Unary NP-hardness of single-machine scheduling to minimize the total tardiness with deadlines

Abstract

We revisit the classical single-machine scheduling problem to minimize total tardiness with deadlines. The problem is binary NP-hard even without the deadline restrictions. It was reported early in Koulamas and Kyparisis (Eur J Oper Res 133:447–453, 2001) that the exact complexity (unary NP-hardness or pseudo-polynomial-time solvability) of the problem is still open. We show that this problem is unary NP-hard.

Appendix: Proof of equations (5) and (6)

We first recall that (i) $\sigma$ is a feasible schedule (subject to the deadlines), (ii) $J_0=J_{\sigma (3t+1)}$, and (iii) $\mathcal{F}^{(i)}_{\sigma }= \{J_{\sigma (1)}, J_{\sigma (2)},\ldots , J_{\sigma (3t)}\}\cap \mathcal{J}^{(i)}$, $i=1, 2, \ldots , t$. For convenience, we write $\mathcal{F}^{(i)}=\mathcal{F}^{(i)}_{\sigma }$ for $i=1,2, \ldots , t$. Then, we have $\mathcal{F}^{(1)}\cup \mathcal{F}^{(2)}\cup \cdots \cup \mathcal{F}^{(t)}= \{J_{\sigma (1)}, J_{\sigma (2)},\ldots , J_{\sigma (3t)}\}$.

Note that $J_0$ satisfies its deadline $\bar{d}_0$ in $\sigma$, and the first 3t normal jobs in $\mathcal{F}^{(1)}\cup \mathcal{F}^{(2)}\cup \cdots \cup \mathcal{F}^{(t)}$ are scheduled before $J_0$ in $\sigma$. Then, we have $p(\mathcal{F}^{(1)})+ p(\mathcal{F}^{(2)})+\cdots +p(\mathcal{F}^{(t)})+p_0\le \bar{d}_0 = L+P= p_0+3t\Delta _1+3\lambda _{t}\Delta _2+\lambda _{t}B$. Consequently,

$$\begin{aligned} p(\mathcal{F}^{(1)}){+} p(\mathcal{F}^{(2)}){+} \cdots {+} p(\mathcal{F}^{(t)}) {\le } 3t\Delta _1+ 3\lambda _{t}\Delta _2{+} \lambda _{t}B. \end{aligned}$$

For each $i\in \{1, 2, \ldots , t-1\}$, we have the following observations:

• All the jobs in $\{J_0\}\cup \mathcal{F}^{(1)}\cup \mathcal{F}^{(2)}\cup \cdots \cup \mathcal{F}^{(i)}$ are completed by the deadline $\bar{d}_0$ of $J_0$. (This statement holds since $\sigma$ is a feasible schedule, $J_0=J_{\sigma (3t+1)}$, and $\mathcal{F}^{(1)}\cup \mathcal{F}^{(2)}\cup \cdots \cup \mathcal{F}^{(i)} \subseteq \{J_{\sigma (1)}, J_{\sigma (2)},\ldots , J_{\sigma (3t)}\}$.)

• For each $i'\in \{i+1, i+2, \ldots , t\}$, all the jobs in $\mathcal{J}^{(i')}$ are completed by their common deadline $L+D^{(i')}$. (This statement holds since $\sigma$ is a feasible schedule.)

• $L+P=\bar{d}_0< L+ D^{(t)}< L+D^{(t-1)}< \cdots < L+D^{(i+1)}$. (This statement holds since, from (1), we have $P< D^{(t)}< D^{(t-1)}< \cdots < D^{(i+1)}$.)

From the above observations, we conclude that all the jobs in $\{J_0\}\cup \mathcal{F}^{(1)}\cup \mathcal{F}^{(2)}\cup \cdots \cup \mathcal{F}^{(i)}$ and $\mathcal{J}^{(t)}\cup \mathcal{J}^{(t-1)}\cup \cdots \cup \mathcal{J}^{(i+1)}$ are completed by time $L+D^{(i+1)}$ in $\sigma$. Then, we have that

$$\begin{aligned}&L+ p(\mathcal{F}^{(1)})+ p(\mathcal{F}^{(2)})+\cdots + p(\mathcal{F}^{(i)}) +p(\mathcal{J}^{(t)})\\&\qquad +\,p(\mathcal{J}^{(t-1)})+ \cdots + p(\mathcal{J}^{(i+1)})\\&\quad =\, p_0 + p(\mathcal{F}^{(1)})+ p(\mathcal{F}^{(2)})+ \cdots \\&\qquad +\,p(\mathcal{F}^{(i)})+ t(P^{(t)}+ P^{(t-1)}+\cdots +P^{(i+1)})\\&\quad \le L+D^{(i+1)}\\&\quad =\, L+ P^{(1)}+ P^{(2)}+ \cdots + P^{(i)}+ t(P^{(t)}\\&\qquad +\, P^{(t-1)}+\cdots +P^{(i+1)})\\&\quad = L+ 3i\Delta _1 +3\lambda _i\Delta _2 +\lambda _iB+ t(P^{(t)}\\&\qquad +\, P^{(t-1)}+\cdots +P^{(i+1)}). \end{aligned}$$

It follows that

$$\begin{aligned}&p(\mathcal{F}^{(1)})+ p(\mathcal{F}^{(2)})+ \cdots + p(\mathcal{F}^{(i)})\\&\quad \le 3i\Delta _1 +3\lambda _i\Delta _2 +\lambda _iB,~ i=1, 2, \ldots , t-1. \end{aligned}$$

This proves Eq. (5).

Since each normal job has a processing time larger than $\Delta _1$ and $3i\Delta _1+ 3\lambda _i\Delta _2+ \lambda _iB< (3i+1)\Delta _1$, from (5), we have

$$\begin{aligned} |\mathcal{F}^{(1)}|{+} |\mathcal{F}^{(2)}|{+} \cdots {+} |\mathcal{F}^{(i)}|{\le } 3i \text{ for } \text{ each } i=1, 2, \ldots , t.\nonumber \\ \end{aligned}$$

(21)

From (4), we have

$$\begin{aligned}&|\mathcal{F}^{(t)}|+ |\mathcal{F}^{(t-1)}|+ \cdots \nonumber \\&\quad +\,|\mathcal{F}^{(i+1)}|\ge 3(t-i) \text{ for } \text{ each } i=1, 2, \ldots , t-1. \end{aligned}$$

(22)

If there is some $x\in \{1, 2, \ldots , t-1\}$ such that $|\mathcal{F}^{(t)}|+|\mathcal{F}^{(t-1)}|+ \cdots + |\mathcal{F}^{(x+1)}|\ge 3(t-x)+1$, then we have

$$\begin{aligned}&|\mathcal{F}^{(1)}|+ 2|\mathcal{F}^{(2)}|+ \cdots + t|\mathcal{F}^{(t)}|\\&\quad =\, \sum _{i=0}^{t-1}(|\mathcal{F}^{(t)}|+\, |\mathcal{F}^{(t-1)}|+ \cdots + |\mathcal{F}^{(i+1)}|)\\&\quad \ge \sum _{i=0}^{t-1}3(t-i)+1= 3\lambda _{t} +1. \end{aligned}$$

Note that each normal job in $\mathcal{F}^{(i)}$ has a processing time larger than $\Delta _1 +i\Delta _2$ for $1\le i\le t$. Then, the completion time of $J_0$ can be estimated by the following way:

$$\begin{aligned}&p(\mathcal{F}^{(1)})+ p(\mathcal{F}^{(2)})+\cdots +p(\mathcal{F}^{(t)})+ p_0\\&\quad>\, |\mathcal{F}^{(1)}|(\Delta _1+ \Delta _2)+ |\mathcal{F}^{(2)}|(\Delta _1+ 2\Delta _2)+ \cdots \\&\qquad +\,|\mathcal{F}^{(t)}|(\Delta _1+ t\Delta _2)+ L\\&\quad = L+ 3t\Delta _1+ (|\mathcal{F}^{(1)}|+ 2|\mathcal{F}^{(2)}|\\&\qquad +\cdots +t|\mathcal{F}^{(t)}|)\Delta _2\\&\quad \ge L+ 3t\Delta _1 +3\lambda _{t}\Delta _2 +\Delta _2 >\, L+ P\\&\quad = \bar{d}_{0}. \end{aligned}$$

This contradicts the feasibility of $\sigma$. Consequently, $|\mathcal{F}^{(t)}|+ |\mathcal{F}^{(t-1)}|+ \cdots + |\mathcal{F}^{(i+1)}|= 3(t-i)$ for each i with $1\le i\le t-1$. This further implies that $|\mathcal{F}^{(i)}|= 3$ for each $i=1, 2, \ldots , t$. Equation (6) follows.