~~\text {for}~~J_{j} \in {\mathcal {J}}_{w}. \end{aligned}$$

(19)

By inequalities (18) and (19),

$$\begin{aligned} \sum _{J_{j} \in {\mathcal {J}}_{w}}E_{j}({\hat{\sigma }}) > (m+1)T. \end{aligned}$$

(20)

Also, by inequalities (9) and (10), and Results 1 and 2, we have

$$\begin{aligned} E_{m+{\hat{a}}_{i}(2)}({\hat{\sigma }}) > l_i T+T^{2},~~ \text {for}~~J_{m+{\hat{a}}_{i}(2)} \in {\mathcal {J}}_{x}, \end{aligned}$$

(21)

and

$$\begin{aligned} E_{2m+{\hat{a}}_{i}(3)}({\hat{\sigma }}) > l_i T+T^{2}+T^{3}, ~~ \text {for}~~J_{2m+{\hat{a}}_{i}(3)} \in {\mathcal {J}}_{y}, \end{aligned}$$

(22)

where $l_i$ is the number of jobs of ${\mathcal {J}}_{w}$ in ${\hat{\pi }}_i$. By Results 1 and 2, and inequalities (20)–(22),

$$\begin{aligned} z({\hat{\sigma }}) \ge \sum _{J_{j} \in {\mathcal {J}}_{w}}E_{j}({\hat{\sigma }})+\sum _{i=1}^{m} \big (E_{m+{\hat{a}}_{i}(2)}({\hat{\sigma }})+E_{2m+{\hat{a}}_{i}(3)}({\hat{\sigma }})\big ) \\ ~~~~~> (m+1)T + \sum _{i=1}^{m} (2l_i T + 2T^{2} + T^{3})>K. \end{aligned}$$

This is a contradiction. Thus, we have the following result.

Result 3:

Exactly one job in ${\mathcal {J}}_{w}$ is processed in ${\hat{\pi }}_i$.

For simplicity, let $J_{{\hat{a}}_{i}(1)}$ be that job. Thus, by Results 1–3, Claim 3 holds. $\square$