Resource-Constrained Scheduling Algorithms for Stochastic Independent Tasks With Unknown Probability Distribution

Abstract

This work introduces scheduling algorithms to maximize the expected number of independent tasks that can be executed on a parallel platform within a given budget and under a deadline constraint. The main motivation for this problem comes from imprecise computations, where each job has a mandatory part and an optional part, and the objective is to maximize the number of optional parts that are successfully executed, in order to improve the accuracy of the results. The optional parts of the jobs represent the independent tasks of our problem. Task execution times are not known before execution; instead, the only information available to the scheduler is that they obey some (unknown) probability distribution. The scheduler needs to acquire some information before deciding for a cutting threshold: instead of allowing all tasks to run until completion, one may want to interrupt long-running tasks at some point. In addition, the cutting threshold may be reevaluated as new information is acquired when the execution progresses further. This work presents several algorithms to determine a good cutting threshold, and to decide when to re-evaluate it. In particular, we use the Kaplan-Meier estimator to account for tasks that are still running when making a decision. The efficiency of our algorithms is assessed through an extensive set of simulations with various budget and deadline values, and ranging over 13 probability distributions. In particular, the AutoPerSurvival(40%,0.005) strategy is proved to have a performance of 77% compared to the upper bound even in the worst case. This shows the robustness of our strategy.

Appendix

1.1 Proof of Theorem 2

We start by recalling some classical results for a unimodal exponential distribution of rate $\lambda$, whose probability density function and cumulative distribution function are respectively denoted by f(x) and F(x) (here we do not add any constant or, in other words, $\delta = 0$):

• Probability that the random variable is no greater than l: $P(X \le l) = F(l) = 1-e^{-\lambda l}$.

• Average value of a random variable which is no greater than l: ${\mathbb {E}}(X \le l) = \int _{x=0}^l x f(x) dx = -l e^{-\lambda l} + \frac{1}{\lambda } \left( 1-e^{-\lambda l}\right)$.

• Average value of the time spent executing a task whose execution time is defined by an exponential distribution of parameter $\lambda$ when the execution is cut at time l if the task has not completed by that time:

  $$\begin{aligned} \left( -l e^{-\lambda l} + \frac{1}{\lambda } \left( 1-e^{-\lambda l}\right) \right) +l(1-F(l)) = \frac{1}{\lambda } \left( 1-e^{-\lambda l}\right) . \end{aligned}$$

Fig. 7

Theoretical yield when varying cutting threshold for each distribution

Fig. 8

Ratio to Oracle of number of tasks successfully executed using different heuristics when varying $p$ for each distribution

Fig. 9

Ratio to Oracle of number of tasks successfully executed using either AutoPerSurvival (for different values of $p_{max}$ and when varying $\epsilon$) or PerSurvival (with different values of $p$)

Fig. 10

Number of successfully executed tasks for the different heuristics with a budget $b =1000$ when task execution times follow a log-normal distribution. Unless otherwise specified, the expectation is $\mu =1$, the standard deviation is $\sigma =3$, and the number of processors is $M =10$

When the cutting threshold l is smaller than or equal to the constant time $\delta$, the yield is obviously null: ${\mathcal {Y}} (l)=0$. Therefore, we only consider the yield for cutting thresholds greater than or equal to $\delta$. To ease the writing, let $y(l)={\mathcal {Y}} (l+\delta )$. Using Eq. 2, we derive the expression of the yield when the cutting threshold is set at $l+\delta$:

$$\begin{aligned} y(l)&= {\mathcal {Y}} (l+\delta )\\&= \frac{p \left( 1-e^{-\lambda l}\right) + (1-p) \left( 1-e^{-\mu l}\right) }{\delta +p\left( \frac{1}{\lambda } \left( 1-e^{-\lambda l}\right) \right) +(1-p)\left( \frac{1}{\mu } \left( 1-e^{-\mu l}\right) \right) }\\&= \frac{p \left( 1-e^{-\lambda l}\right) + (1-p) \left( 1-e^{-\mu l}\right) }{\delta +\frac{p}{\lambda } \left( 1-e^{-\lambda l}\right) +\frac{(1-p)}{\mu } \left( 1-e^{-\mu l}\right) }\\\end{aligned}$$

In order to identify the maximum of the function y(l), we differentiate it:

$$\begin{aligned} y'(l)&\scriptstyle = \frac{ \left( \lambda p e^{-\lambda l} + \mu (1-p)e^{-\mu l}\right) \left( \delta +\frac{p}{\lambda } \left( 1-e^{-\lambda l}\right) +\frac{(1-p)}{\mu } \left( 1-e^{-\mu l}\right) \right) }{\left( \delta +\frac{p}{\lambda } \left( 1-e^{-\lambda l}\right) +\frac{(1-p)}{\mu } \left( 1-e^{-\mu l}\right) \right) ^2}\\&\scriptstyle - \frac{\left( p \left( 1-e^{-\lambda l}\right) + (1-p)\left( 1-e^{-\mu l}\right) \right) \left( p e^{-\lambda l} + (1-p)e^{-\mu l} \right) }{\left( \delta +\frac{p}{\lambda } \left( 1-e^{-\lambda l}\right) +\frac{(1-p)}{\mu } \left( 1-e^{-\mu l}\right) \right) ^2}\\\end{aligned}$$

Because we are only interested by the sign of $y'(l)$, we focus on its numerator, that we denote by g(l):

$$\begin{aligned} g(l)&\scriptstyle = \left( \lambda p e^{-\lambda l} + \mu (1-p)e^{-\mu l}\right) \left( \delta +\frac{p}{\lambda } \left( 1-e^{-\lambda l}\right) +\frac{(1-p)}{\mu } \left( 1-e^{-\mu l}\right) \right) \\&\scriptstyle \quad - \left( p \left( 1-e^{-\lambda l}\right) + (1-p)\left( 1-e^{-\mu l}\right) \right) \left( p e^{-\lambda l} + (1-p)e^{-\mu l} \right) \\&\scriptstyle = \delta \left( \lambda p e^{-\lambda l} + \mu (1-p)e^{-\mu l}\right) \\&\scriptstyle \quad + p(1-p)\left( \left( \frac{\lambda }{\mu }-1\right) e^{-\lambda l} \left( 1-e^{-\mu l}\right) + \left( \frac{\mu }{\lambda }-1\right) e^{-\mu l} \left( 1-e^{-\lambda l}\right) \right) \\&\scriptstyle = e^{-\lambda l} \left( \delta \left( \lambda p + \mu (1-p)e^{-(\mu -\lambda )l}\right) \right. \\&\scriptstyle \qquad \qquad \qquad \left. + p(1-p)\frac{\mu -\lambda }{\lambda \mu }\left( -\lambda \left( 1-e^{-\mu l}\right) + \mu e^{-\mu l} \left( e^{\lambda l}-1\right) \right) \right) . \end{aligned}$$

We focus on the expression

$$\begin{aligned} h(l)&= -\lambda \left( 1-e^{-\mu l}\right) + \mu e^{-\mu l} \left( e^{\lambda l}-1\right) \\&= -\lambda +\lambda e^{-\mu l} + \mu e^{-(\mu -\lambda ) l} - \mu e^{-\mu l}. \end{aligned}$$

We differentiate h (with respect to l):

$$\begin{aligned} h'(l)&= -\lambda \mu e^{-\mu l} - \mu (\mu -\lambda ) e^{-(\mu -\lambda ) l} + \mu ^2 e^{-\mu l}\\&= \mu e^{-\mu l} (\mu -\lambda )\left( 1 - e^{\lambda l} \right) . \end{aligned}$$

Because $\mu -\lambda$, $\lambda$ and l are all nonnegative, $1 - e^{\lambda l} \le 0$ and $h'(l)$ is nonpositive. Therefore, h(l) is a non-increasing function.

Clearly, g(l) and $k(l) = g(l) e^{\lambda l}$ have the same sign. From what precedes, $k(l) = g(l) e^{\lambda l}$ is the sum of two non-increasing functions, the functions $\delta (\lambda p + \mu (1-p)e^{-(\mu -\lambda )l})$ and $p(1-p)\frac{\mu -\lambda }{\lambda \mu } h(l)$. The maximum of k(l) is thus reached for $l=0$ and its minimum is reached when l tends toward $+\infty$:

$$\begin{aligned} k(0)&= \delta \left( \lambda p + \mu (1-p)\right) ; \\ \lim _{l \rightarrow +\infty } k(l)&= \delta \lambda p + p(1-p)\frac{\mu -\lambda }{\lambda \mu }\left( -\lambda \right) \nonumber \\&= \delta \lambda p - p(1-p)\frac{\mu -\lambda }{\mu }\\&= p\left( \delta \lambda - (1-p)\frac{\mu -\lambda }{\mu }\right) . \end{aligned}$$

To conclude, we have several cases to consider:

• 1. $\delta =0$ (arbitrarily small execution times are allowed). Then, $k(0)=0$ and $\lim _{l \rightarrow +\infty } k(l) \le 0$. We have two subcases to consider:

  • (a) $p(1-p)(\mu -\lambda ) = 0$. In this case, we have a monomodal exponential distribution. Then, $y'(l)$ is null, ${\mathcal {Y}} (l)=y(l)$ is constant and equal to $\lambda$. In other words, in this case the yield is optimal whatever the value chosen for the threshold.

  • (b) $p(1-p)(\mu -\lambda ) \ne 0$. In this case, ${\mathcal {Y}} (l)=y(l)$ is a decreasing function and its maximum is achieved when $l=0$, and the optimum yield is then $\lim _{l \rightarrow 0} {\mathcal {Y}} (l)$. To compute this limit we use the equivalent to $e^{-x}$ in 0 which is $1-x$. We obtain:

    $$\begin{aligned} {\mathcal {Y}} (l)&= \frac{p \left( 1-e^{-\lambda l}\right) + (1-p) \left( 1-e^{-\mu l}\right) }{c+\frac{p}{\lambda } \left( 1-e^{-\lambda l}\right) +\frac{(1-p)}{\mu } \left( 1-e^{-\mu l}\right) }\\&\approx \frac{p \left( \lambda l\right) + (1-p)\left( \mu l\right) }{\frac{p}{\lambda } \lambda l+\frac{(1-p)}{\mu } \mu l}\\&= p \lambda + (1-p) \mu . \end{aligned}$$

    Remark: this counter-intuitive result means that the shortest the threshold, the better. It means that, in practice, the scheduler should stop each task as soon as it is started. Obviously, this is not achievable in practice. This peculiar property is a consequence of allowing execution times to be arbitrarily small, as the remainder of this case study will illustrate.

• 2. $\delta > 0$. Once again, we have two subcases to consider:

  • $p(\delta \lambda - (1-p)\frac{\mu -\lambda }{\mu }) \ge 0$. Then k(l), and thus g(l) and $y'(l)$ are nonnegative for all values of l. y and ${\mathcal {Y}}$ are thus increasing and we should never abort the execution of a running task (threshold = $+\infty$). The optimum yield in this case is then:

    $$\begin{aligned} \lim _{l \rightarrow +\infty } {\mathcal {Y}} (l)&= \frac{1}{\delta +\frac{p}{\lambda } +\frac{(1-p)}{\mu } }\cdot \end{aligned}$$

  • $p(\delta \lambda - (1-p)\frac{\mu -\lambda }{\mu }) < 0$ which can be rewritten $\delta < (1-p)\frac{\mu -\lambda }{\lambda \mu }$. In this subcase, we have $y'(l)$ which is a decreasing function, with $y'(0) >0$ and $\lim _{l \rightarrow +\infty } k(l) < 0$ which implies that $y'(l)$ and ${\mathcal {Y}} '(l)$ are negative when l is sufficiently large. Therefore, ${\mathcal {Y}} (l+\delta )$ is first increasing and then decreasing, and has a unique maximum. This maximum is achieved for l satisfying $k(l) = 0$, which could only be solved numerically.

1.2 Additional Graphs and Statistics

We report here the theoretical threshold and the performance for the distributions that were not illustrated in the core of the article. We also report the performance of heuristics when the budget is large with respect to the average task execution time ($b =1000$).