Reducing download delay for cooperative caching in small cell network

Abstract

As mobile devices become more and more popular, mobile traffic demands are increasing exponentially. To deal with such challenge, caching at small cell base stations (SBSs) is one of the most effective techniques for reducing the transmission delay of mobile applications. However, due to the limitation of the SBS storages, the improvement of reducing transmission delay is limited in small cell networks. In this paper, we propose a novel cooperative caching framework in a small cell network in which SBSs are grouped into disjoint clusters and cooperate through backhaul links to make use of SBS storages. A backhaul-based cooperative caching (BCC) scheme is introduced, and the problem for content placement is formulated to minimize the average download delay in small cell networks. Then, we show that the problem is equivalent to the maximization of a monotone submodular function subject to matroid constraints and propose a low-complexity greedy strategy with 1/2 performance guarantee. Simulation results demonstrate that the proposed scheme achieves lower download delay and better cache hit rate than other schemes.

A Derivation for Average Wireless Transmission Rate \({\bar{R}}\)

The file transmission rate can be estimated by applying the Shannon capacity formula in wireless networks as follows:

$$\begin{aligned} R_i = \frac{w}{U_i}\text {log}_2\left( 1 + \text {SINR}_i\right) , \end{aligned}$$

(22)

where w is radio bandwidth available for each SBS and, the serving SBS of user-i is $b_{i,1}$. We assume a busy system that $U_i$ denotes the number of users served by user-i’s $b_{i,1}$. In other words, including user-i, $U_i$ users have the same $b_{i,1}$, and each user shares the bandwidth equally for file delivery.

$$\begin{aligned} \text {SINR}_i = \frac{P_t d_i^{-\alpha }}{\rho + I} \end{aligned}$$

(23)

In the signal-to-interference-plus-noise ratio, $P_t$ is the SBS transmit power in mW. We assume that a user attaches to the nearest SBS, and $d_i$ is the distance between user-i and $b_{i,1}$ in meters. The pass loss exponent is presented as $\alpha$. Let $\rho$ denote the additive Gaussian noise and I denote the inter-cell interference both in mW. Before we obtain the average downloading delay of a request, it is necessary to define the average transmission rate ${\bar{R}}$.

$$\begin{aligned} {\bar{R}}&= \mathbb {E}\left[ R_i\right] \nonumber \\&= \mathbb {E}\left[ \frac{w}{U_i}\text {log}_2\left( 1+\text {SINR}_i\right) \right] \nonumber \\&= \frac{w\mathbb {E}\left[ \text {log}_2\left( 1+\frac{P_td_i^{-\alpha }}{\rho + I}\right) \right] }{\mathbb {E}[U_i]} \end{aligned}$$

(24)

Since the distributions of SBSs and users are modeled as PPP with density $\lambda _b$ and $\lambda _u$ respectively, the average number of users under the SBS is $\mathbb {E}[U_i] = \frac{\lambda _u}{\lambda _b}$, which only depends on the density of SBSs and users and is independent of the transmission distance. To simplify the formula, we derived its lower bound:

$$\begin{aligned}&\mathbb {E}\left[ \text {log}_2\left( 1+\frac{P_td_i^{-\alpha }}{\rho + I}\right) \right] \nonumber \\&\quad \ge \mathbb {E}\left[ \text {log}_2\left( \frac{P_td_i^{-\alpha }}{\rho + I}\right) \right] \nonumber \\&\quad = \text {log}_2\left( \frac{P_t}{\rho + I}\right) - \alpha \mathbb {E}[\text {log}_2d_i]. \end{aligned}$$

(25)

According to [30], the expected value of a logarithm $\mathbb {E}[\text {log}(x)]$ can be approximated in terms of $\mathbb {E}[x]$:

$$\begin{aligned} \mathbb {E}[\text {log}(x)] \approx \text {log}(\mathbb {E}[x]) - \frac{\mathbb {V}[x]}{2\mathbb {E}[x]^2}. \end{aligned}$$

(26)

Thus, the approximation of $\mathbb {E}[\text {log}(d_u)]$ is given by:

$$\begin{aligned} \mathbb {E}[\text {log}(d_i)]&\approx \text {log}(\mathbb {E}[d_i]) - \frac{\mathbb {V}[d_i]}{2\mathbb {E}[d_i]^2} \nonumber \\&= \text {log}(\mathbb {E}[d_i]) - \frac{\mathbb {E}[d_i^2]-\mathbb {E}[d_i]^2}{2\mathbb {E}[d_i]^2}. \end{aligned}$$

(27)

Now we need to solve both $\mathbb {E}[d_i]$ and $\mathbb {E}[d_i^2]$. As SBSs and users are both PPP distributed, we can obtain the probability of the nearest SBS found on the annulus with radius r centered at user-i is $2\lambda _b\pi r e^{-\lambda _b\pi r^2}$ [31]. Hence:

$$\begin{aligned} \mathbb {E}[d_i]&= \int _0^\infty r \cdot 2\lambda _b\pi re^{-\lambda _b\pi r^2}\ dr \nonumber \\&= -\int _0^\infty r \cdot (-2\lambda _b\pi re^{-\lambda _b\pi r^2})\ dr \nonumber \\&\left( \text{ let } u=e^{-\lambda _b\pi r^2},\ \frac{du}{dr}=-2\lambda _b\pi re^{-\lambda _b\pi r^2} \right) \nonumber \\&= -\int _0^\infty r \frac{du}{dr}\ dr = -\left( ru \bigg |_0^\infty - \int _0^\infty u\ dr\right) \nonumber \\&= -\left( \frac{r}{e^{\lambda _b\pi r^2}}\bigg |_0^\infty - \int _0^\infty e^{-\lambda _b\pi r^2}\ dr \right) \nonumber \\&= \int _0^\infty e^{-\lambda _b\pi r^2}\ dr \nonumber \\&\left( \text {Gaussian integral: } \int _{-\infty }^\infty e^{-sx^2}\ dx={\sqrt{\frac{\pi }{s}}} \right) \nonumber \\&=\frac{1}{2\sqrt{\lambda _b}}. \end{aligned}$$

(28)

Then let’s solve $\mathbb {E}[d_i^2]$:

$$\begin{aligned} \mathbb {E}[d_i^2]&= \int _0^\infty r^2 \cdot 2\lambda _b\pi re^{-\lambda _b\pi r^2}\ dr \end{aligned}$$

(29)

Let $u=e^{-\lambda _b\pi r^2},\ \frac{du}{dr}=-2\lambda _b\pi re^{-\lambda _b\pi r^2}, v=r^2$, and $\frac{dv}{dr}=2r$, then

$$\begin{aligned}&\mathbb {E}[d_i^2] = -\int _0^\infty v\ du\nonumber \\&\quad = -\left( vu \bigg |_0^\infty - \int _0^\infty u\ dv\right) \nonumber \\&\quad = -\left( \frac{r^2}{e^{\lambda _b\pi r^2}}\bigg |_0^\infty - \int _0^\infty 2re^{-\lambda _b\pi r^2}\ dr \right) \nonumber \\&\quad = \int _0^\infty 2re^{-\lambda _b\pi r^2}\ dr \nonumber \\&\quad = -\frac{1}{\lambda _b\pi } \int _0^\infty \left( -2\lambda _b\pi re^{-\lambda _b\pi r^2}\right) \ dr \nonumber \\&\quad = -\frac{1}{\lambda _b\pi } \int _0^\infty du \nonumber \\&\quad = -\frac{1}{\lambda _b\pi } \cdot \frac{1}{e^{\lambda _b\pi r^2}}\bigg |_0^\infty = -\frac{1}{\lambda _b\pi } (0-1) \nonumber \\&\quad =\frac{1}{\lambda _b\pi }. \end{aligned}$$

(30)

Substitute the results of Eq. (28) and Eq. (30) into Eq. (27):

$$\begin{aligned} \mathbb {E}[\text {log}d_i]&\approx \text {log}(\mathbb {E}[d_i]) - \frac{\mathbb {E}[d_i^2]-\mathbb {E}[d_i]^2}{2\mathbb {E}[d_i]^2} \nonumber \\&= \text {log}\left( \frac{1}{2\sqrt{\lambda _b}}\right) - \frac{4-\pi }{2\pi }. \end{aligned}$$

(31)

Finally, substitute into Eq. (24), $\bar{R_t}$ is derived as:

$$\begin{aligned} {\bar{R}}&= \mathbb {E}[R_i] = \frac{w\mathbb {E}\left[ \text {log}_2\left( \frac{P_td_i^{-\alpha }}{\rho + I}\right) \right] }{\mathbb {E}[U_i]} \nonumber \\&\quad \approx \frac{w\left[ \text {log}_2\left( \frac{P_t}{\rho + I}\right) - \alpha \mathbb {E}[\text {log}_2d_i]\right] }{\frac{\lambda _u}{\lambda _b}} \nonumber \\&\quad = \frac{\lambda _bw}{\lambda _u}\left\{ \text {log}_2\left( \frac{P_t}{\rho + I}\right) \right. \nonumber \\&\qquad \qquad \left. - \frac{\alpha }{\text {log}2}\left[ \text {log}\left( \frac{1}{2\sqrt{\lambda _b}}\right) - \frac{4-\pi }{2\pi }\right] \right\} \nonumber \\&\quad = \frac{\lambda _bw}{\lambda _u}\left[ \text {log}_2\left( \frac{P_t\left( \frac{1}{2\sqrt{\lambda _b}}\right) ^{-\alpha }}{\rho + I}\right) + \frac{\alpha (4-\pi )}{2\pi \text {log}2}\right] . \end{aligned}$$

(32)