Delay Analysis for Fair Power Allocation Strategyin Single Antenna Broadcast Channel

Alireza Borhani and Soroush AkhlaghiFaculty of Engineering, Shahed University, Tehran, Iran

Emails: {aborhani, akhlaghi}@shahed.ac.ir

Abstract— A single transmit antenna Broadcast Channel (BC)with large number of users in a Rayleigh fading environmentis considered. It is assumed each user either receives apredetermined minimum rate constraint or remains silent.Recently, it is shown that by applying a proper power allocationstrategy, one can maximize the maximum number of activeusers in such channel. The main contribution of this paper isto derive the expected delay of the optimum power allocationstrategy in terms of fairness maximization, where the expecteddelay is defined as the minimum number of channel uses whichguarantees all of the users are serviced at least once. Moreover,for sufficiently large number of channel uses, the averagenumber of services received by a randomly selected user isderived.

Keywords — Broadcast Channel; Power Allocation; DelayAnalysis; Minimum Rate Constraint.

I. INTRODUCTION

Nowadays, Quality of Service (QoS) has been regarded asone of the main concerns in many communication systems.In many wireless applications, the users are required to get aservice with a constant rate and for a certain period of time.In this regard, Andrews et al. [1] proposed an efficient wayto support QoS in wireless channels based on maximizingthe number of users that can be supported with the desiredQoS. Some other works are devoted to the case of delaylimited systems [2], [3]. The traffic rate of wireless networksunder delay constraint with large number of nodes is presentedin [3]. Also, there have been some attempts to make abalance between the system delay and the achieved throughputin wireless networks [4], [5]. The problem of throughput-delay trade off in multicast channels for different schedulingalgorithms is presented in [6].

Accordingly, in [7], it is shown the optimum power al-location strategy in terms of maximizing the sum-rate is tochoosing the best user, meaning to allocate the whole powerto the corresponding codebook of the selected user. Fromnow on, we call this strategy as the Best User SelectionStrategy (BUSS). Although, the BUSS is able to achievethe sum-rate capacity of BC channel, it renders an excessdelay to happen [8], as only one user is being serviced ata given time. This may not be a wise strategy, specifically,for the case of large users. Motivated by this, the followingpaper concerns a variant issue in a Rayleigh fading BroadcastChannel (BC) when there is a base station aims at sendingindividual information to many users [9]. In this regard, anelegant power allocation method is proposed by Keshavarz et

al. [9], which is able to simultaneously achieve the asymptoticsum-rate capacity and to maximize the number of active users,assuming each user is subject to the minimum rate constraintof 𝑅𝑚𝑖𝑛. It is worth mentioning, the optimality of this schemeis limited to the case of large users. However, the expecteddelay of this method yet to be studied. In what follows, wecall this strategy as the Fair Power Allocation Strategy (FPAS).

This motivated us to explore the expected delay of the FPAS.As noting earlier, the main advantage of the aforementionedstrategy is approaching the maximum number of active usersat the same time which dramatically decreases the systemdelay. The system delay is defined as the minimum numberof channel uses which guarantees all of the system users (𝑛)successfully receive at least 𝑝 packets. In [8], it is proved thatin a homogenous single antenna BC and for the case of 𝑝=1,the expected delay of the BUSS, scales as 𝐷=𝑛 log(𝑛).

Similarly, we define the delay as the minimum numberof channel uses in which each user successfully receives atleast one packet. In what follows, we derive the expecteddelay of the FPAS for the case of single antenna BC andmany number of users. In this paper, it is proved that after𝑅𝑚𝑖𝑛

𝑛 log(𝑛)log(log(𝑛)) + 𝜔

(𝑛

log(log(𝑛))

)channel uses, one can claim

that all of the users are serviced at least once. In fact, forthe case of single antenna BC and 𝑛 users, the expecteddelay lies within two extreme points, the result of round-robin scheduling with 𝐷1 = 𝑛, which serves as a lowerbound and 𝐷2 = 𝑛𝑙𝑜𝑔(𝑛) as the upper bound which refersto the case that merely one user is being serviced at a giventime [8]. Furthermore, considering 𝑘 channel uses, it is clearthat some of the users may be serviced more than one time.Subsequently, in this paper, it is shown that the average numberof getting service by a randomly selected user behaves like𝑘 log(log(𝑛))

𝑛𝑅𝑚𝑖𝑛

(1 +𝑂(

√log(𝑘)

𝑘 )).

The remainder of this paper is organized as follows. Sec-tion II presents the system model, formalizes the problem andfinally discusses the Fair Power Allocation Strategy proposedin [9]. Next, the expected delay of this strategy is presented inSection III in details. Section IV shows the simulation results,and finally Section V summarizes findings.

In this paper, according to the Kunth’s notation [10], forany function 𝑓(𝑛) and 𝑔(𝑛), 𝑓(𝑛)=O(𝑔(𝑛)), 𝑓(𝑛)=o(𝑔(𝑛))and 𝑓(𝑛) = 𝜔(𝑔(𝑛)) are equivalent to lim𝑛→∞∣ 𝑓(𝑛)𝑔(𝑛) ∣ < ∞,

lim𝑛→∞∣ 𝑓(𝑛)𝑔(𝑛) ∣=0 and lim𝑛→∞∣ 𝑓(𝑛)𝑔(𝑛) ∣=∞, respectively.

2010 Sixth International Conference on Wireless and Mobile Communications

978-0-7695-4182-2/10 $26.00 © 2010 IEEE

DOI 10.1109/ICWMC.2010.28

151

2010 Sixth International Conference on Wireless and Mobile Communications

978-0-7695-4182-2/10 $26.00 © 2010 IEEE

DOI 10.1109/ICWMC.2010.28

151

II. PROBLEM FORMULATION

We consider a single antenna BC with 𝑛 users, each ofsingle receive antenna. Assuming the transmitted informationat time instant 𝑡 is s(𝑡), thus the received signal at the 𝑖’threceiver can be represented as follows,

𝑦𝑖(𝑡) = ℎ𝑖s(𝑡) + 𝑤𝑖(𝑡) 𝑖 = 1, 2, ..., 𝑛 , (1)

where ℎ𝑖 denotes the channel gain of the 𝑖’th user and isassumed to be available at both the transmitter and the receiversides. Moreover, the channel gains are supposed to be constantacross the coding block, and varies for each block. 𝑤𝑖(𝑡)represents a scaler additive white gaussian noise with variance𝜎2, i.e., 𝑤𝑖(𝑡)∼𝒞𝒩 (0,𝜎2). Further more, it is assumed theaverage transmit power is unit, i.e., 𝐸{∣s(𝑡)∣2} = 1.

It is demonstrated that the single antenna gaussian broadcastchannel is stochastically degraded [7]. As a result, by usingsuperposition coding at the transmitter and applying successiveinterference cancelation at the receivers, one can increasethe received Signal to Interference plus Noise Ratio (SINR)of the 𝑖’th user by canceling those interfering signals forwhich the corresponding channel strength outperforms that ofthe 𝑖’th user in the ordered list. In other words, assumingusers depending on their channel strength are indexed in adescending order, i.e., ∣ℎ1∣2>∣ℎ2∣2>...>∣ℎ𝑛∣2, the interferingsignals arising from users 𝑘 = 1, . . . , 𝑖 − 1 can be readilycanceled at the 𝑖’th receiver. As a result, the sum-rate capacityunder gaussian code book assumption becomes

∑𝑛𝑖=1 𝑅𝑖,

where 𝑅𝑖 is represented as follows,

𝑅𝑖 < log(1 +𝑝𝑖

1/∣ℎ𝑖∣2 +∑𝑖−1

𝑘=1 𝑝𝑘) for 𝑖 = 1, . . . , 𝑛. (2)

where 𝑝𝑖 for 𝑖 = 1, . . . , 𝑛 denotes the power allocated to the𝑖’th user, and it is assumed the received noise power is one.Accordingly, noting (2), the equivalent noise of the 𝑖’th usercan be thought as 𝑁𝑖=

1∣ℎ𝑖∣2 .

It should be noted that if one could find a power allocationstrategy such that the corresponding SINR of each activeuser exceeds 𝛼 − 1, then 𝑅𝑚𝑖𝑛 = log(1 + SINR) = log(𝛼)would be achievable by all active users. Noting above, in [9]for the case of Rayleigh channels, it is proved that 𝑅𝑚𝑖𝑛=log(𝛼) for 𝑚 best users is achievable if 𝑚𝑎𝑥1≤𝑖≤𝑚𝑁𝑖<

1𝛼𝑚 .

Accordingly, noting users are labeled in a descending order,i.e., 𝑁1 < 𝑁2 < . . . < 𝑁𝑚, thus 𝑚 is defined such that𝑁𝑚 < 1

𝛼𝑚 . Moreover, the assigned power to the 𝑖’th useris computed as 𝑝𝑖 =

𝐶𝛼𝑚−𝑖 for 𝑖 = 1, . . . ,𝑚, where 𝐶 is

a constant value and is defined such that the total powerconstraint is satisfied, thus 𝐶 = (1 − 1/𝛼). In this case, itis shown that the maximum number of active users scales as𝑚= log(log(𝑛))

𝑅𝑚𝑖𝑛(for more details refer to [9]). Clearly, as the

minimum rate of each user is 𝑅𝑚𝑖𝑛, the achievable sum-rate ofsuch channel asymptotically approaches the sum-rate capacityof such channel, i.e., log(log(𝑛)).

III. DELAY ANALYSIS

The problem is to find the minimum number of channeluses, namely 𝐷𝑚𝑖𝑛, for which all users are being serviced

at least once with probability approaching one (w.p.1). Inthis correspondence, we define a successful service S as allusers receive at least once. To compute 𝐷𝑚𝑖𝑛, we proposeusing a probabilistic viewpoint to seek for conditions whichleads to 𝑃𝑟{S } → 1. Let S1, ...,S𝑛 denote respectively thetotal number of times for which the users indexed 1 to 𝑛 arebeing serviced. As is noted earlier, the successful event Scorresponds to the event that each user is being serviced atleast once, i.e., S1 ∕= 0, ...,S𝑛 ∕= 0. The following theoremdefines 𝐷 = 𝐷𝑚𝑖𝑛 which guarantees the successful event Sto happen w.p.1 .

Theorem 1: For large number (𝑛) of users, assuming 𝑚users are being serviced simultaneously at a given time, thenecessary and sufficient condition for having 𝑃𝑟{S } → 1 is,

𝐷∼= 𝑛𝑙𝑜𝑔(𝑛)𝑚 + 𝜔( 𝑛

𝑚 ).

Proof: a. Necessary condition:Assuming 𝐷 channel uses are being used, hence, it follows

𝑃𝑟{S } = 𝑃𝑟{S1 ∕= 0, ...,S𝑛 ∕= 0}

= 1− 𝑃𝑟{𝑛∪

𝑖=1

S𝑖 = 0}

≥(𝑎) 1−𝑛∑

𝑖=1

𝑃𝑟{S𝑖 = 0}

=(𝑏) 1− 𝑛(1− 𝑚

𝑛)𝐷

≈(𝑐) 1− 𝑛𝑒−𝑚𝐷𝑛

=(𝑑) 1− 𝑒−𝑚𝐷−𝑛 log(𝑛)

𝑛 (3)

where (𝑎) comes from union upper bound. (𝑏) comes fromthe fact that for each channel use, the probability that a givenuser is being serviced is 𝑚

𝑛 , hence, for 𝐷 channel uses, theprobability of unsuccessful service for a given user becomes(1 − 𝑚

𝑛 )𝐷. (𝑐) is based on using (1 − 𝑚𝑛 )𝑛 = 𝑒−𝑚 for large

value of 𝑛 and finally (𝑑) is based on using 𝑥 = 𝑒log 𝑥. Thus,refering to (3), the necessary condition to have 𝑃𝑟{S } → 1

is 𝑚𝐷𝑚𝑖𝑛∼= 𝑛 log(𝑛)+𝜔(𝑛). As a result, 𝐷𝑚𝑖𝑛

∼= 𝑛 log(𝑛)𝑚 +

𝜔( 𝑛𝑚 ).b. Sufficient condition:As is noted earlier, we have

𝑃𝑟{S }=𝑃𝑟{𝑛∩

𝑖=1

S𝑖 ∕= 0} (4)

Congruent to what is done in [11], it can be argued that (4)can be upper bounded as 𝑃𝑟{S } ≤ ∏𝑛

𝑖=1 𝑃𝑟{S𝑖 ∕= 0}, thuswe arrive at the following

𝑃𝑟{S } ≤𝑛∏

𝑖=1

𝑃𝑟{S𝑖 ∕= 0}

= [1− (1− 𝑚

𝑛)𝐷]𝑛

≤(𝑎)𝑒−𝑛(1−𝑚𝑛 )𝐷

= 𝑒−𝑛𝑒𝐷 log(1−𝑚𝑛

)

(5)

152152

where (𝑎) comes from the fact that (1−𝑥)𝑛≤𝑒−𝑛𝑥. Clearly inorder to have 𝑃𝑟{S }→1, we should have 𝑛𝑒𝐷 log(1−𝑚

𝑛 )→0.Since 𝑛 goes to infinity, we have

𝑛𝑒𝐷 log(1−𝑚𝑛 ) ∼𝑛𝑒−

𝑚𝐷𝑛 (1+O(𝑚

𝑛 ))

= 𝑒−𝑚𝐷−𝑛 log(𝑛)

𝑛 (1+O(𝑚𝑛 )) (6)

Hence by setting 𝐷∼=𝑛 log(𝑛)𝑚 + 𝜔( 𝑛

𝑚 ), the event S happensw.p.1.

Theorem 1 states that in a single antenna BC, for 𝑚 con-current transmissions, the expected delay behaves like 𝑛 log(𝑛)

𝑚 .Referring to the minimum delay Round-Robin scheduling with𝐷1 = 𝑛, the corresponding 𝐷𝑚𝑖𝑛 of the aforementionedscheduling could theoretically approach 𝐷1 if 𝑚 = log(𝑛),which is not feasible as the maximum number of active usersis at most log(log(𝑛))

𝑅𝑚𝑖𝑛. Also, it is interesting to note that having

one active user at a given time corresponds to the case of𝑚 = 1, thus the expected delay becomes 𝑛 log(𝑛) which isin accordance to the result reported in [8]. Finally, referringto theorem 1 and noting the maximum possible value of 𝑚is log(log(𝑛))

𝑅𝑚𝑖𝑛, the expected delay of a single antenna BC be-

comes 𝐷𝑚𝑖𝑛 = 𝑅𝑚𝑖𝑛𝑛 log(𝑛)

log(log(𝑛)) +𝜔(

𝑛log(log(𝑛))

). Accordingly,

the following theorem indicates the average number of servicesreceived by a randomly selected user after 𝑘 transmissions.

Theorem 2: For large (𝑘) number of channel uses, theaverage number of services received by a randomly selected

user behaves like 𝑘𝑚𝑛

(1 +𝑂(

√log(𝑘)

𝑘 ))

w.p.1.Proof: Recall that for each channel use, the probability

that a specific user is being serviced is 𝑚𝑛 . Assuming 𝑥𝑖 is

a binary random variable representing wether the 𝑖’th user isbeing serviced or not,

𝑥𝑖 =

{1 with probability 𝑚

𝑛0 with probability (1− 𝑚

𝑛 )

Clearly, after 𝑘 channel uses, the total number of gettingservice by the aforementioned user can be considered as arandom variable, 𝑋 =

∑𝑛𝑖=1 𝑥𝑖, with Binomial distribution as

B(𝑘, 𝑚𝑛 ), where B(𝑘, 𝑝) denotes a Binomial distribution with

parameters 𝑘 and 𝑝, respectively, as the number of trials andthe probability of success. Using gaussian approximation fora binomial distribution when 𝑘 is large enough, we have

𝑃𝑟{𝑘𝑚𝑛

(1− 𝛿) < S <𝑘𝑚

𝑛(1 + 𝛿)}

≈ 1− 2𝑄( 𝑘𝑚

𝑛 𝛿√𝑘𝑚𝑛 (1− 𝑚

𝑛 )

)(7)

It can be verified that by setting 𝛿=

√2( 𝑛

𝑚−1) log(𝑘)

𝑘 and noting

the approximation 𝑄(𝑥)= 1√2𝜋𝑥

𝑒−𝑥2

2 for 𝑥≫1, it follows,

𝑃𝑟{𝑘𝑚

𝑛

(1− O(

√log(𝑘)

𝑘))<S<

𝑘𝑚

𝑛

(1 + O(

√log(𝑘)

𝑘))}

∼ 1− o(1

𝑘) (8)

It means for sufficient large number of channel uses, theaverage number of services to a specific user, scales as S =𝑘𝑚𝑛

(1 +𝑂(

√log(𝑘)

𝑘 ))

w.p.1.

IV. SIMULATION RESULTS

This section aims at comparing the expected delay of thefollowing scheduling methods in a Rayleigh fading broadcastchannel: (i) the BUSS with 𝑚 = 1 and (ii) the FPAS with𝑚 = log(log(𝑛))

𝑅𝑚𝑖𝑛. Indeed, in the numerical experiment, a

random vector of the channel gains is generated (from thesize of 𝑛). Then, an algorithm finds the position of the 𝑚greatest elements of this vector, which are equivalent to the𝑚 best users among all of the system users. In fact, these 𝑚users are selected to give a service and other 𝑛−𝑚 users areinactive. The aforementioned algorithm is continued until allof the users are selected at least one time.

Fig. 1 compares the system delay of the aforementionedmethods, assuming 𝑅𝑚𝑖𝑛 is set to one. Accordingly, fordifferent number of users ranging from 500 to 1000, theexpected delay of these scheduling methods are simulatedand compared with the theoretical results, showing there isa close agreement between the computed delays derived bythe analytical results to that of the simulations. Referring toFig. 1, there is a major difference between the expected delayof these two methods. Moreover, the small gap between thetheoretical results to that of the simulations is due to thefact that merely the main terms of the analytical results areconsidered, i.e., 𝐷𝑚𝑖𝑛 = 𝑅𝑚𝑖𝑛

𝑛 log(𝑛)log(log(𝑛)) is considered for the

FPAS. As an illustration, considering a system with 600 singleantenna users, when the base station sends the informationbased on the BUSS, it needs around 4000 transmissions toclaim that all of the 600 users receive at least one service.However, it needs only 2000 transmissions for the same claim,when the FPAS is employed at the transmitter.

Also, Fig. 2 compares the average number of getting ser-vices by a randomly selected user for two aforementionedstrategies, where 𝑘 and 𝑅𝑚𝑖𝑛 are set to 𝐷𝐵𝑈𝑆𝑆 = 𝑛 log(𝑛)and 0.1, respectively. Regarding the value of 𝑅𝑚𝑖𝑛, onecan consider a system with channel bandwidth of 200Khz.Consequently, the transmission rate of 0.1 bits per sampleis equivalent to 0.1 × 2𝐵𝑊 = 40K bits per second. Noting𝑘=𝑛 log(𝑛), and referring to theorem 2, the average numberof services to a randomly selected user for the BUSS andthe FPAS scale as S 𝐵𝑈𝑆𝑆 = log(𝑛)

(1 + 𝑂(𝑛− 1

2 ))

andS 𝐹𝑃𝐴𝑆 = log(𝑛) log(log(𝑛))

𝑅𝑚𝑖𝑛

(1 +𝑂(𝑛− 1

2 )), respectively.

Fig. 2 indicates that there is a major gap between two powerallocation strategies when relying on the average number ofgetting services for a randomly selected user.

V. CONCLUSION

In this paper, the expected delay of a Fair Power AllocationStrategies (FPAS) has been studied both analytically andon the basis of simulation results. In analytical study, theresults showed that the system delay is decreased through adouble logarithmic scale when the transmitter uses the FPASinstead of the Best User Selection Strategy (BUSS). In the

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500 550 600 650 700 750 800 850 900 950 10001000

2000

3000

4000

5000

6000

7000

8000

Total number of users

Del

ay (

The

num

ber

of p

acke

ts)

BUSS, Numerical resultFPAS, Numerical resultFPAS, Analytical resultBUSS, Analytical result

Fig. 1. The expected delay of the BUSS as compared to that of the FPAS.

500 550 600 650 700 750 800 850 900 950 10000

20

40

60

80

100

120

140

160

Total number of users

The

ave

rage

num

ber

of s

ervi

ces

to a

spe

cific

use

r

FPAS, Numerical resultFPAS, Analytical resultBUSS, Numerical resultBUSS, Analytical result

Fig. 2. The average number of services for a randomly selected user for twodifferent power allocation strategies, and 𝑅𝑚𝑖𝑛 = 0.1.

simulation study, the results demonstrated that there is a closeagreement between the analytic and numerical results. In fact,the numerical experiment showed that the expected delay ofthe FPAS is substantially less than the delay of the BUSS.Moreover, illustrating the experiment results as well, whenthe FPAS is employed at the transmitter, the average numberof services to a randomly selected user is considerably greaterthan the case of using the BUSS.

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