Minimizing latency in geo-distributed clouds

Abstract

Geo-distributed data-centers in cloud computing are becoming increasingly popular due to their lower end-user perceived latency and increased reliability in distributed applications. The important challenge of resource allocation in cloud management is more pronounced in geo-distributed data-centers compared to traditional data-centers. A geo-distributed cloud manager faces applications whose virtual machines (VMs) are far apart and need to interact with end users, access distributed data and communicate with each other. In such applications, the service level agreement is not met if the communication latency is not bounded. In this paper, we focus on the problem of finding data-centers for hosting VMs while the requested VMs are located in different geo-distributed data-centers and are sensitive to the communication latency. We propose an algorithm to minimize communication latency by taking into account the cloud network topology to be either a tree or that of the Internet. Moreover, our algorithm can utilize user’s locations to find better candidate solutions. In the case of tree topology, we prove that our algorithm finds a solution whose latency is minimum. In addition, we show that our algorithm performs well in the Internet topology, with simulation results indicating that it can reduce the communication latency up to $92~\%$ compared to existing algorithms.

Appendix

Proof of Lemma 1:

Proof

Let Findset(u, v) return the set $S \subset \mathcal {D}$. First, we prove that the latency of each two members of the set S is less than or equal to ltnc(u, v). Second, we prove that the set S is maximal.

Bound on the latency To prove a bound on the latency of the set S, it is enough to verify the latency between two optional leaves of the set S. It is clear from the definition of the set $\mathcal {D}$ that all members of the set S are leaves of a given tree. Since the latency and distance between two nodes are equal in a tree, it suffices to verify the distance, rather than their latency, between two arbitrary nodes a and b of the set S.

Let us consider the nodes a, b, u, $v \in S$ and the paths between them. The tree paths a–u, a–v and u–v have at least a common node. Because if these three paths do not meet at a common node, they build a cycle in the graph and contradict the assumption that the given graph is a tree. In the same way, three paths b–u, b–v and u–v have at least a common node. Let $t_1$ be a common node on paths a–u, a–v and u–v and $t_2$ be a common node on paths b–u, b–v and u–v.

Since there is a unique path between any two distinct nodes on the tree, the path a–u can split in two disjoint paths a–$t_1$ and $t_1$–u such that the following equation on their distances is obtained:

$$\begin{aligned} \mathrm{ltnc}(a,u)=\mathrm{ltnc}(a,t_1)+\mathrm{ltnc}(t_1,u) \end{aligned}$$

(1)

Also, we have the following equations for other mentioned paths:

$$\begin{aligned} \mathrm{ltnc}(v,u)= & {} \mathrm{ltnc}(v,t_1)+\mathrm{ltnc}(t_1,u) \end{aligned}$$

(2)

$$\begin{aligned} \mathrm{ltnc}(b,u)= & {} \mathrm{ltnc}(b,t_2)+\mathrm{ltnc}(t_2,u) \end{aligned}$$

(3)

$$\begin{aligned} \mathrm{ltnc}(v,u)= & {} \mathrm{ltnc}(v,t_2)+\mathrm{ltnc}(t_2,u) \end{aligned}$$

(4)

In addition, we have the following equations from the comparisons made in the FindSet function:

$$\begin{aligned} \mathrm{ltnc}(a,u)\le & {} \mathrm{ltnc}(u,v) \end{aligned}$$

(5)

$$\begin{aligned} \mathrm{ltnc}(a,v)\le & {} \mathrm{ltnc}(u,v) \end{aligned}$$

(6)

$$\begin{aligned} \mathrm{ltnc}(b,u)\le & {} \mathrm{ltnc}(u,v) \end{aligned}$$

(7)

$$\begin{aligned} \mathrm{ltnc}(b,v)\le & {} \mathrm{ltnc}(u,v) \end{aligned}$$

(8)

By knowing ltnc(i, j) equals to ltnc(j, i) and by substituting Eqs. (1) and (2) into Eq. (5), the following equation is obtained:

$$\begin{aligned} \mathrm{ltnc}(a,u)\le & {} \mathrm{ltnc}(u,v) \Rightarrow \nonumber \\ \mathrm{ltnc}(a,t_1) + \mathrm{ltnc}(t_1,u)\le & {} \mathrm{ltnc}(v,t_1) + \mathrm{ltnc}(t_1,u) \nonumber \\ \Rightarrow \mathrm{ltnc}(a,t_1)\le & {} \mathrm{ltnc}(v,t_1) \end{aligned}$$

(9)

Again, by substituting Eqs. (3) and (4) in Eq. (7), the following equation is obtained:

$$\begin{aligned} \mathrm{ltnc}(b,t_2)\le \mathrm{ltnc}(u,t_2) \end{aligned}$$

(10)

We assume in the path from u to v, first we visit $t_1$ and second we visit $t_2$. If it is vice versa, it is enough to rewrite the proof by changing the notations.

The following equation is established by considering the unique path from u to v in the tree:

$$\begin{aligned} \mathrm{ltnc}(v,u) = \mathrm{ltnc}(v,t_1) + \mathrm{ltnc}(t_1,t_2) + \mathrm{ltnc}(t_2,u) \end{aligned}$$

(11)

Adding up Eqs. (9)–(11) leads to

$$\begin{aligned} \mathrm{ltnc}(v,u) \ge \mathrm{ltnc}(a,t_1) + \mathrm{ltnc}(t_1,t_2) + \mathrm{ltnc}(t_2,b) \end{aligned}$$

(12)

Following equation is derived for latencies between the nodes $a\,,b\,,t_1$ and $t_2$:

$$\begin{aligned} \mathrm{ltnc}(a,t_1) + \mathrm{ltnc}(t_1,t_2) + \mathrm{ltnc}(t_2,b)\ge \mathrm{ltnc}(a,b) \end{aligned}$$

(13)

Finally, ltnc$(v,u) \ge \mathrm{ltnc}(a,b)$ is concluded from Eqs. (12) and (13). Because we assume that a and b are two arbitrary nodes of the set S, we can conclude that the distance between any two nodes of the set S is less than or equal to ltnc(u, v). Consequently, ltnc(S) equals ltnc(u, v).

Maximality Let the node d be a member of the set $\mathcal {D}$, which is not in the set S ($d \notin S\,$ and $d \in \mathcal {D}$). According to the definition of the FindSet function, at least one of the inequalities ltnc$(d,u) > \mathrm{ltnc}(u,v)$ or ltnc$(d,v) > \mathrm{ltnc}(u,v)$ is satisfied. Therefore, if the node d is added to the set S, the latency of the set S is increased and is not equal to ltnc(u, v) anymore. Since each optional node $d\notin S\,$ cannot be added to the set S, it is a maximal set.

Proof of Lemma 2:

Proof

Let the set S be the output of the FindSet(u, v) function. Suppose on the contrary that $n_1$ and $n_2$ are two members of the set S such that

$$\begin{aligned} \mathrm{ltnc}(n_1,n_2) > 2 \times \mathrm{ltnc}(u,v) \end{aligned}$$

(14)

Based on the comparisons were made in FindSet(u, v):

$$\begin{aligned} \mathrm{ltnc}(n_1,u)\le & {} \mathrm{ltnc}(u,v) \end{aligned}$$

(15)

$$\begin{aligned} \mathrm{ltnc}(n_2,u)\le & {} \mathrm{ltnc}(u,v) \end{aligned}$$

(16)

In addition, the triangle inequality between latencies states that

$$\begin{aligned} \mathrm{ltnc}(n_1,n_2) \le \mathrm{ltnc}(n_1,u)+ \mathrm{ltnc}(n_2,u) \end{aligned}$$

(17)

Adding up Eqs. (15), (16) and (17) leads to

$$\begin{aligned} \mathrm{ltnc}(n_1,n_2) \le 2 \times \mathrm{ltnc}(u,v) \end{aligned}$$

(18)

Overall, the output latency of FindSet(u, v) cannot be more than $2 \times \mathrm{ltnc}(u,v)$ because Eq. (18) contradicts the Eq. (14).