On-line preemptive machine scheduling with $\ell _p$ norm on two uniform machines

Abstract

One of the basic and fundamental scheduling problems is to minimize the machine completion time vector in the $\ell _p$ norm, a direct extension of the well-studied objective makespan ($l_{\infty }$ norm), on parallel machines. We concentrate on the on-line and preemptive version of this problem where jobs arrive one by one over a list to be allocated to two uniform machines with job preemption permitted. We present a best possible deterministic on-line scheduling algorithm for this problem along with a matching lower bound, generalizing existing results for the identical machines scheduling problem in the literature. One notable feature of this work is the highly involved technicality compared to similar analysis in the existing literature, mainly due to the intrinsic unavailability of a closed-form formula for the competitive ratio.

Appendices

Appendix 1: A proof of the lower on the optimal objective value (3)

We prove a stronger result that the equality holds in (3).

Consider the optimal (off-line) schedule opt before $x$. Let $L$ be the total process time before $x$ and $z$ be any job before $x$ in opt. Denote $C_{\textsc {opt}}$ to be the optimal off-line objective prior to $x$. Note that $\frac{z}{s}\ge \hat{s}\hat{t}L$ if and only if $L-z\le \hat{t}L$ due to (1)–(2). Therefore, the RHS of (3)

$$\begin{aligned} \text {RHS:}&= \root p \of {\left( \max \left\{ \frac{z}{s}, \hat{s}\hat{t}L \right\} \right) ^p + \left( \min \left\{ L-z,\hat{t}L\right\} \right) ^p}\\&= \left\{ \begin{array}{ll} \root p \of {(L-z)^p+\left( \frac{z}{s}\right) ^p},&{}\text { if } \frac{z}{s}\ge \hat{s}\hat{t}L;\\ \root p \of {\left( \hat{s}\hat{t}L\right) ^p+\left( \hat{t}L\right) ^p},&{}\text { otherwise.} \end{array}\right. \end{aligned}$$

Now we show the desired result in two cases.

• Case 1. $\frac{z}{s}\ge \hat{s}\hat{t}L$, which is equivalent to $L-z\le \hat{t}L$. From the case conditions and $\hat{s}\ge 1$, we have that $\frac{z}{s}\ge \hat{s}\hat{t}L\ge \hat{t}L\ge L-z$. Under these conditions, the optimal schedule is to put $z$ entirely on machine 2 and the rest on machine 1, leading to the desired optimal value:

  $$\begin{aligned}&\quad C_{\textsc {opt}}=\root p \of {(L-z)^p+\left( \frac{z}{s}\right) ^p}. \end{aligned}$$

• Case 2. $\frac{z}{s}< \hat{s}\hat{t}L$, which is equivalent to $L-z>\hat{t}L$. In this case, let $y$ be the load assigned to machine 1 in opt. Then the objective value is $c(y)=\root p \of {y^p+(\frac{L-y}{s})^p}$. By taking derivative, we have the stationary point $y^*=\frac{L}{1+\hat{s}^p}=\hat{t}L$. Checking the second derivative convinces us that $c(y)$ achieves its minimum value at $y^*$, which, due to (1)–(2), equals to

  $$\begin{aligned}&\quad \root p \of {\left( \hat{t}L\right) ^p+\left( \frac{L-\hat{t}L}{s}\right) ^p}=\root p \of {\left( \hat{s}\hat{t}L\right) ^p+\left( \hat{t}L\right) ^p}. \end{aligned}$$

  (13)

  Note that the case condition $L-z>\hat{t}L$ implies that we can indeed assign (by preemption) the load $y^*=\hat{t}L$ on machine 1 and the rest on machine 2 to achieve the bound in (13), implying that

  $$\begin{aligned}&\quad C_{\textsc {opt}}=\root p \of {\left( \hat{s}\hat{t}L\right) ^p+\left( \hat{t}L\right) ^p}. \end{aligned}$$

Appendix 2: Competitive ratios for selected $p$ and $s$

See Table 1.

Table 1 Competitive ratios for selected $s$ and $p$