Bounding Queue Length Violation Probability of Joint Channel and Buffer Aware Transmission

Based on Eq. (4), to transmit all the arrival data in each time slot, the average power consumption can be expressed as

To keep this integral finite, the key point is to avoid the influence of $\lim_{x\to 0^{+}}\frac{1}{x}=\infty$. One sufficient condition is to control the integrand not approaching infinity when $x$ approaches 0, i.e. $\lim_{x\to 0^{+}}\frac{g(x)}{x}<\infty$. From Eq. (8), we can find that if $P_{\rm avg}$ is finite, then $\lim_{x\to 0^{+}}g(x)=0$. Since $\lim_{x\to 0^{+}}g(x)=0$, $g(x)$ can then be expanded as

where $\mathcal{O}(x)$ denote the higher-order infinitesimal of $x$. Based on Eq. (9), now $\lim_{x\to 0^{+}}\frac{g(x)}{x}<\infty$ is equivalent to $\lim_{x\to 0^{+}}g^{\prime}(x)<\infty$. Thus, if Eqs. (6) and (7) hold, $P_{\rm avg}$ must be finite. ∎

We start this proof by obtaining the steady state probability of the queue length with the transmission policy. Let $\pi_{k}$ denote the steady state probability of queue length being equal to $k$, $k=0,1,\cdots$. To obtain $\pi_{k}$, we present the steady state equation and normalization condition, which are given by

By substituting the steady state equation into the normalization condition, we obtain $\pi_{0}$ as

Besides, from the steady state equation, we can obtain the local equilibrium equation as

By substituting Eq. (18) into Eq. (19), $\pi_{k}$, $k=1,\cdots$, is given by

For a Rayleigh fading channel with $\mathbb{E}\{|h[n]|^{2}\}=1$, Eq. (15) can be simplified as

From Eq. (21), we have $h_{\rm th}^{k}=-\ln(1-p_{k})$. Let $C_{k}$ denote the average power consumption with the queue length $k$, which is given by

Then, by combining Eqs. (20), (18), and (22), we obtain Eq. (23).

According to [12], the function $E_{1}(x)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt$ has the property that

By combining Eq. (24) with Eq. (23), we obtain Eq. (25).

The Taylor’s expansion of $\ln(1-p_{k})$ can be expressed as $-p_{k}-\frac{1}{2}p_{k}^{2}+O(p_{k}^{2})$. Therefore, we obtain that $\ln(1-p_{i})<-p_{i}$, which indicates $\ln(1-\frac{1}{\ln(1-p_{k})})<\ln(1+\frac{1}{p_{k}})$. Then, Eq. (25) can be further expressed as

Therefore, if the sequence in Eq. (26) is convergent, the average power of this policy is finite. According to D’Alembert’s ratio test [13], we can obtain a sufficient condition of the finite average power consumption as shown in Eq. (16). ∎

If $p_{k}$ approaches 0, then $\ln(1+\frac{1}{p_{k}})\approx\ln(\frac{1}{p_{k}})$. Combined with Eq. (16), the sufficient condition is simplified as Eq. (27). ∎

According to Eqs. (16) and (27), it is easy to verify that, sequences $\{p_{k}=e^{-k},k=0,1,\cdots\},\{p_{k}=e^{-e^{k}},k=0,1,\cdots\},\{p_{k}=e^{-e^{e^{k}}},k=0,1,\cdots\},\cdots$, satisfy Eqs. (16) and (27). We first give a proof on why these sequences can guarantee the finite $P_{\rm avg}$.

Above sequences are all decreasing sequences with $\lim_{k\to\infty}p_{k}=0$. According to Corollary 1, we have to prove that Eq. (27) holds for these sequences. For these sequences, they obey

Besides, with $k\to\infty$, $\ln(p_{k+1})$ and $\ln(p_{k})$ are the same-order infinity. Thus, we obtain

where $C\in\mathbb{R}$ is a constant. According to Eqs. (28) and (29), these two limits have finite values. Therefore, according to the Limit Laws [13], the limit of a product is equal to the product of the limits, which proves that for these sequences,

The length violation probability for the above transmission policy is given by

Since $\mu_{m}<1$, $p_{m}/\mu_{m+1}<1$, and $\pi_{0}<1$, $\varepsilon(q_{\rm th})$ satisfies

For sequences $\{p_{k}=e^{-k},k=0,1,\cdots\},\{p_{k}=e^{-e^{k}},k=0,1,\cdots\},\{p_{k}=e^{-e^{e^{k}}},k=0,1,\cdots\},\cdots$, $\mu_{k}=1-p_{k}\approx 1$ with $k\to\infty$. Thus, according to Eq. (32), the decay rate of the queue length violation probability is dominated by $\sum_{k=q_{\rm th}+1}^{\infty}\prod_{m=0}^{k-1}p_{m}$ It indicates that, by adopting the above transmission policy, the exponent of the queue length violation probability can have an arbitrarily high decay rate with finite $P_{\rm avg}$. ∎

If $m\geq 2$, we obtain

and

According to Lemma 1, we have proved that if $m\geq 2$, the system can achieve zero queue length violation probability with finite $P_{\rm avg}$. Thus, with $m\geq 2$, the system belongs to Scenario 1.

For $m=1$, we have proved in Theorems 1 and 2 that the tail distribution of queue length can have arbitrary-decay-rate exponents with finite $P_{\rm avg}$. Based on this result, we then prove that for $1<m<2$, the system belongs to Scenario 3.

Given $m$, $m^{m}$ and $\Gamma(m)$ are finite positive constants. Intuitively, whether the system belongs to Scenario 3 should be determined by the remaining part in $g(x)$ except these constants. Thus, for the simplicity of expression in the following proof, we define $\ell(x)$ as

Here, we provide a sketch of the proof of the system belonging to Scenario 3 when $1<m<2$. First, we will prove that $\ell(x)$ with $1<m<2$ can ensure that the system belongs to Scenario 3 with the help of Theorems 1 and 2. Based on this conclusion, we will then prove $g(x)=\frac{m^{m}}{\Gamma(m)}\ell(x)$, $1<m<2$, can ensure that the system belongs to Scenario 3.

We see $\ell(x)$ as a function of $m$, which we denote as $w(m)$. The derivative of $w(m)$ is given by

Since for arbitrary $x>0$, $\ln x-x<0$, we have $w^{\prime}(m)<0$ with $1\leq m<2$. Thus, $w(m)$ is a decreasing function on $[1,2)$. Then, for arbitrary $x>0$, we have $w(1)>w(m)$, $m\in(1,2)$, which is equivalent to

Let $h_{\rm th}^{k}(m)$ denote the channel power gain threshold of the heuristic policy under the queue length $k$ given $m$. Then, given the sequence $\{p_{k},k=0,1,\cdots\}$, the relationship between $p_{k}$ and $h_{\rm th}^{k}(m)$ is given by

By combining Eqs. (39) and (40), we obtain that, with given $p_{k}$, $h_{\rm th}^{k}(1)<h_{\rm th}^{k}(m)$, $m\in(1,2)$.

Let $C_{k}(m)$ denote the average power consumption under the queue length $k$ given $m\in(1,2)$. By comparing with Eq. (22), we find that the integrand in the definition of $C_{k}(m)$, i.e., $x^{m-2}e^{-mx}$, is smaller than $e^{-x}/x$ according to Eq. (39). Besides, the integral interval in the definition of $C_{k}(m)$ is $(h_{\rm th}^{k}(m),\infty)$. The length of $(h_{\rm th}^{k}(m),\infty)$ is smaller than $(h_{\rm th}^{k}(1),\infty)$ since $h_{\rm th}^{k}(1)<h_{\rm th}^{k}(m)$, $m\in(1,2)$. Therefore, given the sequence $\{p_{k},k=0,1,\cdots\}$, with smaller integrand and integral interval,

Note that, the steady state probability of queue length under the heuristic policy is determined by the $\{p_{k},k=0,1,\cdots\}$ sequence only. Thus, given the same $\{p_{k},k=0,1,\cdots\}$ sequence, the steady state probability of queue length with $m=1$ is equal to that with $m\in(1,2)$. Moreover, for each queue length $k$, Eq. (41) holds. Thus, we have proved that, with the same sequence $\{p_{k},k=0,1,\cdots\}$, $P_{\rm avg}$ of $m=1$ is larger than $P_{\rm avg}$ of $m\in(1,2)$, which indicates $\ell(x)$ with $1<m<2$ can ensure that the system belongs to Scenario 3.

Then, we present the proof that $g(x)=\frac{m^{m}}{\Gamma(m)}\ell(x)$, $1<m<2$, can ensure that the system belongs to Scenario 3.

Let $h^{k,1}_{\rm th}(m)$ denote the channel power gain threshold of the heuristic policy under the queue length $k$ given $m$ for $\ell(x)$, while $h^{k,2}_{\rm th}(m)$ denote that for $g(x)$. The relationships between $p_{k}$ and $h^{k,1}_{\rm th}(m)$, $h^{k,2}_{\rm th}(m)$ are given by

Since $m^{m}/\Gamma(m)>1$ when $m\in(1,2)$, we obtain $h^{k,1}_{\rm th}(m)>h^{k,2}_{\rm th}(m)$. Let $C_{k}^{(1)}(m)$ and $C_{k}^{(2)}(m)$ denote the average power consumption under the queue length $k$ given $m$ of $\ell(x)$ and $g(x)$, respectively. $C_{k}^{(1)}(m)$ and $C_{k}^{(2)}(m)$ are given by

where $a_{k}$ is given by

Based on Eq. (44) and $h^{k,1}_{\rm th}(m)>h^{k,2}_{\rm th}(m)$, we decompose Eq. (45) as

where the second term of the right side of Eq. (47) is equal to $\frac{m^{m}}{\Gamma(m)}C_{k}^{(1)}(m)$. Let $P_{\rm avg}^{(1)}$ and $P_{\rm avg}^{(2)}$ denote the average power consumption of the heuristic policy with $\ell(x)$ and $g(x)$, respectively. Then, based on Eq. (47), $P_{\rm avg}^{(2)}$ is given by

As we proved before, $P_{\rm avg}^{(1)}$ is finite with $\{p_{k},k=0,1,\cdots\}$ satisfying Eq. (16). Thus, $\frac{m^{m}}{\Gamma(m)}P_{\rm avg}^{(1)}$ is finite with $m\in(1,2)$ with the same $\{p_{k},k=0,1,\cdots\}$ sequence.

For the first term on the right side of Eq. (47), it satisfies

For $m\in(1,2)$, given a decreasing sequence $\{p_{k},k=0,1,\cdots\}$ which satisfies $\lim_{k\to\infty}p_{k}=0$, we obtain that $\lim_{k\to\infty}r_{k}(m)=0$, $m\in(1,2)$. Thus, given sequence $\{p_{k},k=0,1,\cdots\}$ satisfying Eq. (27), the first term on the right side of Eq. (47) is finite. Now we have proved that given sequence $\{p_{k},k=0,1,\cdots\}$ satisfying Eq. (16), $P_{\rm avg}^{(2)}$ is finite, which indicates that $g(x)$ with $1<m<2$ can ensure that the system belongs to Scenario 3.

For $0.5\leq m<1$, we find that

Besides, similar to the proof for $m\in(1,2)$, given the sequence $\{p_{k},k=0,1,\cdots\}$, $P_{\rm avg}$ of $m=1$ is smaller than $P_{\rm avg}$ of $m\in[0.5,1)$. Therefore, we can not ensure whether the system belongs to Scenario 1 or Scenario 3 when $m\in[0.5,1)$. However, at least we can guarantee that the system belongs to Scenario 2 by adopting the fixed power policy. ∎