Low-complexity uplink scheduling algorithms with power control in successive interference cancellation based wireless mud-logging systems

Abstract

Wireless mud-logging systems have been employed in petroleum exploration to guarantee efficiency and safety in well-drilling. For the wireless networks in well-drillings, the tradeoff between throughput and fairness has great relation to the type of wells. The throughput is of the most importance to deep wells, while fairness is of primary concern for shallow wells. In this study, this challenge is addressed by using 2-SIC (successive interference cancellation) techniques. The throughput maximization and the fairness maximization problems are further formulated based on an imperfect 2-SIC model, which differentiates our work from existing ones. In view of the hardness of these problems, a low-complexity approximation algorithm for maximizing throughput and a low-complexity heuristic algorithm for maximizing fairness are proposed. To evaluate the proposed schemes, extensive simulations are performed. According to simulation results, the throughput performance could be improved by 23% using our approximation algorithm of throughput maximization, while the Jain’s fairness index could reach 0.975 using our algorithm of fairness optimization.

Appendices

Appendix 1: Proving Theorem 1

Lemma 5

If$(p_i^*,p_j^*)$is the optimal solution of PCMT-TN, then$p_i^*=p_{max}$or$p_j^*=p_{max}$.

Proof

Assume there is a feasible solution $({p_i'},{p_j'})$. If ${p_i'}\ge {p_j'}$, $(k{p_i'},k{p_j'})$ is still feasible for all $0<k \le \frac{p_{max}}{p_i'}$. Besides, $Th_{i,j}(k{p_i'},k{p_j'})$ is an increasing function of k. So, the optimal solution satisfies $p_i^*=p_{max}$. Similarly, If ${p_i'}<{p_j'}$, the optimal solution satisfies $p_j^*=p_{max}$.$\square$

Proof of Theorem 1

Case 1 Assume $p_j^*=p_{max}$

In this case, PCMT-TN problem is actually

$$\begin{aligned} \mathop {\max }\limits _{{p_i}} T{h_{i,j}}\left( {{p_i},{p_{\max }}} \right) \quad s.t.\quad {G_i}{p_i} \ge {G_j}{p_{\max }}\quad 0 \le {p_i} \le {p_{\max }} . \end{aligned}$$

It can be known that

$$\begin{aligned} \frac{{dT{h_{i,j}}\left( {{p_i},{p_{\max }}} \right) }}{{d{p_i}}} &=\frac{{{\varepsilon ^2}{G_i}^3{p_i}^2 + 2\varepsilon {G_i}^2{N_0}{p_i} + {N_0}^2{G_i}}}{{\left( {{G_j}{p_{\max }} + {N_0}} \right) {{(\varepsilon {G_i}{p_i} + {N_0})}^2}}}\\&\quad+ \frac{{\left( {1 - \varepsilon } \right) {G_i}{G_j}{N_0}{p_{\max }} - \varepsilon {G_i}{G_j}^2{p_{\max }}^2}}{{\left( {{G_j}{p_{\max }} + {N_0}} \right) {{(\varepsilon {G_i}{p_i} + {N_0})}^2}}}. \end{aligned}$$

Since ${\frac{{dTh_{i,j}({p_i,p_{max}})}}{dp_i}} > 0$  for all ${p_i} \ge \frac{{{G_j}{p_{\max }}}}{{{G_i}}}$. $T{h_{i,j}}\left( {{p_i},{p_{\max }}} \right)$ will increase with increasing $p_i$ if ${p_i} \ge \frac{{{G_j}{p_{\max }}}}{{{G_i}}}$. Thus, the maximum is achieved when $p_i=p_{max}$, and the maximum throughput is $T{h_{i,j}}\left( {{p_{\max }},{p_{\max }}} \right)$.

In conclusion, under Cases 1, the solution is always $(p_{max},p_{max})$.

Case 2 Assume $p_j^*<p_{max}$

In this case, $p_i^*=p_{max}$, thus, PCMT-TN problem is

$$\begin{aligned} {\mathop {\max }\limits _{{p_j}} T{h_{i,j}}\left( {{p_{\max }},{p_j}} \right) \quad s.t. \quad 0 \le {p_j} < {p_{\max }}}. \end{aligned}$$

In Case 2, we further divide the problem into two subcases, i.e., Cases 2.1 and 2.2.

Case 2.1  $\sqrt{\varepsilon }{G_i}{p_{\max }} - {N_0} \le 0$. Since

$$\begin{aligned}&\frac{{dT{h_{i,j}}\left( {{p_{\max }},{p_j}} \right) }}{{d{p_j}}}= \frac{{{G_j}({{({G_j}{p_j} + {N_0})}^2} - \varepsilon {G_i}^2{p_{\max }}^2)}}{{\left( {\varepsilon {G_i}{p_{\max }} + {N_0}} \right) {{({G_j}{p_j} + {N_0})}^2}}} , \end{aligned}$$

it can be seen that $\frac{dTh_{i,j} (p_{max},p_j)}{dp_j} \ge 0$ for any $p_j \in [0,p_{max})$. Therefore, the optimal solution in this case is $(p_{max},p_{max}-\delta )$, where $\delta$ is a sufficiently small positive number.

Since the optimal solution in Case 1 is $(p_{max},p_{max})$, the optimal solution is $(p_{max},p_{max})$ if $\sqrt{\varepsilon }{G_i}{p_{\max }} - {N_0} \le 0$.

Case 2.2  $\sqrt{\varepsilon }{G_i}{p_{\max }} - {N_0} > 0$. Note that

$$\begin{aligned} \frac{{dT{h_{i,j}}\left( {{p_{\max }},{p_j}} \right) }}{{d{p_j}}} = {\left\{ \begin{array}{ll}<\!0 &{} \quad 0 \le {p_j} < \frac{{\sqrt{\varepsilon }{G_i}{p_{\max }} - {N_0}}}{{{G_j}}} \\>\!0 &{} \quad {p_j} > \frac{{\sqrt{\varepsilon }{G_i}{p_{max}} - {N_0}}}{{{G_j}}} \\ \end{array}\right. } . \end{aligned}$$

In Case 2.2, we further divide the problem into two cases: Cases 2.2.1 and 2.2.2.

Case 2.2.1  If $(\sqrt{\varepsilon } G_i p_{max}-N_0)/G_j \ge p_{max}$, i.e., $p_{max} \ge \ N_0/(\sqrt{\varepsilon } G_i-G_j)$. At this time, $Th_{i,j}(p_{max},p_j)$ is shown in Fig. 12(a). Obviously, the optimal solution is obtained when $p_j=0$, i.e., the optimal solution in this case is $(p_{max},0)$.

Fig. 12

$Th_{i,j}(p_i^*,p_j)$ when $p_i^*=p_{max}$

Case 2.2.2  If $p_{max}<N_0/(\sqrt{\varepsilon } G_i-G_j )$, $Th_{i,j}(p_{max},p_j)$ in Case 2.2.2 is shown in Fig. 12(b). Obviously, the optimal solution is either $(p_{max},p_{max})$ or $(p_{max},0)$. Specifically, if $Th_{i,j}(p_{max},p_{max}) \ge Th_{i,j}(p_{max},0)$, the optimal solution is $(p_{max},p_{max})$, and it is $(p_{max},0)$ otherwise. $\square$

Appendix 2: Proving Theorem 2

Lemma 6

The optimal solution$(p_i^*,p_j^*)$for PCMF-TN satisfies$p_i^*=p_{max}$or$p_j^*=p_{max}$.

Proof

The proof is similar with that of Lemma 5.$\square$

Lemma 7

The optimal solution$(p_i^*,p_j^*)$for PCMF-TN is shown in (11), where${{\widehat{p}}_i} = \frac{{\sqrt{{N_0}^2 + 4\varepsilon {G_j}^2{p_{\max }}^2 + 4{N_0}\varepsilon {G_j}{p_{\max }}} - {N_0}}}{{2\varepsilon {G_i}}}$, ${{\widehat{p}}_j} = \frac{{\sqrt{{N_0}^2 + 4\varepsilon {G_i}^2{p_{\max }}^2 + 4{N_0}{G_i}{p_{\max }}} - {N_0}}}{{2{G_j}}}$, and $F{a_{i,j}}\left( {{p_i},{p_j}} \right) = \min \left( {1 + \frac{{{G_i}{p_i}}}{{{G_j}{p_j} + {N_0}}},1 + \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_i} + {N_0}}}} \right)$.

Proof

Based on Lemma 6, two possibilities of the optimal solution of PCMF-TN are discussed separately and then merged as follows:

Case 1 Assume $p_i^*=p_{max}$

In this case, PCMF-TN problem is actually formulated as follows:

$$\begin{aligned} \mathop {\max }\limits _{{p_j}} Fa_{i,j}({p_{max}},{p_j}) \quad s.t. \quad 0 \le {p_j} \le {p_{max }} . \end{aligned}$$

Since ${{\widehat{p}}_j}=\frac{{\sqrt{{N_0}^2 + 4\varepsilon {G_i}^2{p_{\max }}^2 + 4{N_0}{G_i}{p_{\max }}} - {N_0}}}{{2{G_j}}}$, it can be easily verified that $\frac{{{G_i}{p_{\max }}}}{{{G_j}{{{\widehat{p}}}_j} + {N_0}}} = \frac{{{G_j}{{{\widehat{p}}}_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}$.

Case 1.1    ${{\widehat{p}}_j}<p_{max}$

Case 1.1.1    For any $p_j \in [0,{{\widehat{p}}_j})$,

since $\frac{{{G_i}{p_{max }}}}{{{G_j}{p_j} + {N_0}}}> \frac{{{G_i}{p_{max}}}}{{{G_j}{{{\widehat{p}}}_j} + {N_0}}} = \frac{{{G_j}{{{\widehat{p}}}_j}}}{{\varepsilon {G_i}p_{max}+N_0}} > \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{max}} + N_0}}$, $Fa_{i,j}({{p_{max}},{p_j}}) = min \left( {1 + \frac{{{G_i}{p_{max}}}}{{{G_j}{p_j} + {N_0}}},1 + \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{max}} + {N_0}}}}\right) = 1 + \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{max}} + {N_0}}} < 1 + \frac{{{G_j}{{{\widehat{p}}}_j}}}{{\varepsilon {G_i}{p_{max}} + {N_0}}} = F{a_{i,j}}({{p_{max}},{{{\widehat{p}}}_j}}).$

Case 1.1.2   For any $p_j \in ({{\widehat{p}}_j},p_{max}]$, since $\frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}> \frac{{{G_j}{{{\widehat{p}}}_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}} = \frac{{{G_i}{p_{\max }}}}{{{G_j}{{{\widehat{p}}}_j} + {N_0}}} > \frac{{{G_i}{p_{\max }}}}{{{G_j}{p_j} + {N_0}}}$,

$$\begin{aligned}&Fa_{i,j}({{p_{max}},{p_j}}< F{a_{i,j}}\left( {{p_{\max }},{{{\widehat{p}}}_j}} \right) . \end{aligned}$$

(10)

In conclusion, for Case 1.1, the optimal solution for PCMF-TN is $(p_{max},{{{\widehat{p}}}_j})$.

Case 1.2   ${{\widehat{p}}_j} \ge p_{max}$

It can be proved that for all $p_j \in [0,{{\widehat{p}}}_j]$, $\frac{{{G_i}{p_{max}}}}{{{G_j}{p_j} + {N_0}}} \ge \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}$ is true. Its proof is as follows: Obviously, $p_j \in [0,{{{\widehat{p}}}_j]}$, $\frac{{{G_i}{p_{max}}}}{{{G_j}{p_j} + {N_0}}} \ge \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}$ is equivalent to ${G_j}^2{p_j}^2 + {N_0}{G_j}{p_j} - \varepsilon {G_i}^2{p_{\max }}^2 - {N_0}{G_i}{p_{\max }} \le 0$. For any $p_j \in [0,{{\widehat{p}}_j}]$, the latter inequality above is true by following the property of quadratic functions.

$$\begin{aligned} ({p_i^*,p_j^*})&={\left\{ \begin{array}{ll} (p_{max},min({{\widehat{p}}_j},p_{max})) &{} \quad Fa_{i,j}(p_{max},min({{\widehat{p}}_j},p_{max})) > Fa_{i,j}(min({{\widehat{p}}_i},p_{max}),p_{max}) \\ (min({{\widehat{p}}_i},p_{max}),p_{max}) &{}\quad Fa_{i,j}(p_{max},min({{\widehat{p}}_j},p_{max})) \le Fa_{i,j}(min({{\widehat{p}}_i},p_{max}),p_{max}) \\ \end{array}\right. } \end{aligned}$$

(11)

$$\begin{aligned} \min \left( {{{{\widehat{p}}}_i},{p_{\max }}} \right) &={\left\{ \begin{array}{ll} {{{\widehat{p}}}_i} &{}\quad \left( {{p_{\max }}< \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j}< {G_i}} \right) \\ p_{max} &{}\quad \left( {\left( {{p_{\max }} \ge \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j} < {G_i}} \right) } \right) \vee \left( {{G_j} = {G_i}} \right) \\ {{{\widehat{p}}}_i} &{}\quad {G_j} \le \sqrt{\varepsilon }{G_i} \\ \end{array}\right. } \end{aligned}$$

(12)

$$\begin{aligned} \min \left( {{{{\widehat{p}}}_j},{p_{\max }}} \right) &={\left\{ \begin{array}{ll} {{{\widehat{p}}}_j} &{} \left( {\left( {{p_{\max }} \ge \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j}< {G_i}} \right) } \right) \vee \left( {{G_j} = {G_i}} \right) \\ p_{max} &{} \left( {{p_{\max }}< \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j} < {G_i}} \right) \\ p_{max} &{} {G_j} \le \sqrt{\varepsilon }{G_i} \\ \end{array}\right. } \end{aligned}$$

(13)

Since $p_{max} \in [0,{{\widehat{p}}_j}]$, it follows $\frac{{{G_i}{p_{\max }}}}{{{G_j}{p_{\max }} + {N_0}}} \ge \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}$.

For any $p_j \in [0,p_{max})$, we can obtain that $F{a_{i,j}}\left( {{p_{\max }},{p_j}} \right) = min\left( {1 + \frac{{{G_i}{p_{\max }}}}{{{G_j}{p_j} + {N_0}}},1 + \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}} \right) = 1 + \frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}< 1 + \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}} = F{a_{i,j}}\left( {{p_{\max }},{p_{\max }}} \right)$, owing to $\frac{{{G_j}{p_j}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}< \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}} \le \frac{{{G_i}{p_{\max }}}}{{{G_j}{p_{\max }} + {N_0}}} < \frac{{{G_i}{p_{\max }}}}{{{G_j}{p_j} + {N_0}}}$.

In conclusion, the optimal solution for PCMF-TN is $(p_{max},p_{max})$ for Case 1.2. Therefore, we can know that the optimal solution for PCMF-TN is $(p_{max},min(p_{max},{{\widehat{p}}_j}))$ in Case 1.

Case 2 Assume $p_j^*=p_{max}$

In this case, PCMF-TN problem can be formulated as follows:

$$\begin{aligned} \mathop {\max }\limits _{{p_i}} Fa_{i,j}({p_{max}},{p_i}) \quad s.t. \quad 0 \le {p_i} \le {p_{max}} \quad G_i p_i \ge G_j p_{max}. . \end{aligned}$$

Since ${{\widehat{p}}_i} = \frac{{\sqrt{{N_0}^2 + 4\varepsilon {G_j}^2{p_{max }}^2 + 4{N_0}\varepsilon {G_j}{p_{max}}} - {N_0}}}{{2\varepsilon {G_i}}}$, it can be easily verified that $\frac{{{G_i}{{{\widehat{p}}}_i}}}{{{G_j}{p_{max }} + {N_0}}} = \frac{{{G_j}{p_{max }}}}{{\varepsilon {G_i}{{{\widehat{p}}}_i} + {N_0}}}$.

Case 2.1    ${{\widehat{p}}_i}<p_{max}$

Case 2.1.1    For any $p_i \in [0,{{\widehat{p}}}_i]$, since $\frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}}< \frac{{{G_i}{{{\widehat{p}}}_i}}}{{{G_j}{p_{\max }} + {N_0}}} = \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{{{\widehat{p}}}_i} + {N_0}}} < \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_i} + {N_0}}}$, $F{a_{i,j}}\left( {{p_i},{p_{\max }}} \right) = \min \left( {1 + \frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}},1 + \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_i} + {N_0}}}} \right) = 1 + \frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}} < 1 + \frac{{{G_i}{{{\widehat{p}}}_i}}}{{{G_j}{p_{\max }} + {N_0}}} = F{a_{i,j}}\left( {{{{\widehat{p}}}_i},{p_{\max }}} \right).$

Case 2.1.2   For any $p_i \in [{{\widehat{p}}}_i, p_{max}]$, since $\frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}}> \frac{{{G_i}{{{\widehat{p}}}_i}}}{{{G_j}{p_{\max }} + {N_0}}} = \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{{{\widehat{p}}}_i} + {N_0}}} > \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_i} + {N_0}}},$

$$\begin{aligned}&F{a_{i,j}}\left( {{p_i},{p_{\max }}} \right) < F{a_{i,j}}\left( {{{{\widehat{p}}}_i},{p_{\max }}} \right) . \end{aligned}$$

(14)

In conclusion, for Case 2.1, the optimal solution for PCMF-TN is $({{{\widehat{p}}}_i},p_{max})$.

Case 2.2    ${{{\widehat{p}}}_i} \ge p_{max}$

Just as that in Case 1.2, it can be easily verified that for any $p_i \in [0, {{{\widehat{p}}}_i}]$, $\frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}} \le \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_i} + {N_0}}}$. Since $p_{max} \in [0,{{{\widehat{p}}}_i}]$, we can thus get $\frac{{{G_i}{p_{\max }}}}{{{G_j}{p_{\max }} + {N_0}}} \le \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_{\max }} + {N_0}}}$.

For any $p_i \in [0, p_{max})$, since $\frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}}< \frac{{{G_i}{p_{\max }}}}{{{G_j}{p_{\max }} + {N_0}}} \le \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_{max}} + {N_0}}} < \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_i} + {N_0}}},$ we can know that $F{a_{i,j}}\left( {{p_i},{p_{\max }}} \right) = \min \left( {1 + \frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}},1 + \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{p_i} + {N_0}}}} \right) = 1 + \frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}}=1 + \frac{{{G_i}{p_i}}}{{{G_j}{p_{\max }} + {N_0}}} < 1 + \frac{{{G_i}{p_{\max }}}}{{{G_j}{p_{\max }} + {N_0}}} = F{a_{i,j}}\left( {{p_{\max }},{p_{\max }}} \right).$

In conclusion, for Case 2.2, the optimal solution for PCMF-TN is $(p_{max},p_{max})$. So, for Case 2, the optimal solution for PCMF-TN is $(min(p_{max},{{{\widehat{p}}}_i}),p_{max})$.

Combining Case 1 with Case 2, we thus get Formula (11). We now try to find $min({{{\widehat{p}}}_j},p_{max})$ and $min({{{\widehat{p}}}_i},p_{max})$ to get a brief expression for the optimal solution to PCMF-TN.$\square$

Lemma 8

If$\left( {\left( {{p_{\max }} \ge \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j} < {G_i}} \right) } \right) \vee \left( {{G_j} = {G_i}} \right)$, then$\min \left( {{{{\widehat{p}}}_i},{p_{\max }}} \right) = {p_{\max }}$and$\min \left( {{{{\widehat{p}}}_j},{p_{\max }}} \right) = {{{\widehat{p}}}_j}$. $min({{{{\widehat{p}}}_i},{p_{max }}}) = {{{\widehat{p}}}_i}$and$min({{{\widehat{p}}}_j},{p_{max}}) = {p_{max}}$in all other cases,.

Proof

Since ${{{\widehat{p}}}_i} = \frac{{\sqrt{{N_0}^2 + 4\varepsilon {G_j}^2{p_{\max }}^2 + 4{N_0}\varepsilon {G_j}{p_{\max }}} - {N_0}}}{{2\varepsilon {G_i}}} < {p_{\max }}$ is equivalent with the inequation $\left( {{G_j}^2 - \varepsilon {G_i}^2} \right) {p_{\max }} < {N_0}({G_i} - {G_j})$, it naturally results in Formula (12). $\square$

Similarly, we can get Formula (13).

So, if $\left( {\left( {{p_{\max }} \ge \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j} < {G_i}} \right) } \right) \vee \left( {{G_j} = {G_i}} \right)$ holds, $\min \left( {{{{\widehat{p}}}_i},{p_{\max }}} \right) = {p_{\max }}$ and $\min \left( {{{{\widehat{p}}}_j},{p_{\max }}} \right) = {{\widehat{p}}_j}$. Otherwise $\min \left( {{{{\widehat{p}}}_i},{p_{\max }}} \right) = {{\widehat{p}}_i}$ and $\min \left( {{{{\widehat{p}}}_j},{p_{\max }}} \right) = {p_{\max }}$.

Proof of Theorem 2

From Lemmas 7 and 8, if $\left( {\left( {{p_{\max }} \ge \frac{{{N_0}\left( {{G_i} - {G_j}} \right) }}{{{G_j}^2 - \varepsilon {G_i}^2}}} \right) \wedge \left( {\sqrt{\varepsilon }{G_i}< {G_j} < {G_i}} \right) } \right) \vee \left( {{G_j} = {G_i}} \right),$$( {p_i^*,p_j^*}) = argmax({F{a_{i,j}}( {{p_{\max }},{p_{\max }}} ), F{a_{i,j}}( {{p_{\max }},{{{\widehat{p}}}_j}} )} )$ where ${{{\widehat{p}}}_j} \le {p_{\max }}$. Based on Formula (10), $F{a_{i,j}}( {{p_{\max }},{{{\widehat{p}}}_j}} )$ is greater than $F{a_{i,j}}( {{p_{\max }},{p_{\max }}} )$ and $F{a_{i,j}}( {{p_{\max }},{{{\widehat{p}}}_j}} ) = 1 + \frac{{{G_i}{p_{\max }}}}{{{G_j}{{{\widehat{p}}}_j} + {N_0}}}$.

$(p_i^*,p_j^* )=argmax( {F{a_{i,j}}( {{{{\widehat{p}}}_i},{p_{\max }}} ), F{a_{i,j}}( {{p_{\max }},{p_{\max }}} )} )$ holds in all other cases, where ${{{\widehat{p}}}_i} \le {p_{\max }}$. Based on Formula (14), $F{a_{i,j}}( {{{{\widehat{p}}}_i},{p_{\max }}} )$ is greater than $F{a_{i,j}}( {{p_{\max }},{p_{\max }}} )$ and $F{a_{i,j}}( {{{{\widehat{p}}}_i},{p_{\max }}} ) = 1 + \frac{{{G_j}{p_{\max }}}}{{\varepsilon {G_i}{{{\widehat{p}}}_i} + {N_0}}}$.

Theorem 2 is thus proved by substituting above results into Formula (11). $\square$