Fair Channel Sharing by Wi-Fi and LTE-U Networks with Equal Priority

Abstract

The paper is concerned with the problem Wi-Fi and LTE-U networks sharing access to a band of communication channels, while also considering the issue of fairness in how the channel is being shared. As a criteria of fairness for such joint access, $\alpha$-fairness and maxmin fairness with regards to expected throughput are explored as fairness metrics. Optimal solutions are found in closed form, and it is shown that these solutions can be either: (a) a channel on/off strategy in which access to the channels is performed sequentially, or (b) a channel sharing strategy, i.e., where simultaneous joint access to the channels is applied. A criteria for switching between these two type of optimal strategies is found, and its robustness on the fairness coefficient is established, as well as the effectiveness of the fairness coefficient to control the underlying protocol of the joint access to the shared resource is managed. Finally, we note that the approach that is explored is general, and it might be adapted to different problems for accessing a sharing resource, like joint sharing of voice and data traffic by cellular carriers.

Appendices

A Appendix I: Proof of Theorem 1

To find the optimal ${{\varvec{q}}}=bq_\alpha$ define the Lagrangian $L_\omega ({{\varvec{q}}})=v_\alpha ({{\varvec{q}}})+\omega (1-q^W-q-q^L).$ Thus, ${{\varvec{q}}}$ is the optimal probability vector then the following conditions has to hold:

$$\begin{aligned} \frac{P^W}{(q^WP^W+qP^W_L)^\alpha } {\left\{ \begin{array}{ll} =\omega ,&{}q^W>0,\\ \le \omega ,&{}q^W=0, \end{array}\right. } \end{aligned}$$

(11)

$$\begin{aligned} \frac{P^L}{(q^LP^L+qP^L_W)^\alpha } {\left\{ \begin{array}{ll} =\omega ,&{}q^L>0,\\ \le \omega ,&{}q^L=0, \end{array}\right. } \end{aligned}$$

(12)

and

$$\begin{aligned} \frac{P^W_L}{(q^WP^W+qP^W_L)^\alpha }+\frac{P^L_W}{(q^LP^L+qP^L_W)^\alpha } {\left\{ \begin{array}{ll} =\omega ,&{}q>0,\\ \le \omega ,&{}q=0. \end{array}\right. } \end{aligned}$$

(13)

Thus, the boundary strategies ${{\varvec{q}}}=(1,0,0)$ and ${{\varvec{q}}}=(0,0,1)$ cannot be optimal.

First we find the condition when the rest boundary strategy ${{\varvec{q}}}=(0,1,0)$ can be optimal. Substituting it into (11)–(13) implies that the following condition has to hold: $\omega =(P^W_L)^{1-\alpha }+(P^L_W)^{1-\alpha }\ge \max \{P^L/(P^L_W)^\alpha ,P^W/(P^W_L)^\alpha \}.$ This condition is equivalent to (8), and (d) follows.

Let us pass to finding channel sharing optimal strategy, i.e., with $q>0$. Then, either $q^W=0$ or $q^L=0$.

Let $q^L=0$. Then, (11)–(13) turn into the following conditions

$$\begin{aligned} P^W/(q^WP^W+qP^W_L)^\alpha =\omega , \end{aligned}$$

(14)

$$\begin{aligned} P^L/(qP^L_W)^\alpha \le \omega \end{aligned}$$

(15)

and

$$\begin{aligned} P^W_L/(q^WP^W+qP^W_L)^\alpha +P^L_W/(qP^L_W)^\alpha =\omega . \end{aligned}$$

(16)

By (14),

$$\begin{aligned} q^WP^W+qP^W_L= \left( P^W/\omega \right) ^{1/\alpha }. \end{aligned}$$

(17)

Since $q^W+q=1$ then $(1-q)P^W+qP^W_L= \left( P^W/\omega \right) ^{1/\alpha }.$ So,

$$\begin{aligned} q=\left( 1 - (P^W)^{(1-\alpha )/\alpha }/\omega ^{1/\alpha }\right) /\left( 1-P^W_L/P^W\right) . \end{aligned}$$

(18)

By (14) and (16),

$$\begin{aligned} \frac{P^W_L}{P^W}\omega +\frac{P^L_W}{(qP^L_W)^\alpha }=\omega . \end{aligned}$$

(19)

Thus,

$$\begin{aligned} q=\frac{(P^L_W)^{(1-\alpha )/\alpha }}{(1-P^W_L/P^W)^{1/\alpha }\omega ^{1/\alpha }}. \end{aligned}$$

(20)

By (18) and (20),

$$\begin{aligned} \frac{1 - (P^W)^{(1-\alpha )/\alpha }/\omega ^{1/\alpha } }{1-P^W_L/P^W}=\frac{(P^L_W)^{(1-\alpha )/\alpha }}{(1-P^W_L/P^W)^{1/\alpha }\omega ^{1/\alpha }}. \end{aligned}$$

(21)

Thus,

$$\begin{aligned} \omega ^{1/\alpha }=(P^W)^{(1-\alpha )/\alpha }+\left( \frac{P^L_W}{1-P^W_L/P^W}\right) ^{(1-\alpha )/\alpha }. \end{aligned}$$

(22)

Thus,

$$\begin{aligned} q=\frac{\displaystyle \frac{(P^L_W)^{(1-\alpha )/\alpha }}{(1-P^W_L/P^W)^{1/\alpha }}}{\displaystyle (P^W)^{(1-\alpha )/\alpha }+\left( \frac{P^L_W}{1-P^W_L/P^W}\right) ^{(1-\alpha )/\alpha }}. \end{aligned}$$

(23)

It is clear that $q>0$. Since, ${{\varvec{q}}}$ is the probability vector we have to find only the condition for q being less or equal to 1. By (23), it is equivalent to

$$\begin{aligned} \frac{(P^L_W)^{(1-\alpha )/\alpha }}{\left( 1- P^W_L/ P^W\right) ^{1/\alpha }} \le (P^W)^{(1-\alpha )/\alpha }+\left( \frac{P^L_W}{1- P^W_L/P^W}\right) ^{(1-\alpha )/\alpha }. \end{aligned}$$

(24)

The last inequality is equivalent to

$$\begin{aligned} (P^L_W)^{1-\alpha }(P^W_L)^{\alpha }\le P^W-P^W_L. \end{aligned}$$

(25)

Finally we have to find the condition that (15) holds. Substituting q from (20) into (15) implies (b).

The case $q^W=0$ as well as the case $q=0$, $q^W>0$ and $q^L>0$ can be considered similarly, and (a) and (c) follow. To deal with (e) denote by ${{\varvec{q}}}_b$ and ${{\varvec{q}}}_c$ the optimal strategies given by (b) and (b). The previous analyze yields that in (d) the optimal strategy is ${{\varvec{q}}}_b$ if $v_\alpha ({{\varvec{q}}}_b)>v_\alpha ({{\varvec{q}}}_c)$, and it is ${{\varvec{q}}}_c$ if $v_\alpha ({{\varvec{q}}}_b)<v_\alpha ({{\varvec{q}}}_c)$.

Note that, by (17) and (19),

$$\begin{aligned} {\begin{matrix} v_\alpha ({{\varvec{q}}}_b)=\frac{1}{1-\alpha }\left( \frac{P^W}{\omega }\right) ^{(1-\alpha )/\alpha }+\frac{1}{1-\alpha }\left( \frac{P^L_W}{\left( 1- P^W_L/P^W\right) \omega }\right) ^{(1-\alpha )/\alpha }. \end{matrix}} \end{aligned}$$

(26)

Substituting (22) into (26) yields

$$ (1-\alpha )v_\alpha ({{\varvec{q}}}_b)=((P^W)^{(1-\alpha )/\alpha }+(P^L_W/(1- P^W_L/P^W))^{(1-\alpha )/\alpha })^\alpha .$$

By symmetry, $v_\alpha ({{\varvec{q}}}_c)$ can be found, and the result follows.    $\blacksquare$

B Appendix II: Proof of Theorem 2

The maxmin problem is equivalent to the following LP problem

$$\begin{aligned} \begin{aligned}&\text{ maximize } \, \nu ,\\&q^W P^W+q P^W_L\ge \nu , q P^L_W+q^L P^L\ge \nu ,\ q^W+q+q^L=1, q^W, q, q^L \ge 0. \end{aligned} \end{aligned}$$

(27)

Let for while the component q of the strategy ${{\varvec{q}}}$ be fixed and optimal, while Then, component $q^W$ and $q^L$ might vary. Since $q^W+q^L=1-q$ the optimal $q^W$ can be found as a solution of the following problem:

$$\begin{aligned} \begin{aligned}&\text{ maximize } \,\nu ,\\&q^W P^W+q P^W_L\ge \nu ,q P^L_W+(1-q)P^L-q^WP^L \ge \nu , q^W\in [0,1-q]. \end{aligned} \end{aligned}$$

(28)

First we look for the optimal channel sharing strategy, i.e., when $q>0$. Figure 5 illustrates that the solution of LP problem (28) can be found as an intersection of the corresponding lines.

Fig. 5.

Solution of the LP problem

Thus, the channel sharing solution holds if and only if $(1-q)P^L+q P^L_W>qP^W_L$ and $(1-q)P^W+q P^W_L>qP^L_W.$ These inequalities are equivalent to

$$\begin{aligned} {\begin{matrix} (1-q)P^L>q(P^W_L-P^L_W) \text{ and } (1-q)P^W>q(P^L_W-P^W_L). \end{matrix}} \end{aligned}$$

(29)

Since either $P^W_L\ge P^L_W$ or $P^L_W>P^W_L$, then one of the conditions (29) always hold. Without loss of generality we assume that

$$\begin{aligned} {\begin{matrix} P^W_L\ge P^L_W. \end{matrix}} \end{aligned}$$

(30)

Then, conditions (29) are equivalent to

$$\begin{aligned} {\begin{matrix} q\le 1/(1+(P^W_L-P^L_W)/P^L). \end{matrix}} \end{aligned}$$

(31)

Let us switch on to finding the optimal $\nu$ and $q^W$. By Fig. 5 and (28),

$$\begin{aligned} {\begin{matrix} q^W P^W+q P^W_L=q P^L_W+(1-q)P^L-q^WP^L =\nu . \end{matrix}} \end{aligned}$$

(32)

Thus,

$$\begin{aligned} {\begin{matrix} q^W=(P^L-(P^L+P^W_L-P^L_W)q)/(P^L+P^W). \end{matrix}} \end{aligned}$$

(33)

Thus, by (30), $q^W$ is decreasing in q, and

$$\begin{aligned} {\begin{matrix} \nu =(P^WP^L+(P^LP^W_L+P^WP^L_W-P^LP^W)q)/(P^W+P^L). \end{matrix}} \end{aligned}$$

(34)

So, (33) and (34) give channel sharing solution of (27) for a fixed q, and (31) is the condition such solution holds. Note that, by (34), $\nu$ is increasing in q if $P^LP^W_L+P^WP^L_W>P^LP^W$, and $\nu$ is decreasing in q otherwise. Thus, if (30) holds, then the channel sharing solution exists ($q>0$) if and only if $P^LP^W_L+P^WP^L_W>P^LP^W,$ and then $q=1/(1+(P^W_L-P^L_W)/P^L).$ Substituting this q into (33) implies $q^W=0$. Thus, $q^L=1-q$. The case of the channel on/off optimal strategy can be considered similarly, and the result follows.    $\blacksquare$