A best possible online algorithm for parallel batch scheduling with delivery times and limited restart

Abstract

We consider the online scheduling on an unbounded (drop-line) parallel batch machine to minimize the time by which all jobs have been delivered. In this paper, all jobs arrive over time and the running batches are allowed limited restart. Here limited restart means that a running batch which contains restarted jobs cannot be restarted again. A drop-line parallel batch machine can process several jobs simultaneously as a batch, and all jobs in a batch start at the same time, and the completion time of a job equals the sum of its starting time and its processing time. Here we consider the restricted model: all jobs have agreeable processing times and delivery times. We provide a best possible online algorithm H with a competitive ratio of $\frac{3}{2}$ for the problem on an unbounded parallel batch machine and the corresponding problem on an unbounded drop-line parallel batch machine, respectively.

Appendix

Proof of Lemma 2.4

Because $S_n<S_{n-1}+p(B_{n-1})$, $B_{n}$ is a restricted batch. Let $p'$ be the longest processing time of jobs arriving in the time interval $(S_{n-1},S_{n}]$ and $q''$ be the largest delivery time of jobs arriving in the time interval $(S_{n-1},S_{n}]$. Among the jobs which arrive in the time interval $(S_{n-1},S_{n}]$ and have the processing time $p'$, select one which arrives latest as the job $J'$. Let $r'$ and $q'$ be the arrival time and the delivery time of $J'$, respectively. Among the jobs which arrive in the time interval $(S_{n-1},S_{n}]$ and have the delivery time $q''$, select one which arrives latest as the job $J''$. Let $r''$ and $q''$ be the arrival time and the delivery time of $J''$, respectively. Thus we have $p'\ge p''$ and $q'\le q''$. Considering that both $J'$ and $J''$ arrive after the time $S_{n-1}$, $p'\ge p''$ and $q'\le q''$, then by Lemma 2.1, we have $L_{{\text{ opt }}}\ge S_{n-1}+p'+q''$. Since $B_{n}$ is a restricted batch, by Step 4.2 and 4.3 of Algorithm H, we have $p(B_{n-1})< p'<\frac{3}{2}p(B_{n-1})$ or $\frac{3}{4}p(B_{n-1})< p'\le p(B_{n-1})$. Note that $r'>S_{n-1}\ge \frac{1}{2}p(B_{n-1})$ by Observation 2.3.

Case 1.$p(B_{n-1})< p'< \frac{3}{2}p(B_{n-1})$.

By Step 4.2 of Algorithm H, we have $p(B_{n})=\max \{p(B_{n-1}), p'\}=p'$, $S_{n}=\max \{p',r'\}$ and $q(B_{n})=\max \{q'',q(B_{n-1})\}$.

As $p(B_{n-1})<p'$, considering that all jobs have agreeable processing times and delivery times, then the largest delivery time of jobs in $B_{n-1}$ is not more than $q'$. Thus $q(B_{n-1})\le q'\le q''$ and $q(B_{n})=\max \{q'',q(B_{n-1})\}=q''$. Now $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=S_{n}+p'+q''$. Now we distinguish the following cases.

If $r'>p'$, then $S_{n}=r'$ and $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=r'+p'+q''$. Note that $q'\le q''$. If $q'=q''$, then $L_{{\text{ opt }}}\ge r'+p'+q'=r'+p'+q''=L_{{\text{ on }}}$, which implies $L_{{\text{ on }}}=L_{{\text{ opt }}}$. If $q'<q''$, noting that all jobs have agreeable processing times and delivery times, then $p'\le p''$. Also considering that $p'\ge p''$, we have $p'=p''$. By the definitions of $J'$ and $J''$ and $p'=p''$, we have $r'\ge r''$. Since $B_{n}$ interrupts $B_{n-1}$ at time $S_{n}=r'$ and $r'>p'$, then by Step 4.2 of Algorithm H, we have $r'<\frac{5}{4}p(B_{n-1})$. Then $r''\le r'<\frac{5}{4}p(B_{n-1})$. If $r''<r'$ and $r''\ge p''$, considering that $p'=p''$ and $r''<\frac{5}{4}p(B_{n-1})$, then by Step 4.2 of Algorithm H, $B_{n}$ will interrupt $B_{n-1}$ at time $r''$, a contradiction to $S_{n}=r'$ and $r''<r'$. If $r''<r'$ and $r''<p''$, considering that $p''=p'$, $r'>p'$ and $p'$ is the longest processing time of jobs arriving in the time interval $(S_{n-1},S_{n}]=(S_{n-1},r']$, then by Step 4.2 of Algorithm H, $B_{n}$ will interrupt $B_{n-1}$ at time $p''=p'$, a contradiction to the fact that $S_{n}=r'$ and $r'>p'$. So the assumption of $r''<r'$ is not right and we have $r''=r'$. So if $q'<q''$, we have $p'=p''$ and $r''=r'$. Then $L_{{\text{ opt }}}\ge r''+p''+q''=r'+p'+q''=L_{{\text{ on }}}$, which implies $L_{{\text{ on }}}=L_{{\text{ opt }}}$.

If $r'\le p'$, then $S_{n}=p'$ and $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=2p'+q''$. Thus $\frac{L_{{\text{ on }}}}{L_{{\text{ opt }}}}\le \frac{2p'+q''}{S_{n-1}+p'+q''}\le \frac{2p'+q''}{\frac{1}{2}p(B_{n-1})+p'+q''}\le \frac{2p'}{\frac{1}{2}p(B_{n-1})+p'}\le \frac{2\times \frac{3}{2}p(B_{n-1})}{\frac{3}{2}p(B_{n-1})+\frac{1}{2}p(B_{n-1})}=\frac{3}{2}$.

Case 2.$\frac{3}{4}p(B_{n-1})< p'\le p(B_{n-1})$.

By Step 4.3 of Algorithm H, we have $p(B_{n})=\max \{p',p(B_{n-1})\}=p(B_{n-1})$, $r'<\frac{5}{4}p(B_{n-1})$ and $S_{n}=\max \{2p(B_{n-1})-p',r'\}$. Note that $S_{n}< \frac{5}{4}p(B_{n-1})$ as $\frac{3}{4}p(B_{n-1})< p'$ and $r'<\frac{5}{4}p(B_{n-1})$. Then $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=S_{n}+p(B_{n-1})+q(B_{n})<\frac{9}{4}p(B_{n-1})+q(B_{n})$.

Subcase 2.1$q''>q(B_{n-1})$.

Then $q(B_{n})=\max \{q(B_{n-1}),q''\}=q''$. Now $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=S_{n}+p(B_{n-1})+q''<\frac{9}{4}p(B_{n-1})+q''$.

Note that $q_{n-1}^{p}$ is the delivery time of $J_{n-1}^{p}$ and $p(B_{n-1})$ is the processing time of $J_{n-1}^{p}$. Considering that $q''>q(B_{n-1})\ge q_{n-1}^{p}$, we have $p''\ge p(B_{n-1})$. Considering that $p''\le p'$, $p'\le p(B_{n-1})$ and $p''\ge p(B_{n-1})$, we have $p''=p'=p(B_{n-1})$. Then $L_{{\text{ opt }}}\ge S_{n-1}+p'+q''\ge \frac{1}{2}p(B_{n-1})+p(B_{n-1})+q''=\frac{3}{2}p(B_{n-1})+q''$ by Observation 2.3. Thus $\frac{L_{{\text{ on }}}}{L_{{\text{ opt }}}}\le \frac{\frac{9}{4}p(B_{n-1})+q''}{\frac{3}{2}p(B_{n-1})+q''}\le \frac{3}{2}$.

Subcase 2.2$q''\le q(B_{n-1})$.

Then $q(B_{n})=\max \{q(B_{n-1}),q''\}=q(B_{n-1})$. Now $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=S_{n}+p(B_{n-1})+q(B_{n-1})$. Since $B_{n}$ interrupts $B_{n-1}$, we have $p'\ge \frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})$ by Step 4.3 of Algorithm H. As $p'\ge \frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})$ and $p'\le p(B_{n-1})$, we have $q(B_{n-1})\le p(B_{n-1})$.

By Lemma 2.2, for the batch $B_{n-1}$, at least one of the jobs $J_{n-1}^{p}$ and $J_{n-1}^{q}$ has the processing time $p(B_{n-1})$ and the delivery time $q(B_{n-1})$ and denote the job by $J_{n-1}^{*}$.

If $S_{n}=2p(B_{n-1})-p'$, then $L_{{\text{ on }}}=S_{n}+p(B_{n-1})+q(B_{n-1})=2p(B_{n-1})-p'+p(B_{n-1})+q(B_{n-1})\le 2p(B_{n-1})-(\frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1}))+p(B_{n-1})+q(B_{n-1})= \frac{9}{4}p(B_{n-1})+\frac{3}{4}q(B_{n-1})$. Note that $J'$ arrives after time $S_{n-1}$, and the processing time of $J_{n-1}^{*}$ is $p(B_{n-1})$, and the delivery time of $J_{n-1}^{*}$ is $q(B_{n-1})$. Thus $L_{{\text{ opt }}}\ge \min \{S_{n-1}+p(B_{n-1})+q(B_{n-1}),p(B_{n-1})+p'\}$ whether $J'$ and $J_{n-1}^{*}$ are in a common batch in the optimal off-line schedule $\pi$ or not. Then $L_{{\text{ opt }}}\ge \min \{S_{n-1}+p(B_{n-1})+q(B_{n-1}),p(B_{n-1})+p'\}\ge \min \{\frac{1}{2}p(B_{n-1})+p(B_{n-1})+q(B_{n-1}),p(B_{n-1})+\frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})\}\ge \min \{\frac{3}{2}p(B_{n-1})+q(B_{n-1}),p(B_{n-1})+\frac{2}{4}p(B_{n-1})+\frac{2}{4}q(B_{n-1})\}=\frac{3}{2}p(B_{n-1})+\frac{1}{2}q(B_{n-1})$. Here the last but one inequality hods as $p(B_{n-1})\ge q(B_{n-1})$. Thus $\frac{L_{{\text{ on }}}}{L_{{\text{ opt }}}}\le \frac{\frac{9}{4}p(B_{n-1})+\frac{3}{4}q(B_{n-1})}{\frac{3}{2}p(B_{n-1})+\frac{1}{2}q(B_{n-1})}=\frac{3}{2}$.

If $S_{n}=r'$, then $r'\ge 2p(B_{n-1})-p'$ and $L_{{\text{ opt }}}\ge r'+p'\ge 2p(B_{n-1})$. Now $L_{{\text{ on }}}=S_{n}+p(B_{n-1})+q(B_{n-1})=r'+p(B_{n-1})+q(B_{n-1})$. Considering that $p'\ge \frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})$ and $q(B_{n-1})\le p(B_{n-1})$, we have $\frac{L_{{\text{ on }}}}{L_{{\text{ opt }}}}\le \frac{r'+p(B_{n-1})+q(B_{n-1})}{r'+p'}=1+\frac{p(B_{n-1})+q(B_{n-1})-p'}{r'+p'}\le 1+\frac{\frac{1}{4}p(B_{n-1})+\frac{3}{4}q(B_{n-1})}{2p(B_{n-1})}\le 1+\frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{3}{2}$. The result follows. $\square$

Proof of Lemma 2.5

As $S_n=S_{n-1}+p(B_{n-1})$, then $B_{n}$ is a free batch and all jobs in $B_{n}$ arrive after the time $S_{n-1}$. Now $L_{{\text{ on }}}=S_{n}+p(B_{n})+q(B_{n})=S_{n-1}+p(B_{n-1})+p(B_{n})+q(B_{n})$. By Lemma 2.2, $L_{{\text{ opt }}}\ge r(B_{n})+p(B_n)+q(B_n)>S_{n-1}+p(B_n)+q(B_n)$. Thus

$$\begin{aligned} L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1}). \end{aligned}$$

(1)

Let $p'$ be the longest processing time of jobs arriving in the time interval $(S_{n-2},S_{n-1}]$. Among the jobs which arrive in the time interval $(S_{n-2},S_{n-1}]$ and have a processing time $p'$, select one which arrives latest as the job $J'$. Let $r'$ and $q'$ be the arrival time and the delivery time of $J'$, respectively. Considering that $B_{n-1}$ is a restricted batch, then by Algorithm H, we have $p(B_{n-2})< p'<\frac{3}{2}p(B_{n-2})$ or $\frac{3}{4}p(B_{n-2})<p'\le p(B_{n-2})$. By Lemma 2.2, at least one of the jobs $J_{n}^{p}$ and $J_{n}^{q}$ has the processing time of $p(B_{n})$ and the delivery time of $q(B_{n})$. Let $J_{n}^{*}$ be one of $J_{n}^{p}$ and $J_{n}^{q}$ whose processing time is $p(B_{n})$ and delivery time is $q(B_{n})$.

Case 1.$p(B_{n-2})< p'<\frac{3}{2}p(B_{n-2})$.

Then $p(B_{n-1})=\max \{p(B_{n-2}),p'\}=p'$. By Step 4.2 of Algorithm H, $S_{n-1}=\max \{p',r'\}$.

If $r'\ge p'$, we have $S_{n-1}=r'$ and $L_{{\text{ opt }}}\ge r'+p'\ge 2p'=2p(B_{n-1})$. Then by the inequality (1), we have $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{1}{2}$.

If $r'<p'$, we have $S_{n-1}=p'=p(B_{n-1})$. If $J'$ starts not before $J_{n}^{*}$ in the optimal off-line schedule $\pi$, considering that $J_{n}^{*}$ arrives after the time $S_{n-1}$, then $L_{{\text{ opt }}}\ge S_{n-1}+p'=2p(B_{n-1})$. Thus we have $L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1})\le \frac{1}{2}L_{{\text{ opt }}}$. If $J'$ starts before $J_{n}^{*}$ in the schedule $\pi$, then $L_{{\text{ opt }}}\ge r'+p'+p(B_{n})+q(B_{n})>S_{n-2}+p(B_{n-1})+p(B_{n})+q(B_{n})$. Considering that $\frac{1}{3}p(B_{n-1})=\frac{1}{3}p'<\frac{1}{2}p(B_{n-2})$ and $S_{n-2}\ge \frac{1}{2}p(B_{n-2})$, we have $S_{n-2}\ge \frac{1}{3}p(B_{n-1})$. Then $L_{{\text{ opt }}}>S_{n-2}+p(B_{n-1})+p(B_{n})+q(B_{n})\ge \frac{1}{3}p(B_{n-1})+p(B_{n-1})=\frac{4}{3}p(B_{n-1})$ and $L_{{\text{ on }}}-L_{{\text{ opt }}}\le S_{n-1}+p(B_{n-1})+p(B_{n})+q(B_{n})-(S_{n-2}+p(B_{n-1})+p(B_{n})+q(B_{n}))=S_{n-1}-S_{n-2}\le p(B_{n-1})-\frac{1}{3}p(B_{n-1})=\frac{2}{3}p(B_{n-1})$. Thus $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{\frac{2}{3}p(B_{n-1})}{\frac{4}{3}p(B_{n-1})}=\frac{1}{2}$.

Case 2.$\frac{3}{4}p(B_{n-2})< p'\le p(B_{n-2})$.

Then $p(B_{n-1})=\max \{p(B_{n-2}),p'\}=p(B_{n-2})$. By Step 4.3 of Algorithm H, we have $S_{n-1}=\max \{2p(B_{n-2})-p',r'\}$. Then $L_{{\text{ on }}}=S_{n-1}+p(B_{n-1})+p(B_{n})+q(B_{n})=S_{n-1}+p(B_{n-2})+p(B_{n})+q(B_{n})$.

If $r'\ge 2p(B_{n-2})-p'$, then $L_{{\text{ opt }}}\ge r'+p'\ge 2p(B_{n-2})=2p(B_{n-1})$. Thus $L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1})\le \frac{1}{2}L_{{\text{ opt }}}$ by the inequality (1).

If $r'<2p(B_{n-2})-p'$, then $S_{n-1}=2p(B_{n-2})-p'<\frac{5}{4}p(B_{n-2})$. Now we distinguish the following cases. If $J_{n-2}^{p}$ or $J'$ starts not before $J_{n}^{*}$ in the schedule $\pi$, considering that $J_{n}^{*}$ arrives after the time $S_{n-1}$, then $L_{{\text{ opt }}}\ge S_{n-1}+\min \{p(B_{n-2}),p'\}=S_{n-1}+p'=2p(B_{n-2})=2p(B_{n-1})$. Thus $L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1})\le \frac{1}{2}L_{{\text{ opt }}}$. If both $J_{n-2}^{p}$ and $J'$ start before $J_{n}^{*}$ in the schedule $\pi$, then $J_{n}^{*}$ starts not earlier than the time $\min \{p(B_{n-2})+p',r'+p(B_{n-2})\}$ in the schedule $\pi$ whether $J_{n-2}^{p}$ and $J'$ are in a common batch or not in the schedule $\pi$. Note that $\min \{p(B_{n-2})+p',r'+p(B_{n-2})\}\ge \min \{p(B_{n-2})+\frac{3}{4}p(B_{n-2}),S_{n-2}+p(B_{n-2})\}\ge \min \{p(B_{n-2})+\frac{3}{4}p(B_{n-2}),\frac{1}{2}p(B_{n-2})+p(B_{n-2})\}= \frac{3}{2}p(B_{n-2})$. Then $J_{n}^{*}$ starts not earlier than the time $\frac{3}{2}p(B_{n-2})$ in the schedule $\pi$ and $L_{{\text{ opt }}}\ge \frac{3}{2}p(B_{n-2})+p(B_{n})+q(B_{n})$. Thus $\frac{L_{{\text{ on }}}}{L_{{\text{ opt }}}}\le \frac{S_{n-1}+p(B_{n-2})+p(B_{n})+q(B_{n})}{\frac{3}{2}p(B_{n-2})+p(B_{n})+q(B_{n})}\le \frac{\frac{5}{4}p(B_{n-2})+p(B_{n-2})+p(B_{n})+q(B_{n})}{\frac{3}{2}p(B_{n-2})+p(B_{n})+q(B_{n})}\le \frac{3}{2}$. The result follows. $\square$

Proof of Lemma 2.6

Since $S_n=S_{n-1}+p(B_{n-1})$, then $B_{n}$ is a free batch, and $r(B_{n})>S_{n-1}$, and $L_{{\text{ on }}}=S_{n-1}+p(B_{n-1})+p(B_{n})+q(B_{n})$. By Lemma 2.2, $L_{{\text{ opt }}}\ge r(B_{n})+p(B_{n})+q(B_{n})>S_{n-1}+p(B_{n})+q(B_{n})$. Then

$$\begin{aligned} L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1}). \end{aligned}$$

(2)

Case 1.$p(B_{n})\ge \frac{3}{2}p(B_{n-1})$.

Note that $L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1})$ and $S_{n-1}\ge \frac{1}{2}p(B_{n-1})$ by Observation 2.3. Then $L_{{\text{ opt }}}>S_{n-1}+p(B_{n})+q(B_{n})\ge \frac{1}{2}p(B_{n-1})+\frac{3}{2}p(B_{n-1})=2p(B_{n-1})$. Thus $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{1}{2}$.

Case 2.$p(B_{n-1})< p(B_{n})< \frac{3}{2}p(B_{n-1})$.

Note that the arrival time of $J_{n}^{p}$ is $r_{n}^{p}$ and the processing time of $J_{n}^{p}$ is $p(B_{n})$. Now we prove that $r_{n}^{p}\ge p(B_{n-1})$. If $r_{n}^{p}<p(B_{n-1})$, noting that $r_{n}^{p}<p(B_{n-1})<p(B_{n})$, then $B_{n}$ will interrupt $B_{n-1}$ at time $p(B_{n})$ by Step 4.2 of Algorithm H, a contradiction. So we have $r_{n}^{p}\ge p(B_{n-1})$. Then $L_{{\text{ opt }}}\ge r_{n}^{p}+p(B_{n})\ge 2p(B_{n-1})$. Thus by the inequality (2), $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{1}{2}$.

Case 3.$p(B_{n})\le p(B_{n-1})$.

If $r_{n}^{p}+p(B_{n})\ge 2p(B_{n-1})$, we have $L_{{\text{ opt }}}\ge r_{n}^{p}+p(B_{n})\ge 2p(B_{n-1})$. Then $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{1}{2}$. Now we assume that $r_{n}^{p}+p(B_{n})<2p(B_{n-1})$, i.e., $r_{n}^{p}< 2p(B_{n-1})-p(B_{n})$.

Subcase 3.1.$p(B_{n})=p(B_{n-1})$.

If $p(B_{n-1})>q(B_{n-1})$, then $\frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})<p(B_{n-1})=p(B_{n})$. Considering that $r_{n}^{p}<2p(B_{n-1})-p(B_{n})=p(B_{n-1})=p(B_{n})$ and $p(B_{n})>\frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})$, then $B_{n}$ will interrupt $B_{n-1}$ at time $2p(B_{n-1})-p(B_{n})=p(B_{n-1})$ by Step 4.3 of Algorithm H, a contradiction. So we have $p(B_{n-1})\le q(B_{n-1})$. By Lemma 2.2, $L_{{\text{ opt }}}\ge p(B_{n-1})+q(B_{n-1})\ge 2p(B_{n-1})$. Then by the inequality (2), we have $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{1}{2}$.

Subcase 3.2.$p(B_{n})<p(B_{n-1})$.

By Lemma 2.2, for the batch $B_{n-1}$, at least one of the jobs $J_{n-1}^{p}$ and $J_{n-1}^{q}$ has the processing time of $p(B_{n-1})$ and the delivery time of $q(B_{n-1})$. So let $J_{n-1}^{*}$ be one of the jobs $J_{n-1}^{p}$ and $J_{n-1}^{q}$ whose processing time is $p(B_{n-1})$ and delivery time is $q(B_{n-1})$. Similarly, let $J_{n}^{*}$ be one of the jobs $J_{n}^{p}$ and $J_{n}^{q}$ whose processing time is $p(B_{n})$ and delivery time is $q(B_{n})$.

If $p(B_{n-1})\le q(B_{n-1})$, by Lemma 2.2, we have $L_{{\text{ opt }}}\ge p(B_{n-1})+q(B_{n-1})\ge 2p(B_{n-1})$. Then $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{2p(B_{n-1})}=\frac{1}{2}$. Now we assume that $p(B_{n-1})>q(B_{n-1})$.

If $\frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})<p(B_{n})<p(B_{n-1})$, considering that $r_{n}^{p}< 2p(B_{n-1})-p(B_{n})<\frac{5}{4}p(B_{n-1})$, then $B_{n}$ will interrupt $B_{n-1}$ by Step 4.3 of Algorithm H, a contradiction. So we have $p(B_{n})\le \frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})$.

If $J_{n-1}^{*}$ starts not earlier than the time $S_{n-1}$ in the schedule $\pi$, then $L_{{\text{ opt }}}\ge S_{n-1}+p(B_{n-1})+q(B_{n-1})$. Note that the processing time of $J_{n}^{q}$ is $p_{n}^{q}$ and the processing time of $J_{n-1}^{p}$ is $p(B_{n-1})$. Then $p_{n}^{q}\le p(B_{n})<p(B_{n-1})$. Considering that $p_{n}^{q}<p(B_{n-1})$, then the delivery time of $J_{n}^{q}$ is not more than the delivery time of $J_{n-1}^{p}$, i.e., $q(B_{n})\le q_{n-1}^{p}$. Then $q(B_{n})\le q_{n-1}^{p}\le q(B_{n-1})$. Thus $L_{{\text{ on }}}-L_{{\text{ opt }}}\le S_{n-1}+p(B_{n-1})+p(B_{n})+q(B_{n})-(S_{n-1}+p(B_{n-1})+q(B_{n-1}))\le p(B_{n})\le \frac{3}{4}p(B_{n-1})+\frac{1}{4}q(B_{n-1})$. Note that $L_{{\text{ opt }}}\ge S_{n-1}+p(B_{n-1})+q(B_{n-1})\ge \frac{1}{2}p(B_{n-1})+p(B_{n-1})+q(B_{n-1})$. Then $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{1}{2}$.

Now we assume that $J_{n-1}^{*}$ starts earlier than the time $S_{n-1}$ in the schedule $\pi$. Considering that all jobs in $B_{n}$ arrive after the time $S_{n-1}$, then $J_{n-1}^{*}$ starts before $J_{n}^{*}$ in the schedule $\pi$. Then $L_{{\text{ opt }}}\ge r(B_{n-1})+p(B_{n-1})+p(B_{n})+q(B_{n})$.

As $B_{n-1}$ is a free batch, by Observation 2.1, we have $S_{n-1}=\frac{1}{2}p(B_{n-1})$, or $S_{n-1}=r(B_{n-1})$, or $S_{n-1}=S_{n-2}+p(B_{n-2})$.

If $S_{n-1}=r(B_{n-1})$ or $S_{n-1}=\frac{1}{2}p(B_{n-1})$, then $L_{{\text{ on }}}-L_{{\text{ opt }}}\le S_{n-1}+p(B_{n-1})+p(B_{n})+q(B_{n})-(r(B_{n-1})+p(B_{n-1})+p(B_{n})+q(B_{n}))=S_{n-1}-r(B_{n-1})\le \frac{1}{2}p(B_{n-1})$ and $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{1}{2}$.

From now on we assume that $S_{n-1}=S_{n-2}+p(B_{n-2})$. So $L_{{\text{ on }}}=S_{n-2}+p(B_{n-2})+p(B_{n-1})+p(B_{n})+q(B_{n})$. Since $B_{n-1}$ is a free batch, we have $r(B_{n-1})>S_{n-2}$. Thus $L_{{\text{ opt }}}\ge r(B_{n-1})+p(B_{n-1})+p(B_{n})+q(B_{n})>S_{n-2}+p(B_{n-1})+p(B_{n})+q(B_{n})$ and $L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-2})$. Noting that $L_{{\text{ on }}}-L_{{\text{ opt }}}\le p(B_{n-1})$, then

$$\begin{aligned} L_{{\text{ on }}}-L_{{\text{ opt }}}\le \min \{p(B_{n-1}),p(B_{n-2})\}. \end{aligned}$$

(3)

By Lemma 2.2, one job of the batch $B_{n-2}$ has the processing time of $p(B_{n-2})$ and the delivery time of $q(B_{n-2})$. Let $J_{n-2}^{*}$ be such a job in $B_{n-2}$ with the processing time $p(B_{n-2})$ and the delivery time $q(B_{n-2})$.

If $B_{n-2}$ is a restricted batch, then we have $S_{n-2}\ge p(B_{n-2})$ by Observation 2.2. Thus $L_{{\text{ opt }}}\ge r(B_{n})>S_{n-1}=S_{n-2}+p(B_{n-2})\ge 2p(B_{n-2})$. Hence $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}} \le \frac{\min \{p(B_{n-1}),p(B_{n-2})\}}{2p(B_{n-2})}\le \frac{1}{2}$ by the inequality (3).

If $B_{n-2}$ is a free batch, we distinguish the following cases. If $p(B_{n-1})\ge \frac{3}{2}p(B_{n-2})$, we have $L_{{\text{ opt }}}\ge r(B_{n-1})+p(B_{n-1})>S_{n-2}+p(B_{n-1})\ge \frac{1}{2}p(B_{n-2})+\frac{3}{2}p(B_{n-2})=2p(B_{n-2})$. Thus $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{\min \{p(B_{n-1}),p(B_{n-2})\}}{2p(B_{n-2})}\le \frac{1}{2}$. If $p(B_{n-2})<p(B_{n-1})< \frac{3}{2}p(B_{n-2})$ or [$\frac{3}{4}p(B_{n-2})<p(B_{n-1})\le p(B_{n-2})$ and $p(B_{n-1})\ge \frac{3}{4}p(B_{n-2})+\frac{1}{4}q(B_{n-2})$], noting that $B_{n-1}$ does not interrupt the free batch $B_{n-2}$, then $r_{n-1}^{p}\ge \frac{5}{4}p(B_{n-2})$ by Step 4.2 and 4.3 of Algorithm H. Note that the arrival time of $J_{n-1}^{p}$ is $r_{n-1}^{p}$ and the processing time of $J_{n-1}^{p}$ is $p(B_{n-1})$. Thus $L_{{\text{ opt }}}\ge r_{n-1}^{p}+p(B_{n-1})\ge \frac{5}{4}p(B_{n-2})+\frac{3}{4}p(B_{n-2})\ge 2\min \{p(B_{n-1}),p(B_{n-2})\}$. So $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{\min \{p(B_{n-1}),p(B_{n-2})\}}{2\min \{p(B_{n-1}),p(B_{n-2})\}}\le \frac{1}{2}$. If [$\frac{3}{4}p(B_{n-2})<p(B_{n-1})\le p(B_{n-2})$ and $p(B_{n-1})<\frac{3}{4}p(B_{n-2})+\frac{1}{4}q(B_{n-2})$] or $p(B_{n-1})\le \frac{3}{4}p(B_{n-2})$, we distinguish the following two cases. If $J_{n-1}^{p}$ and $J_{n-2}^{*}$ are not in a common batch in the schedule $\pi$, $L_{{\text{ opt }}}\ge p(B_{n-1})+p(B_{n-2})\ge 2\min \{p(B_{n-1}),p(B_{n-2})\}$. Thus $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{\min \{p(B_{n-1}),p(B_{n-2})\}}{2\min \{p(B_{n-1}),p(B_{n-2})\}}=\frac{1}{2}$. If $J_{n-1}^{p}$ and $J_{n-2}^{*}$ are in a common batch in the schedule $\pi$, then $L_{{\text{ opt }}}\ge r_{n-1}^{p}+p(B_{n-2})+q(B_{n-2})>S_{n-2}+p(B_{n-2})+q(B_{n-2})\ge \frac{1}{2}p(B_{n-2})+p(B_{n-2})+q(B_{n-2})= \frac{3}{2}p(B_{n-2})+q(B_{n-2})$. Thus $\frac{L_{{\text{ on }}}-L_{{\text{ opt }}}}{L_{{\text{ opt }}}}\le \frac{p(B_{n-1})}{\frac{3}{2}p(B_{n-2})+q(B_{n-2})}\le \frac{\frac{3}{4}p(B_{n-2})+\frac{1}{4}q(B_{n-2})}{\frac{3}{2}p(B_{n-2})+q(B_{n-2})}\le \frac{1}{2}$. The result follows. $\square$