Analyzing the capacity of wireless ad hoc networks

Abstract

In this paper we develop analytical closed form expression for the capacity of a wireless ad hoc network. First, for the general case when nodes can adapt their communication rates to the link quality, a proper formulation for the total network capacity is presented based on the cumulative distribution function (CDF) of the signal to interference power ratio (SIR). Then, a closed form expression for this CDF is analytically derived. This closed form is further studied by fitting it to a normal distribution. Afterwards, the capacity of the network is investigated. By examining the effect of the outage threshold, it is shown that in order to obtain a higher capacity, one may use simple non-adaptive transceivers with higher threshold on the received SIR. These results are obtained by conducting analytical and simulation studies.

Appendix:  Estimating the mean and variance of the received SIR

A Gaussian distribution is fully characterized by its mean and variance [20]. Therefore, in order to approximate the distribution of the SIR at the location with a Gaussian distribution, here its mean and variance are estimated.

According to Eq. (16), we can write:

$$\begin{aligned} \mbox{mean} \{ \mathit{SIR}_{dB} \} &= E \biggl\{ 10\log_{10} \biggl( \frac{1}{d^{\alpha}} \biggr) - I_{dB} \biggr\} \\ & = E \biggl\{ 10\log_{10} \biggl( \frac{1}{ d^{\alpha}} \biggr) \biggr\} - E \{ I_{dB} \} , \end{aligned}$$

(24)

where, E{x} is the expected value of the random variable x. Each term in the right-side of the above equation is calculated separately. The first and second terms are calculated using the probability density function (PDF) of d and I _dB , respectively. These PDFs are the derivative of the corresponding cumulative distribution functions (CDFs) [20] which are given in Eqs. (3) and (15) respectively.

According to Eq. (3) we have:

$$ f_{d}(\rho ) = \frac{dF_{d}(\rho )}{d\rho} = \left \{ \begin{array} {l@{\quad}l} \frac{2\rho}{D^{2} - \varepsilon^{2}}, & \varepsilon \le \rho \le D \\ 0, & \mbox{otherwise}. \end{array} \right . $$

(25)

And, based on Eq. (15) we can write:

$$\begin{aligned} f_{I_{dB}} ( x ) &= \frac{dF_{I_{dB}} ( x )}{dx} \\ &\cong ( N - 1 ) \frac{ ( R^{2} - 10^{ - x / 5\alpha} )^{ ( N - 2 )}}{ ( R^{2} - \varepsilon^{2} )^{ ( N - 1 )}}\frac{ - d ( 10^{ - x / 5\alpha} )}{dx}, \\ &{-} 10\log_{10}R \le x \le - 10\log_{10}\varepsilon. \end{aligned}$$

(26)

Now, the first term in the right-side of Eq. (24) is calculated as follows:

$$\begin{aligned} E \biggl\{ 10\log_{10} \biggl( \frac{1}{ d^{\alpha}} \biggr) \biggr\} &= ( - 10\alpha )E \bigl\{ \log_{10} ( d ) \bigr\} \\ &= ( - 10\alpha )\int_{\varepsilon}^{D} \log_{10} ( d )f_{d} ( x )dx \\ & = \frac{2 ( - 10\alpha )}{\ln 10 \times ( D^{2} - \varepsilon^{2} )} \\ &\quad {}\times \biggl[ \frac{D^{2}\ln D}{2} - \frac{D^{2}}{4} - \frac{\varepsilon^{2}\ln \varepsilon}{2} + \frac{\varepsilon^{2}}{4} \biggr]. \end{aligned}$$

(27)

And, the second term of Eq. (24) is calculated as follows:

$$ E\left\{ I_{dB} \right\} = \int_{ - 10\log_{10}R}^{ - 10\log_{10}\varepsilon} x.f_{I_{dB}}(x)dx. $$

(28)

We define:

$$ y \triangleq = 10^{ - x / 5\alpha}. $$

(29)

Therefore:

$$\begin{aligned} &E \{ I_{dB} \} \\ &\quad = \int_{ - 10\log_{10}R}^{ - 10\log_{10}\varepsilon} x.f_{I_{dB}}(x)dx \\ &\quad =\int_{\varepsilon^{2}}^{R^{2}} \bigl( - 5\alpha \log_{10}(y) \bigr) ( N - 1 )\frac{ ( R^{2} - y )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1}}dy \\ &\quad = \frac{ - 5\alpha ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )}\int_{\varepsilon^{2}}^{R^{2}} \bigl( \ln (y) \bigr) \bigl( R^{2} - y \bigr)^{N - 2}dy \\ &\quad = \frac{ - 5\alpha ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )} \\ &\quad {}\times\int_{\varepsilon^{2}}^{R^{2}} \ln (y)\left [ \sum_{k = 0}^{N - 2} \left \{ \left ( \begin{array} {c} N - 2\\ k \end{array} \right ) \bigl( - R^{2} \bigr)^{N - 2 - k}y^{k} \right \} \right ]dy \\ &\quad = \frac{ - 5\alpha ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )} \\ &\qquad {}\times \sum_{k = 0}^{N - 2} \left \{ \left ( \begin{array} {c} N - 2 \\ k \end{array} \right ) \bigl( - R^{2} \bigr)^{N - 2 - k} \biggl[ \int_{\varepsilon^{2}}^{R^{2}} \ln (y)y^{k}dy \biggr] \right \} \\ &\quad = \frac{ - 5\alpha ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )} \sum_{k = 0}^{N - 2} \biggl\{ \left ( \begin{array} {c} N - 2\\ k \end{array} \right ) \bigl( - R^{2} \bigr)^{N - 2 - k} \\ &\qquad {}\times \biggl[ \frac{R^{2k + 2}\ln R^{2}}{k + 1} - \frac{R^{2k + 2}}{ ( k + 1 )^{2}} \\ &\qquad {}- \frac{\varepsilon^{2k + 2}\ln \varepsilon^{2}}{k + 1} + \frac{\varepsilon^{2k + 2}}{ ( k + 1 )^{2}} \biggr] \biggr\}. \end{aligned}$$

(30)

By replacing (27) and (30) in Eq. (24), the mean of SIR _dB is obtained □.

In order to calculate the variance, we start by Eq. (16) and calculate $E\{\mathit{SIR}_{dB}^{2}\}$.

$$\begin{aligned} E \bigl\{ \mathit{SIR}_{dB}^{2} \bigr\} &= E \biggl\{ \biggl( 10 \log_{10} \biggl( \frac{1}{ d^{\alpha}} \biggr) - I_{dB} \biggr)^{2} \biggr\} \\ &= E \biggl\{ 100 \biggl( \log_{10} \biggl( \frac{1}{d^{\alpha}} \biggr) \biggr)^{2} \biggr\} + E \bigl\{ I_{dB}^{2} \bigr\} \\ &\quad {}- 2E \biggl\{ 10\log_{10} \biggl( \frac{1}{d^{\alpha}} \biggr) \biggr\}E \{ I_{dB} \}. \end{aligned}$$

(31)

The right-side of the above equation consists of 3 terms, among which the third is already known from Eqs. (27) and (30). Therefore, we concentrate on the first and second terms.

$$\begin{aligned} &E \biggl\{ 100 \biggl( \log_{10} \biggl( \frac{1}{d^{\alpha}} \biggr) \biggr)^{2} \biggr\} \\ &\quad = \frac{100\alpha^{2}}{ ( \ln 10 )^{2}}E \bigl\{ ( \ln d )^{2} \bigr\} \\ &\quad = \frac{100\alpha^{2}}{ ( \ln 10 )^{2}}\int_{\varepsilon}^{D} \frac{2x ( \ln x )^{2}}{D^{2} - \varepsilon^{2}}dx \\ &\quad = \frac{200\alpha^{2}}{ ( \ln 10 )^{2} \times ( D^{2} - \varepsilon^{2} )}\int_{\varepsilon}^{D} x ( \ln x )^{2}dx. \end{aligned}$$

To calculate the above integral, we use the following equality:

$$ \begin{aligned} \int x^{n} ( \ln x )^{2}dx &= \frac{x^{n + 1} ( \ln x )^{2}}{n + 1} - \frac{2x^{n + 1}}{ ( n + 1 )^{2}} \biggl( \ln x - \frac{1}{n + 1} \biggr) \\ &\quad {}+ \mathrm{constant}. \end{aligned} $$

We also define:

$$ L(x,n) \buildrel \Delta \over = \frac{x^{n + 1} ( \ln x )^{2}}{n + 1} - \frac{2x^{n + 1}}{ ( n + 1 )^{2}} \biggl( \ln x - \frac{1}{n + 1} \biggr). $$

(32)

Hence:

$$\begin{aligned} &E \biggl\{ 100 \biggl( \log_{10} \biggl( \frac{1}{d^{\alpha}} \biggr) \biggr)^{2} \biggr\} \\ &\quad = \frac{200\alpha^{2}}{ ( D^{2} - \varepsilon^{2} ) ( \ln 10 )^{2}} \bigl( L ( D,1 ) - L ( \varepsilon,1 ) \bigr). \end{aligned}$$

(33)

The second term of Eq. (31) is calculated in a similar way:

$$\begin{aligned} &E \bigl\{ ( I_{dB} )^{2} \bigr\} \\ &\quad = \int_{ - 10\log_{10}R}^{ - 10\log_{10}\varepsilon} x^{2}.f_{I_{dB}}(x)dx \\ &\quad = \int_{\varepsilon^{2}}^{R^{2}} \bigl( - 5\alpha \log_{10}(y) \bigr)^{2} ( N - 1 )\frac{ ( R^{2} - y )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1}}dy, \end{aligned}$$

where, again Eq. (29) has been used to change the integration variable. Therefore:

$$\begin{aligned} E \bigl\{ ( I_{dB} )^{2} \bigr\} &= \frac{25\alpha^{2} ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )^{2}} \\ &\quad {}\times\int _{\varepsilon^{2}}^{R^{2}} \bigl( \ln (y) \bigr)^{2} \bigl( y - R^{2} \bigr)^{N - 2}dy \\ &= \frac{25\alpha^{2} ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )^{2}} \\ &\quad {}\times \int_{\varepsilon^{2}}^{R^{2}} \bigl( \ln (y) \bigr)^{2}\Biggl[ \sum_{k = 0}^{N - 2} \biggl\{ \left ( \begin{array} {c} N - 2 \\ k \end{array} \right ) \\ &\quad {}\times \bigl( - R^{2} \bigr)^{N - 2 - k}y^{k} \biggr\} \Biggr]dy \\ &= \frac{25\alpha^{2} ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )^{2}} \\ &\quad {}\times \sum_{k = 0}^{N - 2} \biggl\{ \left ( \begin{array} {c} N - 2 \\ k \end{array} \right ) \bigl( - R^{2} \bigr)^{N - 2 - k} \\ &\quad {} \times \biggl[ \int_{\varepsilon^{2}}^{R^{2}} \bigl( \ln (y) \bigr)^{2}y^{k}dy \biggr] \biggr\} \\ &= \frac{25\alpha^{2} ( N - 1 ) ( - 1 )^{N - 2}}{ ( R^{2} - \varepsilon^{2} )^{N - 1} ( \ln 10 )^{2}} \\ &\quad {}\times \sum_{k = 0}^{N - 2} \biggl\{ \left ( \begin{array} {c} N - 2 \\ k \end{array} \right ) \bigl( - R^{2} \bigr)^{N - 2 - k} \\ &\quad {} \times \bigl[ L\bigl(R^{2},k\bigr) - L\bigl(\varepsilon^{2},k\bigr) \bigr] \biggr\}, \end{aligned}$$

(34)

where, L(x,n) is defined in Eq. (32).

Equations (33) and (34) calculate the first and second terms of Eq. (31). The third term is obtained from Eqs. (27) and (30). As a result, $E\{\mathit{SIR}_{dB}^{2}\}$ is calculated.

Finally, the variance of the random variable SIR _dB is calculated as follows:

$$ \operatorname{var}\{\mathit{SIR}_{dB} \}= E\bigl\{ \mathit{SIR}_{dB}^{2} \bigr\} - \bigl(E\{\mathit{SIR}_{dB} \}\bigr)^{2}. $$

(35)