Near-linear-time approximation algorithms for scheduling a batch-processing machine with setups and job rejection

Abstract

In this paper we study a single batch-processing machine scheduling model. In our model, a set of jobs having different release dates needs to be scheduled onto a single machine that can process a batch of jobs simultaneously at a time. Each batch incurs a fixed setup time and a fixed setup cost. The decision maker may reject some of the jobs by paying penalty cost so as to achieve a tight makespan, but the total rejection penalty cost is required to be no greater than a given value. Our model extends several existing batch-processing machine scheduling models in the literature. We present efficient approximation algorithms with near-linear-time complexities.

Appendix

Proof of Theorem 1

Consider $\mathcal{A}(\sigma ^*)$, the set of jobs that are accepted in the optimal solution $\sigma ^*$. If $\mathcal{A}(\sigma ^*)=\emptyset$, then $z^*=(1-\alpha )w(\mathcal{J})=z_0$, which indicates $\mathbf{H}_1$ generates an optimal solution in this case (note that we must have $Q_0=R_0-w(\mathcal{J})\ge 0$ in this case). We only need to consider the case when $\mathcal{A}(\sigma ^*)\ne \emptyset$ in the following analysis.

Let $\beta \in \{1,2,\ldots ,n\}$ be the largest job index among all jobs in $\mathcal{A}(\sigma ^*)$. Then, all jobs in set $\mathcal{J}\setminus \{J_{1},\ldots ,J_\beta \}$ are rejected in $\sigma ^*$, and inequality $w(\mathcal{J})-\sum _{j=1}^\beta w_j\le R_0$ holds. Note that $W_i=w(\mathcal{J})-\sum _{j=1}^i w_j$ for $i=1,2,\ldots ,n$. Hence, we must have $Q_\beta =R_0- W_\beta =R_0-w(\mathcal{J})+\sum _{j=1}^\beta w_j\ge 0$. Observe that in $\sigma ^*$ the makespan of the accepted jobs is no less than $r_\beta +p_\beta$, and the total rejection penalty of the rejected jobs is no less than $W_\beta$. This indicates that

$$\begin{aligned} z^*\ge \alpha (r_\beta +p_\beta )+(1-\alpha )(c+W_\beta )=z_\beta . \end{aligned}$$

(12)

Let $\sigma _\eta$ denote the final solution determined in Step 3. By assumption, we have $1\le \eta \le n$. According to the definition of $\eta$, on the one hand, we must have $Q_\eta =R_0-W_\eta \ge 0$, which indicates that it is feasible if all jobs in $\mathcal{J}\setminus \{J_{1},\ldots ,J_\eta \}$ are rejected. On the other hand, we have

$$\begin{aligned}&\alpha (r_\eta +p_\eta )+(1-\alpha )(c+W_\eta )= z_\eta \nonumber \\&\quad =\min _{i=0,1,\ldots ,n}\{z_i\mid Q_i\ge 0\}\le z_\beta . \end{aligned}$$

(13)

Recall that $r_1+p_1\le r_2+p_2 \le \cdots \le r_n+p_n$. We thus have

$$\begin{aligned} \max _{i=1,2,\ldots ,\eta }\{p_i\}\le \max _{i=1,2,\ldots ,\eta }\{r_i+p_i\}=r_\eta +p_\eta \end{aligned}$$

(14)

and

$$\begin{aligned} \max _{i=1,2,\ldots ,\eta }\{r_i\}\le \max _{i=1,2,\ldots ,\eta }\{r_i+p_i\}=r_\eta +p_\eta . \end{aligned}$$

(15)

By the construction of $\sigma _\eta$, the unique job batch starts to be processed at time $\max _{1\le i \le \eta }\{r_i\}$, and thus, its value

$$\begin{aligned} z(\sigma _\eta )&=\alpha \left( \max _{i=1,2,\ldots ,\eta }\{r_i\}\right. \nonumber \\&\quad \left. +\max _{i=1,2,\ldots ,\eta }\{p_i\} \right) +(1-\alpha )(c+W_\eta ). \end{aligned}$$

(16)

By (16), (15), (14), (13) and (12), we have

$$\begin{aligned} z(\sigma _\eta )\le & {} \alpha [(r_\eta +p_\eta )+(r_\eta +p_\eta )]+(1-\alpha )(c+W_\eta ) \\= & {} z_\eta +\alpha (r_\eta +p_\eta )\le 2z_\eta \le 2z_\beta \le 2z^*. \end{aligned}$$

Hence, $\mathbf{H}_1$ is a 2-approximation to problem $\mathcal{P}$.

Consider the time complexity of $\mathbf{H}_1$. Step 1 takes $O(n\log n)$ time. Each of Step 2 and Step 3 only takes O(n) time. Hence, the time complexity of $\mathbf{H}_1$ is $O(n\log n)$.

Finally, we show that the bound is tight. Consider the following instance with 2 jobs: $(r_1, p_1,w_1) = (0, 2, 1)$, $(r_2, p_2,w_2) = (2, 0, 2)$, where $R_0=1$, $\alpha =0.5$ and $c=0$. Then, $r_1+p_1=r_2+p_2=2$, $w_1+w_2 = 3$, $Q_0=-2$, $Q_1=-1$ and $Q_2=1$. Hence, $\eta =2$, and the solution determined by $\mathbf{H}_1$ is as follows: Both $J_1$ and $J_2$ are accepted and processed as a batch starting from time 2. The value of such a solution is equal to $0.5\times (2+2)=2$. However, the optimal schedule is as follows: both $J_1$ and $J_2$ are accepted, $J_1$ starts its processing at time 0, and $J_2$ starts its processing at time 2 (i.e., there are two batches in the optimal solution, each batch contains exactly one job). The value of the optimal solution equals $0.5\times 2=1$. Hence, the bound is tight. This completes the proof. $\square$

Proof of Theorem 2

Consider $\mathcal{A}(\sigma ^*)$, the set of accepted jobs in $\sigma ^*$. If $\mathcal{A}(\sigma ^*)=\emptyset$, then the optimal solution value $z^*=z_0$, which can be obtained by rejecting all the jobs. We only need to consider the case when $\mathcal{A}(\sigma ^*)\ne \emptyset$. According to Step 1 of $\mathbf{H}_1$, we have $r_1\le r_2 \le \cdots \le r_n$ if $p_1=p_2=\cdots =p_n$, and $p_1\le p_2 \le \cdots \le p_n$ if $r_1=r_2=\cdots =r_n$. In either case, we let $\tau$ denote the largest job index among all jobs in $\mathcal{A}(\sigma ^*)$. Then, the optimal solution is as follows: All jobs in $\{J_j\in \mathcal{J}\mid j\le \tau \}$ are accepted and processed as a batch starting from time $r_\tau$, while all other jobs are rejected. The value of the optimal solution is equal to $\alpha (r_\tau +p_\tau )+(1-\alpha )(c+W_{\tau })= z_\tau$. Observe that in $\mathbf{H}_1$, we have $z_i=\alpha (r_i+p_i)+(1-\alpha )(c+W_{i})$ for $i=1,2,\ldots ,n$. As a result, $\eta =\arg \min _{i=1,2,\ldots ,n}\{z_i\mid Q_i\ge 0\} =\tau$ (due to optimality). Hence, $\mathbf{H}_1$ generates an optimal solution to $\mathcal{P}$ when either $p_1=p_2=\cdots =p_n$ or $r_1=r_2=\cdots =r_n$. $\square$

Proof of Lemma 1

Note that the LPT-rule holds for the classical unbounded parallel-batch single-machine scheduling problem $B\mid b= +\infty , r_j\mid C_{\max }$ (see Lee and Uzsoy 1999). Consider $\mathcal{A}(\sigma ^*)$, the set of accepted jobs in the optimal solution $\sigma ^*$. Clearly, in the schedule for the jobs in $\mathcal{A}(\sigma ^*)$, the fixed batch setup cost c does not affect the optimality of the LPT-rule. Hence, the result holds. $\square$

Proof of Lemma 2

Note that $s^*_1\le t_1\Delta$ and $s^*_2-s^*_1\le (t_2-t_1+1)\Delta$. As a result, we have

$$\begin{aligned} B^*_1\subseteq & {} \{J_j\in \mathcal{J}\mid r_j\le s^*_1, p_j\le s^*_2-s^*_1\}\\\subseteq & {} \{J_j\in \mathcal{J}\mid r_j\le t_1\Delta , p_j\le (t_2-t_1+1)\Delta \}=H_1. \end{aligned}$$

We only need to show that $\cup _{\tau =1}^i H_\tau \supseteq \cup _{\tau =1}^i B^*_\tau$ for $i=2,\ldots ,\rho$ if $\rho \ge 2$ in the following.

Suppose that inequality $\cup _{\tau =1}^{\rho '-1} H_\tau \supseteq \cup _{\tau =1}^{\rho '-1} B^*_\tau$ has been shown to be corrected for any $\rho '=2,\ldots ,\rho -1$, and now we would like to prove $\cup _{\tau =1}^{\rho '} H_\tau \supseteq \cup _{\tau =1}^{\rho '} B^*_\tau$. To prove this result, it is equivalent to prove $\cup _{\tau =1}^{\rho '} H_\tau \supseteq B^*_{\rho '}$. By definition, we have

$$\begin{aligned} H_{\rho '}= \{J_j\in \mathcal{J}\setminus (\cup _{\tau =1}^{\rho '-1} H_\tau ) \mid r_j\le t_{\rho '}\Delta , p_j\le (t_{\rho '+1}-t_{\rho '}+1)\Delta \} \end{aligned}$$

and

$$\begin{aligned} B^*_{\rho '}\subseteq \{J_j\in \mathcal{J}\mid r_j\le s^*_{\rho '}, p_j\le s^*_{\rho '+1}-s^*_{\rho '}\}. \end{aligned}$$

Note that $s^*_{\rho '}\le t_{\rho '}\Delta$ and $s^*_{\rho '+1}-s^*_{\rho '}\le (t_{\rho '+1}-t_{\rho '}+1)\Delta$. As a result,

$$\begin{aligned} \cup _{i=1}^{\rho '} H_i\supseteq & {} \{J_j\in \mathcal{J}\mid r_j\le t_{\rho '}\Delta , p_j\le (t_{\rho '+1}-t_{\rho '}+1)\Delta \}\\\supseteq & {} \{J_j\in \mathcal{J}\mid r_j\le s^*_{\rho '}, p_j\le s^*_{\rho '+1}-s^*_{\rho '}\}. \end{aligned}$$

We thus have $\cup _{\tau =1}^{\rho '} H_\tau \supseteq B^*_{\rho '}$. The result holds. $\square$

Proof of Lemma 3

By the construction of $\sigma _\rho$, the starting time of the first batch $H_1$ is no later than $t_1\Delta$, and the setup time plus the processing time of batch $H_1$ is no greater than $(t_2-t_1+1)\Delta$. As a result, the completion time of $H_1$ is no greater than $t_1\Delta +(t_2-t_1+1)\Delta =(t_2+1)\Delta$. Hence, the result holds when $\rho =1$. We consider the case when $\rho \ge 2$ in the following.

Suppose that for any i, $2\le i\le \rho$, the completion time of the $(i-1)$-th batch $H_{i-1}$ is no greater than $(t_{i}+i-1)\Delta$, and we would like to prove that the completion time of the i-th batch $H_i$ is no greater than $(t_{i+1}+i)\Delta$. Note that by (5), we have $r_j\le t_{i}\Delta \le (t_{i}+i-1)\Delta$ for any $J_j\in H_i$, which indicates that the starting time of $H_{i}$ is no later than $(t_{i}+i-1)\Delta$. By (4), the processing time of batch $H_{i+1}$ is no greater than $(t_{i+1}-t_{i}+1)\Delta$. As a result, the completion time of $H_{i}$ is no greater than $(t_{i}+i-1)\Delta +(t_{i+1}-t_{i}+1)\Delta =(t_{i+1}+i)\Delta$. This completes the proof. $\square$

Proof of Lemma 5

By definition, we have $H'_1=H_1$,

$$\begin{aligned} H'_2=\{J_j\in \mathcal{J}\setminus H'_1\mid r_j\le t_2\Delta , p_j\le (t_{\rho +1}-t_2+1)\Delta \}, \end{aligned}$$

and

$$\begin{aligned} H_2= \{J_j\in \mathcal{J}\setminus H_1\mid r_j\le t_2\Delta , p_j\le (t_3-t_2+1)\Delta \}. \end{aligned}$$

As $t_3\le t_{\rho +1}$, we have $(\cup _{\tau =1}^{2}H'_\tau )\supseteq H_{2}$. By Lemma 2, we have $(\cup _{\tau =1}^{2} H'_\tau )\supseteq (\cup _{\tau =1}^{2} H_\tau ) \supseteq (\cup _{\tau =1}^{2} B^*_\tau ).$ Note that $\mathcal{A}(\sigma ^*)=\cup _{i=1}^\rho B_i^*$. Hence, to prove the result, we only need to show $(\cup _{\tau =1}^{3}H'_\tau )\supseteq \cup _{i=3}^\rho B_i^*$. By definition again,

$$\begin{aligned} H'_3=\{J_j\in \mathcal{J}\setminus (\cup _{i=1}^{2} H'_i)\mid r_j\le t_{\rho +1}\Delta , p_j\le \frac{1}{3}t_{\rho +1}\Delta \}. \end{aligned}$$

On important observation is that for any $J_j\in \cup _{i=3}^\rho B_i^*$, we must have

$$\begin{aligned} r_j\le C_{\max }(\sigma ^*)\le t_{\rho +1}\Delta \end{aligned}$$

and

$$\begin{aligned} s+p_j\le \frac{1}{3}C_{\max }(\sigma ^*)\le \frac{1}{3}t_{\rho +1}\Delta \end{aligned}$$

(otherwise, according to the LPT-rule, the total processing time of the first 3 batches in $\sigma ^*$ has been greater than $C_{\max }(\sigma ^*)$, which is a contradiction). This indicates that

$$\begin{aligned} (\cup _{i=3}^\rho B_i^*)\subseteq & {} \{J_j\in \mathcal{J}\mid r_j\le t_{\rho +1}\Delta , p_j\le \frac{1}{3}t_{\rho +1}\Delta \}\\\subseteq & {} (\cup _{\tau =1}^{3}H'_\tau ). \end{aligned}$$

The result holds. $\square$

Proof of Lemma 6

Note that $H'_1=H_1$. By Lemma 3, the completion time of $H'_1$ is bounded by $(t_{2}+1)\Delta$. By definition of $H'_{2}$, on the one hand, $r_j\le t_{2}\Delta \le (t_{2}+1)\Delta$ for any $J_j\in H'_{2}$, which indicates that the starting time of $H'_{2}$ is no later than $(t_{2}+1)\Delta$. On the other hand, the processing time of $H'_{2}$ is no greater than $(t_{\rho +1}-t_{2}+1)\Delta$. Hence, the completion time of $H'_{2}$ is no greater than $(t_{2}+1)\Delta +(t_{\rho +1}-t_{2}+1)\Delta =(t_{\rho +1}+2)\Delta$.

By definition of $H'_{3}$, on the one hand, $r_j\le t_{\rho +1}\Delta \le (t_{\rho +1}+2)\Delta$ for any $J_j\in H'_{3}$, which indicates that the starting time of $H'_{3}$ is no later than $(t_{\rho +1}+2)\Delta$. On the other hand, the processing time of batch $H'_{3}$ is no greater than $\frac{1}{3}t_{\rho +1}\Delta$. As a result, $C_{\max }({\hat{\sigma }}_2)$ the completion time of $H'_{3}$, is no greater than $(t_{\rho +1}+2)\Delta +\frac{1}{3}t_{\rho +1}\Delta =\frac{4}{3}t_{\rho +1}\Delta +2\Delta$. Note that

$$\begin{aligned} (t_{\rho +1}-1)\Delta \le C_{\max }(\sigma ^*) \end{aligned}$$

and

$$\begin{aligned} \frac{10}{3}\Delta =\frac{10\epsilon \cdot Z_0}{21\alpha } \le \frac{10\epsilon \cdot (2z^*)}{21\alpha } \le \frac{z^*\epsilon }{\alpha }. \end{aligned}$$

Hence,

$$\begin{aligned} C_{\max }({\hat{\sigma }}_2)&\le \frac{4}{3}t_{\rho +1}\Delta +2\Delta \le \frac{4}{3}(C_{\max }(\sigma ^*)+\Delta )+2\Delta \\&= \frac{4}{3} C_{\max }(\sigma ^*)+ \frac{10}{3} \Delta \le \frac{4}{3} C_{\max }(\sigma ^*)+\frac{z^*\epsilon }{\alpha }. \end{aligned}$$

The result holds. $\square$

Proof of Lemma 8

After our method to combine jobs is applied, there are no more than $T-2$ jobs remaining in $\mathcal{S}$, and there are no more than $(T-2)\cdot \zeta$ jobs remaining in $\mathcal{L}$. Hence, $|\mathcal{J}'|=(T-2)\cdot \zeta +(T-2)$. To prove the result, we only need to show that $\zeta =O(T\log T)$. By definition,

$$\begin{aligned} \zeta \le \sum _{i=1}^{T-2}\left\lceil \frac{T}{i}\right\rceil \le \sum _{i=1}^{T-2}\left( 1+\frac{T}{i} \right) =T-2+T\sum _{i=1}^{T-2} \frac{1}{i}. \end{aligned}$$

According to the partial sums of the harmonic series, we have

$$\begin{aligned} \sum _{i=1}^{T-2} \frac{1}{i} = \ln (T-1) +\gamma , \end{aligned}$$

where $\gamma \approx 0.5772156649$ is the Euler - Mascheroni constant which is no greater than 1 (see Havil 2009). As a result,

$$\begin{aligned} \zeta\le & {} T-2+T\cdot [\ln (T-1) +1] = 2T-2+T\ln (T-1) \\= & {} O(T\ln T)=O(T\log T). \end{aligned}$$

The result holds. $\square$

Proof of Lemma 10

Let $J_\eta$ be any job in $B'_\mu \setminus \{J_{t_\mu }\}$. By definition, $p_{t_\mu }\ge p_\eta$. If $J_{t_\mu }\in \mathcal{S}$, then ${\tilde{p}}_{t_\mu }={\tilde{p}}_\eta =0$, the result holds. We only need to consider the case when $J_{t_\mu }\in \mathcal{L}$ in the following analysis. Assume that $J_{t_\mu }\in L_{i}^\kappa$ for some integers $i\in \{1,2,\ldots ,T-2\}$ and $\kappa \in \{1,2,\ldots ,\left\lceil T/i\right\rceil \}$. If job $J_\eta$ is also within set $L_{i}^\kappa$, then ${\tilde{p}}_{t_\mu }=\tilde{p}_\eta =i\Delta _1+(\kappa -1)\frac{i}{T}\Delta _1$, the result holds. If job $J_\eta$ is within set $L_{i}^{\kappa '}$ with $1\le \kappa '<\kappa$, then $\tilde{p}_\eta =i\Delta _1+(\kappa '-1)\frac{i}{T}\Delta _1< i\Delta _1+(\kappa -1)\frac{i}{T}\Delta _1={\tilde{p}}_{t_\mu }$, the result holds. If job $J_\eta$ is within set $L_{i'}$ with $1\le i'<i$, then ${\tilde{p}}_\eta \le p_\eta < (i'+1)\Delta _1 \le i\Delta _1\le {\tilde{p}}_{t_\mu }$, the result still holds. This completes the proof. $\square$

Proof of Lemma 11

According to our job partition method, for any $J_j\in \mathcal{J}$, we always have

$$\begin{aligned} p_j-{\tilde{p}}_j\le \Delta _1=\frac{\epsilon Z_0}{5\alpha } \le \frac{2\epsilon z^*}{5\alpha }. \end{aligned}$$

(17)

Hence, the result holds for the case when $\ell =1$. We only need to consider the case when $\ell >1$ in the following analysis.

Remember that $\{J_{t_1},\ldots ,J_{t_{\ell -1}}\}\subseteq \mathcal{L}$, and that $L_i=\{J_j\in \mathcal{L}\mid i\Delta _1< p_j\le (i+1)\Delta _1\}$ for $i=1,2,\ldots ,T-2$. Let $\Phi =\{\tilde{p}_{t_1},{\tilde{p}}_{t_2},\ldots ,{\tilde{p}}_{t_{\ell -1}}\}$. For $i=1,2,\ldots ,T-2$, let $\varphi _i$ be the set of distinct scaled processing times of the jobs in $L_i$, and let $v_i=|\varphi _i\cap \Phi |$ be the number of distinct scaled processing times of the jobs in $L_i$ that are included within $\Phi$. According to our job partition method, we must have

$$\begin{aligned} \sum _{\mu =1}^{\ell -1} {\tilde{p}}_{t_\mu }\ge \sum _{i=1}^{T-2}v_i\cdot i\Delta _1 \end{aligned}$$

(18)

and

$$\begin{aligned} \sum _{\mu =1}^{\ell -1} p_{t_\mu }\le \sum _{i=1}^{T-2}v_i\cdot (i+1)\Delta _1. \end{aligned}$$

(19)

Furthermore, for any $J_j\in L_i$, we have

$$\begin{aligned} p_j-{\tilde{p}}_j\le \frac{i}{T}\Delta _1. \end{aligned}$$

(20)

As a result, by (18), and (19) and (20),

$$\begin{aligned} \sum _{\mu =1}^{\ell -1} (p_{t_\mu }-{\tilde{p}}_{t_\mu })&\le \sum _{i=1}^{T-2}v_i\cdot (\frac{i}{T}\Delta _1) =\frac{1}{T}\sum _{i=1}^{T-2}v_i\cdot i \Delta _1 \nonumber \\&\le \frac{1}{T}\sum _{\mu =1}^\ell {\tilde{p}}_{t_\mu }. \end{aligned}$$

(21)

By Lemma 9 and equality $C_{\max }({\tilde{\sigma }}) =C_{\max }(\pi ^*)$, we have

$$\begin{aligned} \sum _{\mu =1}^{\ell } {\tilde{p}}_{t_\mu }\le C_{\max }({\tilde{\sigma }}) =C_{\max }(\pi ^*)\le \frac{1}{\alpha }z(\pi ^*)\le \frac{1}{\alpha }z^*. \end{aligned}$$

(22)

Hence, by (17),  (21) and (22), we have

$$\begin{aligned} \sum _{\mu =1}^{\ell } (p_{t_\mu }-{\tilde{p}}_{t_\mu })&= (p_{t_\ell }-\tilde{p}_{t_\ell })+\sum _{\mu =1}^{\ell -1} (p_{t_\mu }-{\tilde{p}}_{t_\mu })\nonumber \\&\le \frac{2\epsilon z^*}{5\alpha }+\frac{z^*}{T\alpha } =\frac{3\epsilon }{5\alpha }z^*. \end{aligned}$$

(23)

This completes the proof. $\square$

Proof of Lemma 12

Based on the schedule of ${\tilde{\sigma }}$, we first replace ${\tilde{p}}_j$ by $p_j$ for $j=1,2,\ldots ,n$, and let batches $B'_1,\ldots ,B'_\ell$ start processing as early as possible following this order. After doing so, by Lemma 11, the makespan of the accepted jobs increases by no more than $\frac{3\epsilon z^*}{5\alpha }$. Then, we further replace $\tilde{r}_j$ by $r_j$ for $j=1,2,\ldots ,n$, and let batches $B'_1,\ldots ,B'_\ell$ start processing as early as possible following this order. After doing so, the makespan of the accepted jobs further increases by no more than $\Delta _1=\frac{\epsilon Z_0}{5\alpha }\le \frac{2\epsilon z^*}{5\alpha }$. Hence, the total increment of the makespan of the accepted jobs in $\sigma _f$ is bounded by $\frac{3\epsilon z^*}{5\alpha }+ \frac{2\epsilon z^*}{5\alpha } =\frac{\epsilon z^*}{\alpha }$. This completes the proof. $\square$