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Minimizing Moments of AoI for Both Active and Passive Users through Second-Order Analysis

Siqi Fan Texas A&M University
College Station, USA
[email protected]
   Yuxin Zhong§ Georgia Institute of Technology
Atlanta, USA
[email protected]
   I-Hong Hou Texas A&M University
College Station, USA
[email protected]
   Clement Kam U.S. Naval Research Laboratory
Washington, USA
[email protected]
Abstract

In this paper, we address the optimization problem of moments of Age of Information (AoI) for active and passive users in a random access network. In this network, active users broadcast sensing data while passive users only receive signals. Collisions occur when multiple active users transmit simultaneously, and passive users are unable to receive signals while any active user is transmitting. Each active user follows a Markov process for their transmissions. We aim to minimize the weighted sum of any moments of AoI for both active and passive users in this network. To achieve this, we employ a second-order analysis to analyze the system. Specifically, we characterize an active user’s transmission Markov process by its mean and temporal process. We show that any moment of the AoI can be expressed a function of the mean and temporal variance, which, in turn, enables us to derive the optimal transmission Markov process. Our simulation results demonstrate that this proposed strategy outperforms other baseline policies that use different active user transmission models.

§§footnotetext: Yuxin’s work is done in her study in Texas A&M university.

I Introduction

Recently, there has been a significant increase in the use of time-sensitive applications such as Internet of Things, vehicular networks, sensor networks, and drone systems for surveillance and reconnaissance. While traditional research has focused on optimizing the reliability and throughput of communication, these efforts often fall short in meeting the demands for fresh data. To address this, a performance metric known as age-of-information (AoI) was introduced [1] and has received increasing attention in recent literature [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32].

Previous studies on AoI have commonly considered centralized control over transmission decisions [8, 9, 7, 10, 11, 12, 13] and queuing problems without transmission collisions [15, 18, 16, 19, 17]. However, centralized policies and collisions can lead to large overhead or extra communication, which can be a significant drawback in delay-sensitive applications such as the Internet of Things and transmission-sensitive applications like satellite communications.

To address this issue, we consider a random access network, where active users broadcast sensed data and multiple transmissions at the same time can cause collisions. Since there are no acknowledgements in broadcast, active users do not have any feedback information on collisions. This makes recent feedback-based algorithms for AoI optimization [22, 23, 24, 25, 21, 26] inapplicable.

Additionally, we consider that the network has silent observers, referred to as passive users, who share the same channel as active users. Passive users aim to observe out-of-network radio activities in the same channel but do not make any transmissions. Examples of passive users include sensors detecting malicious jammers, receivers of satellite communications, and radio telescopes tracking celestial objects. The coexistence of passive and active users can improve channel utilization, as passive users can obtain valuable observations when no active user is transmitting.

As AoI depends on previous successful transmissions, memoryless random access algorithms, such as slotted ALOHA, are not suitable for optimizing AoI. Therefore, we consider that each active user follows a Markov process to determine its transmission activities. Furthermore, instead of only considering the mean of AoI, we consider the moments of AoI, which can offer insights into the variance and the maximum value of AoI.

The goal of this paper is to minimize the weighted sum of any moment of AoI for both active users and passive users. There are several challenges that need to be addressed. First, since the transmission strategy of an active user is Markovian instead of i.i.d., the temporal correlation of its transmission activities needs to be explicitly taken into account. Second, active users and passive users have different behavior, which leads to different definitions of AoI between the two types of users. Third, most existing studies on moments of AoI, such as [17, 18, 19], focus only on simple queueing systems and are not applicable to random access networks. Finally, most existing studies on AoI in random access networks only focus on the first moment of AoI and can only obtain the optimal solution in the asymptotic sense.

To address above challenges and solve the optimization problem, we propose to investigate the system behavior through a second-order analysis. We analyze the mean and temporal variance of delivery process to characterize the system. Then, we formulate the expressions of moments of AoI based on these mean and temporal variance, and then further formulate them as functions of state change probabilities in Markov process. By revealing special properties in expressions of moments of AoI for both active and passive users, we find that the optimal action for an active user, under some minor conditions, is to become silent immediately after one attempt of transmission. The simulation results show that our formulation of moments of AoI is valid and our solution outperforms other baseline algorithms that have different active user activity models.

The rest of the paper is organized as follows. Section II introduces our system model and the optimization problem. Section III formulates expressions of moments of AoI based for both active and passive users based on a second-order model. Section IV solves the optimal transmission strategies. Some simulation results are presented in Section V. Finally, Section VI concludes the paper.

II System setting

We consider a wireless system with one channel and two types of users, active users and passive users. Each active user monitors a particular field of interest and uses the wireless channel to transmit its surveillance information to its neighboring receivers. Passive users do not make any wireless transmissions and instead monitor the radio activities in the wireless channel. For example, in a battlefield scenario, an active user can be a surveillance drone that monitors a certain area in the battlefield and broadcasts its video feed to all nearby units. A passive user can be a signal detector that aims to identify and locate enemy jammers or communication devices operating in the same channel.

Next, we discuss the interference relationship among users. We assume that active users are grouped into CC clusters and each cluster has NN (N2N\geq 2) active users. Active users in the same cluster interfere with each other but do not suffer from interference from other clusters. An active user can successfully deliver a packet if it is the only device in the cluster that is transmitting. If multiple active users in the same cluster transmit at the same time, then a collision occurs and none of the packets are received. On the other hand, since passive users need to detect enemy devices that are potentially far away, we assume that a passive user is interfered by all active users and it can only detect radio activities if no active user is transmitting.

The performance of each user is determined by its associated AoI. We assume that time is slotted and the duration of one time slot is the time needed to transmit one packet. Each active user generates a new surveillance packet in each slot. Hence, the AoI of an active user nn at time tt is defined by following recursion:

AoIn,t={1,if users n successfully deliversa packet at time t,AoIn,t1+1,otherwise.\displaystyle AoI_{n,t}=\begin{cases}1,&\text{if users $n$ successfully delivers}\\ &\text{a packet at time $t$,}\\ AoI_{n,t-1}+1,&\text{otherwise}.\end{cases}

Since passive users can only detect radio activities when no active users are transmitting, the AoI of passive user jj at time tt is defined by

AoIj,t={1,if no active user transmits,AoIj,t1+1,otherwise.\displaystyle AoI_{j,t}=\begin{cases}1,\;&\text{if no active user transmits,}\\ AoI_{j,t-1}+1,\;&\text{otherwise}.\end{cases}

We now discuss the transmission strategy of each active user. We assume that the strategy of an active user is governed by a Markov process with two states, a TX state and an Idle state. An active user will transmit a packet at time tt if and only if it is in the TX state. If an active user is in the TX state at time tt, then its state will change to Idle at time t+1t+1 with probability ss. If an active user is in the Idle state at time tt, then its state will change to TX at time t+1t+1 with probability rr. Fig. 1 presents this Markov process. We note that this strategy generalizes the ALOHA network. In particular, if r+s=1r+s=1, then an active user will transmit in each time slot with probability rr independently.

Refer to caption
Figure 1: Markov process model.

We aim to minimize the moments of AoI for both active users and passive users. Specifically, let AoIaAoI_{a} be the random variable of an active user’s AoI and let AoIpAoI_{p} be the random variable of a passive user’s AoI in steady state. For a fixed integer zz and a fixed real number w[0,1]w\in[0,1], we aim to find rr and ss that solve the following optimization problem:

min\displaystyle\min\; F(r,s):=wE[AoIaz]+(1w)E[AoIpz]\displaystyle F(r,s):=wE[AoI_{a}^{z}]+(1-w)E[AoI_{p}^{z}] (1)
s.t. r,s[0,1],\displaystyle r,s\in[0,1], (2)

where E[]E[\cdot] is the expectation function.

III The Second Order Model for AoI

A key challenge in solving the optimization problem (1)-(2) is that it is hard to express E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}] as functions of rr and ss. We propose to adopt the framework of second-order network optimization in Guo et al. [30] to address this challenge. The framework in [30] is only applicable to centrally scheduled systems and the mean of AoI. In this section, we will extend this framework to model random access networks and moments of AoI.

We first define the second-order model that describes the performance of both active and passive users. For an active user nn, Let Sn(t)S_{n}(t) be the indicator function that nn successfully delivers a packet at time tt, that is, nn is the only active user in its cluster that transmits at time tt. We call [Sn(1),Sn(2),][S_{n}(1),S_{n}(2),\dots] the delivery process of nn. Due to symmetry, the delivery processes of all active users follow the same distribution. We then define the mean and temporal variance of the delivery process of an active user by

ma\displaystyle m_{a} :=limTt=1TSn(t)T,\displaystyle:=\lim_{T\xrightarrow{}\infty}\frac{\sum_{t=1}^{T}S_{n}(t)}{T},
va2\displaystyle v_{a}^{2} :=E[(limTt=1TSn(t)TmaT)2],\displaystyle:=E[(\lim_{T\xrightarrow{}\infty}\frac{\sum_{t=1}^{T}S_{n}(t)-Tm_{a}}{\sqrt{T}})^{2}],

respectively, where the expectation is taken over all sample paths.

For a passive user jj, we let Sj(t)S_{j}(t) be the indicator function that no active user transmits, and hence the passive user can monitor radio activities without interference, at time tt. We call [Sj(1),Sj(2),][S_{j}(1),S_{j}(2),\dots] the passive detection process. The mean and temporal variance of the passive detection process are defined as mp:=limTt=1TSj(t)Tm_{p}:=\lim_{T\xrightarrow{}\infty}\frac{\sum_{t=1}^{T}S_{j}(t)}{T} and vp2:=E[(limTt=1TSj(t)TmpT)2]v_{p}^{2}:=E[(\lim_{T\xrightarrow{}\infty}\frac{\sum_{t=1}^{T}S_{j}(t)-Tm_{p}}{\sqrt{T}})^{2}], respectively.

The second-order model uses the means and temporal variances to capture and approximate all aspects of the system. To show its utility, we first derive the closed form expressions of mam_{a}, mpm_{p}, va2v_{a}^{2}, and vp2v_{p}^{2}. Let λ:=rr+s\lambda:=\frac{r}{r+s} and θ=1rs\theta=1-r-s, we have the following theorem:

Theorem 1.
ma\displaystyle m_{a} =λ(1λ)N1,mp=(1λ)CN,\displaystyle=\lambda(1-\lambda)^{N-1},\quad m_{p}=(1-\lambda)^{CN},
va2=\displaystyle v_{a}^{2}= 2k=1((λ+(1λ)θk)(1λ+λθk)N1ma)ma\displaystyle 2\sum_{k=1}^{\infty}((\lambda+(1-\lambda)\theta^{k})(1-\lambda+\lambda\theta^{k})^{N-1}-m_{a})m_{a}
+mama2,\displaystyle+m_{a}-m_{a}^{2},
vp2=\displaystyle v_{p}^{2}= 2k=1((1λ+λθk)CNmp)mp+mpmp2.\displaystyle 2\sum_{k=1}^{\infty}((1-\lambda+\lambda\theta^{k})^{CN}-m_{p})m_{p}+m_{p}-m_{p}^{2}.
Proof.

See Appendix -A. ∎

Next, we show that the moments of AoI of both active users and passive users can be approximated as functions of (ma,va2)(m_{a},v_{a}^{2}) and (mp,vp2)(m_{p},v_{p}^{2}), respectively.

Due to space limitation, we focus on discussing the derivation of moments of AoI of active users.

Let tit_{i} be the ii-th time that an active user nn successfully deliveries its packet, and define ln(i):=titi1l_{n}(i):=t_{i}-t_{i-1}. For active user nn, its AoI from ti1t_{i-1} to ti1t_{i}-1 is from 11 to ln(i)l_{n}(i). So, the summation of zz-th moment of AoI from ti1t_{i-1} to ti1t_{i}-1 is k=1ln(i)kz\sum_{k=1}^{l_{n}(i)}k^{z}. Let LnL_{n} be the total number of deliveries for active user nn. Then, summing over all ln(i)l_{n}(i), we have the sum of all zz-th moment of AoI for active user nn, i=1Lnk=1ln(i)kz\sum_{i=1}^{L_{n}}\sum_{k=1}^{l_{n}(i)}k^{z}. Next, divided by the whole time i=1Lnln(i)\sum_{i=1}^{L_{n}}l_{n}(i) and choosing the expectation, we have E[AoIaz]=E[limLni=1Lnk=1ln(i)kzi=1Lnln(i)]E[AoI_{a}^{z}]=E\Big{[}\lim_{L_{n}\xrightarrow{}\infty}\frac{\sum_{i=1}^{L_{n}}\sum_{k=1}^{l_{n}(i)}k^{z}}{\sum_{i=1}^{L_{n}}l_{n}(i)}\Big{]}. By using the Faulhaber’s formula, we have

E[AoIaz]\displaystyle E[AoI_{a}^{z}] =E[limLni=1Ln1i=1Lnln(i)(1i=1Ln1i=1Ln(ln(i)z+1z+1\displaystyle=E\Big{[}\lim_{L_{n}\xrightarrow{}\infty}\frac{\sum_{i=1}^{L_{n}}1}{\sum_{i=1}^{L_{n}}l_{n}(i)}\Big{(}\frac{1}{\sum_{i=1}^{L_{n}}1}\sum_{i=1}^{L_{n}}\big{(}\frac{l_{n}(i)^{z+1}}{z+1}
+ln(i)z2+k=2zBkk!z!ln(i)zk+1(zk+1)!))]\displaystyle\qquad+\frac{l_{n}(i)^{z}}{2}+\sum_{k=2}^{z}\frac{B_{k}}{k!}\frac{z!l_{n}(i)^{z-k+1}}{(z-k+1)!}\big{)}\Big{)}\Big{]}
=1E[ln(i)](E[ln(i)z+1]z+1+E[ln(i)z]2\displaystyle=\frac{1}{E[l_{n}(i)]}\big{(}\frac{E[l_{n}(i)^{z+1}]}{z+1}+\frac{E[l_{n}(i)^{z}]}{2}
+k=2zBkk!z!E[ln(i)zk+1](zk+1)!),\displaystyle\qquad+\sum_{k=2}^{z}\frac{B_{k}}{k!}\frac{z!E[l_{n}(i)^{z-k+1}]}{(z-k+1)!}\big{)}, (3)

where BkB_{k} is the Bernoulli number.

To approximate E[ln(i)k]E[l_{n}(i)^{k}], we construct an alternative sequence Sn(t)S^{\prime}_{n}(t) that have the same mean and temporal variance as Sn(t)S_{n}(t). Let Dn(t)D_{n}(t) be a Brownian motion random process with mean mam_{a} and variance va2v_{a}^{2}. The sequence {Sn(1),Sn(2),}\{S^{\prime}_{n}(1),S^{\prime}_{n}(2),\dots\} is defined as

Sn(t)={1,if Dn(t)Dn(t)1,0,otherwise,\displaystyle S^{\prime}_{n}(t)=\begin{cases}1,\;&\text{if $D_{n}(t)-D_{n}(t^{-})\geq 1$,}\\ 0,\;&\text{otherwise,}\end{cases} (4)

where t=max{τ|τ<t,Sn(τ)=1}t^{-}=\max\{\tau|\tau<t,S^{\prime}_{n}(\tau)=1\}. Intuitively, we assume that a packet delivery occurs in the alternative sequence every time Dn(t)D_{n}(t) increases by one since the last packet delivery in this sequence. Hence, τ=1tSn(τ)\sum_{\tau=1}^{t}S_{n}^{\prime}(\tau) is close to Dn(t)D_{n}(t), and the process [Sn(1),Sn(2),][S^{\prime}_{n}(1),S^{\prime}_{n}(2),...] has mean mam_{a} and temporal variance va2v_{a}^{2}.

Let ln(i):=tn,itn,i1l_{n}^{\prime}(i):=t^{\prime}_{n,i}-t^{\prime}_{n,i-1}, where tn,it^{\prime}_{n,i} is the ii-th time that Sn(t)=1S^{\prime}_{n}(t)=1. Since Sn(t)S^{\prime}_{n}(t) has the same mean and temporal variance as Sn(t)S_{n}(t), we propose to approximate E[ln(i)k]E[l_{n}(i)^{k}] by E[ln(i)k]E[l^{\prime}_{n}(i)^{k}]. Moreover, ln(i)l^{\prime}_{n}(i) can be regarded as the time needed for Dn(t)D_{n}(t) to increase by 11, which is equivalent to the first-hitting time for the Brownian motion random process with level 11. Thus, ln(i)l_{n}^{\prime}(i) follows the inverse Gaussian distribution IG(1ma,1va2)IG(\frac{1}{m_{a}},\frac{1}{v_{a}^{2}}) ([33, 34]). By using the moment generating function of inverse Gaussian distribution in [35], we have E[ln(i)k]=1makζ=0k1(k1+ζ)!ζ!(k1ζ)!(2mava2)ζE[l^{\prime}_{n}(i)^{k}]=\frac{1}{m_{a}^{k}}\sum_{\zeta=0}^{k-1}\frac{(k-1+\zeta)!}{\zeta!(k-1-\zeta)!}(\frac{2m_{a}}{v_{a}^{2}})^{-\zeta}.

For a passive user jj, let t¯i\bar{t}_{i} be the ii-th time that Sj(t)=1S_{j}(t)=1 and lj(t):=t¯it¯i1l_{j}(t):=\bar{t}_{i}-\bar{t}_{i-1}. By combining the derivations above, we obtain the following:

Theorem 2.

Let BkB_{k} be the Bernoulli number, then

E[AoIaz]=\displaystyle E[AoI_{a}^{z}]= 1E[ln(i)](E[ln(i)z+1]z+1+E[ln(i)z]2\displaystyle\frac{1}{E[l_{n}(i)]}(\frac{E[l_{n}(i)^{z+1}]}{z+1}+\frac{E[l_{n}(i)^{z}]}{2}
+k=2zBkk!z!E[ln(i)zk+1](zk+1)!),\displaystyle+\sum_{k=2}^{z}\frac{B_{k}}{k!}\frac{z!E[l_{n}(i)^{z-k+1}]}{(z-k+1)!}), (5)

and

E[AoIpz]=\displaystyle E[AoI_{p}^{z}]= 1E[lj(i)](E[lj(i)z+1]z+1+E[lj(i)z]2\displaystyle\frac{1}{E[l_{j}(i)]}(\frac{E[l_{j}(i)^{z+1}]}{z+1}+\frac{E[l_{j}(i)^{z}]}{2}
+k=2zBkk!z!E[lj(i)zk+1](zk+1)!),\displaystyle+\sum_{k=2}^{z}\frac{B_{k}}{k!}\frac{z!E[l_{j}(i)^{z-k+1}]}{(z-k+1)!}), (6)

where E[ln(i)k]1makζ=0k1(k1+ζ)!ζ!(k1ζ)!(2mava2)ζE[l_{n}(i)^{k}]\approx\frac{1}{m_{a}^{k}}\sum_{\zeta=0}^{k-1}\frac{(k-1+\zeta)!}{\zeta!(k-1-\zeta)!}(\frac{2m_{a}}{v_{a}^{2}})^{-\zeta} and E[lj(i)k]1mpkζ=0k1(k1+ζ)!ζ!(k1ζ)!(2mpvp2)ζE[l_{j}(i)^{k}]\approx\frac{1}{m_{p}^{k}}\sum_{\zeta=0}^{k-1}\frac{(k-1+\zeta)!}{\zeta!(k-1-\zeta)!}(\frac{2m_{p}}{v_{p}^{2}})^{-\zeta}. \hfill\Box

For example, we have E[AoIa]=12(va2ma2+1ma)+12E[AoI_{a}]=\frac{1}{2}(\frac{v_{a}^{2}}{m_{a}^{2}}+\frac{1}{m_{a}})+\frac{1}{2} and E[AoIa2]=va4ma4+va2ma3+3va2+26ma2+12ma+16E[AoI_{a}^{2}]=\frac{v_{a}^{4}}{m_{a}^{4}}+\frac{v_{a}^{2}}{m_{a}^{3}}+\frac{3v_{a}^{2}+2}{6m_{a}^{2}}+\frac{1}{2m_{a}}+\frac{1}{6}.

With the help of Theorem 2, F(r,s)F(r,s) can now be approximated as a function of (ma,va2,mp,vp2)(m_{a},v_{a}^{2},m_{p},v_{p}^{2}), which, as stated in Theorem 1, can be expressed as functions of rr and ss.

IV AoI Optimization in Second Order Model

In the previous section, we approximate E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}] as functions of rr and ss. In this section, we further study the optimal behavior of active users and solve the optimal rr and ss. We find the interesting property that, when N>C+4N>C+4, the optimal solution requires an active user to the Idle state every time it makes a transmission, i.e., s=1s=1. This result is formally stated below.

Theorem 3.

When N>C+4N>C+4, there exists a λ[0,1N]\lambda\in[0,\frac{1}{N}] such that choosing r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 minimizes F(r,s)F(r,s).

In addition to showing the interesting behavior that, under the optimal solution, an active user will never transmit in two consecutive slots, this theorem also simplifies the search for the optimal solution. In particular, it shows that one only needs to find the optimal λ\lambda, which can be done through a simple line search, in order to obtain the optimal solution.

Recall that λ=rr+s\lambda=\frac{r}{r+s} and θ=1rs\theta=1-r-s, or, equivalently, r=λ(1θ)r=\lambda(1-\theta) and s=(1λ)(1θ)s=(1-\lambda)(1-\theta). Before proving Theorem 3, we first establish some properties about the optimal λ\lambda and θ\theta. We start by the following lemma:

Lemma 1.

For any positive integer zz, E[AoIaz]E[AoI_{a}^{z}] is an increasing function of 1ma\frac{1}{m_{a}} and va2ma2\frac{v_{a}^{2}}{m_{a}^{2}}, and E[AoIpz]E[AoI_{p}^{z}] is an increasing function of 1mp\frac{1}{m_{p}} and vp2mp2\frac{v_{p}^{2}}{m_{p}^{2}}.

Proof.

For active users, we note that each term in (5) is a constant multiplying E[ln(i)k]E[ln(i)]\frac{E[l_{n}(i)^{k}]}{E[l_{n}(i)]}, where k1k\geq 1. For E[ln(i)k]E[ln(i)]\frac{E[l_{n}(i)^{k}]}{E[l_{n}(i)]}, we have

E[ln(i)k]E[ln(i)]=1mak1ζ=0k1(k1+ζ)!ζ!(k1ζ)!(2mava2)ζ,\displaystyle\frac{E[l_{n}(i)^{k}]}{E[l_{n}(i)]}=\frac{1}{m_{a}^{k-1}}\sum_{\zeta=0}^{k-1}\frac{(k-1+\zeta)!}{\zeta!(k-1-\zeta)!}(\frac{2m_{a}}{v_{a}^{2}})^{-\zeta}, (7)

whose ζ\zeta-th term is (k1+ζ)!2ζζ!(k1ζ)!1mak1ζva2ζma2ζ\frac{(k-1+\zeta)!}{2^{\zeta}\zeta!(k-1-\zeta)!}\frac{1}{m_{a}^{k-1-\zeta}}\frac{v_{a}^{2\zeta}}{m_{a}^{2\zeta}}. Since k1ζk-1\geq\zeta, this ζ\zeta-th term in (7) is an increasing function of 1ma\frac{1}{m_{a}} and va2ma2\frac{v_{a}^{2}}{m_{a}^{2}}. Thus, E[ln(i)k]E[ln(i)]\frac{E[l_{n}(i)^{k}]}{E[l_{n}(i)]} is an increasing function of 1ma\frac{1}{m_{a}} and va2ma2\frac{v_{a}^{2}}{m_{a}^{2}}, for any k1k\geq 1. Therefore, E[AoIaz]E[AoI_{a}^{z}] is an increasing function of 1ma\frac{1}{m_{a}} and va2ma2\frac{v_{a}^{2}}{m_{a}^{2}}.

Similarly, E[AoIpz]E[AoI_{p}^{z}] is an increasing function of 1mp\frac{1}{m_{p}} and vp2mp2\frac{v_{p}^{2}}{m_{p}^{2}}. ∎

Based on Lemma  1, we further have

Lemma 2.

For any w[0,1]w\in[0,1], The optimal λ\lambda that minimizes F(r,s)F(r,s) is in range [0,1N][0,\frac{1}{N}].

Sketch of proof.

Recall ma=λ(1λ)N1m_{a}=\lambda(1-\lambda)^{N-1} and mp=(1λ)CNm_{p}=(1-\lambda)^{CN}. When λ>1N\lambda>\frac{1}{N}, it is easy to verify that mam_{a} and mpm_{p} decreases as λ\lambda increases.

In Appendix -B, we further show that, when λ1N\lambda\geq\frac{1}{N}, (va2/ma2)λ0\frac{\partial(v_{a}^{2}/m_{a}^{2})}{\partial\lambda}\geq 0 and (vp2/mp2)λ0\frac{\partial(v_{p}^{2}/m_{p}^{2})}{\partial\lambda}\geq 0.

Therefore, when λ1N\lambda\geq\frac{1}{N}, (1ma\frac{1}{m_{a}}, va2ma2\frac{v_{a}^{2}}{m_{a}^{2}}, 1mp\frac{1}{m_{p}}, vp2mp2)\frac{v_{p}^{2}}{m_{p}^{2}}) all increase as λ\lambda increases. Since E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}] are increasing functions of (1ma,va2ma2)(\frac{1}{m_{a}},\frac{v_{a}^{2}}{m_{a}^{2}}) and (1mp,vp2mp2)(\frac{1}{m_{p}},\frac{v_{p}^{2}}{m_{p}^{2}}) respectively, the optimal λ\lambda is less than or equal to 1N\frac{1}{N}. ∎

Next, we analyze the optimal θ\theta. The lemma below derives the optimal θ\theta when λ\lambda is given and fixed.

Lemma 3.

When λ[0,min{α,β}]\lambda\in[0,\min\{\alpha,\beta\}] is fixed, setting θ=λ1λ\theta=-\frac{\lambda}{1-\lambda}, or, equivalently, setting r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1, minimizes E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}], where α\alpha and β\beta are the smallest positive roots of

hN(y)\displaystyle h_{N}(y) =(N+8)y3(N13)y26y+1,\displaystyle=-(N+8)y^{3}-(N-13)y^{2}-6y+1,
h¯CN(y)\displaystyle\bar{h}_{CN}(y) =(CN10)y3(CN13)y26y+1,\displaystyle=(CN-10)y^{3}-(CN-13)y^{2}-6y+1,

respectively.

Sketch of proof.

When λ\lambda is determined, ma=λ(1λ)N1m_{a}=\lambda(1-\lambda)^{N-1} and mp=(1λ)CNm_{p}=(1-\lambda)^{CN} are determined. Hence, by Theorem 2, E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}] are increasing functions of va2v_{a}^{2} and vp2v_{p}^{2} respectively.

In Appendix -C, we further establish that va2θ0\frac{\partial v_{a}^{2}}{\partial\theta}\geq 0 when λα\lambda\leq\alpha, and vp2θ0\frac{\partial v_{p}^{2}}{\partial\theta}\geq 0 when λβ\lambda\leq\beta. So, when λ[0,min{α,β}]\lambda\in[0,\min\{\alpha,\beta\}], minimum θ\theta will minimize E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}]. In Appendix -C, we also show that α,β<12\alpha,\beta<\frac{1}{2}, and hence the minimum θ\theta is achieved when r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1.

Therefore, when λ[0,min{α,β}]\lambda\in[0,\min\{\alpha,\beta\}] is fixed, choosing r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 minimizes E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}]. ∎

A limitation of Lemma 3 is that it requires λ[0,min{α,β}]\lambda\in[0,\min\{\alpha,\beta\}]. In the following lemma, we establish lower bounds of α\alpha and β\beta.

Lemma 4.

α>1N\alpha>\frac{1}{N} when N>4N>4, and β>1N\beta>\frac{1}{N} when N>C+4N>C+4.

Sketch of proof.

In Appendix -D, we prove that hN(y)h_{N}(y) and h¯CN(y)\bar{h}_{CN}(y) both have only one root in [0,1][0,1].

We further show that hN(1N)>0h_{N}(\frac{1}{N})>0 when N>4N>4, and h¯CN(1N)>0\bar{h}_{CN}(\frac{1}{N})>0 when N>C+4N>C+4, in Appendix -D. Since hN(1)=2N<0h_{N}(1)=-2N<0 and h¯N(1)=2<0\bar{h}_{N}(1)=-2<0, we have α>1N\alpha>\frac{1}{N} when N>4N>4, and β>1N\beta>\frac{1}{N} when N>C+4N>C+4. ∎

We are now ready to prove Theorem 3.

Proof of Theorem 3.

It is shown in Lemma 2 that the optimal λ\lambda is in the range [0,1N][0,\frac{1}{N}], which is covered by the range [0,min{α,β}][0,\min\{\alpha,\beta\}] when N>C+4N>C+4 according Lemma 4. Therefore, the optimal solution in Lemma 3 holds for the optimal λ\lambda when N>C+4N>C+4. Hence, when N>C+4N>C+4, choosing r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 minimizes E[AoIaz]E[AoI_{a}^{z}] and E[AoIpz]E[AoI_{p}^{z}], as well as F(r,s)F(r,s). ∎

V Simulation Results

This section presents simulation results for applying our solution of rr and ss. The performance function is the optimization objective F(r,s)F(r,s).

The results of our solution are compared with three other baseline policies. We provide descriptions of all four policies, along with necessary modifications to fit the simulation setting.

  • Our solution: Based on Theorem 3, we choose the optimal λ\lambda and set r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 for all active users. This optimal λ\lambda is obtained by plugging r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 into Theorem 2 and doing an exhaustive search over λ\lambda for F(r,s)F(r,s) with a precision 0.010.01.

  • Slotted ALOHA: When r+s=1r+s=1, our active user model is reduced to be slotted ALOHA, which is a famous and commonly used model for transmissions with no acknowledgment. According to solutions in [5], we choose r=1Nr=\frac{1}{N} and s=11Ns=1-\frac{1}{N}.

  • Optimal ALOHA: After introducing passive users, the optimal λ\lambda in slotted ALOHA is no longer 1N\frac{1}{N}. Thus, we simulate all possible λ\lambda with precision 0.010.01, and then choose the best λ\lambda to obtain optimal slotted ALOHA. It should be noticed that optimal ALOHA is impossible to implement in real world since its λ\lambda is determined after the whole process is finished.

  • Age Threshold ALOHA (ATA): Age threshold ALOHA is proposed by Yavascan and Uysal [24], which is a slotted ALOHA algorithm with an age threshold. If the AoI of one active user is larger than a threshold, this active user will follow slotted ALOHA transmission with probability rr. Otherwise, this active user will stay silent. Since there is no acknowledgment in the system, we let active users assume their transmissions are always successful. As suggested in [24], r=4.69/Nr=4.69/N and the age threshold is 2.2N2.2N.

We simulate different scenarios with varying values of ww, where ww ranges from 0 to 11. We set N=7N=7, C=2C=2 or 44, and z=1z=1 or 22. Once NN, CC and zz are determined, we can set the values of rr and ss based on policy designs. The system is then simulated for 1010 individual runs, each with 100,000100,000 time slots. The final results are the average of these 1010 runs. We present the value of Actual F(r,s) of one algorithmTheoretical F(r,s) of our solution\frac{\text{Actual }F(r,s)\text{ of one algorithm}}{\text{Theoretical }F(r,s)\text{ of our solution}} for all four algorithms under each setting. Simulation results are shown in Fig. 2. As some values are extremely large, only results that have ratio below 2.42.4 are shown in figures.

Refer to caption
(a) C=2C=2, z=1z=1.
Refer to caption
(b) C=2C=2, z=2z=2.
Refer to caption
(c) C=4C=4, z=1z=1.
Refer to caption
(d) C=4C=4, z=2z=2.
Figure 2: Actual F(r,s) of one algorithmTheoretical F(r,s) of our solution\frac{\text{Actual }F(r,s)\text{ of one algorithm}}{\text{Theoretical }F(r,s)\text{ of our solution}} when N=7N=7

The results show that the average mismatches between theoretical F(r,s)F(r,s) and actual F(r,s)F(r,s) of our solution are about 0.7%0.7\% when z=1z=1 and 12%12\% when z=2z=2. Furthermore, we notice the actual value is always lower than theoretical value. These results indicate that our approximation is useful in practice.

Comparing the performance of different algorithms in Figure 2a and Figure 2b, we can observe that our strategy outperforms other three policies in all settings. Even when compared to the optimal ALOHA, which is not practical in real-world scenarios, our strategy shows a 4%4\% reduction when z=1z=1 and a 9%9\% reduction when z=2z=2. Additionally, the increase in the reduction of our strategy between z=1z=1 and z=2z=2 indicates that our strategy can better optimize the variance of AoI.

In Figure 2c and Figure 2d, we simulate C=4C=4 to evaluate the performance of our strategy when the condition N>C+4N>C+4 is not met. The Results indicate that our strategy outperforms the other three algorithms in nearly all settings, even when the condition N>C+4N>C+4 is not satisfied.

VI Conclusion

This paper studies an optimization problem of moments of AoI for both active and passive users in a random access network. In this network, active users broadcast their transmissions following a Markov process and will interfere with each other. By applying a second-order model, we formulate moments of AoI as functions of state change probabilities in the Markov process. Next, we reveal the special properties of these functions and optimize moments of AoI for both active and passive users. The optimal solution indicates that the best strategy for active users is to become silent immediately after one transmission. Simulations show that this transmission strategy outperforms three baseline algorithms under various settings.

Acknowledgment

This material is based upon work supported in part by NSF under Award Number ECCS-2127721, in part by ARL and ARO under Grant Number W911NF-22-1-0151, and in part by ONR under Contract N00014-21-1-2385.

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-A Proof of Theorem 1

Let xn,tx_{n,t} be the indicator that the active user nn is in TX state in time slot tt.

First, we calculate the mean for active user nn in cluster cic_{i} and passive user jj. We have

ma\displaystyle m_{a} =Prob(xn,t=1)ζn,ζciProb(xζ,t=0)\displaystyle=Prob(x_{n,t}=1)\prod_{\zeta\neq n,\zeta\in c_{i}}Prob(x_{\zeta,t}=0)
=rr+s(sr+s)N1=λ(1λ)N1,\displaystyle=\frac{r}{r+s}\Big{(}\frac{s}{r+s}\Big{)}^{N-1}=\lambda(1-\lambda)^{N-1},

and mp=ζ=1CNProb(xζ,t=0)=(sr+s)CN=(1λ)CNm_{p}=\prod_{\zeta=1}^{CN}Prob(x_{\zeta,t}=0)=(\frac{s}{r+s})^{CN}=(1-\lambda)^{CN}.

Then, we derive the temporal variance for active user nn and passive user jj. According to the central limit theorem of Markov process, we have

va2\displaystyle v_{a}^{2} =Var(Sn(1))+2k=1Cov(Sn(1),Sn(k+1)),\displaystyle=Var(S_{n}(1))+2\sum_{k=1}^{\infty}Cov(S_{n}(1),S_{n}(k+1)),
vp2\displaystyle v_{p}^{2} =Var(Sj(1))+2k=1Cov(Sj(1),Sj(k+1)).\displaystyle=Var(S_{j}(1))+2\sum_{k=1}^{\infty}Cov(S_{j}(1),S_{j}(k+1)).

We first calculate va2v_{a}^{2}. Since Sn(1)S_{n}(1) is a Bernoulli random variable, it has

Var(Sn(1))=E[Sn(1)]E[Sn(1)]2=mama2\displaystyle Var(S_{n}(1))=E[S_{n}(1)]-E[S_{n}(1)]^{2}=m_{a}-m_{a}^{2}
Cov(Sn(1),Sn(k+1))\displaystyle Cov(S_{n}(1),S_{n}(k+1))
=E[Sn(1)Sn(k+1)]E[Sn(1)]E[Sn(k+1)]\displaystyle=E[S_{n}(1)S_{n}(k+1)]-E[S_{n}(1)]E[S_{n}(k+1)]
=(Prob(Sn(k+1)=1|Sn(1)=1)E[Sn(k+1)])E[Sn(1)]\displaystyle=(Prob(S_{n}(k+1)=1|S_{n}(1)=1)-E[S_{n}(k+1)])E[S_{n}(1)]
=(Prob(Sn(k+1)=1|Sn(1)=1)ma)ma.\displaystyle=(Prob(S_{n}(k+1)=1|S_{n}(1)=1)-m_{a})m_{a}.

Assume the active user nn is in cluster cic_{i}, we have Prob(Sn(k+1)=1|Sn(1)=1)=Prob(xn,k+1=1|Sn(1)=1)ζn,ζciProb(xζ,k+1=0|Sn(1)=1)Prob(S_{n}(k+1)=1|S_{n}(1)=1)=Prob(x_{n,k+1}=1|S_{n}(1)=1)\prod_{\zeta\neq n,\zeta\in c_{i}}Prob(x_{\zeta,k+1}=0|S_{n}(1)=1).

Let Gn,n(k)=Prob(xn,k=1|Sn(1)=1)G_{n,n}(k)=Prob(x_{n,k}=1|S_{n}(1)=1) and Gζ,n(k)=Prob(xζ,k=1|Sn(1)=1),ln,lciG_{\zeta,n}(k)=Prob(x_{\zeta,k}=1|S_{n}(1)=1),l\neq n,l\in c_{i}, we have Prob(Sn(k)=1|Sn(1)=1)=Gn,n(k)ζn,ζci(1Gζ,n(k))Prob(S_{n}(k)=1|S_{n}(1)=1)=G_{n,n}(k)\prod_{\zeta\neq n,\zeta\in c_{i}}(1-G_{\zeta,n}(k)) and

va2=\displaystyle v_{a}^{2}= 2k=1(Gn,n(k+1)ζn(1Gζ,n(k+1))ma)ma\displaystyle 2\sum_{k=1}^{\infty}(G_{n,n}(k+1)\prod_{\zeta\neq n}(1-G_{\zeta,n}(k+1))-m_{a})m_{a}
+mama2.\displaystyle+m_{a}-m_{a}^{2}.

Gn,n(k)G_{n,n}(k) can be calculated by

Gn,n(k)\displaystyle G_{n,n}(k) =Prob(xn,k=1|Sn(1)=1)\displaystyle=Prob(x_{n,k}=1|S_{n}(1)=1)
=Gn,n(k1)Prob(xn,k=1|xn,k1=1)\displaystyle=G_{n,n}(k-1)Prob(x_{n,k}=1|x_{n,k-1}=1)
+(1Gn,n(k1))Prob(xn,k=1|xn,k1=0)\displaystyle\quad+(1-G_{n,n}(k-1))Prob(x_{n,k}=1|x_{n,k-1}=0)
=Gn,n(k1)(1s)+(1Gn,n(k1))r\displaystyle=G_{n,n}(k-1)(1-s)+(1-G_{n,n}(k-1))r
=r+θGn,n(k1),\displaystyle=r+\theta G_{n,n}(k-1),

for k>1k>1, and Gn,n(1)=1G_{n,n}(1)=1. By i=2kθkiGn,n(i)\sum_{i=2}^{k}\theta^{k-i}G_{n,n}(i), we have

Gn,n(k)\displaystyle G_{n,n}(k) =i=2kθkir+θk1Gn,n(1)\displaystyle=\sum_{i=2}^{k}\theta^{k-i}r+\theta^{k-1}G_{n,n}(1)
=rr+s+sr+sθk1=λ+(1λ)θk1.\displaystyle=\frac{r}{r+s}+\frac{s}{r+s}\theta^{k-1}=\lambda+(1-\lambda)\theta^{k-1}.

For Gζ,n(k)G_{\zeta,n}(k), we also have

Gζ,n(k)\displaystyle G_{\zeta,n}(k) =Prob(xζ,k=1|Sn(1)=1)\displaystyle=Prob(x_{\zeta,k}=1|S_{n}(1)=1)
=Gζ,n(k1)Prob(xζ,k=1|xζ,k1=1)\displaystyle=G_{\zeta,n}(k-1)Prob(x_{\zeta,k}=1|x_{\zeta,k-1}=1)
+(1Gζ,n(k1))Prob(xζ,k=1|xζ,k1=0)\displaystyle\quad+(1-G_{\zeta,n}(k-1))Prob(x_{\zeta,k}=1|x_{\zeta,k-1}=0)
=Gζ,n(k1)(1s)+(1Gζ,n)(k1)r\displaystyle=G_{\zeta,n}(k-1)(1-s)+(1-G_{\zeta,n})(k-1)r
=r+θGζ,n(k1),\displaystyle=r+\theta G_{\zeta,n}(k-1),

for k>1k>1, and Gζ,n(1)=0G_{\zeta,n}(1)=0. Solving the recursive equation, we have Gζ,n(k)=λλθk1G_{\zeta,n}(k)=\lambda-\lambda\theta^{k-1}. Therefore,

va2=2k=1((λ+(1λ)θk)(1λ+λθk)N1ma)ma\displaystyle v_{a}^{2}=2\sum_{k=1}^{\infty}((\lambda+(1-\lambda)\theta^{k})(1-\lambda+\lambda\theta^{k})^{N-1}-m_{a})m_{a}
+mama2.\displaystyle\quad+m_{a}-m_{a}^{2}.

Now, we calculate vp2v_{p}^{2}. Sine Sj(1)S_{j}(1) is also a Bernoulli random variable in the system, we have

Var(Sj(1))=E[Sj(1)]E[Sj(1)]2=mpmp2\displaystyle Var(S_{j}(1))=E[S_{j}(1)]-E[S_{j}(1)]^{2}=m_{p}-m_{p}^{2}
Cov(Sj(1),Sj(k+1))\displaystyle Cov(S_{j}(1),S_{j}(k+1))
=E[Sj(1)Sj(k+1)]E[Sj(1)]E[Sj(k+1)]\displaystyle=E[S_{j}(1)S_{j}(k+1)]-E[S_{j}(1)]E[S_{j}(k+1)]
=(Prob(Sj(k+1)=1|Sj(1)=1)E[Sj(k+1)])E[Sj(1)]\displaystyle=(Prob(S_{j}(k+1)=1|S_{j}(1)=1)-E[S_{j}(k+1)])E[S_{j}(1)]
=(Prob(Sj(k+1)=1|Sj(1)=1)mp)mp.\displaystyle=(Prob(S_{j}(k+1)=1|S_{j}(1)=1)-m_{p})m_{p}.

Let Gn,j(k)=Prob(xn,k=1|Sj(1)=1)G_{n,j}(k)=Prob(x_{n,k}=1|S_{j}(1)=1), we have Prob(Sj(k+1)=1|Sj(1)=1)=n=1CN(1Gn,j(k))Prob(S_{j}(k+1)=1|S_{j}(1)=1)=\prod_{n=1}^{CN}(1-G_{n,j}(k)) and

vp2\displaystyle v_{p}^{2} =mpmp2+2k=1(n=1CN(1Gn,j(k+1))mp)mp.\displaystyle=m_{p}-m_{p}^{2}+2\sum_{k=1}^{\infty}(\prod_{n=1}^{CN}(1-G_{n,j}(k+1))-m_{p})m_{p}.

Gn,j(k)G_{n,j}(k) can be calculated by

Gn,j(k)\displaystyle G_{n,j}(k) =Prob(xn,k=1|Sj(1)=1)\displaystyle=Prob(x_{n,k}=1|S_{j}(1)=1)
=Gn,j(k1)Prob(xn,k=1|xn,k1=1)\displaystyle=G_{n,j}(k-1)Prob(x_{n,k}=1|x_{n,k-1}=1)
+(1Gn,j(k1))Prob(xn,k=1|xn,k1=0)\displaystyle\quad+(1-G_{n,j}(k-1))Prob(x_{n,k}=1|x_{n,k-1}=0)
=Gn,j(k1)(1s)+(1Gn,j)(k1)r\displaystyle=G_{n,j}(k-1)(1-s)+(1-G_{n,j})(k-1)r
=r+θGn,j(k1),\displaystyle=r+\theta G_{n,j}(k-1),

for k>1k>1, and Gn,j(1)=0G_{n,j}(1)=0. This expression is similar to that of Gζ,n(k)G_{\zeta,n}(k). Thus, solving the recursive equation, we have Gn,j(k)=λλθk1G_{n,j}(k)=\lambda-\lambda\theta^{k-1}. Thus,

vp2=2k=1((1λ+λθk)CNmp)mp+mpmp2.\displaystyle v_{p}^{2}=2\sum_{k=1}^{\infty}((1-\lambda+\lambda\theta^{k})^{CN}-m_{p})m_{p}+m_{p}-m_{p}^{2}.

-B Proof of Lemma 2

First, we show that we only need to consider the case θ0\theta\leq 0. The partial derivative of va2ma\frac{v_{a}^{2}}{m_{a}} with respect to θ\theta when λ\lambda is fixed is

va2θ\displaystyle\frac{\partial v_{a}^{2}}{\partial\theta} =2mak=1[(1λ)kθk1(1λ+λθk)N1\displaystyle=\frac{2}{m_{a}}\sum_{k=1}^{\infty}[(1-\lambda)k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-1}
+(λ+(1λ)θk)(N1)λkθk1(1λ+λθk)N2)]\displaystyle\quad+(\lambda+(1-\lambda)\theta^{k})(N-1)\lambda k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-2})]
=2mak=1[12λ+Nλ2+Nλ(1λ)θk]\displaystyle=\frac{2}{m_{a}}\sum_{k=1}^{\infty}[1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{k}]
×kθk1(1λ+λθk)N2.\displaystyle\quad\times k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-2}.

When θ>0\theta>0, it is easy to verify that the partial derivative is positive. Similarly, we also have the partial derivative of vp2mp\frac{v_{p}^{2}}{m_{p}} with respect to θ\theta is positive when θ>0\theta>0. Thus, the optimal θ\theta is non-positive, and hence we only consider θ0\theta\leq 0 in the following proof.

We first prove (va2/ma2)λ0\frac{\partial(v_{a}^{2}/m_{a}^{2})}{\partial\lambda}\geq 0 when λ1N\lambda\geq\frac{1}{N}. Based on Theorem 1, we have

(va2/ma2)λ=2k=1[(1θk)(1λ+λθk)N1mama2\displaystyle\frac{\partial(v_{a}^{2}/m_{a}^{2})}{\partial\lambda}=2\sum_{k=1}^{\infty}\big{[}\frac{(1-\theta^{k})(1-\lambda+\lambda\theta^{k})^{N-1}m_{a}}{m_{a}^{2}}
+(λ+(1λ)θk)(N1)(1λ+λθk)N2(θk1)mama2\displaystyle+\frac{(\lambda+(1-\lambda)\theta^{k})(N-1)(1-\lambda+\lambda\theta^{k})^{N-2}(\theta^{k}-1)m_{a}}{m_{a}^{2}}
(1Nλ)(1λ)N2(λ+(1λ)θk)(1λ+λθk)N1ma2]\displaystyle-\frac{(1-N\lambda)(1-\lambda)^{N-2}(\lambda+(1-\lambda)\theta^{k})(1-\lambda+\lambda\theta^{k})^{N-1}}{m_{a}^{2}}\big{]}
(1Nλ)(1λ)N2ma2\displaystyle-\frac{(1-N\lambda)(1-\lambda)^{N-2}}{m_{a}^{2}}
=(Nλ1)(1λ)N2ma2+2k=1(1λ)N2(1λ+λθk)N2ma2\displaystyle=\frac{(N\lambda-1)(1-\lambda)^{N-2}}{m_{a}^{2}}+2\sum_{k=1}^{\infty}\frac{(1-\lambda)^{N-2}(1-\lambda+\lambda\theta^{k})^{N-2}}{m_{a}^{2}}
×θk[(N2)λ2(1λ)θk+(N2)λ2+2λ1].\displaystyle\times\theta^{k}\big{[}(N-2)\lambda^{2}(1-\lambda)\theta^{k}+(N-2)\lambda^{2}+2\lambda-1\big{]}.

Let ω(λ,θ,k,N):=(1λ)N2(1λ+λθk)N2θk[(N2)λ2(1λ)θk+(N2)λ2+2λ1]\omega(\lambda,\theta,k,N):=(1-\lambda)^{N-2}(1-\lambda+\lambda\theta^{k})^{N-2}\theta^{k}[(N-2)\lambda^{2}(1-\lambda)\theta^{k}+(N-2)\lambda^{2}+2\lambda-1] be the kk-th term in the above infinite summation. The following part is to prove k=1ξ(λ,θ,k,N)0\sum_{k=1}^{\infty}\xi(\lambda,\theta,k,N)\geq 0 or k=1ξ(λ,θ,k,N)+(Nλ1)0\sum_{k=1}^{\infty}\xi(\lambda,\theta,k,N)+(N\lambda-1)\geq 0.

Based on definition of λ\lambda and θ\theta, we first have (1λ)0(1-\lambda)\geq 0 and (1λ+λθk)0(1-\lambda+\lambda\theta^{k})\geq 0.

Consider the special case θ=1\theta=-1. θ=1\theta=-1 can only be achieved when λ=12\lambda=\frac{1}{2}. Thus, (1λ+λθk)N2(1-\lambda+\lambda\theta^{k})^{N-2} equals to 11 when kk is even, and equals to 0 when kk is odd. In this case, (N2)λ+2λ1>0(N-2)\lambda+2\lambda-1>0. Thus, ω(λ,θ,k,N)0\omega(\lambda,\theta,k,N)\geq 0 for any kk, and hence (va2/ma2)λ0\frac{\partial(v_{a}^{2}/m_{a}^{2})}{\partial\lambda}\geq 0.

Now, we discuss the case θ>1\theta>-1.

We first consider (N2)λ2+2λ10(N-2)\lambda^{2}+2\lambda-1\geq 0, which means λ[N11N2,1]\lambda\in[\frac{\sqrt{N-1}-1}{N-2},1]. In this case, [(N2)λ2(1λ)θk+(N2)λ2+2λ1][(N-2)\lambda^{2}(1-\lambda)\theta^{k}+(N-2)\lambda^{2}+2\lambda-1] is non-negative for all even kk, and is possible negative for odd kk (when kk is larger, equation becomes positive).

Consider kk from 22 to infinite, we compare the value with an even k=2qk=2q and its following odd k=2q+1k=2q+1. As θ(1,0]\theta\in(-1,0], we have |(1λ+λθ2q)N2|>|(1λ+λθ2q+1)N2||(1-\lambda+\lambda\theta^{2q})^{N-2}|>|(1-\lambda+\lambda\theta^{2q+1})^{N-2}| and |θ2q|>|θ2q+1||\theta^{2q}|>|\theta^{2q+1}|. Since θ0\theta\leq 0, when kk is even, we have θk0\theta^{k}\geq 0 and (N2)λ2(1λ)θk0(N-2)\lambda^{2}(1-\lambda)\theta^{k}\geq 0. When kk is odd, we have θk0\theta^{k}\leq 0 and (N2)λ2(1λ)θk0(N-2)\lambda^{2}(1-\lambda)\theta^{k}\leq 0. Hence, θk[(N2)λ2(1λ)θk+(N2)λ2+2λ1]θk[(N2)λ2+2λ1]\theta^{k}\big{[}(N-2)\lambda^{2}(1-\lambda)\theta^{k}+(N-2)\lambda^{2}+2\lambda-1\big{]}\geq\theta^{k}\big{[}(N-2)\lambda^{2}+2\lambda-1\big{]}, for all kk. Let ξ(k,λ,θ,N):=2(1λ+λθk)N2θk((N2)λ2+2λ1)\xi(k,\lambda,\theta,N):=2(1-\lambda+\lambda\theta^{k})^{N-2}\theta^{k}((N-2)\lambda^{2}+2\lambda-1). Based on above results, it is easy to verify ξ(2q,λ,θ,N)+ξ(2q+1,λ,θ,N)>0\xi(2q,\lambda,\theta,N)+\xi(2q+1,\lambda,\theta,N)>0, and hence k=2ξ(k,λ,θ,N)>0\sum_{k=2}^{\infty}\xi(k,\lambda,\theta,N)>0.

The remaining term is ξ(1,λ,θ,N)\xi(1,\lambda,\theta,N) and (Nλ1)(N\lambda-1). Summing them, denote ρ(θ,λ,N):=2θ[(N2)λ2(1+θ)(N2)λ3θ+2λ1]+Nλ1\rho(\theta,\lambda,N):=2\theta[(N-2)\lambda^{2}(1+\theta)-(N-2)\lambda^{3}\theta+2\lambda-1]+N\lambda-1, we have 2(1λ+λθ)N2θ[(N2)λ2(1+θ)(N2)λ3θ+2λ1]+Nλ1ρ(θ,λ,N)2(1-\lambda+\lambda\theta)^{N-2}\theta[(N-2)\lambda^{2}(1+\theta)-(N-2)\lambda^{3}\theta+2\lambda-1]+N\lambda-1\geq\rho(\theta,\lambda,N). The partial derivative of ρ(θ,λ,N)\rho(\theta,\lambda,N) with respect to θ\theta is ρ(θ,λ,N)θ=2[(N2)λ2(1+θ)(N2)λ3θ+2λ1]+2(N2)λ2(1λ)θ\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}=2[(N-2)\lambda^{2}(1+\theta)-(N-2)\lambda^{3}\theta+2\lambda-1]+2(N-2)\lambda^{2}(1-\lambda)\theta, which is an increasing function of θ\theta (convex function in θ(1,0)\theta\in(-1,0)). Choose θ=0\theta=0, ρ(θ,λ,N)θ|θ=0=2(N2)λ2+4λ2>0\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=0}=2(N-2)\lambda^{2}+4\lambda-2>0 when λ[N11N2,1]\lambda\in[\frac{\sqrt{N-1}-1}{N-2},1]. Choosing θ=1\theta=-1, the partial derivative becomes ρ(θ,λ,N)θ|θ=1=4(N2)λ32(N2)λ2+4λ2\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=-1}=4(N-2)\lambda^{3}-2(N-2)\lambda^{2}+4\lambda-2.

If ρ(θ,λ,N)θ|θ=1<0\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=-1}<0, ρ(θ,λ,N)min{ρ(1,λ,N)+12ρ(θ,λ,N)θ|θ=1,ρ(0,λ,N)12ρ(θ,λ,N)θ|θ=0}\rho(\theta,\lambda,N)\geq\min\{\rho(-1,\lambda,N)+\frac{1}{2}\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=-1},\rho(0,\lambda,N)-\frac{1}{2}\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=0}\}. Note ρ(1,λ,N)+12ρ(θ,λ,N)θ|θ=1=2(N2)λ34λ+2+Nλ1+2(N2)λ3(N2)λ2+2λ1=(N2)λ(1λ)0\rho(-1,\lambda,N)+\frac{1}{2}\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=-1}=-2(N-2)\lambda^{3}-4\lambda+2+N\lambda-1+2(N-2)\lambda^{3}-(N-2)\lambda^{2}+2\lambda-1=(N-2)\lambda(1-\lambda)\geq 0 and ρ(0,λ,N)12ρ(θ,λ,N)θ|θ=0=Nλ1(N2)λ22λ+1=(N2)λ(1λ)0\rho(0,\lambda,N)-\frac{1}{2}\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=0}=N\lambda-1-(N-2)\lambda^{2}-2\lambda+1=(N-2)\lambda(1-\lambda)\geq 0. Thus, ρ(θ,λ,N)0\rho(\theta,\lambda,N)\geq 0 in this case.

If ρ(θ,λ,N)θ|θ=1=4(N2)λ32(N2)λ2+4λ2=2(2λ1)((N2)λ2+1)0\frac{\partial\rho(\theta,\lambda,N)}{\partial\theta}|_{\theta=-1}=4(N-2)\lambda^{3}-2(N-2)\lambda^{2}+4\lambda-2=2(2\lambda-1)((N-2)\lambda^{2}+1)\geq 0, the minimum ρ(θ,λ,N)\rho(\theta,\lambda,N) is achieved at smallest θ\theta and λ12\lambda\geq\frac{1}{2}. Since λ12\lambda\geq\frac{1}{2}, θ[λ1λ,0]\theta\in[\frac{\lambda-1}{\lambda},0], and hence ρ(θ,λ,N)ρ(λ1λ,λ,N)=2(λ1)(N2)λ2+6(N2)λ(λ1)2(N4)(λ1)2(λ1)λ+Nλ1\rho(\theta,\lambda,N)\geq\rho(\frac{\lambda-1}{\lambda},\lambda,N)=-2(\lambda-1)(N-2)\lambda^{2}+6(N-2)\lambda(\lambda-1)-2(N-4)(\lambda-1)-\frac{2(\lambda-1)}{\lambda}+N\lambda-1, and 2(λ1)λ2-\frac{2(\lambda-1)}{\lambda}\geq-2. Thus, ρ(θ,λ,N)2λ3(N2)+8(N2)λ2+(207N)λ+2N11\rho(\theta,\lambda,N)\geq-2\lambda^{3}(N-2)+8(N-2)\lambda^{2}+(20-7N)\lambda+2N-11. Taking derivative, we have 6(N2)λ2+16(N2)λ+207N-6(N-2)\lambda^{2}+16(N-2)\lambda+20-7N, which is positive when λ=1\lambda=1. The derivative of 6(N2)λ2+16(N2)λ+207N-6(N-2)\lambda^{2}+16(N-2)\lambda+20-7N is (N2)(1712λ)0(N-2)(17-12\lambda)\geq 0. Combined with λ[12,1]\lambda\in[\frac{1}{2},1], the value of 2λ3(N2)+8(N2)λ2+(207N)λ+2N112\lambda^{3}(N-2)+8(N-2)\lambda^{2}+(20-7N)\lambda+2N-11 is larger than the value of choosing λ=12\lambda=\frac{1}{2} (when N<30)N<30) or the value of λ=12\lambda=\frac{1}{2} plus the derivative at λ=12\lambda=\frac{1}{2} times 12\frac{1}{2} (when N30N\geq 30). The value at λ=12\lambda=\frac{1}{2} is 94N920\frac{9}{4}N-\frac{9}{2}\geq 0. When N30N\geq 30, the value of λ=12\lambda=\frac{1}{2} plus the derivative at λ=12\lambda=\frac{1}{2} times 12\frac{1}{2} is 2N+3>02N+3>0. Thus, ρ(θ,λ,N)0\rho(\theta,\lambda,N)\geq 0 in this case.

As ρ(θ,λ,N)0\rho(\theta,\lambda,N)\geq 0 in above all cases, and hence k=2ξ(k,λ,N)>0\sum_{k=2}^{\infty}\xi(k,\lambda,N)>0.

Next, we consider (N2)λ+2λ1<0(N-2)\lambda+2\lambda-1<0. Solving the inequation, we have λ[1N,N11N2)\lambda\in[\frac{1}{N},\frac{\sqrt{N-1}-1}{N-2}). Note when |(N2)λ2(1λ)θk||(N2)λ2+2λ1||(N-2)\lambda^{2}(1-\lambda)\theta^{k}|\geq|(N-2)\lambda^{2}+2\lambda-1|, θk((N2)λ2(1λ)θk+(N2)λ2+2λ1)0\theta^{k}((N-2)\lambda^{2}(1-\lambda)\theta^{k}+(N-2)\lambda^{2}+2\lambda-1)\geq 0 for both odd and even kk, and hence the summation is nonnegative for these kk. So, we only need to analyze the case |(N2)λ2(1λ)θk||(N2)λ2+2λ1||(N-2)\lambda^{2}(1-\lambda)\theta^{k}|\geq|(N-2)\lambda^{2}+2\lambda-1|, in which ω(λ,θ,k,N)0\omega(\lambda,\theta,k,N)\geq 0 when kk is odd and ω(λ,θ,k,N)0\omega(\lambda,\theta,k,N)\leq 0 when kk is even. Therefore, we consider ω(λ,θ,2q1,N)+ω(λ,θ,2q,N)\omega(\lambda,\theta,2q-1,N)+\omega(\lambda,\theta,2q,N). Note θ2q(N2)λ2+2λ1(N2)λ2(1λ)\theta^{2q}\leq-\frac{(N-2)\lambda^{2}+2\lambda-1}{(N-2)\lambda^{2}(1-\lambda)} in this case.

Recall the function ξ(k,λ,θ,N)\xi(k,\lambda,\theta,N), we have ω(λ,θ,2q1,N)+ω(λ,θ,2q,N)(1λ)N2ξ(2q1,λ,θ,N)+ξ(2q,λ,θ,N)=(1λ)N2θ2q1[(N2)λ2+2λ1][(1λ+λθ2q1)N2+θ(1λ+λθ2q)N2]\omega(\lambda,\theta,2q-1,N)+\omega(\lambda,\theta,2q,N)\geq(1-\lambda)^{N-2}\xi(2q-1,\lambda,\theta,N)+\xi(2q,\lambda,\theta,N)=(1-\lambda)^{N-2}\theta^{2q-1}[(N-2)\lambda^{2}+2\lambda-1][(1-\lambda+\lambda\theta^{2q-1})^{N-2}+\theta(1-\lambda+\lambda\theta^{2q})^{N-2}]. Note θ2q10\theta^{2q-1}\leq 0 and (N2)λ2+2λ10(N-2)\lambda^{2}+2\lambda-1\leq 0. The remaining is to compare (1λ+λθ2q1)N2(1-\lambda+\lambda\theta^{2q-1})^{N-2} and θ(1λ+λθ2q)N2-\theta(1-\lambda+\lambda\theta^{2q})^{N-2}. Divide these two terms, we have (1λ+λθ2q1)N2θ(1λ+λθ2q)N21θ(1λ+λθ11λ+λθ2)N2\frac{(1-\lambda+\lambda\theta^{2q-1})^{N-2}}{-\theta(1-\lambda+\lambda\theta^{2q})^{N-2}}\geq\frac{1}{-\theta}\big{(}\frac{1-\lambda+\lambda\theta^{1}}{1-\lambda+\lambda\theta^{2}}\big{)}^{N-2}.

When N13N\geq 13, we consider the bound θ[λ1λ,0]\theta\in[-\frac{\lambda}{1-\lambda},0] and λ[1N,N11N2]\lambda\in[\frac{1}{N},\frac{\sqrt{N-1}-1}{N-2}], and hence (1λ+λθ2q1)N2θ(1λ+λθ2q)N21λλ(1λλ2/(1λ)1λ+λ3/(1λ)2)N2=1λλ(13λ+2λ213λ+3λ2)N2=1λλ(111/λ23/λ+3)N2\frac{(1-\lambda+\lambda\theta^{2q-1})^{N-2}}{-\theta(1-\lambda+\lambda\theta^{2q})^{N-2}}\geq\frac{1-\lambda}{\lambda}(\frac{1-\lambda-\lambda^{2}/(1-\lambda)}{1-\lambda+\lambda^{3}/(1-\lambda)^{2}})^{N-2}=\frac{1-\lambda}{\lambda}(\frac{1-3\lambda+2\lambda^{2}}{1-3\lambda+3\lambda^{2}})^{N-2}=\frac{1-\lambda}{\lambda}(1-\frac{1}{1/\lambda^{2}-3/\lambda+3})^{N-2}. By doing the partial derivative with respect to λ\lambda, we have (1λ)/λλ=1λ2<0\frac{\partial(1-\lambda)/\lambda}{\partial\lambda}=\frac{-1}{\lambda^{2}}<0 and (1/λ23/λ)λ=1/λ3(3λ2)<0\frac{\partial(1/\lambda^{2}-3/\lambda)}{\partial\lambda}=1/\lambda^{3}(3\lambda-2)<0 when λ[1N,N11N2]\lambda\in[\frac{1}{N},\frac{\sqrt{N-1}-1}{N-2}]. Thus, 1λλ(111/λ23/λ+3)N2\frac{1-\lambda}{\lambda}(1-\frac{1}{1/\lambda^{2}-3/\lambda+3})^{N-2} is a decreasing function of λ\lambda. Taking λ=N11N2\lambda=\frac{\sqrt{N-1}-1}{N-2} into this function, we obtain the lower bound, which is a function of only one variable NN. By doing the partial derivative, it can be verified that it is an increasing function of NN. Taking N=13N=13, we can verify 1λλ(111/λ23/λ+3)N21\frac{1-\lambda}{\lambda}(1-\frac{1}{1/\lambda^{2}-3/\lambda+3})^{N-2}\geq 1. Hence (1λ+λθ2q1)N2θ(1λ+λθ2q)N21\frac{(1-\lambda+\lambda\theta^{2q-1})^{N-2}}{-\theta(1-\lambda+\lambda\theta^{2q})^{N-2}}\geq 1 for N13N\geq 13.

When N<13N<13, we should consider both bound for θ\theta, θ[λ1λ,0]\theta\in[-\frac{\lambda}{1-\lambda},0] and θ2q(N2)λ2+2λ1(N2)λ2(1λ)\theta^{2q}\leq-\frac{(N-2)\lambda^{2}+2\lambda-1}{(N-2)\lambda^{2}(1-\lambda)}, which comes from the inequation condition θk((N2)λ2(1λ)θ2i+(N2)λ2+2λ1)0\theta^{k}((N-2)\lambda^{2}(1-\lambda)\theta^{2i}+(N-2)\lambda^{2}+2\lambda-1)\leq 0. For λ\lambda, we separate its domain to be [1N,Γ][\frac{1}{N},\Gamma] and [Γ,N11N2][\Gamma,\frac{\sqrt{N-1}-1}{N-2}], where Γ=0.031N+0.97N11N2\Gamma=0.03*\frac{1}{N}+0.97*\frac{\sqrt{N-1}-1}{N-2}. It can be verified that, when λ[1N,Γ]\lambda\in[\frac{1}{N},\Gamma], 1λλ(111/λ23/λ+3)N21\frac{1-\lambda}{\lambda}(1-\frac{1}{1/\lambda^{2}-3/\lambda+3})^{N-2}\geq 1 for every NN that N<13N<13, using the same strategy as that for N13N\geq 13 and the only change is the upper bound for λ\lambda. When λ[Γ,N11N2]\lambda\in[\Gamma,\frac{\sqrt{N-1}-1}{N-2}], we analyze the sign of (1λ+λθ2q1)N2+θ(1λ+λθ2q)N2(1-\lambda+\lambda\theta^{2q-1})^{N-2}+\theta(1-\lambda+\lambda\theta^{2q})^{N-2}. Since θ2q(N2)λ2+2λ1(N2)λ2(1λ)\theta^{2q}\leq-\frac{(N-2)\lambda^{2}+2\lambda-1}{(N-2)\lambda^{2}(1-\lambda)}, by doing the partial derivative, the upper bound of θ2q\theta^{2q} is a decreasing function of λ\lambda. Hence, the maximum θ2q\theta^{2q} is achieved when λ=Γ\lambda=\Gamma and then θ2qγ\theta^{2q}\leq-\gamma, where γ=(N2)Γ2+2th1(N2)Γ2(1Γ)\gamma=\frac{(N-2)\Gamma^{2}+2th-1}{(N-2)\Gamma^{2}(1-\Gamma)} is a variable only related to NN. Thus, (1λ+λθ2q1)N2+θ(1λ+λθ2q)N2(1λλγ)N2γ(1λ+λγ)N2(1-\lambda+\lambda\theta^{2q-1})^{N-2}+\theta(1-\lambda+\lambda\theta^{2q})^{N-2}\geq(1-\lambda-\lambda\sqrt{\gamma})^{N-2}-\sqrt{\gamma}(1-\lambda+\lambda\gamma)^{N-2}, which, verified by doing partial derivative, is a decreasing function of λ\lambda. Then, taking λ=N11N2\lambda=\frac{\sqrt{N-1}-1}{N-2}, we can verify that (1λλγ)N2γ(1λ+λγ)N20(1-\lambda-\lambda\sqrt{\gamma})^{N-2}-\sqrt{\gamma}(1-\lambda+\lambda\gamma)^{N-2}\geq 0 for every NN that N<13N<13. Therefore, (1λ+λθ2q1)N2+θ(1λ+λθ2q)N20(1-\lambda+\lambda\theta^{2q-1})^{N-2}+\theta(1-\lambda+\lambda\theta^{2q})^{N-2}\geq 0 when N<13N<13

As (1λ+λθ2q1)N2+θ(1λ+λθ2q)N20(1-\lambda+\lambda\theta^{2q-1})^{N-2}+\theta(1-\lambda+\lambda\theta^{2q})^{N-2}\geq 0 for both N13N\geq 13 and N<13N<13, ω(λ,θ,2q1,N)+ω(λ,θ,2q,N)0\omega(\lambda,\theta,2q-1,N)+\omega(\lambda,\theta,2q,N)\geq 0, and hence k=1ξ(k,λ,θ,N)0\sum_{k=1}^{\infty}\xi(k,\lambda,\theta,N)\geq 0 when (N2)λ+2λ1<0(N-2)\lambda+2\lambda-1<0.

Since k=1ξ(k,λ,θ,N)0\sum_{k=1}^{\infty}\xi(k,\lambda,\theta,N)\geq 0 when (N2)λ+2λ1<0(N-2)\lambda+2\lambda-1<0 and k=1ξ(k,λ,θ,N)+Nλ10\sum_{k=1}^{\infty}\xi(k,\lambda,\theta,N)+N\lambda-1\geq 0 when (N2)λ+2λ10(N-2)\lambda+2\lambda-1\geq 0, we have, when λ1N\lambda\geq\frac{1}{N},

(va2/ma2)λ0,\displaystyle\frac{\partial(v_{a}^{2}/m_{a}^{2})}{\partial\lambda}\geq 0,

Now, we discuss passive users. Based on Theorem 1, we have

(vp2/mp2)λ=2k=1[CN(θk1)(1λ+λθk)CN1mpmp2\displaystyle\frac{\partial(v_{p}^{2}/m_{p}^{2})}{\partial\lambda}=2\sum_{k=1}^{\infty}\Big{[}\frac{CN(\theta^{k}-1)(1-\lambda+\lambda\theta^{k})^{CN-1}m_{p}}{m_{p}^{2}}
+CN(1λ)CN1(1λ+λθk)CNmp2]+CN(1λ)CNmp2\displaystyle+\frac{CN(1-\lambda)^{CN-1}(1-\lambda+\lambda\theta^{k})^{CN}}{m_{p}^{2}}\Big{]}+\frac{CN(1-\lambda)^{CN}}{m_{p}^{2}}
=2CN(1λ)CN1[1λ+k=1(1λ+λθk)CN1θk]mp2.\displaystyle=\frac{2CN(1-\lambda)^{CN-1}[1-\lambda+\sum_{k=1}^{\infty}(1-\lambda+\lambda\theta^{k})^{CN-1}\theta^{k}]}{m_{p}^{2}}.

Since θ(1,0)\theta\in(-1,0) and 1λ+λθk01-\lambda+\lambda\theta^{k}\geq 0 for all kk,

(vp2/mp2)λ\displaystyle\frac{\partial(v_{p}^{2}/m_{p}^{2})}{\partial\lambda}
2CN(1λ)CN1[1λ+k=1(1λ)CN1θk]mp2\displaystyle\geq\frac{2CN(1-\lambda)^{CN-1}[1-\lambda+\sum_{k=1}^{\infty}(1-\lambda)^{CN-1}\theta^{k}]}{m_{p}^{2}}
=CN(1λ)CN1[1λ+(1λ)CN1θ1θ]mp2.\displaystyle=\frac{CN(1-\lambda)^{CN-1}[1-\lambda+(1-\lambda)^{CN-1}\frac{\theta}{1-\theta}]}{m_{p}^{2}}.

When λ[0,12]\lambda\in[0,\frac{1}{2}], θ[λ1λ,0]\theta\in[-\frac{\lambda}{1-\lambda},0], and hence

(vp2/mp2)λ2CN(1λ)CN1[1λ(1λ)CN1λ]mp2\displaystyle\frac{\partial(v_{p}^{2}/m_{p}^{2})}{\partial\lambda}\geq\frac{2CN(1-\lambda)^{CN-1}[1-\lambda-(1-\lambda)^{CN-1}\lambda]}{m_{p}^{2}}
=2CN(1λ)CN1[1(1+(1λ)CN1)λ]mp2\displaystyle=\frac{2CN(1-\lambda)^{CN-1}[1-(1+(1-\lambda)^{CN-1})\lambda]}{m_{p}^{2}}
2CN(1λ)CN1[1(1+1)12]mp2=0.\displaystyle\geq\frac{2CN(1-\lambda)^{CN-1}[1-(1+1)\frac{1}{2}]}{m_{p}^{2}}=0.

When λ[12,1]\lambda\in[\frac{1}{2},1], θ[λ1λ,0]\theta\in[\frac{\lambda-1}{\lambda},0], and then

(vp2/mp2)λ2CN(1λ)CN1[1λ(1λ)CN]mp2\displaystyle\frac{\partial(v_{p}^{2}/m_{p}^{2})}{\partial\lambda}\geq\frac{2CN(1-\lambda)^{CN-1}[1-\lambda-(1-\lambda)^{CN}]}{m_{p}^{2}}
2CN(1λ)CN1[1λ(1λ)]mp2=0.\displaystyle\geq\frac{2CN(1-\lambda)^{CN-1}[1-\lambda-(1-\lambda)]}{m_{p}^{2}}=0.

Therefore, when λ1N\lambda\geq\frac{1}{N},

(vp2/mp2)λ0.\displaystyle\frac{\partial(v_{p}^{2}/m_{p}^{2})}{\partial\lambda}\geq 0.

-C Proof of Lemma 3

The partial derivative of va2v_{a}^{2} with respect to θ\theta when λ\lambda is determined is

va2θ=2mak=1[(1λ)kθk1(1λ+λθk)N1\displaystyle\frac{\partial v_{a}^{2}}{\partial\theta}=2m_{a}\sum_{k=1}^{\infty}[(1-\lambda)k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-1}
+(λ+(1λ)θk)(N1)λkθk1(1λ+λθk)N2)]\displaystyle\quad+(\lambda+(1-\lambda)\theta^{k})(N-1)\lambda k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-2})]
=\displaystyle= 2mak=1[12λ+Nλ2+Nλ(1λ)θk]\displaystyle 2m_{a}\sum_{k=1}^{\infty}[1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{k}]
×kθk1(1λ+λθk)N2.\displaystyle\quad\times k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-2}.

Let ψ(k):=2ma[12λ+Nλ2+Nλ(1λ)θk]kθk1(1λ+λθk)N2\psi(k):=2m_{a}[1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{k}]k\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{N-2} be the kkth term in the above equation.

We first show that α,β<12\alpha,\beta<\frac{1}{2}. Choosing y=0y=0 and y=12y=\frac{1}{2}, we have hN(0)=h¯N(0)=1h_{N}(0)=\bar{h}_{N}(0)=1, hN(12)=38N+112<0h_{N}(\frac{1}{2})=-\frac{3}{8}N+\frac{1}{12}<0 and h¯N(12)=18N<0\bar{h}_{N}(\frac{1}{2})=-\frac{1}{8}N<0. Therefore, there must exists one root that in (0,12)(0,\frac{1}{2}) for both hN(y)h_{N}(y) and h¯N(y)\bar{h}_{N}(y). Thus, λ[0,12)\lambda\in[0,\frac{1}{2}) and hence θ[λ1λ,1]\theta\in[-\frac{\lambda}{1-\lambda},1].

Therefore, [12λ+Nλ2+Nλ(1λ)θk]12×12+Nλ2Nλ(1λ)λ1λ=0[1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{k}]\geq 1-2\times\frac{1}{2}+N\lambda^{2}-N\lambda(1-\lambda)\frac{\lambda}{1-\lambda}=0, and 1λ+λθk1λ+λθ11λλλ1λ>12λ01-\lambda+\lambda\theta^{k}\geq 1-\lambda+\lambda\theta^{1}\geq 1-\lambda-\lambda\frac{\lambda}{1-\lambda}>1-2\lambda\geq 0. So, ψ(k)<0\psi(k)<0 only when θ<0\theta<0 and kk is even.

The expression of ψ(2i1)ψ(2i)\frac{\psi(2i-1)}{\psi(2i)} is

ψ(2i1)ψ(2i)=\displaystyle-\frac{\psi(2i-1)}{\psi(2i)}= 12λ+Nλ2+Nλ(1λ)θ2i112λ+Nλ2+Nλ(1λ)θ2i2i12i\displaystyle-\frac{1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{2i-1}}{1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{2i}}\frac{2i-1}{2i}
×θ2i1θ2i(1λ+λθ2i11λ+λθ2i)N2.\displaystyle\times\frac{\theta^{2i-1}}{\theta^{2i}}\Big{(}\frac{1-\lambda+\lambda\theta^{2i-1}}{1-\lambda+\lambda\theta^{2i}}\Big{)}^{N-2}.

When θ0\theta\geq 0, it is easy to verify that ψ(2i1)ψ(2i)1-\frac{\psi(2i-1)}{\psi(2i)}\geq 1 since θ1\theta\leq 1.

When λ1λθ<0\frac{-\lambda}{1-\lambda}\leq\theta<0, we have

ψ(2i1)ψ(2i)12λ+Nλ2+Nλ(1λ)θ112λ+Nλ2+Nλ(1λ)θ212θ1θ2\displaystyle\frac{\psi(2i-1)}{\psi(2i)}\geq\frac{1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{1}}{1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\theta^{2}}\frac{1}{2}\frac{\theta^{1}}{\theta^{2}}
×(1λ+λθ11λ+λθ2)N2\displaystyle\quad\times\Big{(}\frac{1-\lambda+\lambda\theta^{1}}{1-\lambda+\lambda\theta^{2}}\Big{)}^{N-2}
12λ+Nλ2Nλ(1λ)λ1λ12λ+Nλ2+Nλ(1λ)λ2(1λ)2121λλ\displaystyle\geq\frac{1-2\lambda+N\lambda^{2}-N\lambda(1-\lambda)\frac{\lambda}{1-\lambda}}{1-2\lambda+N\lambda^{2}+N\lambda(1-\lambda)\frac{\lambda^{2}}{(1-\lambda)^{2}}}\frac{1}{2}\frac{1-\lambda}{\lambda}
×(1λλλ1λ1λ+λλ2(1λ)2)N2\displaystyle\quad\times\Big{(}\frac{1-\lambda-\lambda\frac{\lambda}{1-\lambda}}{1-\lambda+\lambda\frac{\lambda^{2}}{(1-\lambda)^{2}}}\Big{)}^{N-2}
=1λ2λ13λ+2λ213λ+(N+2)λ2(1λ213λ+3λ2)N2.\displaystyle=\frac{1-\lambda}{2\lambda}\frac{1-3\lambda+2\lambda^{2}}{1-3\lambda+(N+2)\lambda^{2}}(1-\frac{\lambda^{2}}{1-3\lambda+3\lambda^{2}})^{N-2}.

Define AN(y),B¯(y)A_{N}(y),\bar{B}(y) and CN(y)C_{N}(y) as follows:

AN(y)\displaystyle A_{N}(y) =13y+2y213y+(N+2)y2(1y213y+3y2)N2,\displaystyle=\frac{1-3y+2y^{2}}{1-3y+(N+2)y^{2}}(1-\frac{y^{2}}{1-3y+3y^{2}})^{N-2},
B¯(y)\displaystyle\bar{B}(y) =2y1y,CN(y)=13y(N4)y213y+(N+2)y2.\displaystyle=\frac{2y}{1-y},\quad C_{N}(y)=\frac{1-3y-(N-4)y^{2}}{1-3y+(N+2)y^{2}}.

Then, ψ(2i1)ψ(2i)AN(y)B¯(y)\frac{\psi(2i-1)}{\psi(2i)}\geq\frac{A_{N}(y)}{\bar{B}(y)}.

First, we prove CN(y)B¯(y)C_{N}(y)\geq\bar{B}(y). Subtracting B¯(y)\bar{B}(y) from CN(y)C_{N}(y), we have

CN(y)B¯(y)=13y(N4)y213y+(N+2)y22y1y\displaystyle C_{N}(y)-\bar{B}(y)=\frac{1-3y-(N-4)y^{2}}{1-3y+(N+2)y^{2}}-\frac{2y}{1-y}
=(13y)2(1y)(N4)y22y(N+2)y2(13y+(N+2)y2)(1y)\displaystyle=\frac{(1-3y)^{2}-(1-y)(N-4)y^{2}-2y(N+2)y^{2}}{(1-3y+(N+2)y^{2})(1-y)}
=hN(y)(13y+(N+2)y2)(1y).\displaystyle=\frac{h_{N}(y)}{(1-3y+(N+2)y^{2})(1-y)}.

As α\alpha is the smallest positive root of hN(y)h_{N}(y) and hN(0)=1>0h_{N}(0)=1>0, hN(y)0h_{N}(y)\geq 0 when y[0,α]y\in[0,\alpha]. Note α<12\alpha<\frac{1}{2}.

Now we check whether (13y+(N+2)y2)(1-3y+(N+2)y^{2}) is positive. The derivative of (13y+(N+2)y2)(1-3y+(N+2)y^{2}) is 2(N+2)y32(N+2)y-3 and hence it achieves minimum when y=32(N+2)y=\frac{3}{2(N+2)}. Choosing y=32(N+2)y=\frac{3}{2(N+2)}, we have (13y+(N+2)y2)=192(N+2)+94(N+2)=194(N+2)194(2+2)>0(1-3y+(N+2)y^{2})=1-\frac{9}{2(N+2)}+\frac{9}{4(N+2)}=1-\frac{9}{4(N+2)}\geq 1-\frac{9}{4(2+2)}>0.

Since (13y+(N+2)y2)(1-3y+(N+2)y^{2}), (1y)(1-y) and h¯N(y)\bar{h}_{N}(y) are all non-negative when y[0,α]y\in[0,\alpha], we have

CN(y)B¯(y)=hN(y)(13y+(N+2)y2)(1y)0.\displaystyle C_{N}(y)-\bar{B}(y)=\frac{h_{N}(y)}{(1-3y+(N+2)y^{2})(1-y)}\geq 0.

Next, we prove AN(y)CN(y)A_{N}(y)\geq C_{N}(y).

We know that (1y)N1Ny(1-y)^{N}\approx 1-Ny from binomial approximation. Let j(y)=(1y)N(1Ny)j(y)=(1-y)^{N}-(1-Ny). The derivative of j(y)j(y) is N[1(1y)N1]N[1-(1-y)^{N-1}]. As (1y)N11(1-y)^{N-1}\leq 1 when 0y10\leq y\leq 1, we have j(y)j(y) is increasing when y[0,1]y\in[0,1]. Hence, j(y)j(0)=0j(y)\geq j(0)=0, which means (1y)N1Ny(1-y)^{N}\geq 1-Ny, when y[0,1]y\in[0,1].

Now, we want to find whether y23y23y+1\frac{y^{2}}{3y^{2}-3y+1} is between 0 and 11. The derivative of 3y23y+1y23y^{2}-3y+1-y^{2} is 4y34y-3. Thus, when 0y120\leq y\leq\frac{1}{2}, 3y23y+1y23×143×12+114=03y^{2}-3y+1-y^{2}\geq 3\times\frac{1}{4}-3\times\frac{1}{2}+1-\frac{1}{4}=0. Therefore, 3y23y+1y23y^{2}-3y+1\geq y^{2}. As 3y23y+1>03y^{2}-3y+1>0 and y20y^{2}\geq 0, we have y23y23y+11\frac{y^{2}}{3y^{2}-3y+1}\leq 1 and y23y23y+10\frac{y^{2}}{3y^{2}-3y+1}\geq 0 when y[0,12]y\in[0,\frac{1}{2}].

Thus, when y[0,12]y\in[0,\frac{1}{2}],

(1y23y23y+1)N21(N2)y23y23y+1.\displaystyle(1-\frac{y^{2}}{3y^{2}-3y+1})^{N-2}\geq 1-(N-2)\frac{y^{2}}{3y^{2}-3y+1}.

Then,

AN(y)\displaystyle A_{N}(y) =13y+2y213y+(N+2)y2(1y213y+3y2)N2\displaystyle=\frac{1-3y+2y^{2}}{1-3y+(N+2)y^{2}}(1-\frac{y^{2}}{1-3y+3y^{2}})^{N-2}
13y+2y213y+(N+2)y2[1(N2)y213y+3y2]\displaystyle\geq\frac{1-3y+2y^{2}}{1-3y+(N+2)y^{2}}[1-(N-2)\frac{y^{2}}{1-3y+3y^{2}}]
=13y+2y213y+(N+2)y213y(N5)y213y+3y2.\displaystyle=\frac{1-3y+2y^{2}}{1-3y+(N+2)y^{2}}\frac{1-3y-(N-5)y^{2}}{1-3y+3y^{2}}.

When N2N\geq 2, 13y(N5)y213y+3y21-3y-(N-5)y^{2}\leq 1-3y+3y^{2}, and hence

13y(N5)y213y+3y213y(N4)y213y+2y2.\displaystyle\frac{1-3y-(N-5)y^{2}}{1-3y+3y^{2}}\geq\frac{1-3y-(N-4)y^{2}}{1-3y+2y^{2}}.

Therefore,

AN(y)\displaystyle A_{N}(y) 13y+2y213y+(N+2)y213y(N4)y213y+2y2\displaystyle\geq\frac{1-3y+2y^{2}}{1-3y+(N+2)y^{2}}\frac{1-3y-(N-4)y^{2}}{1-3y+2y^{2}}
=13y(N4)y213y+(N+2)y2=CN(y).\displaystyle=\frac{1-3y-(N-4)y^{2}}{1-3y+(N+2)y^{2}}=C_{N}(y).

So, when y[0,α]y\in[0,\alpha], AN(y)CN(y)B¯(y)A_{N}(y)\geq C_{N}(y)\geq\bar{B}(y). Thus,

ψ(2i1)ψ(2i)AN(λ)B¯(λ)1.\displaystyle\frac{\psi(2i-1)}{\psi(2i)}\geq\frac{A_{N}(\lambda)}{\bar{B}(\lambda)}\geq 1.

Hence, va2θ0\frac{\partial v_{a}^{2}}{\partial\theta}\geq 0 and va2v_{a}^{2} achieves its minimum value when θ\theta is minimized. Thus, choosing r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 minimizes va2v_{a}^{2}.

Since E[AoIaz]E[AoI_{a}^{z}] is an increasing function of va2v_{a}^{2}, choosing r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 also minimizes E[AoIaz]E[AoI_{a}^{z}].

Next, we consider passive users.

Recall the expression of vp2v_{p}^{2},

vp2\displaystyle v_{p}^{2} =2k=1((1λ+λθk)CNmp)mp+mpmp2.\displaystyle=2\sum_{k=1}^{\infty}((1-\lambda+\lambda\theta^{k})^{CN}-m_{p})m_{p}+m_{p}-m_{p}^{2}.

For this expression, it is easy to verify that vp2|θ>0vp2|θ=0v_{p}^{2}|_{\theta>0}\geq v_{p}^{2}|_{\theta=0}. Therefore, vp2v_{p}^{2} reaches its minimum in the region λ1λθ0-\frac{\lambda}{1-\lambda}\leq\theta\leq 0.

The partial derivative of vp2v_{p}^{2} with respect to θ\theta is

vp2θ=2mpk=1CNkλθk1(1λ+λθk)CN1.\displaystyle\frac{\partial v_{p}^{2}}{\partial\theta}=2m_{p}\sum_{k=1}^{\infty}CNk\lambda\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{CN-1}.

Let ϕ(k)=2mpCNkλθk1(1λ+λθk)CN1\phi(k)=2m_{p}CNk\lambda\theta^{k-1}(1-\lambda+\lambda\theta^{k})^{CN-1} be the kthk-th term of the partial derivative above. When θ[λ1λ,0]\theta\in[-\frac{\lambda}{1-\lambda},0], we can find that ϕ(k)0\phi(k)\geq 0 when kk is odd and ϕ(k)<0\phi(k)<0 when kk is even. Thus, vp2θ\frac{\partial v_{p}^{2}}{\partial\theta} is positive if ϕ(2i1)ϕ(2i)>1-\frac{\phi(2i-1)}{\phi(2i)}>1 for all positive integer ii.

Plugging in the formula of ϕ(k)\phi(k),

ϕ(2i1)ϕ(2i)=2i12i1θ(1λ+λθ2i11λ+λθ2i)CN1.\displaystyle-\frac{\phi(2i-1)}{\phi(2i)}=-\frac{2i-1}{2i}\frac{1}{\theta}\Big{(}\frac{1-\lambda+\lambda\theta^{2i-1}}{1-\lambda+\lambda\theta^{2i}}\Big{)}^{CN-1}.

Since θ2i1θλ1λ\theta^{2i-1}\geq\theta\geq-\frac{\lambda}{1-\lambda} and θ2iθ2λ2(1λ)2\theta^{2i}\leq\theta^{2}\leq\frac{\lambda^{2}}{(1-\lambda)^{2}} when θ[λ1λ,0]\theta\in[-\frac{\lambda}{1-\lambda},0], we have

ϕ(2i1)ϕ(2i)\displaystyle-\frac{\phi(2i-1)}{\phi(2i)} ϕ(1)ϕ(2)=121θ(1λ+λθ1λ+λθ2)CN1\displaystyle\geq-\frac{\phi(1)}{\phi(2)}=-\frac{1}{2}\frac{1}{\theta}\Big{(}\frac{1-\lambda+\lambda\theta}{1-\lambda+\lambda\theta^{2}}\Big{)}^{CN-1}
1λ2λ(1λ23λ23λ+1)CN1.\displaystyle\geq\frac{1-\lambda}{2\lambda}\Big{(}1-\frac{\lambda^{2}}{3\lambda^{2}-3\lambda+1}\Big{)}^{CN-1}.

Define A¯CN(y)\bar{A}_{CN}(y) as follows.

A¯CN(y)\displaystyle\bar{A}_{CN}(y) =(1y23y23y+1)CN1.\displaystyle=\Big{(}1-\frac{y^{2}}{3y^{2}-3y+1}\Big{)}^{CN-1}.

When y[0,12]y\in[0,\frac{1}{2}],

A¯CN(y)B¯(y)=(1y23y23y+1)CN12y1y\displaystyle\bar{A}_{CN}(y)-\bar{B}(y)=\Big{(}1-\frac{y^{2}}{3y^{2}-3y+1}\Big{)}^{CN-1}-\frac{2y}{1-y}
1(CN1)y23y23y+12y1y\displaystyle\geq 1-(CN-1)\frac{y^{2}}{3y^{2}-3y+1}-\frac{2y}{1-y}
=h¯CN(y)(3y23y+1)(1y).\displaystyle=\frac{\bar{h}_{CN}(y)}{(3y^{2}-3y+1)(1-y)}.

Note h¯CN(0)=1\bar{h}_{CN}(0)=1, h¯CN(12)=CN8\bar{h}_{CN}(\frac{1}{2})=-\frac{CN}{8}, and β\beta is the smallest positive root of h¯CN(y)\bar{h}_{CN}(y). Thus, h¯CN(y)0\bar{h}_{CN}(y)\geq 0 if y[0,β]y\in[0,\beta]. When y[0,β]y\in[0,\beta], 3y23y+1>03y^{2}-3y+1>0 and 1y>01-y>0, and hence

A¯CN(y)B¯(y)=h¯CN(y)(3y23y+1)(1y)0.\displaystyle\bar{A}_{CN}(y)-\bar{B}(y)=\frac{\bar{h}_{CN}(y)}{(3y^{2}-3y+1)(1-y)}\geq 0.

Therefore, if λ[0,β]\lambda\in[0,\beta],

ϕ(2i1)ϕ(2i)A¯CN(λ)B¯(λ)1.\displaystyle-\frac{\phi(2i-1)}{\phi(2i)}\geq\frac{\bar{A}_{CN}(\lambda)}{\bar{B}(\lambda)}\geq 1.

So, vp2θ>0\frac{\partial v_{p}^{2}}{\partial\theta}>0 and vp2v_{p}^{2} achieves its minimum value when θ\theta is minimized and λ[0,β]\lambda\in[0,\beta]. Thus, choosing r=λ1λr=\frac{\lambda}{1-\lambda} and s=1s=1 minimizes vp2v_{p}^{2}.

-D Proof of Lemma 4

Let us rewrite the expression of hN(y)h_{N}(y).

hN(y)=(N+8)y3(N13)y26y+1.\displaystyle h_{N}(y)=-(N+8)y^{3}-(N-13)y^{2}-6y+1.

The derivative of hN(y)h_{N}(y) is

hN(y)=3(N+8)y22(N13)y6.\displaystyle h^{\prime}_{N}(y)=-3(N+8)y^{2}-2(N-13)y-6.

Let hN(y)=0h^{\prime}_{N}(y)=0, we have y1=2(N13)+4(N13)272(N+8)6(N+8)y_{1}=\frac{2(N-13)+\sqrt{4(N-13)^{2}-72(N+8)}}{-6(N+8)} and y2=2(N13)4(N13)272(N+8)6(N+8)y_{2}=\frac{2(N-13)-\sqrt{4(N-13)^{2}-72(N+8)}}{-6(N+8)}. Thus, the roots of hN(y)h^{\prime}_{N}(y) are either not existing or all negative. Since there is no root when y0y\geq 0 and hN(0)=6<0h^{\prime}_{N}(0)=-6<0, hN(y)h^{\prime}_{N}(y) is always negative when y0y\geq 0. Therefore, hN(y)h_{N}(y) is always decreasing when y0y\geq 0. Considering hN(0)=1>0h_{N}(0)=1>0 and hN(1)=2N<0h_{N}(1)=-2N<0, α\alpha must exist between 0 and 11, and hN(y)>0h_{N}(y)>0 if and only if y<αy<\alpha, when y0y\geq 0.

Let y=1Ny=\frac{1}{N}, we have

hN(1N)\displaystyle h_{N}(\frac{1}{N}) =(N+8)N3(N13)N26N+1\displaystyle=\frac{-(N+8)}{N^{3}}-\frac{(N-13)}{N^{2}}-\frac{6}{N}+1
=N37N2+12N8N3.\displaystyle=\frac{N^{3}-7N^{2}+12N-8}{N^{3}}.

Solving N37N2+12N8=0N^{3}-7N^{2}+12N-8=0, we obtain the only real root, which is approximately 4.87514.8751. Since N37N2+12N8N^{3}-7N^{2}+12N-8\rightarrow\infty when NN\rightarrow\infty, we have N37N2+12N8>0N^{3}-7N^{2}+12N-8>0 when N5N\geq 5.

Therefore, hN(1N)>0h_{N}(\frac{1}{N})>0, and hence α>1N\alpha>\frac{1}{N} when N>4N>4.

Now we proof β>1N\beta>\frac{1}{N} when N>C+4N>C+4. Let us rewrite h¯N(y)\bar{h}_{N}(y) here,

h¯CN(y)=(CN10)y3(CN13)y26y+1.\displaystyle\bar{h}_{CN}(y)=(CN-10)y^{3}-(CN-13)y^{2}-6y+1.

The analysis of h¯CN(y)\bar{h}_{CN}(y) depends on the sign of CN10CN-10.

When CN10=0CN-10=0, h¯CN(y)=3y26y+1\bar{h}_{CN}(y)=3y^{2}-6y+1. There are two roots for h¯CN(y)\bar{h}_{CN}(y), which are 3630.1835\frac{3-\sqrt{6}}{3}\approx 0.1835 and 3+631.8165\frac{3+\sqrt{6}}{3}\approx 1.8165. Note NN must be 1010 when CN=10CN=10 and N>C+4N>C+4. Thus, β0.1835>1N=0.1\beta\approx 0.1835>\frac{1}{N}=0.1.

When CN10<0CN-10<0, the derivative of h¯CN(y)\bar{h}_{CN}(y) is

h¯CN(y)=3(CN10)y22(CN13)y6.\displaystyle\bar{h}^{\prime}_{CN}(y)=3(CN-10)y^{2}-2(CN-13)y-6.

Let h¯CN(y)=0\bar{h}^{\prime}_{CN}(y)=0, we obtain two roots, one is x1=2(CN13)+4(CN13)2+72(CN10)6(CN10)x_{1}=\frac{2(CN-13)+\sqrt{4(CN-13)^{2}+72(CN-10)}}{6(CN-10)}, and the other one is x2=2(CN13)4(CN13)2+72(CN10)6(CN10)x_{2}=\frac{2(CN-13)-\sqrt{4(CN-13)^{2}+72(CN-10)}}{6(CN-10)}. Look at the term in the square root, 4(CN13)2+72(CN10)=4C2N232CN444(CN-13)^{2}+72(CN-10)=4C^{2}N^{2}-32CN-44. It has one negative 4331.19624-3\sqrt{3}\approx-1.1962 and one positive root 4+339.16924+3\sqrt{3}\approx 9.1692. Thus, h¯CN(y)\bar{h}^{\prime}_{CN}(y) does not have real roots when 1<CN<101<CN<10, and hence h¯CN(y)\bar{h}_{CN}(y) is always a decreasing function in this case. Therefore, h¯CN(y)\bar{h}_{CN}(y) only has one root, which is β\beta. In this case, h¯CN(y)>0\bar{h}_{CN}(y)>0 if and only if y<βy<\beta.

When CN10>0CN-10>0, h¯CN()<0\bar{h}_{CN}(-\infty)<0, h¯CN(0)=1>0\bar{h}_{CN}(0)=1>0, h¯CN(1)=2<0\bar{h}_{CN}(1)=-2<0, and h¯CN()>0\bar{h}_{CN}(\infty)>0. Thus, there must exist three roots, one is less than 0, one is between 0 and 11, and one is larger than 11. Then, there is just one root, β\beta, between 0 and 11, and h¯CN(y)>0\bar{h}_{CN}(y)>0 if and only if y<βy<\beta when y(0,1)y\in(0,1).

Therefore, if h¯CN(y1)>0\bar{h}_{CN}(y_{1})>0 for a certain y1(0,1)y_{1}\in(0,1), we have β>y1\beta>y_{1} when 1CN<101\leq CN<10 or CN>10CN>10. Taking y=1Ny=\frac{1}{N} in h¯CN(y)\bar{h}_{CN}(y), we have

h¯CN(1N)\displaystyle\bar{h}_{CN}(\frac{1}{N}) =CN10N3CN13N26N+1\displaystyle=\frac{CN-10}{N^{3}}-\frac{CN-13}{N^{2}}-\frac{6}{N}+1
=CN10CN2+13N6N2+N3N3\displaystyle=\frac{CN-10-CN^{2}+13N-6N^{2}+N^{3}}{N^{3}}
=N3(6+C)N2+(13+C)N10N3.\displaystyle=\frac{N^{3}-(6+C)N^{2}+(13+C)N-10}{N^{3}}.

When N>C+4N>C+4,

h¯CN(1N)\displaystyle\bar{h}_{CN}(\frac{1}{N})\geq (C+5)3(6+C)(C+5)2+(13+C)(C+5)N3\displaystyle\frac{(C+5)^{3}-(6+C)(C+5)^{2}+(13+C)(C+5)}{N^{3}}
10N3=8C+30N3>0.\displaystyle-\frac{10}{N^{3}}=\frac{8C+30}{N^{3}}>0.

Therefore, β>1N\beta>\frac{1}{N} when N>C+4N>C+4.