Priority-based capacity and power allocation in co-located WBANs using Stackelberg and bargaining games

Abstract

To mitigate the interference in co-located wireless body area networks (WBANs), this paper proposes an inter-WBAN priority-based capacity allocation scheme based on the Nash bargaining game, and an intra-WBAN priority-based power control scheme based on the Stackelberg game. Moreover, under the network capacity imposed by the Nash bargaining solution, two pricing mechanisms: non-uniform pricing and uniform pricing, are introduced in the Stackelberg game and the Stackelberg equilibrium under each mechanism is achieved analytically. Additionally, owing to the special features of WBANs, the players priorities indicated by exigency of the sensed data and the energy consumption of sensors are considered in the design of utility functions. Extensive simulations show that the proposed schemes are energy efficient and can improve the network quality of service in terms of real time and reliability of critical data transmission.

Appendix

1.1 A. Proof of Proposition 4.1

Obviously, Problem 4.2 is a convex optimization problem, which can be solved using the Lagrange Multiplier approach. The Lagrange function is given by

$$\begin{aligned} L\left( {{\mathbf{S}},{\varvec{\alpha }},\beta } \right) = \sum \limits _{i = 1}^N {{w_i} \cdot \ln ({S_i} - S_i^{\min })} + \sum \limits _{i = 1}^N {{\alpha _i} \cdot ({S_i} - S_i^{\min })} + \beta \cdot \left( C - \sum \limits _{i = 1}^N {{S_i}}\right) \end{aligned}$$

(31)

Then, the KKT conditions can be written as follows:

$$\begin{aligned} \frac{{\partial L({\mathbf{S}},{\varvec{\alpha }},\beta )}}{{\partial {S_i}}}= & {} 0, \quad \forall i \end{aligned}$$

(32)

$$\begin{aligned} {\alpha _i} \cdot ({S_i} - S_{\min }^i)= & {} 0, \quad \forall i \end{aligned}$$

(33)

$$\begin{aligned} \beta \cdot \left( C - \sum \limits _{i = 1}^N {{S_i}}\right)= & {} 0 \end{aligned}$$

(34)

$$\begin{aligned} {\alpha _i} \ge 0,\mathrm{{ }}\beta \ge 0,\mathrm{{ }}{S_i}\ge & {} S_i^{\min }, \quad \forall i \end{aligned}$$

(35)

$$\begin{aligned} \sum \limits _{i = 1}^N {{S_i}}\le & {} C \end{aligned}$$

(36)

From formulas (31) and (32), we have

$$\begin{aligned} \frac{{\partial L({\mathbf{S}},{\varvec{\alpha }},\beta )}}{{\partial {S_i}}} = \frac{{{w_i}}}{{{S_i} - S_i^{\min }}} + {\alpha _i} - \beta = 0, \quad \forall i \end{aligned}$$

(37)

Thus, the following solution can be obtained

$$\begin{aligned} {S_i} = \frac{{{w_i}}}{{\beta - {\alpha _i}}} + S_i^{\min },\quad \forall i \end{aligned}$$

(38)

Lemma 1

${\alpha _i} = 0, \quad \forall i$

Proof

Suppose that ${\alpha _i} \ne 0$ for some i. it can be deduced from formula (33) that ${S_i} = S_i^{\min }$, which contradicts the meaning of cooperative game and formula (37). Therefore, the assumption that ${\alpha _i} \ne 0$ for some i doesn’t hold, and we have ${\alpha _i} = 0, \ \forall i$. $\square$

Lemma 2

$\sum \limits _{i = 1}^N {{S_i}} = C$

Proof

Suppose that $\sum _{i = 1}^N {{S_i}} \ne C$, according to formula (34), we have $\beta = 0$. Further, it has been proved in Lemma 1 that ${\alpha _i} = 0, \ \forall i$. Then, it can be seen that the equation in formula (37) isn’t true in this case. Thus, the assumption doesn’t hold and we have $\sum _{i = 1}^N {{S_i}} = C$.

According to Lemmas 1 and 2, it can be computed that $\beta = {\left( {\sum _{i = 1}^N {{w_i}} } \right) } \bigg /{\left( {C - \sum _{i = 1}^N {S_i^{\min }} } \right) }$. Substituting it into formula (38), we have

$$\begin{aligned} {S_i} = \frac{{{w_i} \cdot \left( {C - \sum _{i = 1}^N {S_i^{\min }} } \right) }}{{\sum _{i = 1}^N {{w_i}} }} + S_i^{\min },\quad \forall i \end{aligned}$$

Proposition 4.1 is thus proved. $\square$

1.2 B. Proof of Proposition 5.1

The first- and second-order partial derivatives of ${U_k^i}$ with respect to ${p_k^i}$ are given as

$$\begin{aligned} \frac{{\partial U_k^i}}{{\partial p_k^i}}= & {} \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot p_k^i)}} - c_k^i \end{aligned}$$

(39)

$$\begin{aligned} \frac{{{\partial ^2}U_k^i}}{{\partial p{{_k^i}^2}}}= & {} - \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h{{_k^i}^2}}}{{\ln 2 \cdot {{({\sigma ^2} + h_k^i \cdot p_k^i)}^2}}} \end{aligned}$$

(40)

As $\lambda _k^i > 0$ and $\vartheta _k^i \ge 0$, it can be obtained that $\frac{{{\partial ^2}U_k^i}}{{\partial p{{_k^i}^2}}} \le 0$ Thus, ${U_k^i}$ is a concave function of ${p_k^i}$. We can get the optimal solution of Problem 5.1 by setting $\frac{{\partial U_k^i}}{{\partial p_k^i}} = 0$, i.e., $p_k^i = \frac{{\lambda _k^i \cdot \vartheta _k^i}}{{\ln 2 \cdot c_k^i}} - \frac{{{\sigma ^2}}}{{h_k^i}}$. Due to the limited energy, sensor nodes active under the constraint of maximum feasible transmission power ${P_{\max }}$, i.e., $p_k^i = \left( {\frac{{\lambda _k^i \cdot \vartheta _k^i}}{{\ln 2 \cdot c_k^i}} - \frac{{{\sigma ^2}}}{{h_k^i}}} \right) _0^{{P_{\max }}}$. Therefore, we can get

$$\begin{aligned} p{{_k^i}^ * } = \left\{ \begin{array}{l} 0, \quad \mathrm{{ }}c_k^i \ge \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}}\mathrm{{ }} \\ \\ \frac{{\lambda _k^i \cdot \vartheta _k^i}}{{\ln 2 \cdot c_k^i}} - \frac{{{\sigma ^2}}}{{h_k^i}}, \quad \mathrm{{ }}\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot {P_{\max }})}}< c_k^i < \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}} \\ \\ {P_{\max }}, \quad \mathrm{{ }}c_k^i \le \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot {P_{\max }})}}\mathrm{{ }} \\ \end{array} \right. \end{aligned}$$

The solution can also be obtained using the Lagrange Multiplier approach, which is similar to part A of appendix, and it is omitted here. Thus, Proposition 5.1 is proved.

1.3 C. Proof of Proposition 5.2

Problem 5.3 is a convex optimization problem and can be solved using the Lagrange Multiplier method. The Lagrange function is given by

$$\begin{aligned} \begin{array}{l} L({{\mathbf{c}}_k},\alpha ,\varvec{\beta } ,\varvec{\gamma } )= \sum \limits _{i = 1}^m {\left( {\frac{{\lambda _k^i \cdot \vartheta _k^i}}{{\ln 2}} - \frac{{{\sigma ^2}}}{{h_k^i}} \cdot c_k^i} \right) } - \alpha \cdot \left[ {\sum \limits _{i = 1}^m {B_k^i \cdot {{\log }_2}\left( {\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2} \cdot c_k^i}}} \right) } - {S_k}} \right] \\ \\ \quad \mathrm{{ }} - \sum \limits _{i = 1}^m {{\beta _i}} \cdot \left( {\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot {P_{\max }})}} - c_k^i} \right) - \sum \limits _{i = 1}^m {{\gamma _i}} \cdot \left( {c_k^i - \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}}} \right) \\ \end{array} \end{aligned}$$

(41)

Then, the KKT conditions can be written as follows:

$$\begin{aligned}&\frac{{\partial L}}{{\partial c_k^i}} = 0 \end{aligned}$$

(42)

$$\begin{aligned}&\sum \limits _{i = 1}^m {B_k^i \cdot {{\log }_2}\left( {\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2} \cdot c_k^i}}} \right) } \le {S_k}\end{aligned}$$

(43)

$$\begin{aligned}&\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot {P_{\max }})}} < c_k^i,\quad \forall i\end{aligned}$$

(44)

$$\begin{aligned}&c_k^i < \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}},\quad \forall i\end{aligned}$$

(45)

$$\begin{aligned}&{\beta _i} \cdot \left( {\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot {P_{\max }})}} - c_k^i} \right) = 0,\quad \forall i\end{aligned}$$

(46)

$$\begin{aligned}&{\gamma _i} \cdot \left( {c_k^i - \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}}} \right) = 0,\quad \forall i\end{aligned}$$

(47)

$$\begin{aligned}&\alpha \cdot \left[ {\sum \limits _{i = 1}^m {B_k^i \cdot {{\log }_2}\left( {\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2} \cdot c_k^i}}} \right) } - {S_k}} \right] = 0 \end{aligned}$$

(48)

From formulas (44)–(47), it can be seen that ${\beta _i} = {\gamma _i} = 0,\forall i$. Since $\frac{{\partial L}}{{\partial c_k^i}} = - \frac{{{\sigma ^2}}}{{h_k^i}} + {\beta _i} - {\gamma _i} + \frac{{\alpha \cdot B}}{{\ln 2 \cdot c_k^i}} = 0$ , it can be obtained that

$$\begin{aligned} c_k^i = \frac{{\alpha \cdot B \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}}, \quad \forall i \end{aligned}$$

(49)

From formulas (48) and (49), it can be deduced that

$$\begin{aligned} \sum \limits _{i = 1}^m {B_k^i \cdot {{\log }_2}\left( {\frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2} \cdot c_k^i}}} \right) } = {S_k} \end{aligned}$$

(50)

Substituting (49) into (50), we can get that

$$\begin{aligned} \alpha \mathrm{{ = }}{\left( {\frac{{\prod _{i = 1}^m {\frac{{\lambda _k^i \cdot \vartheta _k^i}}{B}} }}{{{2^{{{{S_k}}/ B}}}}}} \right) ^{\frac{1}{m}}} \end{aligned}$$

(51)

Substituting (51) into (49), the price strategy can be obtained as follows:

$$\begin{aligned} c{{_k^i}^ * } = \frac{{{{\left( {\prod _{i = 1}^m {\lambda _k^i \cdot \vartheta _k^i} } \right) }^{\frac{1}{m}}} \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2} \cdot {2^{\frac{{{S_k}}}{{B \cdot m}}}}}}, \quad \forall i \end{aligned}$$

Proposition 5.2 is proved.

1.4 D. Proof of Proposition 5.3.

It is observed that the solution in (19) is the optimal solution of Problem 5.3 if the following condition is satisfied:

$$\begin{aligned} \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot ({\sigma ^2} + h_k^i \cdot {P_{\max }})}}< c_k^i < \frac{{\lambda _k^i \cdot \vartheta _k^i \cdot h_k^i}}{{\ln 2 \cdot {\sigma ^2}}}, \quad \forall i. \end{aligned}$$

Substituting (19) into these inequalities, it can be deduced that

$$\begin{aligned} {S_k} > B \cdot m \cdot {\log _2}\frac{{{A_m}}}{{\lambda _k^i \cdot \vartheta _k^i}}, \quad \forall i \end{aligned}$$

and

$$\begin{aligned} {S_k} < B \cdot m \cdot {\log _2}\left[ {\frac{{{A_m}}}{{\lambda _k^i \cdot \vartheta _k^i}} \cdot \left( {1 + \frac{{{P_{\max }} \cdot h_k^i}}{{{\sigma ^2}}}} \right) } \right] , \quad \forall i. \end{aligned}$$

These inequalities can be transformed as formulas (20) and (21) compactly, respectively. Thus, Proposition 5.3 is proved.