Competitive algorithm for scheduling of sharing machines with rental discount

Abstract

This paper addresses the online parallel machine scheduling problem with machine leasing discount. Rental cost discount is a common phenomenon in the sharing manufacturing environment. In this problem, jobs arrive one by one over-list and must be allocated irrevocably upon their arrivals without knowing future jobs. Each job is with one unit of processing time. One manufacturer leases a limited number of identical machines over a manufacturing resource sharing platform, and pays a rental fee of $\alpha$ per time unit for processing jobs. Especially, when the time duration of a leasing machine reaches the discount time point, the manufacturer will get a discount for further processing jobs on the machine, i.e., the unit time rental cost is $\alpha \beta$, where $\beta =1/2$ is the discount coefficient. The objective function is the sum of makespan and the rental cost of all the sharing machines. When the unit time rental cost $\alpha \ge 2$, we first provide the lower bound of objective value of an optimal schedule in the offline version and prove a lower bound of 1.093 for the problem. Based on the analysis of the offline solution, we present a deterministic online algorithm LS-RD with a tight competitive ratio of 3/2. When $1\le \alpha <2$, we prove that the competitive ratios of algorithm LIST are 1 and 2 for the case of $m=2$ and $m\rightarrow \infty$, respectively. For the general rental discount $0<\beta \le 1$, we give the relevant results for offline and online solutions.

Appendix

Proof for Theorem 8. Similar to the proof of Theorem 3, according to Theorem 7 and Remark 2, we discuss the following six cases.

Case 1: $1\le n\le \mu$.

$$\begin{aligned} \rho _1=\frac{C^{MLS}(I)}{C^*(I)}<\frac{n+\alpha n}{\alpha n}\le \frac{3}{2}. \end{aligned}$$

Case 2: $\mu <n\le 2\alpha \mu (1-\beta )$. If $\alpha (1-\beta )\le \sqrt{2}$, i.e., $2\alpha \mu (1-\beta )\le 4\mu /(\alpha (1-\beta ))$, then

$$\begin{aligned} \rho _2=\frac{C^{MLS}(I)}{C^*(I)}<\frac{n+\alpha \mu +(n-\mu )\alpha \beta }{\alpha n}<\frac{\mu +\alpha \mu }{\alpha \mu }\le \frac{3}{2}. \end{aligned}$$

If $\alpha (1-\beta )>\sqrt{2}$, i.e., $2\alpha \mu (1-\beta )>4\mu /(\alpha (1-\beta ))$, then

$$\begin{aligned} \rho _3=\frac{C^{MLS}(I)}{C^*(I)}\le \frac{n+\alpha \mu +(n-\mu )\alpha \beta }{2\sqrt{n\mu \alpha (1-\beta )}+n\alpha \beta }. \end{aligned}$$

(7)

Because the rightmost part of (7) is monotonically decreasing when $n\in (\mu , \alpha \mu (1-\beta ))$ and increasing when $n\in [\alpha \mu (1-\beta ),2\alpha \mu (1-\beta )]$. Thus it achieves the maximum at $2\alpha \mu (1-\beta )$. Hence, we have

$$\begin{aligned} \rho _3=\frac{C^{MLS}(I)}{C^*(I)}\le \frac{2\alpha \mu (1-\beta )+\alpha \mu +(2\alpha \mu (1-\beta )-\mu )\alpha \beta }{2\sqrt{2\alpha ^2\mu ^2(1-\beta )^2}+2\alpha ^2\mu \beta (1-\beta )}=\frac{3+2\alpha \beta }{2\sqrt{2}+2\alpha \beta }<\frac{11}{10} \end{aligned}$$

Case 3: $\alpha \mu k(k-1)(1-\beta )<n\le \alpha \mu k(k-1)(1-\beta )+\mu$. If $4\mu /(\alpha (1-\beta ))>\alpha \mu k(k-1)(1-\beta )$ then

$$\begin{aligned} \rho _4&=\frac{C^{MLS}(I)}{C^*(I)}<\frac{n\alpha +\alpha \mu (1-\beta )(2k-1)-(k^2-k)\alpha ^2(1-\beta )^2\mu }{\alpha n}\\ {}&<1+\frac{2k-1-(k^2-k)\alpha (1-\beta )}{\alpha k(k-1)}\le \frac{3}{2\alpha }+\beta \le \frac{5}{4} \end{aligned}$$

where the second inequality holds because $n>\alpha \mu k(k-1)(1-\beta )$.

If $4\mu /(\alpha (1-\beta ))<\alpha \mu k(k-1)(1-\beta )+\mu$ then

$$\begin{aligned} \rho _5&=\frac{C^{MLS}(I)}{C^*(I)}\le \frac{n\alpha +\alpha \mu (1-\beta )(2k-1)-(k^2-k)\alpha ^2(1-\beta )^2\mu }{2\sqrt{n\mu \alpha (1-\beta )}+n\alpha \beta }\\&<\frac{n\alpha (1-\beta )+\alpha \mu (1-\beta )(2k-1)-(k^2-k)\alpha ^2(1-\beta )^2\mu }{2\sqrt{n\mu \alpha (1-\beta )}}\\&\le \frac{k\alpha (1-\beta )}{\sqrt{\alpha ^2k(k-1)(1-\beta )^2+(1-\beta )\alpha }}\le \frac{k}{\sqrt{k(k-1)}}\le \sqrt{2} \end{aligned}$$

where the third inequality holds because $n\le \alpha \mu k(k-1)(1-\beta )+\mu$.

Case 4: $\alpha \mu k(k-1)(1-\beta )+\mu <n\le \alpha \mu k^2(1-\beta )$. If $4\mu /(\alpha (1-\beta ))>\alpha \mu k(k-1)(1-\beta )+\mu$. Then

$$\begin{aligned} \rho _6&=\frac{C^{MLS}(I)}{C^*(I)}<\frac{2\alpha \mu k(1-\beta )+n\alpha \beta }{\alpha n}<\beta +\frac{2\mu k(1-\beta )}{\alpha \mu k(k-1)(1-\beta )+\mu }\\&\le \beta +\frac{2k(1-\beta )}{2k(k-1)(1-\beta )+1}\le \beta +\frac{4(1-\beta )}{4(1-\beta )+1}\le \frac{7}{6}. \end{aligned}$$

If $4\mu /(\alpha (1-\beta ))\le \alpha \mu k(k-1)(1-\beta )+\mu$. Then

$$\begin{aligned} \rho _7=\frac{C^{MLS}(I)}{C^*(I)}\le \frac{2\alpha \mu k(1-\beta )+n\alpha \beta }{2\sqrt{n\alpha \mu (1-\beta )}+n\alpha \beta }<\frac{k\alpha (1-\beta )}{\sqrt{\alpha ^2k(k-1)(1-\beta )^2+(1-\beta )\alpha }}\le \sqrt{2} \end{aligned}$$

Case 5: $\alpha \mu k^2(1-\beta )<n\le \alpha \mu k(k+1)(1-\beta )$.

$$\begin{aligned} \rho _8&=\frac{C^{MLS}(I)}{C^*(I)}<\frac{n/k+1+\alpha \mu k(1-\beta )+n\alpha \beta }{2\sqrt{n\mu \alpha (1-\beta )}+n\alpha \beta }<\frac{n/k+1+\alpha \mu k(1-\beta )}{2\sqrt{n\mu \alpha (1-\beta )}}\\&\le \frac{\alpha \mu (k+1)(1-\beta )+1+\alpha \mu k(1-\beta )}{2\sqrt{\mu ^2\alpha ^2 k(k+1)(1-\beta )^2}}\le \frac{2k(1-\beta )+(1-\beta )+1/4}{2(1-\beta )\sqrt{k(k+1)}}\\&\le \frac{4k+3}{4\sqrt{k(k+1)}}<\frac{11}{9} \end{aligned}$$

where the third inequality holds since $\frac{n/k+1+\alpha \mu k(1-\beta )}{2\sqrt{n\mu \alpha (1-\beta )}}$ is monotonically decreasing when $n\in (\alpha \mu k^2(1-\beta ), k+\alpha \mu k^2(1-\beta ))$ and monotonically increasing when $n\in [k+\alpha \mu k^2(1-\beta ), \alpha \mu k(k+1)(1-\beta )]$, and achieves the maximum at $\alpha \mu k(k+1)(1-\beta )$.

Case 6: $\alpha \mu m^2(1-\beta )< n$. $\rho _9=\frac{C^{MLS}(I)}{C^*(I)}=1$.