New approximation algorithms for machine scheduling with rejection on single and parallel machine

Abstract

In this paper we consider three machine scheduling problems with the special feature that jobs may be rejected at a certain penalty. There are n jobs which are characterized by a release date, a processing time and a penalty. Each job is either accepted and then processed by one machine, or rejected and then a rejection penalty is paid. The objective is to minimize the maximum completion time of all accepted job plus the total penalties of all rejected jobs. When jobs have identical release dates, we present a ($\frac{3}{2}-\frac{1}{2m}$)-approximation algorithm for the parallel machine problem. When jobs have general release dates, we propose a $\frac{4}{3}$-approximation algorithm for the single machine problem and a ($1+\max \{0.618,1-\frac{1}{m}\}$)-approximation algorithm for the parallel machine problem, respectively.

Appendices

Appendix A

Claim

For the relaxation problem (2), there exists an optimal solution that satisfies

• (1) If $x_{[j]}>0$ for some j, then for each $i<j$, $x_{[i]}=1$.

• (2) If $x_{[j]}<1$ for some j, then for each $l>j$, $x_{[l]}=0$.

Proof

Let ${x_{[i]}|(i=1,\dots ,n)}$ be an optimal solution for the relaxation problem. Suppose for some j, $x_{[j]}>0$, there exist some $i$ $(i<j)$ such that $x_i<1$. Let $\delta =\min \{(1-x_{[i]})p_{[i]}, x_{[j]}p_{[j]}\}$. We increase $x_{[i]}$ to $x_{[i]}+\delta /p_{[i]}$ and decrease $x_{[j]}$ to $x_{[j]}-\delta /p_{[j]}$. This will enable us to obtain a new solution and the the objective value will not increase. Therefore the new solution is also optimal. By repeating the above operation, we can obtain an optimal solution which satisfies property (1). Similarly, we can show that there exists an optimal solution that satisfies property (2). $\square$

Proof of Lemma 2:

If $M(\overline{A}_k)<p_k$ and $M(U_k)\ge p_k$, by Claim 1, we know that there exists some integer j such that $x_{[i]}=1$ for each $i<j$ and $x_{[i]}=0$ for each $i>j$ in the optimal solution. Let $Z^R$ be the optimal value for the relaxation problem. Thus

$$\begin{aligned} Z^R=\max \{p_{k}, \frac{1}{m}(p_k+\sum _{i=1}^{j-1}p_{[i]}+x_{[j]}p_{[j]})\}+(1-x_{[j]})w_{[j]}+ \sum _{i=j+1}^{n-1}w_{[i]} \end{aligned}$$

Recall that $M(\overline{A}_k)<p_k$, $M(U_k)\ge p_k$, $\frac{1}{m}p_i>w_i$ for each $J_i\not \in \overline{A}_k$ and and $J_{[j]}$ is the j-th job from $A_k\cup R_k$ in non-increasing order of $\frac{w_j}{p_j}$. By calculating the value of $Z^R$, we know that the relaxation problem (2) obtains its minimum value when $j=v$ and

$$\begin{aligned} Z^{OPT}\ge Z^R= \frac{1}{m}(p_k+\sum _{i=1}^{v-1}p_{[i]}+\alpha p_{[v]}) + \sum _{i=v+1}^{n-1}w_{[i]}+(1-\alpha )w_{[v]} \end{aligned}$$

where v is the first integer such that $\sum _{i=1}^vp_{[i]}\ge (m-1)p_k$ and $\alpha$ is a number such that $\alpha p_{[v]}= x_{[v]}p_{[v]} =(m-1)p_{k}-\sum _{i=1}^{v-1}p_{[i]}$. $\square$

Appendix B

He et al. (2016) revisited $1|r_j,rej|C_{max}+\sum w_j$. In their manuscript, they present an improved approximation algorithm with a worst case ratio of 5/4. In the appendix, we will give an instance to show that its worst case ratio is greater than 5/4 indeed. Firstly, we will describe the algorithm presented by He et al. (2016).

For the job set $J=\{j|1\le j\le n\}$, assume that there are m different release dates $(r_{i_1}, r_{i_2}, \dots , r_{i_m} )$ with $r_{i_1}<r_{i_2}< \dots <r_{i_m}$. Let $J(i_k) := \{j | r_j = r_{i_k}$ and $w_j/p_j > 1\}$ and $J^\prime (i_k) := \{ j | r_j = r_{i_k}$ and $w_j/p_j \le 1\}$.

A procedure named Weighted Shortest Remaining Processing Time(WSRPT) is introduced firstly.

Procedure WSRPT

• Step 1. Let $A0 := \{j | 1 \le j \le n$ and $w j/p j > 1 = \{ j_1, j_2, \dots , j_{m}\}$, where $r_{j1}< r_{j2}<\dots < r_ {j_m}$ .

• Step 2. Schedule all jobs in $A_0$ as early as possible according to the ERD rule.

• Step 3. Schedule the jobs in $\bigcup _{k=1}^{m-1}J^\prime (i_k)$ preemptively in the idle intervals in the interval $[r_{i_1}, +\infty )$ as early as possible. When there are several jobs available (i.e., jobs that have been released but not yet been completed), schedule the job with the largest $w_j/p_j^\prime$ first, where $p_j^\prime$ is the current remaining processing time of job j.

Algorithm AR

• Call Procedure WSRPT to obtain a schedule $\pi$.

• For each $j_k$, $1 \le k \le m$, define $A(j_k) = \{i : C_i(\pi ) \le C_{j_k}(\pi )$ or $r_i < r_{j_k}$ and $C_i(\pi ) > C_{j_k}(\pi )$ and $w_i/p_i^\prime > 1\}$, where $p_i^\prime$ is the remaining processing time of job i at time $t = C_{j_k}(\pi )$. Let $R(j_k) = J\backslash A(j_k)$, and let $\pi (j_k)$ be the schedule with accepted job set $A(j_k)$ and rejected job set $R(j_k)(1 \le k \le m)$.

• Let $\pi ^*$ be the schedule with the smallest value of the objective function among the schedules $\pi (j_0), \pi (j_1), \pi , \pi (j_m)$, where $\pi (j_0)$ is the schedule in which all jobs are rejected.

Now, we present an instance to show that the worst case ratio of Algorithm AR is greater than 5/4. There are $n+2$ jobs and w is a constant number with $0\le w \le 1$. $J_{i}(0\le i\le n-1)$ is released at time i/n with $p_i=(1-w)/n$ and $w_i=w(1-w)/(n-i(1-w))-w/n^2$. $J_n$ is released at time 0 with $p_n=1, w_n=w$. $J_{n+1}$ is released at time $1-w$ with $p_{n+1}=w$, $w_{n+1}=w-w/n$. Algorithm AR will accept $J_n$ and reject all other jobs. The total penalties will be

$$\begin{aligned} RC =\sum _{j=0}^{n-1}w_j+w_{n+1}=\sum _{j=0}^{n-1}\{\frac{w(1-w)}{n-i(1-w)}-\frac{w}{n^2}\}+w(1-\frac{1}{n}) \end{aligned}$$

When n tends to $\infty$,

$$\begin{aligned} RC= & {} \lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}\{\frac{w(1-w)}{n-i(1-w)}-\frac{w}{n^2}\}+w\\= & {} w\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}\frac{1}{1-\frac{i(1-w)}{n}}\frac{1-w}{n}+\lim _{n\rightarrow \infty }\frac{w}{n}+w\\= & {} w\int _0^{1-w}\frac{1}{1-x}\mathrm {d}x+w\\= & {} -w\ln w+w \end{aligned}$$

Thus the objective value is $Z=1-w\ln w+w$.

While an optimal solution will accept all jobs except $J_n$ and the optimal value is $Z^*=1+w$.

Let $w=1/e$ where $e\approx 2.7183$ is Euler Number. Then the ratio

$$\begin{aligned} \frac{Z}{Z^*}=\frac{1-w\ln w+w}{1+w}=\frac{1+\frac{2}{e}}{1+\frac{1}{e}}\approx 1.2689 \end{aligned}$$

which is greater than 5/4.