Speed scaling scheduling of multiprocessor jobs with energy constraint and makespan criterion

Abstract

We are given a set of parallel jobs that have to be executed on a set of speed-scalable processors varying their speeds dynamically. Running a job at a slower speed is more energy-efficient, however, it takes a longer time and affects the performance. Every job is characterized by the processing volume and the number or the set of the required processors. Our objective is to minimize the maximum completion time so that the energy consumption is not greater than a given energy budget. For various particular cases, we propose polynomial-time approximation algorithms, consisting of two stages. At the first stage, we give an auxiliary convex program. By solving this problem, we get processing times of jobs and a lower bound on the makespan. Then, at the second stage, we transform our problem into the corresponding scheduling problem with the constant speed of processors and construct a feasible schedule. We also obtain an “almost exact” solution for the preemptive settings based on a configuration linear program.

Appendix

Appendix contains a polynomial time algorithm solving the lower bound model (1)–(5). We consider the following two subproblems.

Subproblem (P1)

$$\begin{aligned}&\frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_jp_j\rightarrow \min , \\&\quad \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j^{\alpha }p_j^{1-\alpha }\le E, \\&\quad p_j\ge 0,\ j\in {\mathcal {J}}. \end{aligned}$$

Subproblem (P2)

$$\begin{aligned}&\max _{j\in {\mathcal {J}}}p_j\rightarrow \min , \\&\quad \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j^{\alpha }p_j^{1-\alpha }\le E, \\&\quad p_j\ge 0,\ j\in {\mathcal {J}}. \end{aligned}$$

Problems (P1) and (P2) are relaxations of the problem (1)–(5). Therefore, if an optimal solution of (P1) or (P2) is a feasible solution of (1)–(5), it is also an optimal solution of (1)–(5).

Our algorithm for (1)–(5) consists of three steps. At the first two steps, problems (P1) and (P2) are solved, and at the third step, a combination of (P1) and (P2) is formed by equating the values of the objective functions.

1.1 The first step

Before solving problem (P1), we check the following condition

$$\begin{aligned} \max _{j\in {\mathcal {J}}}{W_j}\le \frac{1}{m}\sum _{j\in {\mathcal {J}}}{\textit{size}_jW_j}. \end{aligned}$$

(46)

If inequality (46) takes place, then the optimal solution of problem (P1) is an optimal solution of problem (1)–(5), otherwise, go to the second step. Indeed, problem (P1) corresponds to the case when each rigid job j is replaced by $\textit{size}_j$ single-processor jobs of the same duration $p_j$, and it is required to schedule them on one processor. So, in an optimal solution of (P1) all jobs have the same speed s, which can be found through equation

$$\begin{aligned} \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_js^{\alpha -1}= E. \end{aligned}$$

The processing times of jobs are

$$\begin{aligned} p_i=\frac{W_i}{s}=\frac{W_i\left( \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j\right) ^{\frac{1}{\alpha -1}}}{E^{\frac{1}{\alpha -1}}},\ i\in {\mathcal {J}}. \end{aligned}$$

(47)

The condition (46) guarantees that the inequality $\max _{j\in {\mathcal {J}}}p_j\le \frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_jp_j$ holds for job durations (47).

1.2 The second step

Initially we check the condition

$$\begin{aligned} \sum _{j\in {\mathcal {J}}} \textit{size}_j\le m, \end{aligned}$$

(48)

which indicates that the optimal solution of problem (P2) is an optimal solution of problem (1)–(5). Problem (P2) corresponds to simultaneous execution of the jobs. Thus, all jobs should have identical processing times

$$\begin{aligned} p_i=\frac{\left( \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW^{\alpha }_j \right) ^{\frac{1}{\alpha -1}}}{E^{\frac{1}{\alpha -1}}},\ i\in {\mathcal {J}}. \end{aligned}$$

(49)

The condition (48) guarantees that $\frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_j p_j \le \max _{j\in {\mathcal {J}}}p_j$ for job durations (49). However, if the total number of processors required by all jobs is more then m, we go to the third step.

1.3 The third step

If neither inequality (46) nor inequality (48) is satisfied, according to the Karush–Kuhn–Tucker necessary and sufficient conditions (see the last section in “Appendix”), the following equality should hold for an optimal solution of problem (1)–(5):

$$\begin{aligned} \frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_j p_j = \max _{j\in {\mathcal {J}}}p_j. \end{aligned}$$

As a result we have problem (P3):

$$\begin{aligned}&\sum _{j\in {\mathcal {J}}}{} \textit{size}_jp_j\rightarrow \min , \\&\quad \max _{j\in {\mathcal {J}}}p_j - \frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_jp_j=0, \\&\quad \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j^{\alpha }p_j^{1-\alpha }\le E, \\&\quad p_j\ge 0,\ j\in {\mathcal {J}}. \end{aligned}$$

Solving (P3) begins with the optimal solution of (P1), where the processing times of jobs are equal to

$$\begin{aligned} p_i=\frac{W_i\left( \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j\right) ^{\frac{1}{\alpha -1}}}{E^{\frac{1}{\alpha -1}}},\ i\in {\mathcal {J}}. \end{aligned}$$

As we can see, job durations $p_j$ are proportional to works $W_j$. We order jobs in non-increasing work-values $W_1\ge W_2\ge \cdots W_n$. Let l denote the job with a minimal processing volume such that

$$\begin{aligned} W_l> \frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_j W_j. \end{aligned}$$

(50)

This condition is equivalent to the condition $p_l> \frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_j p_j$.

Because the objective of (P3) is a linear function, a necessary condition for the optimal solution of (P3) is

$$\begin{aligned} p_j= \frac{1}{m}\sum _{i\in {\mathcal {J}}}{} \textit{size}_i p_i,\ j=1,\dots ,l. \end{aligned}$$

Then we formulate a new problem (P4):

$$\begin{aligned}&\sum _{j\in {\mathcal {J}}}{} \textit{size}_jp_j\rightarrow \min , \\&\quad p_j = \frac{1}{m}\sum _{i\in {\mathcal {J}}}{} \textit{size}_ip_i,\ j=1,\dots ,l, \\&\quad \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j^{\alpha }p_j^{1-\alpha }\le E, \\&\quad p_j\ge 0,\ j\in {\mathcal {J}}. \end{aligned}$$

In order to find an optimal solution of problem (P4) we use the Lagrangian method constructing the Lagrangian function

$$\begin{aligned} L(p_j,\lambda _j)= & {} \sum _{j\in {\mathcal {J}}}{} \textit{size}_jp_j+ \sum _{j=1}^l \lambda _j \left\{ p_j- \sum _{i\in {\mathcal {J}}}\frac{\textit{size}_ip_i}{m} \right\} + \\&\quad \lambda _{l+1}\left( \sum _{j\in {\mathcal {J}}}{} \textit{size}_jW_j^{\alpha }p_j^{1-\alpha }- E\right) \end{aligned}$$

We calculate the partial derivatives, equate them to zero and find the processing times of jobs

$$\begin{aligned} p_j= & {} \frac{\sum _{i=l+1}^{n}{} \textit{size}_iW_i}{(m-\sum _{i=1}^l \textit{size}_i)}C,\ j=1,\dots ,l, \end{aligned}$$

(51)

$$\begin{aligned} p_j= & {} W_jC,\ j=l+1,\dots ,n, \end{aligned}$$

(52)

where

$$\begin{aligned} C=\left( \frac{E}{\sum _{j=1}^l\frac{W_j^{\alpha }{} \textit{size}_j (\sum _{i=l+1}^{n}W_i \textit{size}_i)^{1-\alpha } }{(m-\sum _{i=1}^l \textit{size}_i)^{1-\alpha }} +\sum _{j=l+1}^nW_j \textit{size}_j} \right) ^{\frac{1}{1-\alpha }}. \end{aligned}$$

Note that $m>\sum _{i=1}^l \textit{size}_i$ due to condition (50) for choosing index l.

If the following inequality is satisfied for the obtained durations of jobs $l+1,\dots ,n$

$$\begin{aligned} \max _{j=l+1,\dots ,n}p_j\le \frac{1}{m}\sum _{j\in {\mathcal {J}}}{} \textit{size}_j p_j, \end{aligned}$$

(53)

then we have an optimal solution of problem (1)–(5). Otherwise, we go to solving the problem (P4) with new value of l.

Inequality (53) can be rewritten in the following form using the presented durations of jobs (51), (52)

$$\begin{aligned} \max _{j=l+1,\dots ,n}W_j=W_{l+1}\le \frac{1}{(m-\sum _{i=1}^l \textit{size}_i)}\sum _{j=l+1}^{n}{} \textit{size}_j W_j. \end{aligned}$$

(54)

So, instead of solving problem (P4) several times we only need to find such value of l for which condition (54) is satisfied. This can be done in $O(\min \{n,m\})$ steps. After that, we use formulas (51), (52) to calculate processing times and obtain an optimal solution of the initial problem (1)–(5).

At each of the three steps, the formulas for processing times of jobs include exponentiation and root extraction operations. Therefore $p_j,\ j\in {\mathcal {J}},$ may be irrational and computed with any desired accuracy $\varepsilon '>0$ in $O\left( n\mathrm {log}\left( \frac{1}{\varepsilon '} n m W_{\max }\right) \right)$ time, where $W_{\max }=\max _{j\in {\mathcal {J}}}W_j$ (see Newton’s method or dichotomy method in [32]). If we set $\varepsilon '=\frac{\varepsilon }{n}$, then we obtain an $OPT+\varepsilon$ solution of problem (1)–(5) as the objective is the weighted sum of job durations.

1.4 Karush–Kuhn–Tucker conditions

Consider the following convex program:

$$\begin{aligned}&f(x) \rightarrow \min \\&\quad g_i(x)\le 0,\ i=1,\dots ,m, \\&\quad x\in R^n, \end{aligned}$$

where all functions $g_i$ are differentiable, and there is an interior point $x'$ such that $g_i(x')<0$ for all $i=1,\dots ,m$.

Let $\lambda _i$ be the dual variable associated with the constraint $g_i(x)\le 0$. The Karush–Kuhn–Tucker conditions are:

$$\begin{aligned} g_i(x)\le 0,\ i= & {} 1,\dots ,m, \\ \lambda _i\ge 0,\ i= & {} 1,\dots ,m, \\ \lambda _i g_i(x)= 0,\ i= & {} 1,\dots ,m, \\ \nabla L(x,\lambda )= & {} 0, \end{aligned}$$

where $L(x,\lambda )=f(x)+\sum _{i=1}^n \lambda _i g_i(x)$ is the Lagrangian function. Karush–Kuhn–Tucker conditions are necessary and sufficient for solutions $x\in R^n$ and $\lambda \in R^m$ to be primal and dual optimal [11].