Quick resource allocation in heterogeneous networks

Abstract

In this paper, an innovative technique is given to reduce the implementation time taken for doing the seamless communication in heterogeneous networks. When a new user arrives into the locality of one of the heterogeneous networks, resource allocation is to be done fastly by that heterogeneous network so as not to interrupt the data transfer to the newly arrived user. To do this, heterogeneous network allots optimal powers to S channels for maximizing the capacity of the newly arrived user while confirming to the interference bounds of the other users of the heterogeneous network and at the same time preserving the power (to save the energy) in heterogeneous network. To resolve this power allocation problem, the proposed algorithm finds out the tight lower bound and tight upper bound to the number of positive powers ($=N$) first and then finds out the indices of positive power allotted channels thereby discerning N itself (so as to calculate N powers exclusively). This is the unprecedented method as existing algorithms are iterative algorithms that deal with all of the assigned S channels. Hence, the proposed algorithm reduces the worst-case computational complexity of the erstwhile algorithms by the factor of $O(S^2)$.

Appendix

1.1 Proof for Proposition 5:

Proof

The covariance relation between two random variables A and B can be described as [39]

$$\begin{aligned} E\left( A B \right) = E\left( A \right) E\left( B\right) + Cov\left( A , B \right) \end{aligned}$$

(52)

wherein $E\left( A\right) = \frac{\sum _{v=1}^{V} a_v }{V}$ is mean of A random variable and $Cov\left( A , B \right)$ is covariance of A and B random variables. (52) is used in the following proof.

The below given proof is comprised of three segments.

Part 4: Finding bound for the water level ${\frac{1}{ E\left( \gamma _v\right) }}$

Figure 8 gives the water level for vth user in OPAN. In Fig. 8, the noise levels (${R}^{*}_{mv}$, m = 1, 2, $\ldots$) are arranged as bottom steps of a container. When water is poured above the bottom steps (or noise levels) upto the water level ($= \frac{1}{\gamma _{v}}$), the amount of water that is there above the mth noise level (or power allotted to the mth channel) is ${P}^{*}_{mv} = \frac{1}{\gamma _{v}}-{R}^{*}_{mv}$. The area of occupied water (or slashed lines part in Fig. 8) is $I_v$.

Fig. 8

Water level for vth user in OPAN

From Fig. 8, it can be grasped that the water level $\frac{1}{\gamma _v}$ $\propto$ $I_v$ because the height of water level is directly dependent on the total amount (or area) of occupied water. Hence, the covariance between $\gamma _v$ and $I_v$ is negative. Applying this in (52) for $\gamma _v$ and $I_v$ variables, we get

$$\begin{aligned}&E\left( \gamma _v I_v \right) = E\left( \gamma _v \right) E\left( I_v\right) + Cov\left( \gamma _v , I_v \right) \end{aligned}$$

(53)

$$\begin{aligned}&\Rightarrow E\left( \gamma _v I_v \right) < E\left( \gamma _v \right) E\left( I_v\right) \end{aligned}$$

(54)

In (53),

$$\begin{aligned} E\left( I_v\right) = \frac{\sum _{v=1}^{V} I_v}{V} \, and \, E\left( \gamma _v\right) = \frac{\sum _{v=1}^{V} \gamma _v}{V}. \end{aligned}$$

(55)

Likewise, the water level ($\frac{1}{\gamma _v}$) > the noise levels $\left( {R}^{*}_{mv}, m \in X_N \right)$ since power ($P^{*}_{mv} = \frac{1}{\gamma _v}$ - ${R}^{*}_{mv}$) > 0, m $\in X_N$ [7]. In Fig. 8, $X_N$ is given as {1,2, $\ldots$, N} for better understanding of the concepts. This further implies that the height of water level ($\frac{1}{\gamma _v}$) $\propto {R}^{*}_{mv}, m \le N$. Consecutively, the covariance between $\gamma _v$ and ${R}^{*}_{mv}$ is negative. Applying this concept in (52) for $\gamma _v$ and ${R}^{*}_{mv}$ variables, we get

$$\begin{aligned}&E\left( \gamma _v {R}^{*}_{mv} \right) = E\left( \gamma _v \right) E\left( {R}^{*}_{mv}\right) + Cov\left( \gamma _v , {R}^{*}_{mv} \right) \end{aligned}$$

(56)

$$\begin{aligned}&\Rightarrow E\left( \gamma _v {R}^{*}_{mv} \right) < E\left( \gamma _v \right) E\left( {R}^{*}_{mv}\right) \end{aligned}$$

(57)

(9)–(10) is again given here for convenience.

$$\begin{aligned}&\sum _{v=1}^{V} \gamma _v \, C_{v}= \sum _{m \in X_N } q_m \end{aligned}$$

(58)

$$\begin{aligned}&\, {\mathrm {with}}: C_{v} = I_{v} + \sum _{m \in X_N } q_m \, {R}^{*}_{mv} \end{aligned}$$

(59)

Substituting (59) in (58); and then simplifying it further, we obtain

$$\begin{aligned} \sum _{v=1}^{V} \gamma _v \, I_{v} + \sum _{m \in X_N } q_m \, \sum _{v=1}^{V} \gamma _v \, {R}^{*}_{mv} = \sum _{m \in X_N } q_m \end{aligned}$$

(60)

After using the definition of (55) in (60), (60) becomes

$$\begin{aligned} V E\left( \gamma _v \, I_{v}\right) + \sum _{m \in X_N } q_m \, V \, E \left( \gamma _v \, {R}^{*}_{mv} \right) = \sum _{m \in X_N } q_m \end{aligned}$$

(61)

Applying (54) and (57) in (61) , we procure

$$\begin{aligned}&\sum _{m \in X_N } q_m < V \, E\left( \gamma _v\right) \, E\left( I_{v}\right) + \, V \sum _{m \in X_N } q_m \, E\left( \gamma _v\right) \, E\left( {R}^{*}_{mv}\right) \end{aligned}$$

(62)

$$\begin{aligned}&\Rightarrow \frac{1}{ E\left( \gamma _v\right) } < \frac{ \sum _{v=1}^{V} I_{v} + \,\sum _{m \in X_N } q_m \, \sum _{v=1}^{V} {R}^{*}_{mv} }{\sum _{m \in X_N } q_m } \end{aligned}$$

(63)

Part 5: Procuring an additional bound for water level ${\frac{1}{ E\left( \gamma _v\right) }}$

This part of the proof is given by contradiction. That is, the relation for water level for $m \notin X_N$ is derived, and then, contradictory relation is applied for the water level for $m \in X_N$.

(17) identifies that the channels that are assigned with zero powers have

$$\begin{aligned} \sum _{v=1}^{V} \gamma _v \, {R}^{*}_{mv} = 1, m \notin X_N \end{aligned}$$

(64)

Applying the definition of (55) in (64) to get

$$\begin{aligned} V \, E\left( \gamma _v \, {R}^{*}_{mv} \right) = 1, m \notin X_N \end{aligned}$$

(65)

Figure 8 points out that ${R}^{*}_{(N+1)v}>$ the water level ($\frac{1}{\gamma _v}$) [7]. Then, it is evident that as water level ($\frac{1}{\gamma _v}$) increases, ${R}^{*}_{(N+1)v}$ also increases. In other words, ${R}^{*}_{(N+1)v} \propto \frac{1}{\gamma _v}$. But, $(N+1)\,\notin X_N$. Therefore, (57) can be applied as well to $m \notin X_N$. That is,

$$\begin{aligned} E\left( \gamma _v {R}^{*}_{mv} \right) < E\left( \gamma _v \right) E\left( {R}^{*}_{mv}\right) , m \notin X_N. \end{aligned}$$

(66)

Substituting (66) in (65), we attain

$$\begin{aligned}&1 < V \, E\left( \gamma _v \right) E\left( {R}^{*}_{mv}\right) , m \notin X_N \end{aligned}$$

(67)

$$\begin{aligned}&\Rightarrow \frac{1}{ E\left( \gamma _v \right) } < \,\sum _{v=1}^{V} {R}^{*}_{mv}, m \notin X_N \end{aligned}$$

(68)

Contrary to the above, the opposite to (68) occurs for $m \in X_N$. That is,

$$\begin{aligned} \frac{1}{ E\left( \gamma _v \right) } > \,\sum _{v=1}^{V} {R}^{*}_{mv}, m \in X_N \end{aligned}$$

(69)

Along with (15), one more similarity can be observed in between WFP and OPAN now. For vth exclusive WFP, $\frac{1}{\gamma _{_{v,WFP}}} > {R}^{*}_{Nv}$; whereas for OPAN, $\frac{1}{ E\left( \gamma _v \right) } > \,\sum _{v=1}^{V} {R}^{*}_{Nv}$ with $N = |X_N|$.

Part 6: Merging the above two proof parts to get the upper bound ${U_{N1}}$

From (63) and (69), we obtain

$$\begin{aligned}&\sum _{v=1}^{V} {R}^{*}_{mv}< \frac{1}{ E\left( \gamma _v\right) } < \left\{ \frac{1}{\gamma _{_{sum}}} \triangleq \right. \nonumber \\&\quad \left. \frac{ \sum _{v=1}^{V} I_{v} + \,\sum _{m \in X_N } q_m \, \sum _{v=1}^{V} {R}^{*}_{mv} }{\sum _{m \in X_N } q_m } \right\} \end{aligned}$$

(70)

Water level for vth exclusive WFP is [7]

$$\begin{aligned} \frac{1}{\gamma _{_{v, WFP}}} = \frac{C_{v}}{\sum _{m \in X_N } q_m} = \frac{I_{v} + \sum _{m \in X_N } q_m \, {R}^{*}_{mv} }{\sum _{m \in X_N } q_m} \end{aligned}$$

(71)

Likewise, the water level calculated for sum of noise levels $\left( {R}^{*}_{m} = \sum _{v=1}^{V} {R}^{*}_{mv}\right)$ is :

$$\begin{aligned} \frac{1}{\gamma _{_{sum}}} \triangleq \frac{ \sum _{v=1}^{V} I_{v} + \,\sum _{m \in X_N } q_m \, \sum _{v=1}^{V} {R}^{*}_{mv} }{\sum _{m \in X_N } q_m }. \end{aligned}$$

(72)

The above is obtained from (70). Water level calculated for the sum of noise levels is shown in Fig. 9. Comparing both the water levels from (71) and (72), it can be perceived here that

• 1. Water level for vth exclusive WFP ($=\frac{1}{\gamma _{_{v, WFP}}}$) is calculated for the individual noise levels ${R}^{*}_{mv}$ where as $\frac{1}{\gamma _{_{sum}}}$ is calculated for the sum of noise levels ${R}^{*}_{m}$ = $\sum _{v=1}^{V} {R}^{*}_{mv}$.

• 2. The area of water occupied in vth exclusive WFP is $I_{v}$; whereas the area of water occupied for the water level of $\frac{1}{\gamma _{_{sum}}}$ is $\sum _{v=1}^{V} I_{v}$.

• 3. For vth exclusive WFP, $\frac{1}{\gamma _{_{v, WFP}}}$ > ${R}^{*}_{mv}$, m $\in X_N$ where as $\frac{1}{\gamma _{_{sum}}} > \sum _{v=1}^{V} {R}^{*}_{mv}$, m $\in X_N$ from (70).

(69) infers that ‘$\frac{1}{ E\left( \gamma _v\right) } > \sum _{v=1}^{V} {R}^{*}_{mv}$’ is followed for $X_N$ values of m. Further, ‘$\frac{1}{\gamma _{_{sum}}} > \frac{1}{ E\left( \gamma _v\right) }$’ from (70). Then, it can be inferred that ‘$\frac{1}{\gamma _{_{sum}}} > \sum _{v=1}^{V} {R}^{*}_{mv}$’ is fulfilled either for $X_N$ values of m (or) for more than $X_N$ values of m (i.e. for $U_{N1} > |X_N|$).

Fig. 9

Water level for sum of noise levels ${R}^{*}_{m} = \sum _{v=1}^{V} {R}^{*}_{mv}$ in OPAN

Part 7: Obtaining another upper bound ${U_{N2}}$

‘$\frac{1}{ E\left( \gamma _v\right) }$’ represents the water level for the mean of $\gamma _v$’s. Taking into account that OPAN includes vth exclusive WFP, upper bound for the number of positive powers can also be obtained from all the V exclusive WFPs. Subsequently, the upper bound $U_{N2}$ becomes equivalent to the mean of $X_{_{v,WFP}}$, $v \le V$ wherein $X_{_{v,WFP}}$ is attained as the ‘m’ values for which $\{ \frac{1}{\gamma _{_{v, WFP}}} \ge {R}^{*}_{mv}, \hbox { m } \le S \}$ is fulfilled.

Part 8: Upper bound from ${U_{N1}}$ and ${U_{N2}}$

Finally, the upper bound for the number of positive powers in OPAN is $U_{N}$ = maximum of $U_{N1}$ and $U_{N2}$.

Hence, the Proposition is proved.$\square$