Optimal Power Control and Scheduling for Real-Time and Non-Real-Time Data
aa r X i v : . [ c s . I T ] M a y Optimal Power Control and Scheduling forReal-Time and Non-Real-Time Data
Ahmed Ewaisha, Cihan Tepedelenlio ˘glu
School of Electrical, Computer, and Energy Engineering, Arizona State University, USA.Email: { ewaisha, cihan } @asu.edu Abstract —We consider a joint scheduling-and-power-allocation problem of a downlink cellular system. The systemconsists of two groups of users: real-time (RT) and non-real-time (NRT) users. Given an average power constraint on the basestation, the problem is to find an algorithm that satisfies the RThard deadline constraint and NRT queue stability constraint. Wepropose two sum-rate-maximizing algorithms that satisfy theseconstraints as well as achieving the system’s capacity region. Inboth algorithms, the power allocation policy has a closed-formexpression for the two groups of users. However, interestingly,the power policy of the RT users differ in structure from thatof the NRT users. The first algorithm is optimal for the on-offchannel model with a polynomial-time scheduling complexity inthe number of RT users. The second, on the other hand, works forany channel fading model which is shown, through simulations,to have an average complexity that is close-to-linear. We alsoshow the superiority of the proposed algorithms over existingapproaches using extensive simulations.
I. I
NTRODUCTION
Quality-of-service-based scheduling has received much at-tention recently. It is shown in [2] and [3] that quality-of-service-aware scheduling results in a better performance inLTE systems compared to best-effort techniques. Dependingon the application, quality-of-service (QoS) metrics capturelong-term throughput [4], short-term throughput [5], per-useraverage delay [6], average number of packets missing aspecific deadline [7], or the average time a user waits to receiveits data [8]. Real-time audio and video applications requirealgorithms that take hard deadlines into consideration. This isbecause if a real-time packet is not transmitted on time, thecorresponding user might experience intermittent connectivityof its audio or video.The problem of scheduling for wireless systems under hard-deadline constraints has been widely studied in the literature(see, e.g., [9] and [10] for a survey). In [7] the authorsconsider binary erasure channels and present a sufficient andnecessary condition to determine if a given problem is feasible.The work is extended in three different directions. The firstdirection studies the problem under delayed feedback [3]. Thesecond considers general channel fading models [11]. Thethird studies multicast video packets that have strict deadlinesand utilize network coding to improve the overall networkperformance [12], [13]. Unlike the time-framed assumption inthe previous works, the authors of [14] assume that arrivalsand deadlines do not have to occur at the edges of a time
The work in this paper has been supported by NSF Grant ECCS-1307982.Parts of this work have been accepted in IEEE WCNC 2017 conference[1]. frame. They present a scheduling algorithm under the on-offchannel fading model and present its achievable region undergeneral arrivals and deadline patterns but with a fixed powertransmission. In [15] the authors study the scheduling problemin the presence of real-time and non-real-time data. Unlikereal-time data, non-real-time data do not have strict deadlinesbut have an implicit stability constraint on the queues. Usingthe dual function approach, the problem was decomposedinto an online algorithm that guarantees network stability andsatisfies the real-time users’ constraint.Power allocation has not been considered for RT users inthe literature, to the best of our knowledge. In this paper,we study a throughput maximization problem in a downlinkcellular system serving RT and NRT users simultaneously.We formulate the problem as a joint scheduling-and-power-allocation problem to maximize the sum throughput of theNRT users subject to an average power constraint on thebase station (BS), as well as a delivery ratio requirementconstraint for each RT user. The delivery ratio constraintrequires a minimum ratio of packets to be transmitted by a harddeadline, for each RT user. Perhaps the closest to our work arereferences [15] and [16]. The former does not consider powerallocation, while the latter assumes that only one user can bescheduled per time slot. The contributions in this paper are asfollows: • We present two scheduling-and-power-allocation algo-rithms. The first is for the on-off channel fading modelwhile the second is for the continuous channel fadingmodel. • We show that both algorithms are optimal. That is,both satisfy the average power constraint, the deliveryratio requirement constraint, in addition to achieving thecapacity region. However, the complexity of the first ispolynomial in the number of users, while the second isshown to have an average complexity that is close-to-linear. • We present closed-form expressions for the power alloca-tion policy used by both algorithms. It is shown that thepower allocation expressions for the RT and NRT usershave a different structure. • Through simulations, we show the complexity andthroughput performances of the proposed algorithms overbaseline ones.The rest of this paper is organized as follows. In Section IIwe present the system model and the underlying assumptions.The problem is formulated in Section III. For the on-off channel model, the proposed power-allocation and schedulingalgorithm as well as its optimality is presented in SectionIV. In Section V we present the optimal algorithm for thecontinuous channel model as well as another optimal algorithmwith a lower complexity. The capacity region of the problemis presented in Section VI. Simulation results and comparisonswith baseline approaches is presented in Section VII. Finally,the paper is concluded in Section VIII.II. S
YSTEM M ODEL
We assume a time slotted downlink system with slot du-ration T seconds. The system has a single base station (BS)having access to a single frequency channel. The interferencecoming from all other neighboring BSs is assumed to betreated as noise. There are N users in the system indexedby the set N , { , · · · , N } . The set of users is dividedinto the RT users N R , { , · · · , N R } , and NRT users N NR , { N R + 1 , · · · , N } with N R and N NR , N − N R denoting the number of RT and NRT users, respectively.We model the channel between the BS and the i th useras a fading channel with power gain γ i ( k ) . The distributionand statistics of γ i ( k ) is arbitrary and need not be known tothe BS nor to any of the users. In this paper, we present theproblem for the on-off channel fading case in Sections III andIV and then we generalize this to the continuous fading casein Section V. The on-off model [7] corresponds to the well-known binary erasure channel model and models whether thechannel is in outage or not. While the continuous fading modelis more general and captures all independent and identicallydistributed channel distributions its solution, as will be seen,has a higher complexity.For the on-off channel model, if channel i is in a non-outagestate during the k th slot then γ i ( k ) = 1 , otherwise γ i ( k ) =0 . Channel gains are fixed over the whole slot and changeindependently in subsequent slots and are independent acrossusers. Hence, the channel gain follows a Bernoulli process.Channels with a more general fading model will be discussedin Section V. Moreover, γ i ( k ) is known to the BS, for all i ∈ N , at the beginning of the each slot. A. Packet Arrival Model
Let a i ( k ) ∈ { , } be the indicator of a packet arrival foruser i ∈ N at the beginning of the k th slot. { a i ( k ) } is assumedto be a Bernoulli process with rate λ i packets per slot andassumed to be independent across all users in the system.Packets arriving at the BS for the RT users are called real-time packets. RT packets have a strict transmission deadline.If an RT packet is not transmitted by this deadline, this packetis dropped out of the system and does not contribute towardsthe throughput of the user. However, RT user i is satisfied ifit receives, on average, more than q i % of its total number ofpackets. We refer to this constraint as the QoS constraint foruser i . Here we assume that real-time packets arriving at thebeginning of the k th slot have their deadline at the end of thisslot.On the other hand, packets arriving to the BS for the NRTusers can be transmitted at any point in time. Thus, packets for Fig. 1. In the k th time slot, the BS chooses N k users to be scheduled. Alltime slots have a fixed duration of T seconds. NRT user i are stored, at the BS, at user i ’s (infinite-sized [17])buffer and served on a first-come-first-serve basis. Since thearrival rate λ i , for NRT user i , might be higher than what thesystem can support, we define r i ( k ) as an admission controllerfor user i at slot k . At the beginning of slot k , the BS sets r i ( k ) to if the BS decides to admit user i ’s arrived packetto the buffer, and to otherwise. The time-average number ofpackets admitted to user i ’s buffer is A i , lim sup K →∞ K K X k =1 E [ r i ( k )] , i ∈ N NR . (1)And the queue associated with NRT user i is given by Q i ( k + 1) = ( Q i ( k ) + Lr i ( k ) − µ i ( k ) R i ( k )) + , i ∈ N NR , (2)where r i ( k ) is the admission control decision variable for NRTuser i at the beginning of slot k . We note that no admissioncontroller is defined for the RT users since their buffers cannotbuild up due to the presence of a deadline. B. Service Model
Following [11] we assume that more than one user can bescheduled in one time slot. However, due to the existence ofa single frequency channel in the system, the BS transmits tothe scheduled users sequentially as shown in Fig. 1. At thebeginning of the k th slot, the BS selects a set of RT usersdenoted by S R ( k ) ⊆ N R and a set of NRT users S NR ( k ) ⊆N NR to be scheduled during slot k . Thus a total of N k , |N k | users are scheduled at slot k where N k , S R ( k ) ∪ S NR ( k ) (Fig. 1). Moreover, the BS assigns an amount of power P i ( k ) for every user i ∈ N k . This dictates the transmission rate foreach user according to the channel capacity given by R i ( k ) = log (1 + P i ( k ) γ i ( k )) . (3)Finally, the BS determines the duration of time, out of the T seconds, that will be allocated for each scheduled user. Wedefine the variable µ i ( k ) to represent the duration of time,in seconds, assigned for user i ∈ N during the k th slot (Fig.1). Hence, µ i ( k ) ∈ [0 , T ] for all i ∈ N . The BS decides thevalue of µ i ( k ) for each user i ∈ N at the beginning of slot k . Since RT users have a strict deadline, then if an RT user isscheduled at slot k , then it should be allocated the channel fora duration of time that allows the transmission of the wholepacket. Thus we have µ i ( k ) = (cid:26) LR i ( k ) if i ∈ S R ( k )0 if i ∈ N R \S R ( k ) , (4) where L is the number of bits per packet, that is assumedto be fixed for all packets in the system. The extension tomultiple packet types of different lengths will be addressedin Section V-C. Equation (4) means that, depending on thetransmission power, if RT user i is scheduled at slot k , then it isassigned as much time as required to transmit its L bits. Hence,unlike the NRT users that have µ i ( k ) ∈ [0 , T ] , µ i ( k ) is furtherrestricted to the set { , L/R i ( k ) } for the RT users. For easeof presentation, we denote Q ( k ) , [ Q ( k ) , · · · , Q N NR ( k )] T .In the next section we present the problem formally.III. P ROBLEM F ORMULATION FOR O N -O FF C HANNELS
We are interested in finding the scheduling and powerallocation algorithm that maximizes the sum-rate of all NRTusers subject to the system constraints. In this paper we restrictour search to slot-based algorithms which, by definition, takethe decisions only at the beginning of the time slots.Now define the average rate of user i ∈ N NR to be R i , lim inf K →∞ P Kk =1 µ i ( k ) R i ( k ) / ( LT K ) packets perslot. Thus the problem is to find the scheduling, powerallocation and packet admission decisions at the beginning ofeach slot, that solve the following problemmaximize { µ ( k ) , P ( k ) , r ( k ) } ∞ k =1 X i ∈N NR R i , (5)subject to r i ( k ) ≤ a i ( k ) , ∀ i ∈ N NR , (6) lim sup k →∞ E [ Q i ( k )] < ∞ ∀ i ∈ N NR , (7) R i ≥ λ i q i , ∀ i ∈ N R , (8) lim sup K →∞ KT K X k =1 X i ∈N P i ( k ) µ i ( k ) ≤ P avg , (9) ≤ P i ( k ) ≤ P max , ∀ i ∈ N , (10) X i ∈N µ i ( k ) = T ∀ k ≥ , (11) ≤ µ i ( k ) ≤ T ∀ i ∈ N , (12)where µ ( k ) , [ µ i ( k )] i ∈N , P ( k ) , [ P i ( k )] i ∈N while r ( k ) , [ r i ( k )] i ∈N NR . Constraint (6) says that no packets should beadmitted to the i th buffer if no packets arrived for user i .Constraint (7) means that the queues of the NRT users have tobe stable. Constraint (8) indicates that the resources allocatedto a RT user i need to be such that the fraction of packetstransmitted by the deadline are greater than the required QoS q i . Constraint (9) is an average power constraint on the BStransmission power. Finally constraint (11) guarantees that thesum of durations of transmission of all scheduled users doesnot exceed the slot duration T . In this paper, we assume thatthe NRT user with the longest queue has enough packets,at each slot, to fit the whole slot duration which is a validassumption in the heavy traffic regime. It will be clear thatthe generalization to the non-heavy traffic regime is possibleby allowing multiple NRT users to be scheduled but this isomitted for brevity. IV. P ROPOSED A LGORITHM FOR O N -O FF C HANNELS
We use the Lyapunov optimization technique [18] to findand optimal algorithm that solves (5). We do this on four steps:i) We define, in Section (IV-A) a “virtual queue” associatedwith each average constraint in problem (5). This helps indecoupling the problem across time slots. ii) In Section IV-B,we define a Lyapunov function, its drift and a, per-slot, rewardfunction. The latter is proportional to the objective of (5). iii)Based on the virtual queues and the Lyapunov function, weform an optimization problem, for each slot k , that minimizesthe drift-minus-reward expression the solution of which isthe proposed power allocation and scheduling algorithm. InSection IV-C, we propose an efficient way to solve thisproblem optimally. iv) Finally, we show that this minimizationguarantees reaching an optimal solution for (5), in SectionIV-D. A. Problem Decoupling Across Time Slots
We define a virtual queue associated with each RT user asfollows Y i ( k + 1) = ( Y i ( k ) + a i ( k ) q i − i ( k )) + , i ∈ N R , (13)where i ( k ) , ( µ i ( k )) with ( · ) = 1 if its argument isnon-zero and ( · ) = 0 otherwise. For notational conveniencewe denote Y ( k ) , [ Y ( k ) , · · · , Y N R ( k )] T . Y i ( k ) is a measureof how much constraint (8) is violated for user i . We willlater show a sufficient condition on Y i ( k ) for constraint (8)to be satisfied. Hence, we say that the virtual queue Y i ( k ) isassociated with constraint (8). Similarly, we define the virtualqueue X ( k ) , associated with constraint (9), as X ( k + 1) = (cid:18) X ( k ) + P i ∈N P i ( k ) µ i ( k ) T − P avg (cid:19) + . (14)To provide a sufficient condition on the virtual queues tosatisfy the corresponding constraints, we use the followingdefinition of mean rate stability of queues [18, Definition 1]to state the lemma that follows. Definition 1.
A random sequence { Y i ( k ) } ∞ k =0 is said to bemean rate stable if and only if lim sup K →∞ E [ Y i ( K )] /K = 0 holds. Lemma 1.
If, for some i ∈ N NR , { Y i ( k ) } ∞ k =0 is mean ratestable, then constraint (8) is satisfied for user i .Proof. Proof follows along the lines of Lemma 3 in [18].Lemma 1 shows that when the virtual queue Y i ( k ) is meanrate stable, then constraint (8) is satisfied for user i ∈ N NR .Similarly, if { X ( k ) } ∞ k =0 is mean rate stable, then constraint(9) is satisfied. Thus, our objective would be to devise analgorithm that guarantees the mean rate stability of Y i ( k ) forall RT users as well as the mean rate stability for X ( k ) . B. Applying the Lyapunov Optimization
The quadratic Lyapunov function is defined as L yap ( U ( k )) , X i ∈N R Y i ( k ) + 12 X i ∈N NR Q i ( k ) + 12 X ( k ) , (15) where U ( k ) , ( Y ( k ) , Q ( k ) , X ( k )) , and the Lyapunov driftas ∆( k ) , E U ( k ) [ L k +1 ( U ( k + 1)) − L yap ( U ( k ))] where E U ( k ) [ x ] , E [ x | U ( k )] is the conditional expectation of therandom variable x given U ( k ) . Squaring (2), (13) and (14)taking the conditional expectation then summing over i , thedrift becomes bounded by ∆( k ) ≤ C + Ψ( k ) , (16)where C , P i ∈N R (cid:0) q i + 1 (cid:1) + P + P + N NR (cid:2) L + T R (cid:3) (17)and we use R max , log (1 + P max ) , while Ψ( k ) , X i ∈N R E U ( k ) [ Y i ( k ) ( λ i q i − i ( k ))]+ X ( k ) X i ∈N E U ( k ) [ µ i ( k ) P i ( k )] T − P avg ! + X i ∈N NR Q i ( k ) (cid:0) E U ( k ) [ Lr i ( k ) − µ i ( k ) R i ( k )] (cid:1) . (18)We define B max as an arbitrarily chosen positive controlparameter that controls the performance of the algorithm.We shall discuss the tradeoff on choosing B max later on.Since E U ( k ) [ Lr i ( k )] represents the average number of bitsadmitted to NRT user i ’s buffer at slot k , we refer to B max P i ∈N NR E U ( k ) [ Lr i ( k )] as the “reward term”. We sub-tract this term from both sides of (16), then use (18) andrearrange to bound the drift-minus-reward term as ∆( k ) − B max X i ∈N NR E U ( k ) [ Lr i ( k )] ≤ C − X ( k ) P avg + E U ( k ) " X i ∈N R Ψ R ( i, k ) + E U ( k ) " X i ∈N NR Ψ NR ( i, k ) µ i ( k ) + E U ( k ) " X i ∈N NR ( Q i ( k ) − B max ) Lr i ( k ) + X i ∈N R Y i ( k ) λ i q i , (19)where Ψ R ( i, k ) and Ψ NR ( i, k ) are given by Ψ R ( i, k ) , (cid:18) Y i ( k ) − LT R i ( k ) X ( k ) P i ( k ) (cid:19) i ( k ) , i ∈ N R , (20) Ψ NR ( i, k ) , Q i ( k ) R i ( k ) − X ( k ) P i ( k ) T , i ∈ N NR , (21)respectively, where we used (4) in (20). The proposed algo-rithm schedules the users, allocates their powers and controlsthe packet admission to minimize the right-hand-side of (19)at each slot. Since the only term in right-hand-side of (19)that is a function in r i ( k ) ∀ i ∈ N NR is the fourth term,we can decouple the admission control problem from thejoint scheduling-and-power-allocation problem. Minimizingthis term results in the following admission controller: set r i ( k ) = a i ( k ) if Q i ( k ) < B max and otherwise. Minimizing the remaining terms yieldsmaximize P ( k ) , µ ( k ) P i ∈S R ( k ) Ψ R ( i, k ) + P i ∈N NR Ψ NR ( i, k ) µ i ( k ) subject to (10) , (11) and (12) . (22)This is a per-slot optimization problem the solution of whichis an algorithm that minimizes the upper bound on the drift-minus-reward term defined in (19). Next we show how to solvethis problem in an efficient way. C. Efficient Solution for the Per-Slot Problem
We first solve for the NRT variables then use its result tosolve for the RT variables.
1) NRT variables:
To solve this problem optimally, we firstfind the optimal power-allocation-and-scheduling policy forthe NRT users through the following lemma.
Lemma 2.
If an NRT user i is scheduled to transmit any of itsNRT data during the k th slot, then the optimum power levelfor this NRT with respect to (w.r.t.) problem (22) is given by P i ( k ) = min (cid:18) T Q i ( k ) X ( k ) − (cid:19) + , P max ! . (23) Moreover, in the heavy traffic regime, the optimum NRTuser to be scheduled, if any, w.r.t. problem (22) is i ∗ NR , arg max i ∈N NR Ψ ∗ NR ( i, k ) , where Ψ ∗ NR ( i, k ) comes by substi-tuting (23) in (21) .Proof. We observe that, for any i ∈ N NR , the only term in(22) that is a function in P i ( k ) is Ψ NR ( i, k ) . Differentiating(21) w.r.t. P i ( k ) for all i ∈ N NR , equating the results to0 and noting the minimum and maximum power constraints(10), we get the water-filling power allocation formula (23).This completes the first part of the lemma.To prove the second part, we substitute by (23) in (21) to get Ψ ∗ NR ( i, k ) . We continue the proof by contradiction. Supposethat the optimal scheduled NRT set is given by S ∗ NR ( k ) = { i ∗ NR , j } where j = i ∗ NR and Ψ ∗ NR ( j, k ) < Ψ ∗ NR ( i ∗ NR , k ) .Thus, there exists some values α > and β > suchthat the corresponding scheduler would be µ i ∗ NR ( k ) = α and µ j ( k ) = β , while µ l k ( k ) = 0 for all l k / ∈ { i ∗ NR , j } . Inother words, α seconds are assigned to i ∗ NR and β secondsassigned to j . However, if user i ∗ NR has enough backloggeddata, which happens in the heavy traffic regime, then we canincrease its assigned duration to µ i ∗ NR = α + β and thusset µ j ( k ) = 0 , to get an increase in the objective of (22)by β (Ψ ∗ NR ( i ∗ NR , k ) − Ψ ∗ NR ( j, k )) > which contradicts withthe optimality of S ∗ NR ( k ) and completes the proof of thelemma.Lemma 2 provides the optimal scheduling policy for theNRT users, at the k th slot, as well as the optimal powerallocation w.r.t. problem (22). The lemma shows that if any ofthe NRT users is going to be scheduled in the k th slot, thenonly one of them is going to be scheduled. This means thatthe scheduling policy for the NRT users is µ i ( k ) = (cid:26) T − P i ∈S ∗ R ( k ) µ i ( k ) i = i ∗ NR N NR \{ i ∗ NR } (24) which is a manipulation of (11). Substituting (24) and Ψ NR ( i, k ) in (22), the latter becomesmaximize µ i ∗ NR ( k ) , [ µi ( k ) ,Pi ( k ) ] i ∈N NR X i ∈S R ( k ) Ψ R ( i, k ) + Ψ ∗ NR ( i ∗ NR , k ) µ i ∗ NR ( k ) (25)subject to (12) , (10) and µ i ∗ NR ( k ) = T − X i ∈S R ( k ) LR i ( k ) , which is simpler than (22) since it is not a function in theNRT variables except µ i ∗ NR ( k ) . Finding the optimal value of µ i ∗ NR ( k ) solves the NRT scheduling problem. We will firstsolve for µ i ( k ) for all RT users then use (24) to find µ i ∗ NR ( k ) .
2) RT Variables:
To find the scheduler of the RT users thatis optimal w.r.t. problem (25), we first solve for [ P i ( k )] i ∈N R given a fixed set S R ( k ) , then we discuss the scheduling policythat solves for this set. To solve for [ P i ( k )] i ∈N R , we presentthe following definition then present a theorem that discussesthe optimum power allocation policy for the RT users. Definition 2.
We define the Lambert power allocation policyfor the RT users as P i ( k ) = min T Ψ ∗ NR ( i ∗ NR ,k ) X ( k ) − W (cid:16)h Ψ ∗ NR ( i ∗ NR ,k ) TX ( k ) − i e − (cid:17) − , P max , (26) i ∈ S R ( k ) , where W ( z ) is the principle branch of theLambert W function [19] while Ψ ∗ NR ( i, k ) is defined in Lemma2. Theorem 1.
Given any set S R ( k ) , if the Lambert power policyresults in P i ∈S R ( k ) L/ log(1 + P i ( k )) ≤ T , then it is theoptimum RT-users’ power allocation policy given that S R ( k ) is the scheduling set at slot k . Otherwise, the optimum powerallocation policy is given by P i ( k ) = exp P i ∈S R ( k ) LT − , i ∈ S R ( k ) . (27) Proof.
We prove this theorem by applying the Lagrangeoptimization [20, Ch. 5] technique to problem (25) then usethe complementary slackness condition.Since µ i ( k ) ≥ for all i ∈ N R (see (4)), then we havethe constraint µ i ∗ NR ( k ) ≤ T always holds from (24). Thus wedefine the Lagrange multiplier φ to be the multiplier associatedwith the constraint µ i ∗ NR ( k ) ≥ . The Lagrangian becomes L agr , X i ∈S R ( k ) Ψ R ( i, k ) + (Ψ ∗ NR ( i ∗ NR , k ) + φ ) × T − X i ∈S R ( k ) L log (1 + P i ( k ) γ i ( k )) (28)Differentiating (28) with respect to P i ( k ) and equating to gives log (1 + P i ( k ) γ i ( k )) X ( k ) LT − ( X ( k ) P i ( k ) /T + Ψ ∗ NR ( i ∗ NR , k ) + φ ) γ i ( k )1 + P i ( k ) γ i ( k ) = 0 . (29)After some manipulations and denoting ˜ φ , (Ψ ∗ NR ( i ∗ NR , k ) + φ ) T /X ( k ) (30)we get log (1 + P i ( k ) γ i ( k )) = 1 + ˜ φγ i ( k ) −
11 + P i ( k ) γ i ( k ) , P . (31)Thus we get ˜ P e ˜ P = (cid:16) ˜ φγ i ( k ) − (cid:17) e − which has twosolutions in ˜ P (see [19]), one of them yields a negative valuefor P i ( k ) . Hence, with the help of W ( · ) , which is the inversefunction of xe x , we can write a unique solution for (29) as P i ( k ) = 1 γ i ( k ) ˜ φγ i ( k ) − W (cid:16)h ˜ φγ i ( k ) − i e − (cid:17) − , i ∈ S R ( k ) . (32)To calculate (32), we need to find the value of φ satisfyingthe complementary slackness condition φµ i ∗ NR ( k ) = 0 . Hencewe have one of the two following possibilities might yield theoptimal solution: 1) setting φ = 0 and thus µ i ∗ NR ( k ) ≥ , or 2)setting µ i ∗ NR ( k ) = 0 and thus φ ≥ . If setting φ = 0 yields P i ∈S R ( k ) L/ log(1 + P i ( k )) ≤ T then the Lambert powerallocation policy in (26) is optimum since there exists noother non-negative value for φ that yields P i ∈S R ( k ) L/ log(1+ P i ( k )) = T while satisfying µ i ∗ NR ( k ) = 0 (to satisfy thecomplementary slackness). On the other hand, if setting φ = 0 yields P i ∈S R ( k ) L/ log(1 + P i ( k )) > T , then φ cannot be . Thus we have µ i ∗ NR ( k ) = 0 , which means that the timeslot will be allocated for RT users only. The correspondingvalue of φ should satisfy P i ∈S R ( k ) L/ log(1 + P i ( k )) = T .From (32), we observe that P i ( k ) = P j ( k ) for all i, j ∈S R ( k ) because γ i ( k ) = 1 for all i ∈ S R ( k ) . Thus wehave L |S R ( k ) | / log(1 + P i ( k )) = T . This yields the powerallocation policy (27) and completes the proof.Theorem 1 gives closed-form expressions for the powerfunction of the RT users given any scheduling set S R ( k ) . Tofind the optimum scheduling set S R ( k ) that solves problem(25), we present the following definition then mention atheorem that decreases the complexity of this search. Definition 3.
At slot k , the set S R ( k ) is said to be a“candidate” set if and only if Y i ( k ) ≥ Y j ( k ) for all i ∈ S R ( k ) and all j / ∈ S R ( k ) . Otherwise it is called a “non-candidate”set. We note that the definition of candidate sets assumes that allRT users have γ i ( k ) = 1 . If this assumption does not hold atsome time slot k , then we eliminate the users with γ i ( k ) = 0 from the system for this time slot and consider only those with γ i ( k ) = 1 . Theorem 2.
The optimal RT set that solves (25) is one of thecandidate sets.Proof.
We prove this theorem by contradiction. Suppose that S ∗ R ( k ) is the optimal set and that it is not a candidate set.That is, ∃ i ∈ S R ( k ) and j / ∈ S R ( k ) such that Y i ( k ) < Y j ( k ) .It is easy to show that the Lambert power policy results inthe fact that P i ( k ) depends on |S R ( k ) | and not on S R ( k ) for any i ∈ S R ( k ) and any S R ( k ) . Thus, replacing user i with user j results in having P j ( k ) = P i ( k ) which meansthat X ( k ) P j ( k ) µ j ( k ) = X ( k ) P i ( k ) µ i ( k ) holds. But since Y i ( k ) < Y j ( k ) , swapping the two users increases the objectivefunction of (25) and results in a candidate set. This contradictswith the fact that S R ( k ) is optimal while being non-candidate.Theorem 2 says that there will be no scheduled RT usershaving a value of Y j ( k ) smaller than any of the unscheduledRT users. This theorem suggests an algorithm to reduce thecomplexity of scheduling the RT users from O (cid:0) N R (cid:1) to O ( N R ) . This algorithm is to list the RT users in a descendingorder of their Y i ( k ) . Without loss of generality, in the remain-ing of this paper, we will assume that Y > Y · · · > Y N R .We now propose Algorithm 1 which is the scheduling andpower allocation algorithm for problem (5). Algorithm 1 isexecuted at the beginning of the k th slot and, without lossof generality, it assumes: 1) all RT users in the system havereceived a packet at the beginning of the k th slot, 2) all usersin the system have an “on” channel. If, at some slot, any ofthese assumptions does not hold for some users, these usersare eliminated from the system for this slot. That is, they willnot be scheduled. In addition, we assume heavy traffic regime,thus the NRT user with the longest queue has enough data tofill the entire time slot. We define the set S RT to be the set ofall candidate sets. D. Optimality of Proposed Algorithm
We first define R (opt) i to be the throughput of NRT user i under the optimal algorithm that solves (5). We define thisalgorithm to be the one that sets, at each time slot k , thevariables P i ( k ) , µ i ( k ) , i ( k ) and R i ( k ) to the values ˜ P i ( k ) , ˜ µ i ( k ) , ˜ i ( k ) and ˜ R i ( k ) , respectively, where the latter valuessatisfy lim sup K →∞ K K − X k =0 E (cid:2) ˜ i ( k ) (cid:3) ≥ λ i q i , ∀ i ∈ N R , (33) lim sup K →∞ K K − X k =0 X i ∈N E " ˜ µ i ( k ) ˜ P i ( k ) T ≤ P avg , (34) lim sup K →∞ K K − X k =0 E " ˜ µ i ( k ) ˜ R i ( k ) L = R (opt) i , ∀ i ∈ N NR , (35)where R (opt) i is the optimal rate for user i ∈ N NR withrespect to solving (5). The following theorem gives a boundon the performance of Algorithm 1 compared to the optimalalgorithm that has a genie-aided knowledge of R (opt) i which, Algorithm 1
Scheduling and Power Allocation Algorithm Define the auxiliary functions Ψ X ( · ) : S RT → R + and P X ( · , · ) : S RT × N R → R + . Initialize P X ( S , i ) = 0 for all S ∈ S RT and all i ∈ N R . Sort the RT users in a descending order of Y i ( k ) . Withoutloss of generality, assume that Y > Y · · · > Y N R . Find the user i ∗ NR with longest queue Q i ( k ) and set S R ( k ) to be an empty set. while i ≤ N R do S R ( k ) = S R ( k ) ∪ { i } and set the power according to(26) ∀ i ∈ S R ( k ) . Calculate µ i ( k ) and µ i ∗ NR ( k ) according to (4) and (24),respectively. if µ i ∗ NR ( k ) < then Set µ i ( k ) = 0 for all i ∈ N NR and set the powerallocation for all i ∈ S R ( k ) according to (27) andrecalculate µ i ( k ) according to (4). end if Set Ψ X ( S R ( k )) = P i ∈S R ( k ) ( Y i ( k ) − X i ( k ) µ i ( k )) +Ψ ∗ NR ( i ∗ NR , k ) µ i ∗ NR ( k ) . Set P X ( S R ( k ) , i ) = P i ( k ) , ∀ i ∈ S R ( k ) . i ← i + 1 . end while Set the optimum scheduling set S ∗ R ( k ) =arg max S R ( k ) Ψ X ( S R ( k )) . Set P ∗ i ( k ) = P X ( S ∗ R ( k ) , i ) for all i ∈ N R , and set theNRT scheduler according to (24). For each i ∈ N NR , set r i ( k ) = a i ( k ) if Q i ( k ) < B max and otherwise. Update equations (2), (13) and (14) at the end of the k thslot.we show that, due to this knowledge it can solve the problemoptimally. Theorem 3.
For the on-off channel model, if problem (5) is feasible, then for any B max > Algorithm 1 results insatisfying all constraints in (5) and achieves an average ratesatisfying X i ∈N NR R i ≥ X i ∈N NR R (opt) i − C LB max . (36) Proof.
See Appendix ATheorem 3 says that Algorithm 1 yields an objective func-tion (5) that is arbitrary close to the performance of the optimalalgorithm that solves (5).V. E
XTENSIONS TO C ONTINUOUS F ADING C HANNELS
In the case of continuous fading, i.e. γ i ( k ) ∈ [0 , γ max ] where γ max < ∞ is the maximum channel gain that γ i ( k ) cantake, we expect the power allocation to depend on the channelgain. An algorithm that solves this case is a generalization ofAlgorithm 1 that assumes γ i ( k ) ∈ { , } . However, as will bedemonstrated later, the scheduling algorithm of the RT usershas a higher complexity order than the special case of on-offchannel gains. We adopt the same model as in Section II except that weallow γ i ( k ) to take any value in the interval [0 , γ max ] , for all i ∈ N . The transmission rate for this case is still given by (3),and the optimization problem is the same as (5) with a newassumption for γ i ( k ) . A. Derivation of the Algorithm
Algorithm 2 is based on the same Lyapunov optimizationprocedure as explained in Section IV. Following this pro-cedure, we reach optimization problem (25) with the newdefinition of γ i ( k ) . We now present the solution for the NRTusers followed by that of the RT users. Lemma 3.
If user i ∈ N NR is scheduled to transmit any of itsNRT data during the k th slot, then the optimum power levelfor this NRT w.r.t. problem (25) in the continuous fading caseis given by P i ( k ) = min (cid:18) Q i ( k ) X ( k ) − γ i ( k ) (cid:19) + , P max ! . (37) Moreover, in the heavy traffic regime, the scheduled NRT user,if any, that optimally solves problem (5) is given by i ∗ NR = arg max i ∈N NR Ψ ∗ NR ( i, k ) , (38) with ties broken randomly uniformly, while Ψ ∗ NR ( i, k ) comesby substituting (37) in (21) .Proof. The proof is similar to that of Lemma 2 and is omittedfor brevity.Lemma 3 presents the optimal power and scheduling policyfor the NRT users. To solve for the RT users, we assumea fixed subset S R ( k ) ⊆ N R of RT users to be scheduledduring the k th slot and find the power allocation of theseusers. Consequently, the optimum set S ∗ R ( k ) is the one thatmaximizes (25). In Section V-B, we present an algorithm thatfinds this optimum set as well as discussing the complexity ofthis algorithm.Assuming that the users in the set S R ( k ) are scheduled atthe k th slot, the problem is to find the transmission powerlevels for all the users in this set. We answer this question inthe following theorem. Theorem 4.
In the continuous-fading channel model, givensome non-empty set S R ( k ) , the power allocation policy P i ( k ) = min γ i ( k ) ˜ φγ i ( k ) − W (cid:16)h ˜ φγ i ( k ) − i e − (cid:17) − , P max , (39) i ∈ S R ( k ) , with ˜ φ , (Ψ ∗ NR ( i ∗ NR , k ) + φ ) T /X ( k ) and Ψ ∗ NR ( i, k ) defined in Lemma 3, is optimal w.r.t. (25) when φ is set to a non-negative value that satisfies (11) .Proof Sketch: The proof is similar to that of Theorem 1.The only difference is that we have to obtain the optimumvalue of φ satisfying (11). We note that instead of finding φ > using a 1-dimensional grid search, we can use thebisection method [21, Ch.9] which requires the monotonicity γ i ( k ) P o w e r P i ( k ) Water−FillingLambert Power Policy
Fig. 2. The Lambert power policy decreases with the channel gain, while thewater-filling policy increases with the gain. of the left-hand-side of (11), a fact that can be shown easilyby showing that the derivative, of this left-hand-side, withrespect to φ is always negative. Moreover, since the bisectionalgorithm needs a bracketing interval, it can be easily shownthat the optimum φ satisfies φ ≤ φ max , − Ψ ∗ NR ( i ∗ NR , k ) +exp( L |S R ( k ) | /T ) L |S R ( k ) | X ( k ) P max / (exp( L |S R ( k ) | /T ) − .It is clear that the Lambert power policy in (39) has adifferent structure than the water-filling policy in (37). Thereason is because the former is for the RT users while thelater is for the NRT users. We plot the two policies in Fig.2 with L = 1 , T = 1 , P max = 20 while Q i ( k ) /X ( k ) = 15 .The Lambert policy is plotted assuming a single RT user isscheduled at slot k while the water-filling policy is plottedassuming a single NRT user is scheduled at slot k . We notethat when a RT user i is the only scheduled user, (39) isequivalent to P i ( k ) = min (cid:18) e L/T − γ i ( k ) , P max (cid:19) , (40)We contrast the fact that, while the water-filling is an increas-ing function in the channel gain, the Lambert is a decreasingfunction in the channel gain. This is because the RT user has asingle packet of a fixed length to be transmitted. If the channelgain increases, then the power decreases to keep the sametransmission rate resulting in the same transmission durationof one slot. This result holds when multiple RT users arescheduled as well as demonstrated in the following theorem. Theorem 5.
Let S R ( k ) be some scheduling RT set at slot k .The power P i ( k ) given by (39) is monotonically decreasingin γ i ( k ) ∀ i ∈ S R ( k ) .Proof Sketch: Proof follows by differentiating (39) withrespect to γ i ( k ) for some user i , while having φ satisfying(11), and showing that the resulting derivative is always non-positive for γ i ( k ) ≥ . The optimum scheduling algorithm for the RT users is tofind, among all subsets of the set N R , the set that gives thehighest objective function of (25). B. Proposed Algorithm and Proof of Optimality
The exhaustive approach to the scheduling problem is toevaluate the objective function of (25) for all N R possible setsand choose the set that gives the highest objective function.This may be not practical when the number of RT users islarge. Observing the approach in the special on-off case andinspired by Theorem 2 that reduces the search space, weprovide here a similar approach. We first provide the followingdefinition which is analogous to Theorem 2. Theorem 6.
At slot k , for any set S R ( k ) , if there exists some i / ∈ S R ( k ) and some j ∈ S R ( k ) such that Y i ( k ) > Y j ( k ) and γ i ( k ) > γ j ( k ) , then S R ( k ) cannot be an optimal RT set, withrespect to problem (25) , for the continuous channel model.Proof Sketch: The proof is carried out by contradiction. Wecan show that if Y i ( k ) > Y j ( k ) and γ i ( k ) > γ j ( k ) for some i / ∈ S R ( k ) and some j ∈ S R ( k ) , then we could form anotherset S ′ ( k ) by swapping users i and j and thus increase theobjective function of (25).This theorem provides a sufficient condition for non-optimality. In other words, we can make use of this theoremto restrict our search algorithm to the sets that do not satisfythis property. Before presenting the proposed algorithm, wedefine the set S RT as the set of all possible subsets of the set N R . Theorem 7.
For the continuous channel model, if problem 5is feasible, then for any B max > and any ǫ ∈ (0 , thereexists some finite constant C such that Algorithm 2 satisfiesall constraints in (5) and achieves an average sum throughputsatisfying X i ∈N NR R i ≥ X i ∈N NR R ∗ i − C LB max , (41) where R ∗ i is the optimal rate for user i w.r.t. (5) .Proof. The proof is similar to that of Theorem 3 and C isdefined as C but with R max , log (1 + P max γ max ) . We omitthe proof for brevity.Due to the problem being a combinatorial problem with ahuge amount of possibilities, we could not reach a closed-form expression for the complexity order of this algorithm.However, simulations will show its complexity improvementover the exhaustive search algorithm. C. Extensions to Packets with Different Lengths
Let L i ( k ) be the length of user i ’s packet at slot k , i ∈ N . Itcan be easily shown that the power-allocation-and-schedulingfor the NRT users and the power allocation for the RT users,namely Lemma 3 and Theorem 4 do not change. That is,replacing L with L i ( k ) in (27) yields an optimal solutionas well. However, there are two possible extensions for thescheduling algorithm for the RT users, namely Algorithm 2.We discuss them next. Algorithm 2
Lambert-Strict Algorithm Define the auxiliary functions Ψ X ( · ) : S RT → R + and P X ( · , · ) : S RT × N R → R + . Initialize P X ( S , i ) = 0 for all S ∈ S RT and all i ∈ N R . Find the user i ∗ NR given in (38) and calculate its powergiven by (37). for S ∈ S RT do if ∃ some i / ∈ S and some j ∈ S such that Y i ( k ) >Y j ( k ) and γ i ( k ) > γ j ( k ) then Set Ψ X ( S ) = −∞ . Skip this iteration and go to step 4 to continue withthe next set in S RT . end if φ ← φ max + ∆ φ while φµ i ( k ) = 0 do φ ← φ − ∆ φ Calculate P i ( k ) given by (39) for all i ∈ S and set µ i ∗ NR ( k ) = T − P i ∈S µ i ( k ) . end while Set Ψ X ( S ) = P i ∈S ( Y i ( k ) − X i ( k ) µ i ( k )) +Ψ ∗ NR ( i ∗ NR , k ) µ i ∗ NR ( k ) and P X ( S , i ) = P i ( k ) , i ∈S . Set Ψ X ( S ) = P i ∈S ( Y i ( k ) − X i ( k ) µ i ( k )) +Ψ ∗ NR ( i ∗ NR , k ) µ i ∗ NR ( k ) . Set P X ( S , i ) = P i ( k ) , ∀ i ∈ S . i ← i + 1 . end for Set the optimum scheduling set S ∗ R ( k ) =arg max S Ψ X ( S ) . Set P ∗ i ( k ) = P X ( S ∗ R ( k ) , i ) for all i ∈ N R , and set theNRT scheduler according to (24). For each i ∈ N NR , set r i ( k ) = a i ( k ) if Q i ( k ) < B max and otherwise. Update equations (2), (13) and (14) at the end of the k thslot.
1) Homogeneous RT Users:
This is where all packets ofall RT users have the same lengths at slot k but they change(randomly and independently) from a slot to the other. Thatis, L i ( k ) = L j ( k ) for all i and j , but L i ( k ) and L i ( k ) need not be the same for k = k . This could be the caseif all RT users are streaming the same information from thesame server, or if their packet lengths change from a slot tothe other but are highly correlated across users in the sensethat L i ( k ) = L j ( k ) is a valid approximation. In this case,Algorithm 2 is still optimal since it solves problem (25) whichis a per-slot optimization problem, namely, it is not affectedwith the packet lengths at preceding and succeeding time slots.
2) Heterogeneous RT Users:
This is where the packetlength changes significantly from a user to the other in additionto its change (randomly and independently) from a slot tothe other. In this case, the scheduling algorithm of the RTusers proposed in Algorithm 2 is suboptimal. In order for thealgorithm to be optimal, Steps 5 through 8 of the algorithmneed to be removed. That is, the algorithm goes over all subsetsof the set N R . The complexity of the optimal algorithm isexponential in the number of RT users. However, suboptimal algorithms could still be developed. One example is to modifyStep 3 of Algorithm 1 by sorting the users according toa decreasing order of Y i ( k ) γ i ( k ) /L i ( k ) . Consequently, thisyields an algorithm of a linear complexity in N R . The sortingaccording to Y i ( k ) γ i ( k ) /L i ( k ) stems from the fact that RTusers with higher Y i ( k ) and γ i ( k ) and lower L i ( k ) should bemore favored to be scheduled.VI. C APACITY R EGION
In Section V, Algorithm 2 is shown to maximize theNRT sum-throughput subject to the system constraints. Inthis section we want to study the stability of the system.Specifically, we are interested to answer the following twoquestions:1) What is the capacity region of the system under thecontinuous fading model?2) What scheduling and power-allocation algorithms canachieve this capacity region?Studying the system’s capacity region means that we needto find all arrival rate vectors λ NR under which the NRTusers’ queues are stable (i.e. have a stationary distribution).This needs to be studied assuming that all arriving packets areadmitted to their respective buffers. Hence we first eliminatethe admission controller r ( k ) by replacing the queue equation(2) with Q i ( k + 1) = ( Q i ( k ) + La i ( k ) − µ i ( k ) R i ( k )) + . (42)More formally, the first question now becomes: what is theclosure of all admissible arrival rate vectors? An admissiblearrival rate vector is defined next. Definition 4.
An arrival rate vector λ NR , [ λ i ] i ∈N NR issaid to be admissible if there exists a power-allocation andscheduling algorithm under which constraints (7) and (8) aresatisfied given the power and scheduling constraints (9) - (12) . For simplicity we henceforth assume that the channelgain γ i ( k ) ∈ M where M is a discrete finite set, theelements of which are in the range [0 , γ max ] . With a slightabuse in notation, we define γ i ( m ) , γ i ( k ) to be thegain of user i when the channel is in fading state m , [ γ ( m ) , · · · , γ N ( m )] T ∈ M N during slot k . We also define µ i ( m , k ) and P i ( m , k ) to be, respectively, the duration andpower allocated to user i ∈ N when the channel is in fadingstate m , [ γ ( m ) , · · · , γ N ( m )] T ∈ M N during slot k , and π m to be the probability of occurrence of fading state m . Wenow mention the following definition then state Theorem 8that answers the first question. Definition 5.
An arrival rate vector λ NR is said to belongto the “Lambert Region” R Lamb if and only if there existsa sequence of time duration vectors { µ ( k ) } and a powerallocation policy { P ( k ) } that make λ NR satisfy λ i = 1 L X m ∈M N µ i ( m , k ) log (1 + P i ( m , k ) γ i ( m )) π m , (43) i ∈ N NR , while having { µ ( k ) } and { P ( k ) } satisfy q i λ i ≤ X m ∈M N µ i ( m , k ) log (1 + P i ( m , k ) γ i ( m )) , i ∈ N R , (44) X i ∈N µ i ( m , k ) ≤ T, ∀ k ≥ , m ∈ M N , (45) lim sup K →∞ K K X k =1 X i ∈N X m ∈M µ i ( m , k ) P i ( m , k ) ≤ P avg , (46) µ i ( m , k ) ≥ , i ∈ N , ∀ k ≥ , m ∈ M N , (47) P i ( m , k ) ≥ , i ∈ N , ∀ k ≥ , m ∈ M N . (48) Theorem 8. If λ NR (1+ ǫ ) ∈ R Lamb then Algorithm 2 satisfies (7) - (12) . Otherwise, then problem (5) is infeasible.Proof. See Appendix BTheorem 8 says that R Lamb is in fact the system’s capacityregion. This answers the first question. Moreover, the secondquestion is answered in the proof, as shown in AppendixB. In the proof, we show that with a simple modificationto Algorithm 2 we can achieve this capacity region. Themodification is by setting r i ( k ) = a i ( k ) for all i ∈ N NR .VII. S IMULATION R ESULTS
We simulate the system for the on-off channel model as wellas the continuous channel model. For both models, we assumethat all channels are statistically homogeneous, i.e. γ i = γ forall i ∈ N where γ is a fixed constant. Moreover, all RT usershave homogeneous delivery ratio requirements, thus q i = q forall i ∈ N R for some parameter q . All parameter values usedin the simulations are: L = 1 bits, P max = 20 and γ i = 1 .We compare the throughput of the RT users, which is theobjective of problem (5), to that of a simple power allocationand scheduling algorithm that we call “FixedP” algorithm. Inthe FixedP algorithm, all scheduled users transmit with themaximum power, i.e. P i ( k ) = P max for all i ∈ N and all k ≥ , while the scheduling policy is to flip a biased coin andchoose to schedule either the NRT users or the RT users. Thecoin is set to schedule the RT users with probability q (thedelivery ratio requirement for all users), at which case the RTusers are sorted according to Y i ( k ) and scheduled one by oneuntil the current slot ends. On the other hand, when the coinchooses the NRT users, the FixedP policy assigns the entiretime slot to the NRT user with the longest queue. A. On-Off Channel Model
We assume that we have N = 20 users that is split equallybetween the RT and NRT users, i.e. N R = N NR = 20 . Fig. 3shows a substantial increase in the average rate of the proposedalgorithm over the FixedP algorithm with over at low P avg values and at high P avg values. We simulated thesystem with B max = 10 , T = 1 and q = 0 . .In Fig. 4, the sum of average NRT users’ throughput isplotted while keeping P avg = 10 but changing q . We cansee that the FixedP algorithm results in a large degradation inthe throughput compared to Algorithm 1 which allocates the P avg N R T A v g R a t e P i ∈ N R T ¯ R i ( b i t s / c h a n - u s e ) Algorithm 1FixedP
Fig. 3. Sum of average throughput for all NRT users. The FixedP algorithmassigns a fixed power to all users set at P max . Delivery Ratio Requirement q i = q N R T A v g R a t e P i ∈ N R T ¯ R i ( b i t s / c h a n - u s e ) Algorithm 1FixedP
Fig. 4. As q increases, the RT users are assigned the channel more frequently.This comes at the expense of the NRT’s throughput. However, the proposedalgorithm outperforms the FixedP algorithm. power and schedules the users optimally with respect to (5).The decrease in the throughput observed in both curves of Fig.4 is due to the increase in the parameter q . This increase makesconstraint (8) more stringent and thus decreases the feasibleregion decreasing the throughput.In Fig. 5 we show the effect of increasing the number ofusers on the system’s throughput. As the number of usersincrease, more RT users have to be scheduled. This comesat the expense of the time allocated to the NRT users thusdecreasing the throughput for the two plotted algorithms. B. Continuous Channel Fading Model
In this simulation setup, we assume the channels are fadingaccording to a Rayleigh fading model with avg power gain
13 14 15 16 17 18 19 200.911.11.21.31.41.5
Number of RT/NRT users N R = N NR N R T A v g R a t e P i ∈ N R T ¯ R i ( b i t s / c h a n - u s e ) Algorithm 1FixedP
Fig. 5. As N increases, the RT users are allocated the channel more at theexpense of the NRT users’ throughput. of γ = 1 . In Fig. 6, we plot the complexity of the Lambert-Strict algorithm as well as the exhaustive search algorithmwith exponential complexity versus the number of users N R .The complexity is measured in terms of the average numberof iterations, per-slot, where we have to evaluate the objectivefunction of (25). Since this complexity changes from a slotto the other, we plot the average of this complexity. As thenumber of users increases, the Lambert-Strict algorithm has anaverage complexity close to linear. However, the there is nosacrifice in the throughput of the NRT users. This is shown inFig. 7. The reason stems from the optimality of the Lambert-Strict algorithm that does not eliminate any RT users fromscheduling unless it is a suboptimal user. We note that wesimulated this system with B max = 100 , T = 5 , P avg = 10 and q = 0 . . VIII. C ONCLUSIONS
We discussed the problem of throughput maximization indownlink cellular systems in the presence of RT and NRTusers. We formulated the problem as a joint power-allocation-and-scheduling problem. Using the Lyapunov optimizationtheory, we presented two algorithms to optimally solve thethroughput maximization problem. The first algorithm is forthe on-off channel fading model while the second is forthe continuous channel fading model. The power allocationsfor both algorithms are in closed-form expressions for theRT as well as the NRT users. We showed that the NRTpower allocation is water-filling-like which is monotonicallyincreasing in the channel gain. On the other hand, the RTpower allocation has a totally different structure that we callthe “Lambert Power Allocation”. It is found that the latter isa decreasing function in the channel gain.The two algorithms differ in the complexity of the adoptedscheduling policies. The first algorithm has a linear complexitywhile the second is shown, through simulations, to have aclose-to-linear complexity. We presented the capacity region of Number of RT/NRT users N R = N NR A v e r ag e C o m p l e x i t y p e r s l o t( c a nd i d a t e s e t s ) Exhaustive SearchLambert−StrictLinear Complexity
Fig. 6. As the number of NRT users in the system increase the complex-ity increases exponentially for exhaustive search and nearly linear for theLambert-Strict algorithm.
Number of RT/NRT users N R = N NR N R T A v g R a t e P i ∈ N R T ¯ R i ( b i t s / c h a n - u s e ) Exhaustive SearchLambert-Strict Alg.
Fig. 7. The Lambert-Strict Algorithm yields the same throughput as theexhaustive search algorithm but with a lower average complexity. the problem and showed that the proposed algorithms achievethis region. A
PPENDIX AP ROOF OF T HEOREM Proof.
We divide the proof into two parts. First, we show thatthe virtual queues are mean rate stable. This proves that con-straints (8) and (9) are satisfied. Second, through the Lyapunovoptimization technique we show that the drift-minus-rewardterm is within a constant gap from the performance of theoptimal, genie-aided algorithm [22], [23].
1) Mean Rate Stability:
According to (22), Algorithm 1minimizes Ψ( k ) where the minimization is taken over allpossible scheduling and power allocation algorithms includingthe optimal algorithm that solves (5). We define Ψ ∗ ( k ) , min Ψ( k ) . Thus we can write Ψ ∗ ( k ) ≤ ˜Ψ( k ) where ˜Ψ( k ) isthe value of Ψ( k ) evaluated at the optimal algorithm and isgiven by ˜Ψ( k ) , X i ∈N R E U ( k ) (cid:2) Y i ( k ) (cid:0) λ i q i − ˜ i ( k ) (cid:1)(cid:3) + X ( k ) X i ∈N E U ( k ) h ˜ µ i ( k ) ˜ P i ( k ) i T − P avg + X i ∈N NR Q i ( k ) (cid:16) E U ( k ) h LR (opt) i − ˜ µ i ( k ) ˜ R i ( k ) i(cid:17) , (49)where ˜ P i ( k ) , ˜ µ i ( k ) , ˜ i ( k ) and ˜ R i ( k ) satisfy (33), (34) and(35). Taking E [ · ] to (49), summing over k = 0 · · · K − ,dividing by K , taking the limit as K → ∞ and using (33),(34) and (35) gives lim sup K →∞ K K − X k =0 E h ˜Ψ( k ) i ≤ (50)Evaluating by Algorithm 1 in the right-side of (16), and taking E [ · ] with respect to U ( k ) to both sides gives X i ∈N R E (cid:2) Y i ( k ) (cid:3) + 12 X i ∈N NR E (cid:2) Q i ( k ) (cid:3) + 12 E (cid:2) X ( k ) (cid:3) ≤ C + E [Ψ ∗ ( k )] . (51)Removing the two summations on the left-side of (51), sum-ming over k = 0 · · · K − , dividing by K then taking thelimit as K → ∞ yields lim sup K →∞ E (cid:2) X ( K ) (cid:3) K ≤ C + lim K →∞ K K − X k =0 E [Ψ ∗ ( k )] ( a ) ≤ C + lim K →∞ K K − X k =0 E h ˜Ψ( k ) i ( b ) ≤ C . (52)where inequalities (a) and (b) in (52) follow from the inequal-ity Ψ ∗ ( k ) ≤ ˜Ψ( k ) and (50), respectively. Jensen’s inequalitysays that E [ X ( K )] ≤ E (cid:2) X ( K ) (cid:3) . Dividing by K , takingthe square root, passing K → ∞ and using (52) completesthe mean rate stability proof. Similarly we can show the meanrate stability of Y i ( k ) .
2) Objective Function Optimality:
Evaluating the right-hand-side of (19) at the optimal policy that has a genie-aided knowledge of the optimum reward r i ( k ) = R (opt) i weget ∆( k ) − B max P i ∈N NR E U ( k ) [ Lr i ( k )] ≤ C + ˜Ψ( k ) − B max P i ∈N NR R (opt) i which is similar to equation (20) in [18].The optimality proof continues along the lines of Theorem 2in [18]. A PPENDIX BP ROOF OF T HEOREM Proof.
We divide our proof into two parts. In the first part(Achievability), we show that if λ NR is strictly within theregion R Lamb , then the queues can be stabilized. And thealgorithm that stabilizes these queues is a modified version ofAlgorithm 2. We show this using the Lyapunov optimizationtechnique [24, pp.120]. In the second part (Converse), we showthat if λ NR / ∈ R Lamb , then there exists no algorithm thatguarantees the stability of the NRT queues.
1) Achievability:
We will show here that the followinginequality holds under Algorithm 2 which is the key to theproof. X i ∈N NR λ i Q i ( k ) + X i ∈N R λ i q i Y i ( k ) − X i ∈N X ( k ) P avg ≤ E U ( k ) " X i ∈N NR Q i ( k ) D i ( k ) + X i ∈N R Y i ( k ) D i ( k ) − E U ( k ) "X i ∈N X ( k ) µ i ( k ) P i ( k ) T , (53)where B i ( k ) , µ i ( k ) log (1 + P i ( k ) γ i ( k )) . Once this in-equality is proven, the rest of the achievability proof workssimilar to Theorem 5.3.2 in [24, pp.120]. Since λ NR (1 + ǫ ) ∈R Lamb , to prove (53) we multiply (43) by λ i , (44) by λ i , and(46) by ( − P avg ) , then add the three inequalities after summingthe first over i ∈ N NR and the second over i ∈ N R yielding X i ∈N NR λ i Q i ( k ) + X i ∈N R λ i q i Y i ( k ) − X i ∈N X ( k ) P avg ≤ X m ∈M N X i ∈N NR Q i ( k ) D i ( m , k ) + X i ∈N R Y i ( k ) D i ( m , k ) (54) − X i ∈N X ( k ) µ i ( m , k ) P i ( m , k ) T ! π m (55) ≤ X m ∈M N " X i ∈N NR Ψ ∗ NR ( i, k ) + X i ∈N R Ψ ∗ R ( i, k ) π m , (56)where D i ( m , k ) , µ i ( k ) log (1 + P i ( k ) γ i ( k )) while in-equality (55) follows since the objective of problem (22) isan upper bound on (55). But since the right-hand-side of (53) can be manipulated to give E U ( k ) " X i ∈N NR Q i ( k ) D i ( m , k ) + X i ∈N R Y i ( k ) D i ( m , k ) (57) − X i ∈N X ( k ) µ i ( k ) P i ( k ) T (58) = E U ( k ) " X i ∈N NR (cid:18) Q i ( k ) D i ( m , k ) − X ( k ) µ i ( k ) P i ( k ) T (cid:19) (59) + X i ∈N R (cid:18) Y i ( k ) D i ( m , k ) − X ( k ) µ i ( k ) P i ( k ) T (cid:19) (60) = X m ∈M N " X i ∈N NR Ψ ∗ NR ( i, k ) + X i ∈N R Ψ ∗ R ( i, k ) π m ≥ (61) X i ∈N NR λ i Q i ( k ) + X i ∈N R λ i q i Y i ( k ) − X i ∈N X ( k ) P avg (62)where the left side of the inequality in (62) follows byevaluating (60) at Algorithm 2 while its right side followsfrom (56) which completes the proof of (53).
2) Converse:
The converse is done by showing that theupper bound of the sum of the number bits served from allNRT buffers under the best, possibly genie-aided, policy isless than the sum of bits arriving to the NRT buffers if thearrival rate does not satisfy (44) through (48).From the strict separation theorem [24, pp.10], if λ / ∈ R
Lamb then there exists a vector β , [ β , · · · β N NR ] T ∈ R N NR anda constant δ > such that for any vector x ∈ R Lamb thefollowing holds X i ∈N NR β i λ i ≥ X i ∈N NR β i x i + δ (63)Define H ( k + 1) = H ( k ) + P i ∈N NR β i ( La i ( k ) − B i ( k )) asthe weighted sum of the queues where B i ( k ) , µ i ( k ) R i ( k ) is the number of bits transmitted to user i at slot k . Hence wehave H ( K ) = K − X k =0 X i ∈N NR β i ( La i ( k ) − B i ( k )) . (64)Define the set K K ( l ) , { k : m ( k ) = l, ≤ k < K } we canbound the second term in (64) as follows X i ∈N NR β i lim sup K →∞ K − X k =0 B i ( k ) K ≤ X i ∈N NR β i lim sup K →∞ M X l =1 X k ∈K K ( l ) ˜ B i ( k ) |K K ( l ) | |K K ( l ) | K (65) = X i ∈N NR β i M X l =1 ˜ B ( l ) i π l = X i ∈N NR β i M X l =1 Lx i π l . (66) Adding Lδ to both sides of (66) and using (63) yields X i ∈N NR β i lim sup K →∞ K − X k =0 B i ( k ) K + Lδ ≤ L M X l =1 π l X i ∈N NR β i x i + δ ! ≤ X i ∈N NR β i Lλ i = lim K →∞ K − X k =0 La i ( k ) K . (67)Combining (67) and (64) we conclude that lim sup K →∞ H ( K ) = ∞ which means that the weightedsum of the queues is unbounded, under the best possiblepolicy, when λ NR / ∈ R Lamb which completes the proof.R
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