On-line Scheduling with a Monotonous Subsequence Constraint

Abstract

In this paper, we study a new on-line scheduling problem that each server has to process a monotonous request subsequence. The customer requests are released over-list, and the operator has to decide whether or not to accept the current request and arrange it to a server immediately. The goal of this paper is to find a strategy which accepts the maximal requests. When the number of servers k is less than that of the request types m, we give several lower bounds for this problem. Also, we present the optimal strategy for $k=1$ and $k=2$ respectively.

A Appendix

The analysis of ratio $\frac{OPT(R)}{A(R)}$ in Theorem 1. Case 2.2.

As we already know, $A(R)= 2k_{1}+2max\{k_{2},k_{3}\} +m \cdot (k- k_{1}-max\{k_{2},k_{3}\} )$ and $OPT(R)= k\cdot m$ for all $m>k>2$, where $k_{2}+k_{3}+2k_{1}=k_{sum}$ $(k< k_{sum}\le 2k)$. To make the ratio of $\frac{OPT(R)}{A(R)}$ easier to solve, we analyse

$$\begin{aligned}&\frac{A(R)}{OPT(R)}=\frac{2k_{1}+2max\{k_{2},k_{3}\} +m\cdot (k- k_{1}-max\{k_{2},k_{3}\} )}{k\cdot m} \end{aligned}$$

(5.1)

$$\begin{aligned}&\nonumber =\frac{k_{1}\cdot (2-m)+max\{k_{2},k_{3}\}\cdot (2-m) +m\cdot k}{k\cdot m}\\&\nonumber =1+C \\&C= \frac{k_{1}\cdot (2-m)+max\{k_{2},k_{3}\}\cdot (2-m)}{k\cdot m} . \end{aligned}$$

(5.2)

We analyse the change of C with respect to $k_{1}$, $max\{k_{2},k_{3}\}$ and $k_{1}+ max\{k_{2},k_{3}\}$, respectively.

$$\begin{aligned}&\frac{\partial C}{\partial k_{1}}=\frac{\partial C}{\partial max\{k_{2},k_{3}\}}=\frac{\partial C}{\partial k_{1}+max\{k_{2},k_{3}\}}= \frac{(2-m)\cdot k\cdot m}{(k\cdot m)^{2}} \\&\nonumber <0 \text { (Because } m>k>2 ). \end{aligned}$$

(5.3)

Because $k_{2}+k_{3}+2k_{1}=k_{sum}$ $(k< k_{sum}\le 2k)$ and 5.3, we have:

$$\begin{aligned}&C \le \frac{\frac{k+1}{2}\cdot (2-m)}{m\cdot k} \nonumber = -\frac{1}{3} \text { (Because } m>k>2 ), \end{aligned}$$

and

$$\begin{aligned}&\frac{OPT(R)}{A(R)}\ge \frac{1}{1-\frac{1}{3}}\\&\nonumber \ge \frac{3}{2} \text { (Because } m>k>2 ), \end{aligned}$$

(5.4)