Fair Scheduling of Two-Hop Transmission with Energy Harvesting

Abstract

In this paper, we consider a two-hop network with a source node (SN) and a relay node (RN) who want to communicate data to a destination node (DN). The SN cannot be directly connected to the DN, but rather is connected only via the RN. The RN does not have an external source of energy, and thus needs to harvest energy from the SN to communicate, while the SN has an external source of energy and can harvest energy straight from it. Thus, a dilemma for the SN arises: how much to share harvested energy with the RN to make it relay the SN’s data to the DN. Fair performing of their communication tasks is considered as an incentive for the SN and the RN to cooperate. The optimal $\alpha$ fair schedule is found for each $\alpha$. It is shown that an altruistic strategy for one of the nodes comes in as a part of the cooperative solution (corresponding $\alpha =0$), while the maxmin strategy (corresponding $\alpha$ tending to infinity) is proved to be egalitarian. Using Nash bargaining over the obtained continuum of fair solutions, we design a trade-off strategy.

A Appendix

Proof of Theorem

1 . By (1) and (3), $T_s\le p_h T/(p_h+p_s)$. Then, due to v given by (4) is increasing on $T_s$, (5) follows.    $\blacksquare$

Proof of Theorem

2 . To find the optimal $p_s$ we have to find derivation of v on $p_s$: $d\,v(p_s)/d\, p_s=\left( h_s(p_h+p_s)/(1+h_sp_s)-\ln (1+h_sp_s) \right) T p_h/(p_h+p_s)^2.$ Thus, $dv/dp_s\{>,=,<\} 0$ if and only if $1+a/x-\ln (x) \{>,=,< \} 0$, where $x=1+h_sp_s$ and $a=h_sp_h-1$. It is clear that $a>-1$ and $x\ge 1$. For a fixed $a>-1$ the function $1+a/x-\ln (x)$ is decreasing on $x\ge 1$. Moreover, the equation $1+a/x-\ln (x)=0$ has the unique root $x=\exp (\text{ LambertW }\left( a/e\right) +1)$, and the result follows.    $\blacksquare$

Proof of Theorem

3 . Since $v(\varvec{T})$ is concave, to find the optimal $\varvec{T}$ the KKT Theorem can be applied. First we define Lagrange function $L_{\omega _1,\omega _2,\omega _3}(\varvec{T})$ with $\omega _1$, $\omega _2$ and $\omega _3$ are Lagrange multipliers as follows:

$$\begin{aligned} \begin{aligned} L_{\omega _1,\omega _2,\omega _3}(\varvec{T})&=\frac{(T_{rd}L_r)^{1-\alpha }}{1-\alpha }+\frac{(T_{sd}L_s)^{1-\alpha }}{1-\alpha }+\omega _1(T-T_h-T_{hr}-T_{sd}-T_{rd})\\&\quad +\omega _2(p_hT_h-T_{hr}p_s-T_{sd}p_s)+\omega _3(\gamma h_{sr}p_sT_{hr}-T_{rd}p_r). \end{aligned} \end{aligned}$$

(22)

Then, for $\varvec{T}$ to be optimal, besides of conditions (15b)–(15d), the following relations have to hold:

$$\begin{aligned} \partial L/\partial T_{rd}=L_r^{1-\alpha }/(T_{rd})^{\alpha }-\omega _1-p_r\omega _3{\left\{ \begin{array}{ll} =0,&{} T_{rd}>0,\\ \le 0,&{} T_{rd}=0, \end{array}\right. } \end{aligned}$$

(23)

$$\begin{aligned} \partial L/\partial T_{sd}=L_s^{1-\alpha }/(T_{sd})^{\alpha }-\omega _1-p_s\omega _2{\left\{ \begin{array}{ll} =0,&{} T_{sd}>0,\\ \le 0,&{} T_{sd}=0, \end{array}\right. } \end{aligned}$$

(24)

$$\begin{aligned} \partial L/\partial T_{h}=-\omega _1+p_h\omega _2=0, \end{aligned}$$

(25)

$$\begin{aligned} \partial L/\partial T_{hr}=-\omega _1-p_s\omega _2 +\gamma h_{sr}p_s\omega _3 =0. \end{aligned}$$

(26)

By (25), we have that

$$\begin{aligned} \omega _2=\omega _1/p_h. \end{aligned}$$

(27)

By (26) and (27), we have that $\omega _3=(1+p_s/p_h)\omega _1/(\gamma h_{sr}p_s).$ Then, (17) and (23) yield that:

$$\begin{aligned} T_{rd}=\frac{L_r^{1/\alpha -1}}{(\omega _1+p_r\omega _3)^{1/\alpha }}= \frac{L_r^{1/\alpha -1}}{(1+P_r(1+P_s))^{1/\alpha } \omega _1^{1/\alpha }}=\frac{A_r^{1/\alpha -1} }{(1+P_r(1+P_s))\omega _1^{1/\alpha }}. \end{aligned}$$

(28)

By (24), in notation (17), we have that

$$\begin{aligned} T_{sd}=\frac{L_s^{1/\alpha -1}}{(\omega _1+p_s\omega _2)^{1/\alpha }}= \frac{L_s^{1/\alpha -1}}{(1+P_s)^{1/\alpha }\omega _1^{1/\alpha }}= \frac{A_s^{1/\alpha -1}}{(1+P_s)\omega _1^{1/\alpha }}. \end{aligned}$$

(29)

By (15d), (17) and (28) the following relation holds

$$\begin{aligned} T_{hr}=T_{rd} p_r/(\gamma h_{sr}p_s)=P_rT_{rd}=P_rA_r^{1/\alpha -1}/((1+P_r(1+P_s))\omega _1^{1/\alpha }). \end{aligned}$$

(30)

By (15c), (29) and (30), using notation (17) we have that

$$\begin{aligned} {\begin{matrix} T_h&=\frac{p_s}{p_h}\left( T_{hr}+T_{sd}\right) =P_s\left( T_{hr}+T_{sd}\right) =\frac{P_sP_rA_r^{1/\alpha -1} }{(1+P_r(1+P_s))\omega _1^{1/\alpha }}+\frac{P_sA_s^{1/\alpha -1}}{(1+P_s)\omega _1^{1/\alpha }}. \end{matrix}} \end{aligned}$$

(31)

Then, summing (28)–(31) and taking into account (15c) imply that $\omega _1^{1/\alpha }=(A_s^{1/\alpha -1}+A_r^{1/\alpha -1})/T.$ Substituting this $\omega _1$ into (28)–(31) implies (16). Then,

$$\begin{aligned} v_{s,\alpha }=L_s T_{su,\alpha }=T A_s^{1/\alpha }/(A_s^{1/\alpha -1}+A_r^{1/\alpha -1}), \end{aligned}$$

(32)

$$\begin{aligned} v_{r,\alpha }=L_r T_{ru,\alpha }=T A_r^{1/\alpha }/(A_s^{1/\alpha -1}+A_r^{1/\alpha -1}). \end{aligned}$$

(33)

Dividing (32) by (33) yields the first relation in (18). Note that

$$\begin{aligned} \frac{A_s^{1/\alpha }}{A_s^{1/\alpha -1}+A_r^{1/\alpha -1}}=A_s\frac{A_s^{1/\alpha -1}}{A_s^{1/\alpha -1}+A_r^{1/\alpha -1}}= A_s\left( 1- \frac{1}{A_r}\frac{A_r^{1/\alpha }}{A_s^{1/\alpha -1}+A_r^{1/\alpha -1}}\right) . \end{aligned}$$

This, jointly with (32) and (33), implies the second relation in (18), and the result follows.    $\blacksquare$

Proof of Theorem

4 : By (21), two cases arise: $A_s>A_r$ and $A_s<A_r$. Let $A_s>A_r$. Then, by (18), (19) and (21), $NP(v_s,v_r)=(v_s-v_{\infty })v_r=(TA_s- A_s v_r/A_r-v_{\infty })v_r= A_s(TA_s/(A_s+A_r)- v_r/A_r)v_r.$ Thus, the $(TA_s- A_s v_{\infty }/(2A_r),v_{\infty }/2)$ is the NBS, and the result follows from (19) and the first relation in (18).    $\blacksquare$