Cloud service provisioning in two types of DCN with awareness of delay and link failure probability

Abstract

Cloud service based on data center network (DCN) has become an attractive choice for various applications. Traditionally, multiple DCs are distributed at different nodes across a given optical network, and users access DCs through predefined routes (this architecture is named as Multiple Independent DCN, MI-DCN). However, as there exist transmission delay and failure probability on each link, such a network may not be a good choice for the service providers from the perspective of service reliability and cost. Therefore, we propose the idea of regrouping all the racks and distributing each rack group on a special node, where there exists a gateway (this architecture is named as Integrated Distributed DCN, ID-DCN). As each group can provide service independently, by properly grouping and routing, the whole network can work more efficiently with lower cost and higher reliability. In this paper, we study the service provision in the above two types of DCN. With the given failure probability and transmission delay on each link, we aim to minimize the total service cost and design the access routes for the demands originated from each node. To integrate the system cost, we introduce two cost scaling factors for delay and failure probability, which can be flexibly adjusted to control their relative importance (i.e., the weights). Based on mathematical approximation, a novel method is proposed to compute the failure probabilities of individual service paths. This translates our objective function into a linear expression. Then, we formulate two integer linear programs (ILP) to compare the solutions of the two scenarios. Via extensive numerical experiments, the performance of the two schemes is properly verified.

Appendix proof of the theorem

Theorem

When we take the mathematic approximation in (3), it has an exact boundary which can be seen as inequation (5).

Proof

First, we list the Taylor expantion when getting Lagrange remainder term as below:

$$\begin{aligned} \hbox {e}^{x}=1+x+\hbox {e}^{\theta x}\frac{x^{2}}{2!}\ \theta \in \left( {0,1} \right) \end{aligned}$$

(22)

After moving the terms on both sides, we can have this:

$$\begin{aligned} 1-\hbox {e}^{x}=-x-\hbox {e}^{\theta x}\frac{x^{2}}{2!}\ \theta \in \left( {0,1} \right) \end{aligned}$$

(23)

Based on (2–3) and (23), we have:

$$\begin{aligned} \varepsilon _\mathcal{L}= & {} \mathop {\tilde{f}_\mathcal{L} } -f_\mathcal{L}\nonumber \\= & {} \frac{1}{2}\hbox {e}^{\theta \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \ln \left( {1-f_{uv} } \right) }\left[ {\mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \ln \left( {1-f_{uv} } \right) } \right] ^{2} \theta \in \left( {0,1} \right) \nonumber \\ \end{aligned}$$

(24)

Moreover, because the following inequation holds:

$$\begin{aligned} \frac{-x}{1-x}\le \ln \left( {1-x} \right) \le -x \end{aligned}$$

(25)

we have:

$$\begin{aligned}&\frac{-f_{uv} }{1-f_{uv} }\le \ln \left( {1-f_{uv} } \right) \le -f_{uv} \end{aligned}$$

(26)

$$\begin{aligned}&\mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \frac{-f_{uv} }{1-f_{uv} }\le \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \ln \left( {1-f_{uv} } \right) \le \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} -f_{uv} \end{aligned}$$

(27)

We take $f_\mathcal{L}^*$ to denote the maximum failure probability of all the links along $\mathcal{L}$. $f_\mathcal{L}^*$ can be expressed as this:

$$\begin{aligned} f_\mathcal{L}^*=\max \left\{ {f_{uv} ,\left( {u,v} \right) \in \mathcal{L}} \right\} \end{aligned}$$

(28)

Then, we have:

$$\begin{aligned} \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \frac{-f_{uv} }{1-f_\mathcal{L}^*}\le \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \ln \left( {1-f_{uv} } \right) \le \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} -f_{uv} \end{aligned}$$

(29)

Assume $r_\mathcal{L} =\mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} f_{uv}$, we have:

$$\begin{aligned} \frac{-1}{1-f_\mathcal{L}^*}r_\mathcal{L} \le \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \ln \left( {1-f_{uv} } \right) \le -r_\mathcal{L} \end{aligned}$$

(30)

Let $r_{\mathcal{L}^{*}} =\mathop {\min }\limits _\mathcal{L} \left\{ {r_\mathcal{L} } \right\}$ and $r_{\mathcal{L}^{+}} =\mathop {\max }\limits _\mathcal{L} \left\{ {r_\mathcal{L} } \right\}$, then we have:

$$\begin{aligned} \frac{-r_{\mathcal{L}^{+}} }{1-f_{\mathcal{L}^{+}}^*}\le \mathop \sum \limits _{\left( {u,v} \right) \in \mathcal{L}} \ln \left( {1-f_{uv} } \right) \le -r_{\mathcal{L}^{*}} \le 0 \end{aligned}$$

(31)

Therefore, from (24) and (31), we have:

$$\begin{aligned} \frac{1}{2}\hbox {e}^{\theta \frac{-r_{\mathcal{L}^{+}} }{1-f_{\mathcal{L}^{+}}^*}}\cdot r_{\mathcal{L}^{*}}^2 \le \varepsilon _\mathcal{L} \le -r_\mathcal{L}^*\le \frac{1}{2}\hbox {e}^{\theta r_{\mathcal{L}^{*}} }\left( {\frac{-r_{\mathcal{L}^{+}} }{1-f_{\mathcal{L}^{+}}^*}} \right) ^{2} \end{aligned}$$

(32)

Let $\theta =1$, we can get the result as in equation (5).

This proves the theorem. $\square$