Performance analysis of HARQ protocols with link adaptation on fading channels

Abstract

This work considers hybrid automatic repeat request (HARQ) protocols on a fading channel with Chase combining and deals with both Rayleigh and Nakagami-m fading. We derive the packet loss probability and the throughput for HARQ both for a slow-varying and a fast-varying channel. We then consider link adaptation with complete channel state information (CSI) for which the instantaneous signal-to-noise ratio (SNR) is known and with incomplete CSI for which only the average SNR is known. We derive analytical formulae of the long-term throughput. These formulae are simple enough to be used for higher level simulations. We show that the throughput is slightly higher on a slow-varying channel but at the expense of a higher loss probability.

Appendices

Appendix A: Long-term throughput of ARQ protocols

In [6] the following formula is given for throughput T _K (γ):

$$ T_K(\gamma) = \sum\limits_{k=1}^{K} \left(\varphi_{k-1}(\gamma) - \varphi_{k}(\gamma)\right) \frac{R}{k}\mbox{.} $$

(29)

The rationale behind the formula is that the throughput is divided by k when there is k transmissions of the same packet. Authors of [6] show that the throughput derived from Eq. 29 is higher than the one derived from Eq. 5. Formula 29 can be used when considering the expected throughput for one epoch of an ARQ round. However, the channel is immediately used for a new transmission (see Fig. 8). In other words, when considering the long-term throughput, the average must be calculated over the time. Equation 29 has to be modified to include a weight that takes into account the duration of the ARQ round:

$$ T_K(\gamma) = {\sum\limits_{k=1}^{K} w_k \left(\varphi_{k-1}(\gamma) - \varphi_{k}(\gamma)\right) \frac{R}{k} } $$

(30)

with

$$ w_k = \frac{kD}{\sum\limits_{i=1}^{K} i D \left(\varphi_{i-1}(\gamma) - \varphi_{i}(\gamma)\right) + K D \varphi_{k}(\gamma)} $$

(31)

where D is the transmission duration of a packet. As $\sum_{k=1}^{K} (\varphi_{k-1}(\gamma) - \varphi_{k}(\gamma)) = 1-\varphi_{K}(\gamma)$, then

$$ T_K(\gamma) = R \frac{1-\varphi_{K}(\gamma)}{N_K(\gamma)} $$

(32)

which is the same as Eq. 32. That is why we do not consider Eq. 29 in Section 2.3 to derive the throughput.

Fig. 8

Evolution of the throughput with time with permanent packet transmission

Appendix B: Detailed derivation of the packet error probability on a fast-varying channel

Let define $x=m \gamma /\overline{\gamma}$, $X=m\gamma_M /\overline{\gamma}$ and $\alpha=g \overline{\gamma}/m$. We call x the “normalized SNR”. Note that f(x) = 1 if x ≤ X and f(x) = e ^{α(x − X)} if x ≥ X and $p(x)=\frac{x^{m-1}}{(m-1)!}e^{-x}$.

We use mathematical induction to compute the loss probability. Let us assume that there is already some accumulated normalized SNR at the receiver at the beginning of the transmission. Let x _A be this accumulated normalized SNR. Let Φ _k (x _A ) be the probability to have at least k transmissions given that the accumulated normalized SNR before the first transmission is x _A . Probability Φ _k (x _A ) depends on the average SNR but for the sake of simplicity we omit parameter $\overline{\gamma}$. Probability Φ_1 + k (x _A ) may be computed by considering that both the SNR at the first transmissions and x _A impact the 1 + k receptions of the packet:

$$ \Phi_{1+k}\left(x_A\right) = \int_{0}^{\infty} p(x) f\left(x_A+x\right) \Phi_{k}\left(x_A+x\right) {\rm d}x\mbox{.} $$

(33)

General formula of the loss probability.

We want to prove that

$${\begin{array}{rll} &&\Phi_{k}(x_A) \notag\\ && \;\;\; = \left\{ \begin{array}{l@{\quad}l} \begin{array}{l} 1- e^{-(X-x_A)} \sum_{j=0}^{k-1} \sum_{l=0}^{m-1}\\ {\kern6pt} \times\left[1 - B_{k-j}^l C_{k-j}^m \right] \dfrac{\left(X-x_A\right)^{jm+l}}{(jm+l)!}\\ \end{array} & \mbox{if $x_A \le X$} \\[24pt] e^{-k\alpha(x_A-X)} C_{k}^m& \mbox{if $x_A>X$}\\ \end{array} \right. \end{array}} $$

(34)

with B _n  = 1 + nα and $C_{n}= \prod_{j=1}^{n} \frac{1}{1 +j\alpha}$. We use the following conventions: $\sum_{i=0}^{-1}x_i=0$ for any x _i . We can re-write Eq. 34 as

$$ \begin{array}{rll} &&\Phi_{k}(x_A) \notag\\ && \;\;\; = \left\{ \begin{array}{@{}l@{\,\,\,\,}l} 1- e^{-(X-x_A)} \sum_{i=0}^{km-1}A_{k,i}\frac{\left(X-x_A\right)^{i}}{i!} & \mbox{if $x_A \le X$} \\ e^{-k\alpha(x_A-X)} C_{k}^m& \mbox{if $x_A>X$} \end{array} \right. \end{array} $$

(35)

with $A_{k,i}=\big(1 {\kern-0.6pt} - {\kern-0.6pt} B_k^i C_{k}^m \big)$ if 0 ≤ i ≤ m  − 1 and $A_{k,i}{\kern-0.6pt}={\kern-0.6pt}\big(1 -$ $B_{k-j}^{i-mj} C_{k-j}^m \big)$ if m ≤ i < mk with $j=\lfloor i/m \rfloor$. Note that A _k + 1,i  = A _{k,i − m} for i ≥ m.

Loss probability for SNR higher than X.

For k = 0 Eq. 35 is equivalent to Φ₀(x _A ) = 1, which is correct. The formula is then checked for k = 0. Using $f(x+x_A)=e^{\alpha(x+x_A-X)}$ and combining Eq. 33 and the lower part of Eq. 35 for x _A  ≥ X gives:

$$ \begin{array}{rll} \Phi_{k+1}\left(x_A\right) &=& C_{k}^m e^{-\alpha(k+1)(x_A-X)}\nonumber\\ &&\times \, \frac{1}{(m-1)!} \int_{0}^{\infty} x^{m-1} e^{-(\alpha(k+1)+1)x} {\rm d}x \end{array} $$

(36)

and then

$$ \Phi_{k+1}\left(x_A\right) = C_{k}^m \frac{1}{(1+\alpha(k+1))^{m}} e^{-\alpha(k+1)(x_A-X)}\mbox{.} $$

(37)

As $C_{k+1}= C_{k} \frac{1}{(1+\alpha(k+1))}$ we have then

$$ \Phi_{k+1}\left(x_A\right) = C_{k+1}^m e^{-\alpha(k+1)(x_A-X)} $$

(38)

which is Eq. 35 for k + 1.

Loss probability for SNR lower than X.

We now consider the case where x _A  < X. We easily check that Eq. 34 is valid for k = 0. Assuming Eq. 35 is valid for k, we have

$$ \begin{array}{rll} &&{\kern-6pt} \Phi_{k+1}(x_A) \\ &&= \int_{0}^{X-x_A} \frac{1}{(m-1)!}x^{m-1} e^{-x}\\ &&{\kern6pt} \times\left(1- e^{-(X-x_A-x)} \sum\limits_{i=0}^{mk-1} A_{k,i} \frac{\left(X-x_A - x\right)^i }{i!} \right){\rm d}x \\ &&{\kern6pt}+\, \int_{X-x_A}^{\infty} \frac{1}{(m-1)!}x^{m-1} \\ &&{\kern6pt}\times\, e^{-x} e^{-\alpha(x + x_A-X)} e^{-k\alpha(x + x_A-X)} C_{k}^m ~{\rm d}x\mbox{.} \end{array} $$

Let x _D  = X − x _A , we have

$$ \begin{array}{rll} \Phi_{k+1}(x_A) &=& \frac{1}{(m-1)!} \int_{0}^{x_D} x^{m-1} e^{-x} {\rm d}x - \frac{1}{(m-1)!} e^{-x_D}\\ && \times\,\sum\limits_{i=0}^{mk-1} \frac{A_{k,i}}{i!} \int_{0}^{x_D} x^{m-1} {(x_D - x)^i } ~{\rm d}x \\ &&+\, \frac{1}{(m-1)!}C_{k}^m e^{(k+1)\alpha x_{D}}\\ &&\times\, \int_{x_D}^{\infty} x^{m-1} e^{-((k+1)\alpha+1)x} ~{\rm d}x\mbox{.} \end{array} $$

As $\int_{0}^{x_D} x^{m-1} {(x_D - x)^i } {\rm d}x = x_D^{m+i} i!(m-1)!/(m+i)!$, it comes

$$ \begin{array}{rll} \Phi_{k+1}(x_A) &=& 1 - e^{-x_D} \sum\limits_{i=0}^{m-1}\frac{x_D^i}{i!} - e^{-x_D} \sum\limits_{i=0}^{mk-1} A_{k,i} \frac{x_D^{m+i}}{(m+i)!} \\ &&+\, C_{k+1}^m e^{-x_D} \sum\limits_{i=0}^{m-1}((k+1)\alpha+1)^i \: \frac{x_{D}^i}{i!} \end{array} $$

and then

$${\begin{array}{rll} \Phi_{k+1}(x_A) &=& 1 - e^{-x_D} \sum\limits_{i=0}^{m-1}\left(1-C_{k+1}^m B_{k+1}^i\right)\frac{x_D^i}{i!}\notag\\ && -\, e^{-x_D} \sum\limits_{i=m}^{m(k+1)-1} A_{k,i-m} \frac{x_D^{i}}{i!}\mbox{.} \end{array}} $$

(39)

We can see that Eq. 39 is equivalent to Eq. 35 for k + 1. Hence, for all values of k and x _A , Eqs. 35 and 34 are valid. Choosing x _A  = 0 we find Eq. 9.