Hierarchical coded caching with heterogeneous cache sizes

Abstract

In this paper, we delve into the intricacies of a network architecture enabled by two layers of caches, where users receive their requested content via intermediate helpers connected to a central server. While coded caching in a two-layer hierarchical model has previously demonstrated its potential to enhance data rates when cache capacities are uniform and no coordination exists between users, our work takes a leap further. We introduce the dimension of heterogeneous cache sizes among both the helpers and users, addressing scenarios where the number of popular files can be either less or more than the number of users within the network. Leveraging a recently proposed modified coded caching scheme, combined with a zero-padding technique, we present novel results on data rates, supported by an illustrative example. Our contribution extends to the formulation of two distinct coded schemes within the hierarchical scenario. Furthermore, we optimize the proportions of files and memories allocated to each scheme, facilitating data transfer efficiency and then derive the lower bound for the total rate. Moreover, we demonstrate that the total rate achieved by the proposed heterogeneous approach is lower than that of a homogeneous network with caches equivalent to the minimum cache size present in the homogeneous network. However, it is more than a homogeneous network with the similar average cache size. In addition, we illustrate by proper selection of the proportions of files and memories allocated to each scheme, we can decrease the performance degradation due to the heterogeneity of the network.

Appendices

Appendices

1.1 Appendix A: Proof of Theorem 1

In this section, we derive the approximate optimum value of $\alpha$ and $\beta$ to minimize $R_T = R_1+k_1 R_2$. Consider two cases.

A. Case $N\ge k_1 k_2$: In $R_T$, we can ignore $R_1^A$ and $R_1^B$ compare $k_1 R_2$ because $k_1 R_2$ has the coefficient of $k_1$ and for $R_1^A$ which has the coefficient of $\alpha k_2$, we have

$$\begin{aligned} \sum _{i=1}^{k_1} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) \ll \sum _{i=1}^{k_2} \prod _{j=1}^{i} \left( 1-\frac{\beta M_U^j}{\alpha N}\right) , \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^{k_1} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) \ll \sum _{i=1}^{k_2} \prod _{j=1}^{i} \left( 1-\frac{(1-\beta )M_U^j}{(1-\alpha )N}\right) . \end{aligned}$$

This is because the memory sizes of the helpers are much more than the memory sizes of the users. So we have $\min {R_T} \approx \min k_1R_2.$

By setting $\alpha$ = $\beta$ in $\frac{\partial R_2(\alpha ,\beta )}{\partial \beta }$, we have $\frac{\partial R_2(\alpha ,\beta )}{\partial \beta } =0$ and by setting $\alpha$ = $\beta$ in $\frac{\partial ^2 R_2(\alpha ,\beta )}{\partial \beta ^2}$, we have $\frac{\partial ^2 R_2(\alpha ,\beta )}{\partial \beta ^2} \ge 0.$ By setting $\alpha = \beta$ in $R_T = R_1+ k_1R_2$, we have

$$\begin{aligned} R_T(\alpha )= & {} \alpha k_2 \sum _{i=1}^{k_1} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) \\{} & {} +(1-\alpha ) \sum _{i=1}^{k_1 k_2} \prod _{j=1}^{i} \left( 1-\frac{M_U^{j-\lfloor \frac{j}{k_2} \rfloor k_2}}{N}\right) \\{} & {} + k_1 \sum _{i=1}^{k_2} \prod _{j=1}^{i} \left( 1-\frac{M_U^j}{N}\right) \\= & {} R_{T,1}(\alpha ) + R_{T,2}(\alpha ) +R_{T,3}. \end{aligned}$$

Notice $R_{T,3}$ is constant, so we calculate $\min ( R_{T,1}(\alpha ) + R_{T,2}(\alpha )).$ Values of $R_{T,1}(\alpha )$ are zero in the interval of $\alpha \in [0, \frac{M_H^1}{N}]$ and then increases to $k_2 \sum _{i=1}^{k_1} \prod _{j=1}^{i} (1-\frac{M_H^j}{N})$. On the other hand, $R_{T,2}(\alpha )$ decreases linearly, so in the interval of $\alpha \in [0, \frac{M_H^1}{N}]$, the minimum of $R_{T,1}(\alpha ) + R_{T,2}(\alpha )$ occurs in the point $\alpha =\frac{M_H^1}{N}$. For the interval of $\alpha \in [\frac{M_H^1}{N}, 1]$, the linear approximation of $R_{T,1}(\alpha ) + R_{T,2}(\alpha )$ can be written and the gradient of the line or

$$\begin{aligned} \frac{k_2 \sum _{i=1}^{k_1} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{N}\right) - \left( 1-\frac{M_H^1}{N}\right) \sum _{i=1}^{k_1 k_2} \prod _{j=1}^{i} \left( 1-\frac{M_U^{j- \lfloor \frac{j}{k_2} \rfloor k_2}}{N}\right) }{1-\frac{M_H^1}{N}} \end{aligned}$$

is positive, so in the interval of $\alpha \in [\frac{M_H^1}{N}, 1]$, we have an increasing function which has the minimum value in $\frac{M_H^1}{N}$.

B. Case $N<k_1 k_2$: Among different parts, $R_1^A$ and $k_1 R_2$ have the factor of $\alpha N$ and $k_1$, respectively, so we can ignore $R_1^B$ compared to them. On the other hand, $R_1^A$ includes $\sum _{i=1}^{N} \prod _{j=1}^{i} (1-\frac{M_H^j}{\alpha N})$ and $k_1 R_2$ includes $\sum _{i=1}^{N} \prod _{j=1}^{i} (1-\frac{\beta M_U^j}{\alpha N})$ or $\sum _{i=1}^{N} \prod _{j=1}^{i} (1-\frac{(1-\beta )M_U^j}{(1-\alpha )N})$. We know that the memory sizes of the helpers are much more than the memory sizes of the users. So

$$\begin{aligned} \sum _{i=1}^{N} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) \ll \sum _{i=1}^{N} \prod _{j=1}^{i} \left( 1-\frac{\beta M_U^j}{\alpha N}\right), \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^{N} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) \ll \sum _{i=1}^{N} \prod _{j=1}^{i} \left( 1-\frac{(1-\beta )M_U^j}{(1-\alpha )N}\right) . \end{aligned}$$

Hence, we can ignore $R_1^A$ compared to $k_1 R_2$ and $\min {(R_T)} \approx \min ({k_1 R_2}).$ Setting $\alpha = \beta$ in $\frac{\partial R_2(\alpha ,\beta )}{\partial \beta }$, we have $\frac{\partial R_2(\alpha ,\beta )}{\partial \beta } =0,$ and setting $\alpha$ = $\beta$ in $\frac{\partial ^2 R_2(\alpha ,\beta )}{\partial \beta ^2}$, we have $\frac{\partial ^2 R_2(\alpha ,\beta )}{\partial \beta ^2} \ge 0.$ By setting $\alpha =\beta$ in $R_T$ and setting $\alpha =\frac{M_H^1}{N}$ in $\frac{\partial R_T(\alpha )}{\partial \alpha }$, we have

$$\begin{aligned} \frac{\partial R_T\left( \frac{M_H^1}{N}\right) }{\partial \alpha }= 0 + N -\sum _{i=1}^{N} \prod _{j=1}^{i} \left( 1-\frac{M_U^j}{N}\right) . \end{aligned}$$

Because a large value of N, $\sum _{i=1}^{N} \prod _{j=1}^{i} (1-\frac{M_U^j}{N})$ tends to be N, so we have $\frac{\partial R_T(\alpha )}{\partial \alpha }= 0.$

1.2 Appendix B: Proof of Theorem 2

In this section, we derive the information-theoretic lower bound on $R_T$ under the heterogeneous cache sizes which is independent of any specific scheme.

In the first layer, the total memories of the helpers, the total memories of the users, and the total information signals which are transmitted by the server must be at least equal to the size of different files which are reconstructed by the users. Let $s_1 \in \{1,2,....., \min \{N,k_1\}\}$, $s_2 \in \{1,2,....., min\{N,k_2\}\}$ and the set of users (i, j) which $i \in \{1,2,....., s_1\}$ and $j \in \{1,2,....., s_2\}$. Then we have the following cut-set bound for $R_1$:

$$\begin{aligned}{} & {} \max _{\begin{array}{c} U_1 \subset \{1,2,....., min\{N,k_1\}\} \\ |U_1|=s_1 \end{array}} \left\{ s_1 \sum _{\begin{array}{c} l \in U_1 \end{array}} M_H^l \right\} \\{} & {} \quad + \max _{\begin{array}{c} U_2 \subset \{1,2,....., \min \{N,k_2\}\} \\ |U_2|=s_2 \end{array}} \left\{ s_1s_2 \sum _{\begin{array}{c} l \in U_1 \end{array}} M_U^l \right\} \\{} & {} \quad + \left\lfloor \frac{N}{s_1 s_2} \right\rfloor R^*_1 \ge s_1s_2 \left\lfloor \frac{N}{s_1 s_2} \right\rfloor \end{aligned}$$

so we have

$$\begin{aligned} R^*_1\ge & {} s_1s_2\left( 1- \frac{\max _{\begin{array}{c} U_1 \subset \{1,2,....., min\{N,k_1\}\} \\ |U_1|=s_1 \end{array}} \left\{ s_1 \sum _{\begin{array}{c} l \in U_1 \end{array}} M_H^l \right\} }{N-s_1s_2}\right. \\{} & {} + \left. \frac{\max _{\begin{array}{c} U_2 \subset \{1,2,....., \min \{N,k_2\}\} \\ |U_2|=s_2 \end{array}} \left\{ s_1s_2 \sum _{\begin{array}{c} l \in U_1 \end{array}} M_U^l \right\} }{N-s_1s_2}\right) \end{aligned}$$

which can be rewritten as

$$\begin{aligned} R^*_1&\ge \max _{\begin{array}{c} s_1 \in \{1,2,....., \min \{N,k_1\}\} \\ s_2 \in \{1,2,....., \min \{N,k_2\}\} \end{array}}\\&\quad \left\{ s_1s_2 \left( 1- \frac{\max _{\begin{array}{c} U_1 \subset \{1,2,....., min\{N,k_1\}\} \\ |U_1|=s_1 \end{array}} \left\{ s_1 \sum _{\begin{array}{c} l \in U_1 \end{array}} M_H^l \right\} }{N-s_1s_2}\right. \right. \\&\quad \left. \left. + \frac{\max _{\begin{array}{c} U_2 \subset \{1,2,....., \min \{N,k_2\}\} \\ |U_2|=s_2 \end{array}} \left\{ s_1s_2 \sum _{\begin{array}{c} l \in U_1 \end{array}} M_U^l \right\} }{N-s_1s_2 } \right) \right\} \\&\triangleq R^{lb}_1 (\overrightarrow{\textbf{M}_{\textbf{H}}}, \overrightarrow{\textbf{M}_{\textbf{U}}}) \end{aligned}$$

We go through the same way for the rate between the helpers and their users. We have

$$\begin{aligned}{} & {} \max _{\begin{array}{c} U_2 \subset \{1,2,....., \min \{N,k_2\}\} \\ |U_2|=s_2 \end{array}} \\{} & {} \quad \left\{ s_2 \sum _{\begin{array}{c} l \in U_2 \end{array}} M_U^l \right\} + \left\lfloor \frac{N}{s_2} \right\rfloor R^*_2 \ge s_2 \left\lfloor \frac{N}{s_2} \right\rfloor \end{aligned}$$

so we have

$$\begin{aligned} R^*_2\ge s_2\left( 1- \frac{\max _{\begin{array}{c} U_2 \subset \{1,2,....., \min \{N,k_2\}\} \\ |U_2|=s_2 \end{array}} \left\{ s_2 \sum _{\begin{array}{c} l \in U_2 \end{array}} M_U^l \right\} }{N-s_2}\right) \end{aligned}$$

and finally

$$\begin{aligned} R^*_2&\ge \max _{\begin{array}{c} s_2 \in \{1,2,....., min\{N,k_1\}\} \end{array}}\\&\quad \left\{ s_2\left( 1- \frac{\max _{\begin{array}{c} U_2 \subset \{1,2,....., min\{N,k_2\}\} \\ |U_2|=s_2 \end{array}} \left\{ s_2 \sum _{\begin{array}{c} l \in U_2 \end{array}} M_U^l \right\} }{N-s_2} \right) \right\} \\&\triangleq R^{lb}_2 (\overrightarrow{\textbf{M}_{\textbf{H}}}, \overrightarrow{\textbf{M}_{\textbf{U}}}) \end{aligned}$$

Now, by obtaining the lower bounds for $R^*_1$ and $R^*_2$, the lower bound for $R_T = R_1+k_1 R_2$ is derived as the above cut-set bound.

1.3 Appendix C: Comparing the rate of the proposed heterogeneous method and the rate of a homogeneous model with minimum cache size

The first rate between the server and the helpers when $N \ge k_1 k_2$ is

$$\begin{aligned} R_1^{Het}(\alpha ,\beta )= & {} R_1^{Het-A}(\alpha ,\beta ) +R_1^{Het-B}(\alpha ,\beta ) \\= & {} \alpha k_2 \sum _{i=1}^{k_1} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) \\{} & {} + (1-\alpha )\sum _{i=1}^{k_1 k_2} \prod _{j=1}^{i} \left( 1 - \frac{(1-\beta )M_U^{j}}{(1-\alpha )N}\right) . \end{aligned}$$

If the memory sizes of all helpers are equivalent and equal to the minimum size, and also the memory sizes of all users are equivalent and equal to the minimum size, we have:

$$\begin{aligned} R_1^{Hom}(\alpha ,\beta )= & {} R_1^{Hom-A}(\alpha ,\beta ) +R_1^{Hom-B}(\alpha ,\beta ) \\= & {} \alpha k_2 \sum _{i=1}^{k_1} \left( 1-\frac{M_H^1}{\alpha N}\right) ^i \\{} & {} + (1-\alpha )\sum _{i=1}^{k_1 k_2} \left( 1 - \frac{(1-\beta )M_U^{1}}{(1-\alpha )N}\right) ^i. \end{aligned}$$

Taking into account the summation of the geometric progression, we obtain:

$$\begin{aligned}{} & {} R_1^{Hom}(\alpha ,\beta ) \\{} & {} \quad = R_1^{Hom-A}(\alpha ,\beta ) +R_1^{Hom-B}(\alpha ,\beta ) \\{} & {} \quad =\alpha k_2 \left[ \left( 1 -\left( 1-\frac{M_H^1}{\alpha N}\right) ^{k_1}\right) \left( \frac{\alpha N}{M_H^1}-1\right) \right] \\{} & {} \qquad + (1-\alpha ) \left[ \left( 1- \left( 1-\frac{(1-\beta )M_U^1}{(1-\alpha ) N}\right) ^{k_1 k_2}\right) \right. \\{} & {} \qquad \qquad \quad \left. \left( \frac{(1-\alpha ) N}{(1-\beta )M_U^1}-1\right) \right] , \end{aligned}$$

which is the rate of the first layer in the homogeneous cache size scheme presented in [4]. On the other hand, we assume that $M_H^1< M_H^2<... < M_H^{k_1}$, so we have $(1-\frac{M_H^1}{\alpha N})> (1-\frac{M_H^2}{\alpha N}) > . . . (1-\frac{M_H^{k_1}}{\alpha N}).$ As $(1-\frac{M_H^i}{\alpha N}) < 0$, for each i we have $(1-\frac{M_H^1}{\alpha N})^i > \prod _{j=1}^{i} (1-\frac{M_H^i}{\alpha N}),$ so

$$\begin{aligned} \alpha k_2 \sum _{i=1}^{k_1} \left( 1-\frac{M_H^1}{\alpha N}\right) ^i >\alpha k_2 \sum _{i=1}^{k_1} \prod _{j=1}^{i} \left( 1-\frac{M_H^j}{\alpha N}\right) , \end{aligned}$$

which leads to

$$\begin{aligned} R_1^{Hom-A}(\alpha ,\beta ) > R_1^{Het-A}(\alpha ,\beta ). \end{aligned}$$

With the same argument, we have $R_1^{Hom-B}(\alpha ,\beta ) > R_1^{Het-B}(\alpha ,\beta )$ and therefore

$$\begin{aligned} R_1^{Hom}(\alpha ,\beta ) > R_1^{Het}(\alpha ,\beta ). \end{aligned}$$

Similar to the above argument, we have

$$\begin{aligned} R_2^{Hom}(\alpha ,\beta ) > R_2^{Het}(\alpha ,\beta ). \end{aligned}$$

Now since $R_T(\alpha ,\beta ) = R_1(\alpha ,\beta )+ k_1 R_2(\alpha ,\beta )$, we have

$$\begin{aligned} R_T^{Hom}(\alpha ,\beta ) > R_T^{Het}(\alpha ,\beta ). \end{aligned}$$