Achieving arbitrary throughput-fairness trade-offs in the inter cell interference coordination with fixed transmit power problem

Abstract

We study the problem of inter cell interference coordination (ICIC) with fixed transmit power in OFDMA-based cellular networks, in which each base station (BS) needs to decide as to which subchannel, if any, to allocate to each of its associated mobile stations (MS) for data transmission. In general, there exists a trade-off between the total throughput (sum of throughputs of all the MSs) and fairness under the allocations found by resource allocation schemes. We introduce the concept of $\tau -\alpha -$fairness by modifying the concept of $\alpha -$fairness, which was earlier proposed in the context of congestion control for packet-switched networks. The concept of $\tau -\alpha -$fairness allows us to achieve arbitrary trade-offs between total throughput and fairness by selecting an appropriate value of $\alpha$ in $[0,\infty )$. We show that for every $\alpha \in [0,\infty )$ and every $\tau > 0$, the problem of finding a $\tau -\alpha -$fair allocation is NP-Complete. Further, we show that for every $\alpha \in [0, \infty )$, there exist thresholds such that if the potential interference levels experienced by each MS on every subchannel are above the threshold values, then the problem can be optimally solved in polynomial time by reducing it to the bipartite graph matching problem. Also, we propose a simple, distributed subchannel allocation algorithm for the ICIC problem, which is flexible, requires a small amount of time to operate, and requires information exchange among only neighboring BSs. We investigate via simulations as to how the algorithm parameters should be selected so as to achieve any desired trade-off between the total throughput and fairness. Finally, we compare the performance of the proposed algorithm with those of the simulated annealing based and opportunistic subchannel allocation algorithms in terms of the total throughput, fairness index and computational complexity.

Appendices

Appendix

Proof of Lemma 1

First, we will show that the function f(x) is quasi-convex on the domain $x \ge 1$.

property 1

A function Q(.) is quasi-convex if $Q''(z)>0$ whenever $Q'(z)=0$ [48].

Let

$$\begin{aligned} y=(x-1)\beta + \frac{1}{\eta }. \end{aligned}$$

(47)

Then,

$$\begin{aligned} f(x)=g(y), \end{aligned}$$

(48)

where $g(y)=$

$$\begin{aligned} \left( \frac{y}{\beta }+1-\frac{1}{\eta \beta }\right) \frac{\left( \tau +\log \left( 1 + \frac{1}{y} \right) \right) ^{1-\alpha }}{1-\alpha }-(\frac{y}{\beta }-\frac{1}{\eta \beta })\frac{\tau ^{1-\alpha }}{1-\alpha }. \end{aligned}$$

Let $p= \left( \tau +\log \left( 1 + \frac{1}{y} \right) \right)$. Then,

$$\begin{aligned} g'(y)= & {} \frac{1}{\beta }\left( \frac{p^{1-\alpha }}{1-\alpha }-\frac{(y+\beta -\frac{1}{\eta })\left( p \right) ^{-\alpha }}{y(y+1)}-\frac{\tau ^{1-\alpha }}{1-\alpha } \right) . \end{aligned}$$

(49)

$$\begin{aligned} g'(y)= & {} 0 \Leftrightarrow y+\beta -\frac{1}{\eta }= \frac{y(y+1)(p-p^{\alpha }\tau ^{1-\alpha })}{1-\alpha }. \end{aligned}$$

(50)

Further, $g''(y)$

$$\begin{aligned} =\frac{1}{\beta y(y+1)p^{\alpha }}\left( -2+ \frac{y+\beta -\frac{1}{\eta }}{y(y+1)p^{\alpha }}\left( (2y+1)p^{\alpha }-\alpha p^{\alpha -1} \right) \right) \end{aligned}$$

So $g''(y) > 0$

$\Leftrightarrow py(2(\beta -\frac{1}{\eta })-1)-\alpha (y+\beta -\frac{1}{\eta })+p(\beta -\frac{1}{\eta })) > 0$

Substituting from (50) in the above inequality, we get:

$py(2(\beta -\frac{1}{\eta })-1)-\alpha y(y+1)\frac{p-p^{\alpha }\tau ^{1-\alpha }}{1-\alpha }+p(\beta -\frac{1}{\eta })) > 0$

$$\begin{aligned} \Leftrightarrow y(2(\beta -\frac{1}{\eta })-1)-\alpha y(y+1)\frac{1-\left( 1+\frac{\log (1+\frac{1}{y})}{\tau }\right) ^{\alpha -1}}{1-\alpha }+(\beta -\frac{1}{\eta }) > 0 \end{aligned}$$

(51)

Now, we find sufficient conditions for (51) to hold for three different values of $\alpha$:

• (a) $\alpha <1:$

  Because $(1+x)^r \ge 1+ rx \;\; \forall x \ge -1, \;r \in \; \mathscr {R} \backslash (0,1)$ [49], a sufficient condition for (51) to hold is

  $y(2(\beta -\frac{1}{\eta })-1)-\alpha \frac{y(y+1)}{1-\alpha }[1-\{1+(\alpha -1)\frac{\log (1+\frac{1}{y})}{\tau }\}]+(\beta -\frac{1}{\eta })) > 0$.

  Because $\log (1+x) \le x \; \forall x \ge -1$ [50], a sufficient condition for the above inequality to hold is

  $y\tau (2(\beta -\frac{1}{\eta })-1)-\alpha (y+1)+\tau (\beta -\frac{1}{\eta })) > 0.$

  $\Leftrightarrow y(2\tau (\beta -\frac{1}{\eta })-\tau -\alpha )-\alpha +\tau (\beta -\frac{1}{\eta })) > 0$.

A sufficient condition for the above inequality to hold is

$$\begin{aligned} \beta \ge \frac{\alpha }{\tau }+\frac{1}{\eta }\;\; \text{ and } \; \;\tau < \alpha . \end{aligned}$$

(52)

Hence, when (52) holds and $\alpha < 1$, then $g''(y)> 0$ when $g'(y) = 0$.

• (b) $1< \alpha < 2:$ From (51),

  $y(2(\beta -\frac{1}{\eta })-1)+\alpha y(y+1)\frac{1-\left( 1+\frac{\log (1+\frac{1}{y})}{\tau }\right) ^{\alpha -1}}{\alpha -1}+(\beta -\frac{1}{\eta }) > 0.$

  Because $(1+x)^r \le 1+ rx \;\; \forall x \ge -1, \;r \in \; (0,1)$ [49], a sufficient condition for the above inequality to hold is

  $y(2(\beta -\frac{1}{\eta })-1)+\alpha \frac{y(y+1)}{\alpha -1}[1-\{1+(\alpha -1)\frac{\log (1+\frac{1}{y})}{\tau }\}]+(\beta -\frac{1}{\eta })) > 0.$

  Because $\log (1+x) \le x \; \forall x \ge -1$ [50], a sufficient condition for the above inequality to hold is

  $y\tau (2(\beta -\frac{1}{\eta })-1)-\alpha (y+1)+\tau (\beta -\frac{1}{\eta })) > 0$

  $\Leftrightarrow y(2\tau (\beta -\frac{1}{\eta })-\tau -\alpha )-\alpha +\tau (\beta -\frac{1}{\eta })) > 0.$

A sufficient condition for the above inequality to hold is

$$\begin{aligned} \beta \ge \frac{\alpha }{\tau }+\frac{1}{\eta }\;\; \text{ and } \; \; \tau < \alpha , \end{aligned}$$

which is the same as (52). Hence, when (52) holds and $1< \alpha < 2$, then $g''(y)> 0$ when $g'(y) = 0$.

(c) $\alpha \ge 2:$ From (51),

$$\begin{aligned}&y\left( 2\left( \beta -\frac{1}{\eta }\right) -1\right) +\alpha y(y+1)\frac{1-\left( 1+\frac{\log \left( 1+\frac{1}{y}\right) }{\tau }\right) ^{\alpha -1}}{\alpha -1}\nonumber \\&\quad +(\beta -\frac{1}{\eta }) > 0, \end{aligned}$$

(53)

because $(1+x)^r \le 1+ \left( 2^r-1\right) x \;\; \forall x \in [0,1], \;r \in \; \mathscr {R} \backslash (0,1)$ [49].

Now,

$$\begin{aligned} 0 \le \log (1+\eta ) \le \tau \Rightarrow \log (1+\frac{1}{y}) \le \tau \;(\text{ since }\; y \ge \frac{1}{\eta }). \end{aligned}$$

(54)

Next, if (54) holds, a sufficient condition for (53) to hold is

$y\left( 2\left( \beta -\frac{1}{\eta }\right) -1\right) +\alpha y(y+1)\frac{1-\left( 1+\left( 2^{\left( \alpha -1\right) }-1\right) \frac{\log \left( 1+\frac{1}{y}\right) }{\tau }\right) }{\alpha -1}+\left( \beta -\frac{1}{\eta }\right) > 0$

Using the fact that $\log (1+x)\le x \;\;\forall x \ge -1$ [50], a sufficient condition for the above inequality to hold is

$y\tau \left( 2(\beta -\frac{1}{\eta })-1\right) - \frac{\alpha (y+1)\left( 2^{(\alpha -1)}-1\right) }{\alpha -1}+\tau (\beta -\frac{1}{\eta })> 0$

$\Leftrightarrow y\left( 2\tau \left( \beta -\frac{1}{\eta }\right) -\tau -\frac{\alpha (2^{(\alpha -1)}-1)}{\alpha -1}\right) - \frac{\alpha (2^{(\alpha -1)}-1)}{\alpha -1}+\tau \left( \beta -\frac{1}{\eta }\right) > 0$.

The above inequality holds if the following two inequalities hold:

$$\begin{aligned} \beta -\frac{1}{\eta }>\frac{\alpha \left( 2^{(\alpha -1)}-1\right) }{\tau (\alpha -1)} \end{aligned}$$

(55)

and $2\tau \left( \beta -\frac{1}{\eta }\right) -\tau -\frac{\alpha \left( 2^{(\alpha -1)}-1\right) }{\alpha -1}>0$.

Using (55), a sufficient condition for the above inequality to hold is

$$\begin{aligned}&2\tau \frac{\alpha \left( 2^{\left( \alpha -1\right) }-1\right) }{\tau \left( \alpha -1\right) }-\tau -\frac{\alpha \left( 2^{\left( \alpha -1\right) }-1\right) }{\alpha -1}>0.\nonumber \\&\quad \Leftrightarrow \frac{\alpha \left( 2^{\left( \alpha -1\right) }-1\right) }{\tau (\alpha -1)}>1. \end{aligned}$$

(56)

Hence, when (54), (55) and (56) hold and $\alpha \ge 2$, then $g''(y)> 0$ when $g'(y) = 0$. Therefore, it follows from Property 1 that g(.) is quasi-convex when (52) holds (respectively, (54), (55) and (56) hold) and $\alpha \in [0,2)\backslash \{1\}$ (respectively, $\alpha \ge 2$).

Now, it follows from (47) and (48) that $f'(x)=\beta g'(y)$ and $f''(x)=\beta ^2 g''(y).$ Therefore, f(.) also satisfies the condition in Property 1 whenever g(.) satisfies it. Hence, f(.) is quasi-convex when (52) holds (respectively, (54), (55) and (56) hold) and $\alpha \in [0,2)\backslash \{1\}$ (respectively, $\alpha \ge 2$).

Also, $\lim _{x \rightarrow \infty } f(x)$

$$\begin{aligned}&= \lim _{x \rightarrow \infty } \frac{\tau ^{1-\alpha }}{1-\alpha }\left[ x\left( 1+\frac{\log \left( 1 + \frac{\eta }{(x-1)\eta \beta + 1} \right) }{\tau }\right) ^{1-\alpha }-(x-1) \right] \nonumber \\&= \lim _{x \rightarrow \infty } \frac{\tau ^{1-\alpha }}{1-\alpha }\left[ \frac{\left( \left( 1+\frac{\log \left( 1 + \frac{\eta }{(x-1)\eta \beta + 1} \right) }{\tau }\right) ^{1-\alpha }-1\right) }{\frac{1}{x}}+1 \right] \end{aligned}$$

(57)

Using L’Hopital’s rule,

$$\begin{aligned} \lim _{x \rightarrow \infty } f(x) = \frac{\tau ^{-\alpha }\left( 1+\beta \tau -\alpha \right) }{\beta (1-\alpha )}. \end{aligned}$$

(58)

Now, let

$$\begin{aligned} \beta > \frac{1-\alpha }{\frac{\left( \tau +\log \left( 1 + \eta \right) \right) ^{1-\alpha }}{\tau ^{-\alpha }}-\tau }. \end{aligned}$$

(59)

From (58) and (59), it is easy to show that for $\alpha \in [0, \infty ) \backslash \{1\}$:

$$\begin{aligned} \lim _{x \rightarrow \infty } f(x) < f(1) . \end{aligned}$$

(60)

Now, consider the sublevel set:

$$S = \{x > 1 | f(x) < f(1) \}.$$

By (60), there exists $x_0 > 1$ such that $x \in S$ for all $x > x_0$. Also, clearly $1 \in S$. Since $f(\cdot )$ is quasi-convex, the set S is convex [48]; so $x \in S$ for all $x > 1$. That is, when (54), (55), (56) and (59) hold (respectively, (52) and (59) hold) for $\alpha \ge 2$ (respectively, $\alpha \in [0,2) \backslash \{1\}$), then $f(x) < f(1)$ for all $x > 1$ and the result follows.

Proof of Lemma 2

First, we will show that the function $f_1(x)$ is quasi-convex on the domain $x \ge 1$. Let

$$\begin{aligned} y=(x-1)\beta + \frac{1}{\eta }. \end{aligned}$$

(61)

Then,

$$\begin{aligned} f_1(x)=g_1(y). \end{aligned}$$

(62)

where $g_1(y)=$

$$\begin{aligned} \left( \frac{y}{\beta }+1-\frac{1}{\eta \beta }\right) \log \left( \tau +\log \left( 1 + \frac{1}{y} \right) \right) -(\frac{y}{\beta }-\frac{1}{\eta \beta })\log \tau . \end{aligned}$$

Now, $g_1'(y)= \frac{1}{\beta }\left[ \log p-\frac{y+\beta -\frac{1}{\eta }}{y(y+1)p}-\log \tau \right] ,$

where, $p= \left( \tau +\log \left( 1 + \frac{1}{y} \right) \right) .$

$$\begin{aligned} g_1'(y)=0 \Leftrightarrow y+\beta -\frac{1}{\eta }= y(y+1)p \log \frac{p}{\tau }. \end{aligned}$$

(63)

Now, $g_1''(y)$

$=\frac{1}{\beta }\left[ -\frac{1}{y(y+1)p}-\left( \frac{y(y+1)p-\left( y+\beta -\frac{1}{\eta }\right) \left( (2y+1)p-1\right) }{\left( y(y+1)p\right) ^{2}}\right) \right]$

$$\begin{aligned} =\frac{1}{\beta \left( y(y+1)p\right) ^{2}}\left( -yp-y+2yp\left( \beta -\frac{1}{\eta }\right) +(p-1)\left( \beta -\frac{1}{\eta }\right) \right) . \end{aligned}$$

So $g_1''(y) > 0$

$\Leftrightarrow py\left( 2\left( \beta -\frac{1}{\eta }\right) -1\right) -\left( y+\beta -\frac{1}{\eta }\right) +p\left( \beta -\frac{1}{\eta }\right) ) > 0.$

Substituting from (63) in the above inequality, we get:

$py\left( 2\left( \beta -\frac{1}{\eta }\right) -1\right) -y(y+1)p \log \left( 1+\frac{\log \left( 1+\frac{1}{y}\right) }{\tau }\right) +\left( p\left( \beta -\frac{1}{\eta }\right) \right) > 0$

As $\log (1+x) \le x \;\;\forall x \ge -1$ [50], a sufficient condition for the above inequality to hold is

$y\left( 2\left( \beta -\frac{1}{\eta }\right) -1\right) -y(y+1)\frac{1}{y\tau }+\left( \left( \beta -\frac{1}{\eta }\right) \right) > 0.$

$\Leftrightarrow y\left( 2\tau \left( \beta -\frac{1}{\eta }\right) -1-\tau \right) -1+\left( \tau \left( \beta -\frac{1}{\eta }\right) \right) > 0.$

A sufficient condition for the above inequality to hold is

$$\begin{aligned} \beta \ge \frac{1}{\tau }+\frac{1}{\eta }\;\; \text{ and } \; \; \tau < 1. \end{aligned}$$

(64)

Hence, under the condition in (64), $g_1''(y)> 0$ when $g_1'(y) = 0$ for $\alpha =1.$ Therefore, it follows from Property 1 that $g_1(.)$ is quasi-convex.

Now, it follows from (61) and (62) that $f_1'(x)=\beta g_1'(y)$ and $f_1''(x)=\beta ^2 g_1''(y).$ Therefore, $f_1(.)$ also satisfies the condition in Property 1 whenever $g_1(.)$ satisfies it. Hence, $f_1(.)$ is quasi-convex when (64) holds.

Also, $\lim _{x \rightarrow \infty } f_1(x)$

$$\begin{aligned}&= \lim _{x \rightarrow \infty } x\log \left[ \tau \left( 1+ \frac{\log \left( 1+\frac{\eta }{(x-1)\eta \beta +1}\right) }{\tau } \right) \right] -x\log \tau +\log \tau \nonumber \\&= \lim _{x \rightarrow \infty } x\log \left( 1+ \frac{\log \left( 1+\frac{\eta }{(x-1)\eta \beta +1}\right) }{\tau } \right) +\log \tau \nonumber \\&= \lim _{x \rightarrow \infty } \frac{\log \left( 1+ \frac{\log \left( 1+\frac{\eta }{(x-1)\eta \beta +1}\right) }{\tau } \right) }{\frac{1}{x}} +\log \tau . \end{aligned}$$

(65)

Using L’Hopital’s rule,

$$\begin{aligned} \lim _{x \rightarrow \infty } f_1(x) = \frac{1}{\beta \tau }+\log \tau . \end{aligned}$$

(66)

Now, let

$$\begin{aligned} \beta > \frac{1}{\tau \log \left( 1+ \frac{\log \left( 1+\eta \right) }{\tau }\right) }. \end{aligned}$$

(67)

From (66) and (67), it is easy to show that

$$\begin{aligned} \lim _{x \rightarrow \infty } f_1(x) < f_1(1). \end{aligned}$$

(68)

Now, consider the sublevel set:

$$S = \{x > 1 | f_1(x) < f_1(1) \}.$$

By (68), there exists $x_0 > 1$ such that $x \in S$ for all $x > x_0$. Also, clearly $1 \in S$. Since $f_1(\cdot )$ is quasi-convex, the set S is convex [48]; so $x \in S$ for all $x > 1$. That is, when (64) and (67) hold, then $f_1(x) < f_1(1)$ for all $x > 1$ and the result follows.