Correlation adaptive task scheduling

Abstract

The Internet of Things offers a vast infrastructure where numerous devices interact to collect data or perform processing activities (tasks). These devices are, usually, equipped with sensors, software and storage capabilities being able to process the collected data. Task scheduling in edge computing environments has gained considerable attention lately due to the fact that the Edge computing provides lower latency compared to the Cloud. The main challenge is to find a way to maximize the utilization of limited resources available in the edge compared to the Cloud and minimize response time. Many research efforts have been published in order to overcome this challenge. The main limitation of these efforts is the fact that they do not account for task requirements or task correlation that originates from these requirements. In this paper, we focus on the development of a mechanism that utilizes correlation between tasks and takes task requirements into consideration in order to provide efficient task scheduling. Our vision is to minimize task failures and maximize resource utilization with great benefits for the efficient management of the limited resources.

Appendix A

1.1 Proof 1

We provide the proof for Lemma 1. As mentioned before $tfv_i[l], tfm_i[l], trov_i[l], NFV_j[l], NFM_j[l],NROV_j[l]$ are independent between them, so we can calculate the initial probability separately for each pair of statistical measure (variance, mean and range of values).

${\textbf{E}}(f(i,j,l)) = 1*[P(f(i,j,l)=1] + 0*[P(f(i,j,l)=0] = [P(f(i,j,l)=1] = P[tfv_i[l] \cap NFV_j[l] \ne \emptyset \vee tfv_i[l] = 0] *P[tfm_i[l] \cap NFM_j[l] \ne \emptyset \vee tfm_i[l] = 0] * P[trov_i[l] \cap NROV_j[l] \ne \emptyset \vee trov_i[l] = 0]$

To avoid confusion, we perform further calculations separately. Each task requirement has a probability p of existing. This leads to the following:

$P[tfv_i[l] \cap NFV_j[l] \ne \emptyset \vee tfv_i[l] = 0] = p*P[tfv_i[l] \cap NFV_j[l] \ne \emptyset ] + (1-p) = p*P[NFV_j[l] \in [c,d]] + (1-p) = p*P[c \le NFV_j[l] \le d] + (1-p) = p* (P[NFV_j[l] \le d] - P[NFV_j[l]< c]) + (1-p) P[tfm_i[l] \cap NFM_j[l] \ne \emptyset \vee tfm_i[l] = 0] = p*P[tfm_i[l] \cap NFM_j[l] \ne \emptyset ] + (1-p) = p*P[NFM_j[l] \in [e,f]] + (1-p) = p*P[e \le NFM_j[l] \le f] + (1-p) = p * (P[NFM_j[l] \le f] - P[NFM_j[l] < e]) + (1-p)$

$P[trov_i[l] \cap NROV_j[l] \ne \emptyset \vee trov_i[l] = 0] = p * P[trov_i[l] \cap NROV_j[l] \ne \emptyset ] + (1-p)$

where

$P(trov_i[l] \cap NROV_j[l] \ne \emptyset ) = P[[g,h] \cap [NROV_j[l][1],NROV_j[l][2]] \ne \emptyset ) = 1 - P([g,h] \cap [NROV_j[l][1],NROV_j[l][2]]) = \emptyset ) = 1- P( (NROV_j[l][2]< g) \cup ([NROV_j[l][1]> h) ) = 1 - (P(NROV_j[l][2]< g) + P( (NROV_j[l][1] > h) ) = 1 - ( P(NROV_j[l][2]< g) + 1 - P( (NROV_j[l][1] \le h) ) = 1 - P(NROV_j[l][2]< g) - 1 + P( (NROV_j[l][1] \le h) ) = P( (NROV_j[l][1] \le h) ) - P(NROV_j[l][2] < g)$

1.2 Proof 2

We provide the proof for Lemma 2.

${\textbf{E}}(su_i[j]) = 1 * P(\sum _{l=1}^{r}f(i,j,l) =r) + 0 * P(\sum _{l=1}^{r}f(i,j,l) \ne r) = P(\sum _{l=1}^{r}f(i,j,l) =r) = P(f(i,j,1) = 1 \cap f(i,j,2) = 1 \cap \ldots \cap f(i,j,r) = 1 ) = \prod _{l=1}^{r} P(f(i,j,l) = 1) = \prod _{l=1}^{r} {\textbf{E}}(f(i,j,l))$