IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012 571
Broadcasting with anEnergy Harvesting Rechargeable Transmitter
Jing Yang, Member, IEEE, Omur Ozel, Student Member, IEEE, and Sennur Ulukus, Member, IEEE
AbstractβIn this paper, we investigate the transmission com-pletion time minimization problem in an additive white Gaussiannoise (AWGN) broadcast channel, where the transmitter is ableto harvest energy from the nature, using a rechargeable battery.The harvested energy is modeled to arrive at the transmitterduring the course of transmissions. The transmitter has afixed number of packets to be delivered to each receiver. Theobjective is to minimize the time by which all of the packets aredelivered to their respective destinations. To this end, we optimizethe transmit powers and transmission rates in a deterministicsetting. We first analyze the structural properties of the optimaltransmission policy in a two-user broadcast channel via the dualproblem of maximizing the departure region by a fixed time π .We prove that the optimal total transmit power sequence hasthe same structure as the optimal single-user transmit powersequence in [1], [2]. In addition, the total power is split optimallybased on a cut-off power level; if the total transmit power islower than this cut-off level, all transmit power is allocated tothe stronger user; otherwise, all transmit power above this levelis allocated to the weaker user. We then extend our analysis toan π -user broadcast channel. We show that the optimal totalpower sequence has the same structure as the two-user case andoptimally splitting the total power among π users involves πβ1cut-off power levels. Using this structure, we propose an algorithmthat finds the globally optimal policy. Our algorithm is based onreducing the broadcast channel problem to a single-user problemas much as possible. Finally, we illustrate the optimal policy andcompare its performance with several suboptimal policies underdifferent settings.
Index TermsβEnergy harvesting, rechargeable wireless net-works, broadcast channels, transmission completion time mini-mization, throughput maximization.
I. INTRODUCTION
WE consider a wireless communication network whereusers are able to harvest energy from nature using
rechargeable batteries. Such energy harvesting capabilities willmake sustainable and environmentally friendly deploymentof wireless communication networks possible. While energy-efficient scheduling policies have been well-investigated in
Manuscript received October 13, 2010; revised March 14, 2011 andSeptember 19, 2011; accepted October 31, 2011. The associate editor co-ordinating the review of this paper and approving it for publication was C.Xiao.
This work was supported by NSF Grants CCF 04-47613, CCF 05-14846,CNS 07-16311, CCF 07-29127, CNS 09-64632 and presented in part at theIEEE International Conference on Communications (ICC), Kyoto, Japan, May,2011.
J. Yang was with the Department of Electrical and Computer Engineering,University of Maryland, College Park, MD 20742, USA. She is now with theDepartment of Electrical and Computer Engineering, University of Wisconsin-Madison, WI 53706, USA (e-mail: [email protected]).
O. Ozel and S. Ulukus are with the Department of Electrical and ComputerEngineering, University of Maryland, College Park, MD 20742 (e-mail:{omur, ulukus}@umd.edu).
Digital Object Identifier 10.1109/TWC.2011.120911.101813
traditional battery powered (un-rechargeable) systems [3]β[8],energy-efficient scheduling in energy harvesting networks withnodes that have rechargeable batteries has only recently beenconsidered [1], [2]. References [1], [2] consider a single-usercommunication system with an energy harvesting transmitter,and develop a packet scheduling scheme that minimizes thetime by which all of the packets are delivered to the receiver.
In this paper, we consider a multi-user extension of the workin [1], [2]. In particular, we consider a wireless broadcastchannel with an energy harvesting transmitter. As shown inFig. 1, we consider a broadcast channel with one transmitterand π receivers, where the transmitter node has π + 1queues. The π data queues store the data arrivals intendedfor the individual receivers, while the energy queue stores theenergy harvested from the environment. The π receivers havedifferent channel gains, and the broadcast channel is degraded[9]. Our objective is to adaptively change the transmissionrates that go to the users according to the instantaneous dataand energy queue sizes, such that the transmission completiontime is minimized.
In [1], [2], we prove that the optimal scheduling policyin a single-user energy harvesting communication systemhas a βmajorizationβ structure. The transmit power is keptconstant between energy harvests, the sequence of transmitpowers increases monotonically, and only changes when theenergy constraint is tight in the optimal schedule. The opti-mal transmit power sequence is the most majorized energy-feasible power sequence during the transmission duration. Wedevelop, in [1], [2], an algorithm to obtain the optimal off-linescheduling policy based on these properties. Reference [10]extends [1], [2] to the case where rechargeable batteries havefinite storage capacities. We extend [1], [2] in [11] to a fadingchannel, and develop optimal off-line and on-line schedulingpolicies under stochastic fading and energy arrival processes.References [12], [13] address single-user energy harvestingrechargeable systems as well, with a slotted time systemmodel. Although the slotted time system model can model amore practical scenario, e.g., a scenario where block encodingis used, a continuous time system model is more general in thesense that the continuous time model can be used to analyzea slotted time system after proper rearrangements. Therefore,we consider a continuous time energy harvesting system modelas in our previous work [1], [2], [11]. We focus on the off-line problem in this paper with the goal of determining thestructural properties of the optimum broadcast scheduler, anddeveloping an iterative optimal off-line scheduling algorithm.We also provide benchmark on-line scheduling algorithms forperformance comparison, while we leave the solution of theoptimal on-line policy for future work.
1536-1276/12$31.00 cβ 2012 IEEE
572 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012
energy queue
Ei
B1
TXB2
BM
RX 1
RX M
RX 2
data queues
Fig. 1. An energy harvesting π -user broadcast channel.
Another line of research in wireless communications withenergy harvesting nodes has been presented in [14]β[16] whichconsiders the simultaneous transmission of information andenergy. While this is a novel and interesting direction, in thispaper, we do not consider the transmission of energy. We onlyconsider the transmission of data. In particular, we addressthe issue of the replenishment of the transmission energyduring the course of the communication session, and obtainthe optimal off-line adaptation of the transmit power to thefluctuating levels of energy in a broadcasting scenario.
References [10], [11] investigate two related problems insingle-user communication with an energy harvesting trans-mitter. The first problem is to maximize the throughput(number of bits transmitted) with a given deadline constraint,and the second problem is to minimize the transmissioncompletion time with a given number of bits to transmit.These two problems are dual: given the energy arrival profile,if the maximum number of bits that can be sent by π isπ΅β in the first problem, then the minimum time to transmitπ΅β bits in the second problem must be π , and the optimaltransmission policies for these two problems must be identical.In this paper, we follow the same dual problem approach tosolve the transmission completion time minimization problemin the broadcast channel. An alternative solution method forthe second problem in the broadcast channel is providedin the independent and concurrent work [17]. This methoddivides the problem into local sub-problems that consider onlytwo energy arrival epochs at a time. Iterations on the localproblems are shown to converge to the global optimum.
We first analyze the structural properties of the optimalpolicy for the first problem in a two-user broadcast channel.Our goal is to determine the maximum departure region witha given deadline constraint π . The maximum departure regionis defined as the set of all (π΅1, π΅2) that can be transmitted tousers reliably with a given deadline π . By using the convexityof the departure region and a Lagrangian approach, we showthat the optimal total transmit power policy is independentof the operating point on the boundary, and has the sameβmajorizationβ structure as the single-user solution. As forthe way of splitting the total transmit power between the twousers, we prove that there exists a cut-off power level for thestronger user, i.e., only the power above this cut-off powerlevel is allocated to the weaker user. We then investigate themaximum departure region in an π -user broadcast channel.We show that there exist π β 1 cut-off power levels and the
total power is split according to these cut-off power levels andthe hierarchy among the channel gains of the users.
Next, we consider the second problem, where our goal isto minimize the time, π , by which a given number of bitsare delivered to their intended receivers. Due to the dualitybetween these problems, the optimal policy in the secondproblem has the same structural properties as those in thefirst problem. Using these optimal structural properties, wedevelop an iterative algorithm that finds the optimal scheduleefficiently. In particular, we start with the two-user case and weobtain the optimal total power in the first epoch, π1. Given thefact that there exists a cut-off power level, ππ, for the strongeruser, the optimal policy depends on whether π1 is smaller orlarger than ππ, which, at this point, is unknown. Performingiterations on ππ that alternates according to whether ππ shouldbe increased or decreased, we approach the optimal policyiteratively. At each iteration, single-user problems as in [1],[2] are solved. The algorithm naturally extends to the π -user broadcast channel. We perform alternating iterations onthe cut-off level for the strongest user and determine whetherit should be decreased or increased in a similar fashion. Wealso discuss the computational requirement of the proposedalgorithm. Finally, we provide numerical illustrations and per-formance analysis for the optimal off-line policy. Specifically,we compare the performance of the optimal off-line policywith that of three practical semi-on-line sub-optimal policieswhich require no or partial off-line knowledge about theenergy harvesting process.
II. SYSTEM MODEL AND PROBLEM FORMULATION
The system model is as shown in Figs. 1 and 2. Thetransmitter has an energy queue and π data queues (Fig. 1).The physical layer is modeled as an AWGN broadcast channel,where the received signals at the π receivers are
ππ = π + ππ, π = 1, . . . ,π (1)
where π is the transmit signal, and ππ is a Gaussiannoise with zero-mean and variance π2
π, and without loss ofgenerality π2
1 β€ π22 β€ . . . β€ π2
π . Therefore, the first useris the strongest and the π th user is the weakest user in ourbroadcast channel. Next, for clarity of exposition we writethe capacity region for this broadcast channel for π = 2,and subsequently generalize it to π users. Assuming that thetransmitter transmits with power π , the capacity region forthe two-user AWGN broadcast channel is [9]
π1 β€ 1
2log2
(1 +
πΌπ
π21
)(2)
π2 β€ 1
2log2
(1 +
(1β πΌ)π
πΌπ + π22
)(3)
where πΌ is the fraction of the total power spent for themessage transmitted to the first user. Let us denote π(π) β12 log2 (1 + π) for future use. Then, the capacity region is
π1 β€ π(πΌππ21), π2 β€ π
((1βπΌ)ππΌπ+π2
2
). Working on the boundary
of the capacity region, we have
π = π212
2(π1+π2) + (π22 β π2
1)22π2 β π2
2 (4)
β π(π1, π2) (5)
YANG et al.: BROADCASTING WITH AN ENERGY HARVESTING RECHARGEABLE TRANSMITTER 573
. . .
t
(B1, . . . , BM)0 s1 s2 sK
E1 E2 EKE0
T
Fig. 2. System model. (π΅1, . . . , π΅π ) bits to be transmitted to the users areavailable at the transmitter at the beginning. Energies arrive (are harvested) atpoints denoted by β. π denotes the transmission completion time by whichall of the bits are delivered to their respective destinations.
Therefore, π is the smallest power necessary to transmit atrates π1 and π2 in this broadcast channel. We note that π(π1, π2)is a strictly convex function of (π1, π2).
The rate region for the π -user broadcast channel is ob-tained using an π -level superposition code [9]. The capacityregion for the π -user AWGN broadcast channel is the set ofrate vectors (π1, . . . , ππ ):
ππ =1
2log2
(1 +
πΌππβπ<π πΌππ + π2
π
), π = 1 . . . ,π (6)
where πΌπ β₯ 0 andβ
π πΌπ = 1. Similar to the derivation in (3)-(5), we obtain the minimum power to achieve the rate vector(π1, . . . , ππ ) by a recursive formula as follows:
π(π)(π1, . . . , ππ ) = 22ππ π(πβ1)(π1, . . . , ππβ1)
+ π2π (22ππ β 1) (7)
where π(π)(π1, . . . , ππ ) is the minimum power for the π -user AWGN broadcast channel. Note that (7) reduces to (5)for π = 2 where π(1)(π1) = π2
1(22π1 β 1). Also note
that π(π)(π1, . . . , ππ ) is strictly convex in (π1, . . . , ππ ) byinduction.
As shown in Fig. 1, the transmitter has π΅π bits destinedto the πth receiver. Energy is harvested at times π π withamounts πΈπ, π β₯ 1. πΈ0 denotes the initial energy availablein the battery before the communication starts. Our goal is toselect a transmission policy that minimizes the time, π , bywhich all of the bits are delivered to their intended receivers.The transmitter adapts its transmit power and the portionsof the total transmit power used to transmit signals to theπ users according to the available energy level and theremaining number of bits. The energy consumed must satisfythe causality constraints, i.e., at any given time π‘, the totalamount of energy consumed up to time π‘ must be less than orequal to the total amount of energy harvested up to time π‘.
Before we proceed to give a formal definition of theoptimization problem and its solution, we start with the dualproblem of finding the maximum departure region of the bitsthe transmitter can deliver to the users by any fixed time π . Aswe will observe in the next section, solving the dual problemenables us to identify the optimal structural properties forthe original problem, and these properties help us reduce theoriginal problem into simple scenarios, which can be solvedefficiently. In the next section, we consider the two-user case,and generalize it to the most general π -user case in SectionIV.
III. CHARACTERIZING π(π ): LARGEST (π΅1, π΅2) REGION
FOR A GIVEN π
In this section, our goal is to characterize the maximumdeparture region for a given deadline π . Let us denote thenumber of bits transmitted to the πth user by time π‘ via a ratefunction ππ(π), in 0 β€ π β€ π‘, as π΅π(ππ(π‘)) =
β« π‘0ππ(π)ππ ,
π = 1, 2.
Definition 1 For any fixed transmission duration π ,the maximum departure region, denoted as π(π ),is the union of (π΅1, π΅2) under any feasible rateallocation policy over the duration [0, π ), i.e.,π(π ) =
βͺπ1(π),π2(π)
(π΅1(π1(π )), π΅2(π2(π ))), subject to the
energy causality constraintβ« π‘0π(π1, π2)(π)ππ β€ β
π:π π<π‘πΈπ,for 0 β€ π‘ β€ π .
We call any policy which achieves the boundary of π(π )to be optimal. In the single-user scenario in [1], [2], wefirst examined the structural properties of the optimal policy.Based on these properties, we developed an algorithm tofind the optimal scheduling policy. In this broadcast scenarioalso, we first analyze the structural properties of the optimalpolicy, and then obtain the optimal solution based on thesestructural properties. The following lemma which was provedfor a single-user problem in [1], [2] was also proved for thebroadcast problem in [17].
Lemma 1 Under the optimal policy, the transmission rateremains constant between energy harvests, i.e., the rate onlypotentially changes at an energy harvesting instant.
Proof: We prove this using the strict convexity of π(π1, π2).If the transmission rate for any user changes between twoenergy harvesting instants, then, we can always equalize thetransmission rate over that duration without contradicting withthe energy constraints. Based on the convexity of π(π1, π2),after equalization of rates, the energy consumed over thatduration decreases, and the saved energy can be allocated toboth users to increase the departures. Therefore, changing ratesbetween energy harvests is sub-optimal. β
Therefore, in the following, we only consider policies wherethe rates are constant between any two consecutive energyarrivals. We denote the rates that go to both users as (π1π, π2π)over the duration [π πβ1, π π). An illustration of the maximumdeparture region is shown in Fig. 3.
Lemma 2 π(π ) is a convex region.
Proof: Proving the convexity of π(π ) is equivalent to provingthat, given any two achievable points (π΅1, π΅2) and (π΅β²
1, π΅β²2)
in π(π ), any point on the line between these two points is alsoachievable, i.e., in π(π ). Assume that (π΅1, π΅2) and (π΅β²
1, π΅β²2)
can be achieved with rate allocation policies (r1, r2) and(rβ²1, r
β²2), respectively. Consider the policy (πr1 + οΏ½ΜοΏ½rβ²1, πr2 +
οΏ½ΜοΏ½rβ²2), where οΏ½ΜοΏ½ = 1β π. Then, the energy consumed up to π π
574 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012
(B1,B2)
B1
B2
Fig. 3. The maximum departure region and trajectories to reach the boundary.Dotted trajectory is not possible.
isπβ
π=1
π(ππ1π + οΏ½ΜοΏ½πβ²1π, ππ2π + οΏ½ΜοΏ½πβ²2π)ππ
β€ π
πβπ=1
π(π1π, π2π)ππ + οΏ½ΜοΏ½
πβπ=1
π(πβ²1π, πβ²2π)ππ (8)
β€ π
πβ1βπ=0
πΈπ + οΏ½ΜοΏ½
πβ1βπ=0
πΈπ =
πβ1βπ=0
πΈπ (9)
Therefore, the energy causality constraint is satisfied for anyπ β [0, 1], and the new policy is energy-feasible. Any pointon the line between (π΅1, π΅2) and (π΅β²
1, π΅β²2) can be achieved.
When π β= 0, 1, the inequality in (8) is strict. Therefore, wesave some amount of energy under the new policy, which canbe used to increase the throughput for both users. This impliesthat π(π ) is strictly convex. β
In order to simplify the notation, for any given π , weassume that there are π β 1 energy arrivals over (0, π ). Wedenote the last energy arrival time before π as π πβ1, andπ π = π . We use ππ to denote the length of the durationbetween two consecutive energy arrival instances π π and π πβ1,i.e., ππ = π π β π πβ1, with π1 = π 1 and ππ = π β π πβ1, asshown in Fig. 4.
Since π(π ) is a strictly convex region, its boundary can becharacterized by solving the following optimization problemfor all π1, π2 β₯ 0,
maxr1,r2
π1
πβπ=1
π1πππ + π2
πβπ=1
π2πππ
s.t.πβ
π=1
π(π1π, π2π)ππ β€πβ1βπ=0
πΈπ, 0 < π β€ π (10)
where r1 and r2 denote the rate sequences π1π and π2π forusers 1 and 2, respectively. For π1 = 0 or π2 = 0, the problemin (10) reduces to the throughput maximization problem forthe user which has the non-zero coefficient. The solution ofthis single-user problem is provided in [11]. We will referto this problem as the single-user problem and its solutionas the single-user solution in the rest of the paper. Due tothe duality between solving the throughput maximization andtransmission completion time minimization problems in thesingle-user scenario, we also refer to the solution in [1], [2]as the single-user solution.
(r12, r22)
T
E1
Β· Β· Β·
Β· Β· Β·
sNβ1
l1 l2 l3 lN
(r11, r21) (r1N , r2N)(r13, r23)
Β· Β· Β·
Β· Β· Β·t
(B1, B2)
E0 ENβ1E2
0 s1 s2
Fig. 4. Rates (π1π, π2π) and corresponding durations ππ with a givendeadline π .
The problem in (10) is a convex optimization problem witha convex cost function and a convex constraint set, therefore,the unique global solution should satisfy the extended KKTconditions. The Lagrangian is
β(r1, r2,π,πΈ) =π1
πβπ=1
π1πππ + π2
πβπ=1
π2πππ
βπβπ=1
ππ
(πβ
π=1
π(π1π, π2π)ππ βπβ1βπ=0
πΈπ
)
+
πβπ=1
πΎ1ππ1π +
πβπ=1
πΎ2ππ2π (11)
Taking the derivatives with respect to π1π and π2π, and settingthem to zero, we have the optimality conditions
π1 + πΎ1π β( πβ
π=π
ππ
)π212
2(π1π+π2π) = 0 (12)
π2 + πΎ2π β( πβ
π=π
ππ
)(π212
2(π1π+π2π) + (π22 β π2
1)22π2π
)= 0
(13)
for π = 1, . . . , π , with the complementary slackness condi-tions
ππ
( πβπ=1
π(π1π, π2π)ππ βπβ1βπ=0
πΈπ
)= 0, π = 1, . . . , π
(14)
πΎ1ππ1π = 0, πΎ2ππ2π = 0, π = 1, . . . , π (15)
Based on (12)-(15), we first prove an important property ofthe optimal policy.
Lemma 3 The optimal total transmit power of the transmitteris independent of the values of π1, π2, and it is the same asthe single-user optimal transmit power. Specifically,
ππ = arg minππβ1<πβ€π
{βπβ1π=ππβ1
πΈπ
π π β π ππβ1
}(16)
ππ =
βππβ1π=ππβ1
πΈπ
π ππ β π ππβ1
(17)
i.e., at π‘ = π ππ , ππ switches to ππ+1.
The proof of Lemma 3 is provided in Appendix A.
Since the total transmit power can be obtained irrespectiveof the values of π1, π2, the optimization problem in (10) isseparable over each duration [π πβ1, π π). Specifically, for 1 β€
YANG et al.: BROADCASTING WITH AN ENERGY HARVESTING RECHARGEABLE TRANSMITTER 575
π β€ π , the local optimization problem becomes
maxπ1π,π2π
π1π1π + π2π2π
s.t. π(π1π, π2π) β€ ππ (18)
We relax the power constraint to be an inequality to makethe constraint set convex. Thus, this becomes a convex opti-mization problem. This does not affect the solution since theobjective function is always maximized on the boundary of itsconstraint set, i.e., the capacity region defined by the transmitpower ππ.
We first note that due to the degradedness of the seconduser, when π2
π1β€ 1, the total power ππ is allocated to the first
user only and no bits are transmitted to the second user. When1 < π2
π1, we define
ππ β(π1π
22 β π2π
21
π2 β π1
)+
(19)
After a first order derivative analysis, we find the solution of(18) in terms of ππ as follows
π1π =1
2log2
(1 + min{ππ, ππ}
)(20)
π2π =1
2log2
(1 +
(ππ β ππ)+
ππ + π22
)(21)
In the optimal solution, power allocated to the first user canbe at most ππ, and the remaining power is allocated to thesecond user. Hence, we call ππ the cut-off power level.
Lemma 4 For fixed π1, π2, under the optimal power policy,there exists a constant cut-off power level, ππ, for the firstuser. If the total power level is below ππ, then, all the power isallocated to the first user; if the total power level is higher thanππ, then, all the power above ππ is allocated to the seconduser.
In the proof of Lemma 3 in Appendix A, we note that theoptimal power ππ monotonically increases in π. CombiningLemma 3 and Lemma 4, we illustrate the structure of theoptimal policy in Fig. 5. Moreover, the optimal way of splittingthe power in each epoch is such that both usersβ shares of thepower monotonically increase in time. In particular, the seconduserβs share is monotonically increasing in time. Hence, thepath followed in the (π΅1, π΅2) plane is such that it changesdirection to get closer to the second userβs departure axis asshown in Fig. 3. The dotted trajectory cannot be optimal, sincethe path does not get closer to the second userβs departure axisin the middle (second) power epoch.
Corollary 1 Under the optimal policy, the transmission ratefor the first user, {π1π}ππ=1, is either a constant sequence (zeroor a positive constant), or an increasing sequence. Moreover,before π1π achieves the value at which it stays constant, wehave π2π = 0; and after π1π achieves the value at whichit stays constant, π2π becomes a monotonically increasingsequence.
Based on Lemma 3, we observe that for fixed π , π1 andπ2, the optimal total power allocation is unique, i.e., does notdepend on π1 and π2. However, the way the total power is
E0
sK
Β· Β· Β·
Β· Β· Β·
EK
t0
E1 E2 E3
s2 s3 s4
E4
s1
(B1, B2)
T
P
Pc
P3
P1
P2
si1 si2
Fig. 5. Optimally splitting the total power between the signals that go tothe two users.
split between the two users depends on π1, π2. In fact, the cut-off power level ππ varies depending on the value of π2/π1.For different values of π2/π1, the optimal policy achievesdifferent boundary points on the maximum departure region,and varying the value of π2/π1 traces the boundary of thisregion.
IV. π(π ) FOR AN π -USER BROADCAST CHANNEL
The maximum departure region π(π ) in the π -user broad-cast channel is defined similar to the two-user case as the unionof achievable (π΅1, . . . , π΅π ) pairs where π΅π =
β« π0 ππ(π)ππ
and the instantaneous rates are subject to the energy causalityconstraintβ« π‘
0
π(π)(π1, . . . , ππ )(π)ππ β€βπ:π π<π‘
πΈπ, 0 β€ π‘ β€ π (22)
The structures of the optimal policy that achieves the boundaryof π(π ) established in Lemmas 3-4 and Corollary 1 naturallyextend to the π -user case. First, note that Lemmas 1 and 2immediately extend to the π -user case as π(π)(π1, . . . , ππ )is a strictly convex function. Hence, the boundary of π(π ) isachieved by the unique solution of the following optimizationproblem for all π1, . . . , ππ β₯ 0,
maxr1,...,rπ
π1
πβπ=1
π1πππ + . . .+ ππ
πβπ=1
πππππ
s.t.πβ
π=1
π(π)(π1π, . . . , πππ)ππ β€πβ1βπ=0
πΈπ, βπ (23)
where rπ denotes the rate sequence {πππ}ππ=1 for user πand π is the number of epochs in [0, π ]. The correspondingLagrangian is
β(r1, . . . , rπ ,π,πΈ) = π1
πβπ=1
π1πππ + . . .+ ππ
πβπ=1
πππππ
βπβπ=1
ππ
(πβ
π=1
π(π)(π1π, . . . , πππ)ππ βπβ1βπ=0
πΈπ
)
+
πβπ=1
πΎ1ππ1π + . . .+
πβπ=1
πΎπππππ (24)
Taking the derivatives with respect to πππ for all π,π andsetting them to zero, we get the necessary KKT optimalityconditions. By using the recursive formula in (7) and the
576 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012
KKT optimality conditions, we can show that the optimal totalpower allocation is
π(π)(πβ1π, . . . , π
βππ) = max
π
{ππβππ=π ππ
β π2π
}(25)
As the complementary slackness conditions in (14)-(15) holdin the π -user case as well, we recover Lemma 3, i.e., theoptimal total power sequence is exactly the same sequence asin (17) for the π -user case irrespective of the values of ππ,π = 1, . . . ,π .
Splitting the total power among π users requires a cut-offpower structure as in the two-user case that is stated in Lemma4. Since the optimal total power is obtained irrespective of thevalues of ππ, the optimization problem in (23) is separableover each duration [π πβ1, π π). Specifically, for 1 β€ π β€ π ,the corresponding local optimization problem is
maxπ1π,...,πππ
π1π1π + . . .+ πππππ
s.t. π(π)(π1π, . . . , πππ) β€ ππ (26)
Whenever ππ β€ ππ for any 1 β€ π < π β€ π , i.e., whenevera degraded user has a smaller coefficient, the solution of (26)is such that πβππ = 0 for any value of ππ. Hence, we removethose users. The remaining π β€ π users are such that π2
1 β€π22 β€ . . . β€ π2
π with π1 < π2 < . . . < ππ . Using a first orderdifferential analysis, the optimal cut-off power levels for theremaining π users must satisfy the following equations (seeAppendix B): For π = 1, . . . , π β 1
πππ = max
{(πππ
2οΏ½ΜοΏ½ β ποΏ½ΜοΏ½π
2π
ποΏ½ΜοΏ½ β ππ
)+
, ππ(πβ1)
}(27)
where by convention, we set ππ0 = 0, πππ = β and οΏ½ΜοΏ½is the smallest user index with πποΏ½ΜοΏ½ > πππ. We note thatππ0 β€ ππ1 β€ . . . β€ ππ(π β1) β€ πππ . For given ππ, the optimalsolution is
π1π =1
2log
(1 +
min{ππ, ππ1}π21
)(28)
π2π =1
2log
(1 +
min{(ππ β ππ1)+, ππ2 β ππ1}
ππ1 + π22
)(29)
...
ππ π =1
2log
(1 +
(ππ β ππ(π β1))+
ππ(π β1) + π2π
)(30)
We show the structure of optimally splitting the total poweramong the users for π =π in Fig. 6. Note that the hierarchyamong the channels of the users can be directly observedin Fig. 6. The top portion of the total power is allocatedto the user with the worst channel and the power below itis interference for this user. The bottom portion of the totalpower is allocated to the user with the best channel and thisuser experiences no interference. We also note that the timesequence of the total power is the same as in the two-user caseand the cut-off power levels are independent of the valuesof the varying total power levels. We also remark that theπ β 1 cut-off power levels are not necessarily distinct. Whenππ(π+1) = πππ for some 1 β€ π β€ π β 2, we must haveπ(π+1)π = 0 for all epochs π.
E0 E3 E4 E5 E6E2E1 . . .
Pc(Mβ1)
T
...
Pc3
Pc2
Pc1
P
P3 P4 P5
P2P1
share of user M
share of user 1
(B1, . . . , BM)
share of user 2
T
Fig. 6. Optimally splitting the total power for π users.
V. MINIMIZING THE TRANSMISSION COMPLETION TIME πFOR A GIVEN (π΅1, . . . , π΅π )
In this section, our goal is to minimize the transmissioncompletion time of (π΅1, . . . , π΅π ) bits. We start with the two-user case, and then generalize the algorithm to the π -userscenario.
A. Two-User Scenario
We formulate the optimization problem as follows:
minr1,r2
π
s.t.πβ
π=1
π(π1π, π2π)ππ β€πβ1βπ=1
πΈπ, 0 < π β€ π(π )
π(π )βπ=1
π1πππ = π΅1,
π(π )βπ=1
π2πππ = π΅2 (31)
where π(π )β 1 is the number of energy arrivals (excludingπ‘ = 0) over (0, π ), and ππ(π ) = π β π π(π )β1. Since π(π )depends on π , the optimization problem in (31) is not a convexoptimization problem in general. Therefore, we cannot solveit using standard convex optimization tools.
We first note that this is exactly the dual problem ofmaximizing the departure region for fixed π . They are dual inthe sense that, if the minimum transmission completion timefor (π΅1, π΅2) is π , then (π΅1, π΅2) must lie on the boundaryof π(π ), and the transmission policy should be exactly thesame for some (π1, π2). This is based on the fact thatπ(π ) β π(π β²) for any π < π β². Assume (π΅1, π΅2) does notlie on the boundary of π(π ). Then, either (π΅1, π΅2) cannotbe achieved by π or (π΅1, π΅2) is strictly inside π(π ) andhence (π΅1, π΅2) can be achieved by π β² < π . Therefore, if(π΅1, π΅2) does not lie on the boundary of π(π ), then π cannotbe the minimum transmission completion time. We have thefollowing lemma.
Lemma 5 When π΅1, π΅2 β= 0, under the optimal policy, thetransmissions to both users must be finished at the same time.
Proof: This lemma can be proved based on Corollary 1. If thetransmission completion time for both users is not the same,then over the last duration, we transmit only to one of theusers, while the transmission rate to the other user is zero. Thiscontradicts with the monotonicity of the transmission rates for
YANG et al.: BROADCASTING WITH AN ENERGY HARVESTING RECHARGEABLE TRANSMITTER 577
both users. Therefore, under the optimal policy, the transmittermust finish transmitting to both users at the same time. β
Lemma 5 is proved in [17] also, by using a differentapproach. The authors prove it in [17] mainly based on theconvexity of the capacity region of the broadcast channel.
According to Lemma 5, the problem of optimal selectionof ππ requires solving a fixed point equation. In particular, ππ
must be chosen such that the resulting transmission completiontime for the first and second user are equal. Therefore, wepropose the following algorithm to solve the transmissioncompletion time minimization problem.
First, we aim to identify π1, the first total transmit powerstarting from π‘ = 0 in the system. This is exactly the same asidentification of π1 in the corresponding single-user problem.For this, as in [1], [2], we treat the energy arrivals as if theyhave arrived at time π‘ = 0, and obtain a lower bound for thetransmission completion time as in [1], [2]. We compute theminimum amount of energy required to finish (π΅1, π΅2) by π 1.This requires to transmit at constant rates (π΅1
π 1, π΅2
π 1), and the
amount of energy is equal to π(π΅1
π 1, π΅2
π 1
)π 1, denoted as π΄1.
Then, we compare π΄1 with πΈ0. If πΈ0 is greater than π΄1, thisimplies that the transmitter can finish the transmission beforeπ 1, and future energy arrivals are not needed. In this case, theminimum transmission completion time is the solution of thefollowing equation
π
(π΅1
π,π΅2
π
)π = πΈ0 (32)
If π΄1 is greater than πΈ0, this implies that the final transmissioncompletion time is greater than π 1, and some of the future en-ergy arrivals must be utilized to complete the transmission. Wecalculate the amount of energy required to finish (π΅1, π΅2) byπ 2, π 3, . . . , and denote them as π΄2, π΄3, . . . , and compare thesewith πΈ0 +πΈ1,
β2π=0 πΈπ ,
β3π=0 πΈπ , . . . , until the first π΄π that
becomes smaller thanβπβ1
π=0πΈπ . We denote the corresponding
time index as οΏ½ΜοΏ½1. Then, we assume that we can useβοΏ½ΜοΏ½1β1
π=0 πΈπ
to transmit (π΅1, π΅2) at constant rates. And, the correspondingtransmission completion time is the solution of the followingequation
π
(π΅1
π,π΅2
π
)π =
οΏ½ΜοΏ½1β1βπ=0
πΈπ (33)
We denote the solution of (33) as π , and the correspondingpower as π1. From our analysis, we know that the solution tothis equation is a lower bound for the minimum transmissioncompletion time. We check whether this constant power π1 isfeasible, when the actual energy arrival times are imposed. Ifit is feasible, it gives us the minimal transmission completiontime; otherwise, we get π1 by selecting the minimal slopeaccording to (17). That is to say, we draw all of the linesfrom π‘ = 0 to the corner points of the energy arrival instancesbefore π , and choose the line with the smallest slope. Wedenote by π π1 the corresponding duration associated with π1.Please see [2, Fig. 8] for a visualization of the algorithm.
Once π1 is selected, it is the optimal total transmit powerover the duration [0, π π1) in our broadcast channel problem.This is due to Lemma 3 and the fact that (π΅1, π΅2) must
lie on the boundary of the departure curve at the minimumtransmission completion time. We defer the rigorous proofof optimality to Theorem 1 which immediately follows thealgorithm. Next, we need to divide this total power betweenthe signals transmitted to the two users. Based on Lemma 4and Corollary 1, if the cut-off power level ππ is higherthan π1, then, the transmitter spends all π1 for the strongeruser; otherwise, the first user finishes its transmission with aconstant power ππ.
We will first determine whether ππ lies in [0, π1] or itis higher than π1. Assume ππ = π1. The transmissioncompletion time for the first (stronger) user is
π1 =π΅1
π(π1)(34)
Next, we calculate the maximum number of bits departed fromthe second user by π1 given the set ππ, and denote it asπ·2(π1, ππ):
π·2(π1, ππ) =
π(π1)βπ=1
1
2log
(1 +
[π βπ β ππ]
+
ππ + π2
)(π π β π πβ1)
(35)
where π β1 , π
β2 , . . . , π
βπ(π1)
is the optimal total power allo-
cation by deadline π1. {π βπ }π(π1)
π=1 is found via single-useroptimal power allocation (c.f. Lemma 3) by deadline π1 as in[11]. Note that π β
1 = π1.If π·2(π1, π1) is smaller than π΅2, we need to decrease the
rate for the first user. In this case, the transmission power forthe first user is constant ππ β [0, π1] throughout the entireduration. In particular, ππ is the unique solution of
π΅2 = π·2
(π΅1
π(ππ), ππ
)(36)
Note that π·2
(π΅1
π(ππ), ππ
)is a continuous, strictly monotoni-
cally decreasing function of ππ, hence the solution for ππ in(36) exists and it is unique. We can use bisection method onππ to solve (36).
If π·2(π1, ππ) is larger than π΅2, that implies π2 < π1, andwe need to increase the power allocated for the first user,i.e., ππ > π1. Therefore, according to Lemma 4, over theduration [0, π π1), the optimal policy is to allocate the entireπ1 to the first user only. We allocate π1 to the first user,calculate the number of bits departed for the first user, andremove them from π΅1. This simply reduces the problem to thatof transmitting (π΅β²
1, π΅2) bits starting at time π‘ = π π1 , whereπ΅β²
1 = π΅1 β π(π1)π π1 . The process is illustrated in Fig. 7.Then, the minimum transmission completion time is
π = π ππΏ +π΅1 β
βπΏπ=1 π(ππ)(π ππ β π ππβ1
)
π(ππ)(37)
where πΏ is the number of recursions needed to get ππ. Inboth scenarios, we reduce the problem into a simple form,and obtain the final optimal policy. We state our algorithmformally as Algorithm 1 below.
We have the following theorem which proves the optimalityof the proposed algorithm. We provide the proof of thistheorem in Appendix C.
578 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012
sK
Β· Β· Β·
Β· Β· Β·
EK
t0
E1 E2 E3
s2 s3 s4
E4
s1
(B1, B2)
T
P
P3
P1
P2
si1 si2
Pc
E0
Fig. 7. Search for the cut-off power level ππ iteratively.
Algorithm 1 The algorithm to minimize the transmissioncompletion time for a given (π΅1, π΅2)
Initialization: Set π = 0, π π0 = 0, π0 = 0.while π΅1 > 0 doπ = π+ 1;Determine ππ and π ππ through the single-user method in[1], [2].Set ππ = ππ, π1 = π΅1
π(ππ).
Calculate π·2(π1, ππ), the maximum departures from thesecond user by π1 + π ππβ1 given ππ.if π΅2 β₯ π·2(π1, ππ) then
Allocate ππ to the first user over [π ππβ1 , π ππ), updateπ΅1.
elseSearch for ππ β [ππβ1, ππ] s.t. π·2(π1, ππ) = π΅2
through bisection method.end if
end while
Theorem 1 The algorithm is feasible and optimal.
B. Generalizing the Algorithm for π Users
In the π -user case, there are π β 1 cut-off power levelsas shown in Fig. 6. Using this structure, we generalize thealgorithm to find the minimum π . We first determine the totalpower level at the first epoch using the same approach withπ(π)(π1, . . . , ππ ) function. In particular, we assume that theenergies are available at time π‘ = 0 and calculate the energynecessary to send (π΅1, . . . , π΅π ) by time π‘ = π 1. Comparingthis energy with πΈ0, we decide if minimum π is smaller orlarger than π 1 and proceed similarly to the two-user case todetermine the initial total power level π1.
Having determined π1, we now decide whether ππ1 > π1
or otherwise. We set ππ1 = π1 and calculate
π1 =π΅1
π(ππ1)(38)
We need to determine whether the remaining bits(π΅2, . . . , π΅π ) can be sent by π1. Because of the cut-off structure of the optimal policy in the π -user scenario,we search for the cut-off power level for users 2, 3, . . ., πin a sequential way. The cut-off power levels are selected toensure that the maximum number of departures from the πthuser equals π΅π, π β₯ 2. If such feasible cut-off power levels
for all of the remaining π β 1 users can be obtained over[0, π1), it implies that the minimum transmission completiontime π β€ π1, and ππ1 β€ π1. Otherwise, once we find thatthe cut-off power level for a user is infeasible, it implies thatnot all of the users can be served by π1, and ππ1 > π1.
First, we obtain the optimal total power allocation bydeadline π1, denoted as π β
2 , . . . , πβπ(π1)
. We set ππ2 = π β2 . If
π(ππ2βππ1
ππ1+π22)(π1 β π π1) > π΅2, the optimal cut-off power level
ππ2 β [ππ1, πβ2 ] and it satisfies π(ππ2βππ1
ππ1+π22)(π1 β π π1) = π΅2.
Otherwise, we set ππ2 = π β3 and repeat the same procedure
until the optimal ππ2 is achieved. The remaining cut-off powerlevels can be determined in a similar manner. If any of theseoptimal power levels becomes infeasible, i.e., takes a valueabove π β
π(π1), we get back to ππ1 and adjust it accordingly.
The feasibility and the optimality of the algorithm can beproved through similar steps in Appendix C and is omittedfor the brevity of the paper.
C. Computation Requirement of the Proposed Algorithm
In a real life implementation of the algorithm, the iterationsof the algorithm are stopped when sufficient accuracy isreached. For given ππ1, the main computational block of thealgorithm is composed of two stages: In the first stage, wecalculate the single-user optimal solution, i.e., the optimalpower sequence π β
1 , . . . , πβπ(π1)
given the deadline π1. Inthe second stage, we determine whether the remaining bits(π΅2, . . . , π΅π ) can be sent by π1. The optimal power sequenceπ β1 , . . . , π
βπ(π1)
can be found as in [1], [2] by a geometricframework which requires drawing lines to the energy arrivalpoints and selecting the line with the minimum slope. Denot-ing the number of epochs as π , calculation of the minimum ofπ(π) elements requires π(π) comparisons and we repeat itπ(π) times. Hence, the first stage of the main computationalblock requires π(π2) operations. We can perform the single-user optimization also by the directional water-filling algo-rithm in [11] which also has π(π2) complexity. In the secondstage, the algorithm goes through π β
π , calculates the remainingcut-off power levels and determines whether the remainingbits (π΅2, . . . , π΅π ) can be sent by π1. Since the optimal cut-off power level increases as the user index increases, and wealways start with the lowest possible power level, this stagerequires π(π + π) operations in the worst case. Hence,the main block of the algorithm requires π(π2 + π + π)computations.
The main block is run each time the first userβs cut-offpower level ππ1 is updated. In the algorithm that we presented,ππ1 is updated recursively as in (37). After deciding the rangethat ππ1 lies, we perform bisection method until the desiredaccuracy is reached. The number of recursions πΏ in (37) isπ(π) in the worst case and the bisection search is at mostπ(log(1π )) where π is the desired accuracy. Therefore, the totalnumber of iterations scale as π(π) and the algorithm requiresπ((π2+π+π)(πβlog π)) operations overall. Here, the totalnumber of epochs πΎ upper bounds the number of epochs π .Therefore, the overall number of operations needed is upperbounded by π((πΎ2 +π +πΎ)(πΎ β log π)).
YANG et al.: BROADCASTING WITH AN ENERGY HARVESTING RECHARGEABLE TRANSMITTER 579
VI. SIMULATIONS
We consider a band-limited AWGN broadcast channel withπ = 3 users. The bandwidth is π΅π = 1 MHz and the noisepower spectral density is π0 = 10β19 W/Hz. We assume thatthe path losses between the transmitter and the receivers are100 dB, 105 dB and 110 dB.
π1 = π΅π log2
(1 +
πΌ1πβ1π0π΅π
)= log2
(1 +
πΌ1π
10β3
)Mbps (39)
π2 = π΅π log2
(1 +
πΌ2πβ2πΌ1πβ2 +π0π΅π
)= log2
(1 +
πΌ2π
πΌ1π + 10β2.5
)Mbps (40)
π3 = π΅π log2
(1 +
(1β πΌ1 β πΌ2)πβ2(πΌ1 + πΌ2)πβ2 +π0π΅π
)= log2
(1 +
(1β πΌ1 β πΌ2)π
(πΌ1 + πΌ2)π + 10β2
)Mbps (41)
Therefore, we have
π(π1, π2, π3) = 10β32π1+π2+π3 + (10β2.5 β 10β3)2π2+π3
+ (10β2 β 10β2.5)2π3 β 10β2 W (42)
A. Deterministic Energy Arrivals
In this subsection, we illustrate the off-line optimal pol-icy in a deterministic energy arrival sequence setting. Inparticular, we assume that at times t = [0, 5, 6, 8, 9, 11] s,energies with the following amounts are harvested: E =[20, 10, 3.5, 8, 10, 10] mJ.
We first study the two-user broadcast channel by removingthe third user, i.e., setting π΅3 = 0. We find the maximumdeparture region of the two-user broadcast channel π(π ) forπ = 6, 8, 9, 10 s, and plot them in Fig. 8. Note that themaximum departure regions are convex, and as π increases,π(π ) monotonically expands.
We next consider the same energy arrival sequence with(π΅1, π΅2) = (21, 2) Mbits. We have the optimal transmissionpolicy, as shown in Fig. 9. In this example, the cut-off poweris less than π1 and hence the power share (and the rate) ofthe first user is constant throughout the interval in which thebits are transmitted. The transmitter finishes its transmissionby time π = 9.28 s, and the last energy harvest is not used.Note that (21, 2) Mbits point (marked with *) in Fig. 8 is notincluded in π(π ) at π = 9 s while it is strictly included inπ(π ) at π = 10 s.
Finally, we consider the same energy arrival sequence with(π΅1, π΅2, π΅3) = (12, 6, 3) Mbits and the optimal policy isshown in Fig. 10. We calculate the cut-off power levels asππ1 = 0.963 mW and ππ2 = 2.619 mW. The bits of all threeusers are always transmitted throughout the communication.The last energy arrival is used in this case and the transmissionis finished by π = 12.33 s.
B. Stochastic Energy Arrivals
In this subsection, we consider stochastic energy arrivalsin the two-user case, i.e., we set π΅3 = 0. We compare
0 5 10 15 20 25 300
2
4
6
8
10
12
14
B1 (Mbits)
B2 (
Mbi
ts)
T=6 sT=8 sT=9 sT=10 s
Fig. 8. The maximum departure region of the broadcast channel for variousπ .
20
50 86
10
9
10
t
(B1, B2) = (21, 2)
P
11
10 3.5 8
Pc
T = 9.28
44.5
8
35.5
T
Fig. 9. Cut-off power ππ = 3.798 mW. Optimal trans-mit rates r1 = [2.262, 2.262, 2.262, 2.262] Mbps and r2 =[0.041, 0.1386, 0.6814, 2.4723] Mbps, with durations l = [5, 3, 1, 0.28] s.
the performance of the off-line optimal policy with those ofthree suboptimal policies. These policies are inspired by theoptimal off-line policy while they require partial or no off-lineknowledge of the energy arrivals.
1) Constant Power Constant Share (CPCS) Policy: Thispolicy transmits with constant power equal to the averagerecharge rate, π = πΌ[πΈ], whenever the battery energy is non-zero and the transmitter is silent otherwise. In addition, thestrong userβs power share is constant whenever the transmitteris non-silent. In particular, the constant power share is set toπΌβ which is the solution of the following equation:
π΅1
π΅2=
log2
(1 + πΌπΌ[πΈ]
π21
)log2
(1 + (1βπΌ)πΌ[πΈ]
πΌπΌ[πΈ]+π22
) (43)
Note that CPCS does not require off-line or on-line knowledgeof the energy arrivals. It requires the first order statistics, i.e.,the mean, of the energy arrival process.
2) Greedy Power Constant Share (GPCS) Policy: Thispolicy also keeps the total transmit power share constant equalto the solution of (43) throughout the transmission. However,different from the CPCS policy, the power is greedily updated.The policy assumes knowledge of the time instant at which thenext energy arrival occurs and at the start of the πth epoch, theavailable energy is allocated to the next epoch only and hencethe power is set to ππ =
πΈπβ1
βπ. Note that GPCS requires partial
580 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012
20
50 86
10
9
10
t
(B1, B2, B3) = (12, 6, 3)
P
T = 12.33
11
4.5
Pc1
4
Pc2
10 3.5 8
6
7.5
T
Fig. 10. Cut-off power levels ππ1 = 0.963 mW and ππ2 =2.619 mW. Optimal transmit rates r1 = [0.9731, 0.9731, 0.9731, 0.9731]Mbps, r2 = [0.4869, 0.4869, 0.4869, 0.4869] Mbps and r3 =[0.1498, 0.2005, 0.3425, 0.4718] Mbps with durations l = [5, 3, 3, 1.33] s.
off-line knowledge of the energy arrivals as well as the firstorder statistics, i.e., the mean, of the energy arrival process.
3) Greedy Power Dynamic Share (GPDS) Policy: Thispolicy allocates power greedily in each epoch and the powershares are dynamically updated. In particular, the policy as-sumes knowledge of the time instant at which the next energyarrival occurs and at the start of the πth epoch, the arrivingenergy πΈπβ1 is spread over βπ. For ππ =
πΈπβ1
βπ, πΌβ
π is calculatedas the solution of the following equation:
π΅1π
π΅2π=
log2 (1 + πΌππ)
log2
(1 + (1βπΌ)ππ
πΌππ+π2
) (44)
where π΅1π and π΅2π are the remaining bits of user 1 and user 2,respectively, at the beginning of epoch π. Note that this policyessentially performs the initialization in [17] and it requirespartial off-line knowledge of the energy arrivals as well ason-line knowledge of the data backlog.
In the simulations, we consider a compound Poisson energyarrival process. The average inter-arrival time is 1 s and thearriving energy is a random variable which is distributeduniformly in [0, 2πππ£π] mJ, where πππ£π is the average rechargerate. The performance metric of the policies is the averagetransmission completion time over 1000 realizations of thestochastic energy arrival process. We first set the π = π΅1
π΅2ratio
constant, i.e., π΅1 = ππ΅2. We plot the performances for π = 3and πππ£π = 1 mJ/s with varying π΅2 in Fig. 11. We observethat the average transmission completion times of the policiesincrease with the number of bits. It is notable that CPCS policyperforms better with respect to the greedy power policies eventhough greedy power policies use partial off-line information.Hence, transmitting with constant power proves to be usefulas observed in the single-user case in [1], [2], [11]. We alsoobserve that dynamically varying the power shares of the usersyields better performance compared to the constant case. Next,we plot the variation of the average transmission completiontime with respect to the average recharge rate in Fig. 12.CPCS policy performs better with respect to the greedy powerpolicies for small average recharge rates; however, GPDSpolicy outperforms CPCS in the high average recharge rateregime. Therefore, adapting the power share according to userloads proves to be useful in the high average recharge rateregime.
1 2 3 4 5 6 7 8 9 100
10
20
30
40
50
60
70
B2 (Mbits)
aver
age
tran
smis
sion
com
plet
ion
time
(s)
Optimal offβlineCPCSGPDSGPCS
Fig. 11. Average transmission completion time versus π΅2 when π = 3 andπππ£π = 1 mJ/s.
1 2 3 4 5 6 7 8 9 105
10
15
20
25
30
average recharge rate (mJ / s)
aver
age
tran
smis
sion
com
plet
ion
time
(s)
Optimal offβlineCPCSGPDSGPCS
Fig. 12. Average transmission completion time versus average recharge ratewhen π΅1 = 12 Mbits and π΅2 = 4 Mbits.
VII. CONCLUSIONS
We investigated the transmission completion time mini-mization problem in an energy harvesting broadcast channel.We first analyzed the structural properties of the optimaltransmission policy, and proved that the optimal total transmitpower has the same structure as in the single-user channel. Wealso proved that there exists a cut-off power for the strongeruser. If the optimal total transmit power is lower than thiscut-off level, all power is allocated to the stronger user, andotherwise, all power above this level is allocated to the weakeruser. We then extended our results to the π -user broadcastchannel for which total power sequence has the same structureas the two-user case and the optimal splitting of the total powerrequires π β 1 cut-off levels. Based on the structure of theoptimal policy, we developed an iterative algorithm to obtainthe globally optimal off-line transmission policy. Finally, weprovided an extensive numerical analysis of the optimal policyand compared its performance with suboptimal policies underdifferent settings.
YANG et al.: BROADCASTING WITH AN ENERGY HARVESTING RECHARGEABLE TRANSMITTER 581
APPENDIX APROOF OF LEMMA 3
According to the expression of π(π1π, π2π) in (5) and theKKT conditions in (12)-(13), we have
π(π1π, π2π) =π2 + πΎ2πβπ
π=π ππβ π2
2 (45)
β₯ π21
(22(π1π+π2π) β 1
)(46)
=π1 + πΎ1πβπ
π=π ππβ π2
1 (47)
β₯ π1βππ=π ππ
β π21 (48)
where (46) becomes an equality when π2π = 0. Therefore,when π2π > 0, (45)-(48) imply
π(π1π, π2π) =π2βππ=π ππ
β π22 >
π1βππ=π ππ
β π21 (49)
When π2π = 0, we must have π1π > 0. Otherwise, ifπ1π = 0, we can always let the weaker user transmit with somepower over this duration without contradicting with any energyconstraints. Since there is no interference from the strongeruser, the departure from the weaker user can be improved,thus it contradicts with the optimality of the policy. Therefore,when π2π = 0, πΎ1π = 0, and (45)-(48) imply
π(π1π, π2π) =π1βππ=π ππ
β π21 >
π2βππ=π ππ
β π22 (50)
Therefore, we can express π(π1π, π2π) in terms of theLagrange multipliers as follows:
π(π1π, π2π) = max
{π1βππ=π ππ
β π21 ,
π2βππ=π ππ
β π22
}(51)
If π2βππ=π ππ
β π22 >
π1βππ=π ππ
β π21 for some οΏ½ΜοΏ½, then, we have
π2 β π1βππ=π ππ
β₯ π2 β π1βππ=οΏ½ΜοΏ½ ππ
> π22 β π2
1 , βπ > οΏ½ΜοΏ½ (52)
where the first inequality follows from ππ β₯ 0 for π =1, 2, . . .π . Therefore, we conclude that there exists an integerοΏ½ΜοΏ½, 0 β€ οΏ½ΜοΏ½ β€ π , such that, when π β€ οΏ½ΜοΏ½, π2π = 0; and whenπ > οΏ½ΜοΏ½, π2π > 0.
Furthermore, (49)-(50) imply that the energy constraint atπ‘ = π οΏ½ΜοΏ½ must be tight. Otherwise, by the complementaryslackness conditions in (14), ποΏ½ΜοΏ½ = 0, and (50) implies
π(π1οΏ½ΜοΏ½, π2οΏ½ΜοΏ½) =π1βπ
π=οΏ½ΜοΏ½+1 ππβ π2
1
>π2βπ
π=οΏ½ΜοΏ½+1 ππβ π2
2 = π(π1,οΏ½ΜοΏ½+1, π2,οΏ½ΜοΏ½+1) (53)
which contradicts with (49). Therefore, in the following, whenwe consider the energy constraints, we only need to considertwo segments [0, π οΏ½ΜοΏ½) and [π οΏ½ΜοΏ½+1, π π) separately.
When π < οΏ½ΜοΏ½, based on (49), if ππ = 0, we haveπ(π1π, π2π) = π(π1,π+1, π2,π+1). Starting from π = 1,π(π1π, π2π) remains constant until an energy constraint be-comes tight. Therefore, between any two consecutive epochs,
when the energy constraints are tight, the power level remainsconstant. Similar arguments hold when π β₯ οΏ½ΜοΏ½. Thus, thecorresponding power level is
ππ =
βππβ1π=ππβ1
πΈπ
π ππ β π ππβ1
(54)
where π ππβ1 and π ππ are two consecutive epochs with tightenergy constraint.
Finally, we need to determine the epochs when the energyconstraint becomes tight. We observe that π(π1οΏ½ΜοΏ½, π2οΏ½ΜοΏ½) mustmonotonically increase in π, as ππ β₯ 0. Hence, the individualterms in the max{., .} function in (51) are monotonicallyincreasing. In addition, both terms in the max{., .} functionstrictly increases when energy constraint becomes tight. There-fore, we conclude that
ππ = arg minππβ1<πβ€π
{βπβ1π=ππβ1
πΈπ
π π β π ππβ1
}(55)
This completes the proof.
APPENDIX BTHE CUT-OFF POWER LEVELS
We solve the π -variable local optimization problem in (26)by using a Lagrangian analysis. Specifically, the Lagrangianfunction is
π»(π1π, . . . , πππ,πΆ, πΎ) =
πβπ=1
πππππ +
πβπ=1
πΌππππ
β πΎ[π(π)(π1π, . . . , πππ)β ππ]
where the Lagrange multipliers πΎ and πΆ = [πΌ1, . . . , πΌπ ] sat-isfy πΎ, πΌπ β₯ 0 with the complimentary slackness conditions
πΎ[π(π)(π1π, . . . , πππ)β ππ] = 0, πΌππππ = 0 (56)
Taking the derivative of the Lagrangian π» with respect toπππ, and setting them to zero, we have
ππ + πΌπ β πΎβ²22βπ
π=π πππ(π(πβ1) + π2π) = 0, (57)
for π = 1, 2, . . . ,π , where πΎβ² = (2 ln 2)πΎ. Because of thenonnegativity of ππ and πΌπ, in order to have a solutionsatisfying the KKT conditions, we must have πΎβ² > 0. Then,considering two consecutive equations, we have
ππ + πΌπ
ππβ1 + πΌπβ1=
π(πβ1) + π2π
π(πβ2) + π2πβ1
β 1
22π(πβ1)π
=π(πβ1) + π2
π
π(πβ2) + π2πβ1
β 1
1 +π(πβ1)βπ(πβ2)
π(πβ2)+π2πβ1
=π(πβ1) + π2
π
π(πβ1) + π2πβ1
(58)
If πΌπ = 0, i.e., π(π) > π(πβ1), we have
π(πβ1) = max
{(ππβ1π
2π β πππ
2πβ1
ππ β ππβ1
)+
, π(πβ2)
}
582 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 11, NO. 2, FEBRUARY 2012
If πΌπ β= 0, i.e., π(π) = π(πβ1), and πβ 1 is the smallestuser index with π(πβ1) > π(πβ1), we have
ππβ1
ππβ1 + πΌπβ1=π(πβ1β1) + π2
πβ1
π(πβ2) + π2πβ1
β 1
22π(πβ1)π
=π(πβ1) + π2
πβ1
π(πβ1) + π2πβ1
(59)
where (59) follows from the fact that π(πβ1β1) = π(πβ1).Thus,
π(πβ1) = max
{(ππβ1π
2πβ1
β ππβ1π2πβ1
ππβ1 β ππβ1
)+
, π(πβ2)
}Therefore, the cut-off power levels are determined in thegeneral form as in (27).
APPENDIX CPROOF OF THEOREM 1
Before we proceed to prove the optimality of the algorithm,we introduce the following lemma first, which is useful in theproof of the optimality of the algorithm.
Lemma 6 For any πΌ β [0, 1], π(
πΌπΈ/π(1βπΌ)πΈ/π+π2
)π monoton-
ically increases in π .
Proof: The monotonicity can be verified by taking derivatives.We have(
π
(πΌπΈ/π
(1 β πΌ)πΈ/π + π2
)π
)β²
=1
2log2
(π2 + πΈ/π
)β 1
2log2
(π2 + (1β πΌ)πΈ/π
)β πΈ
2 ln 2
πΈ
πΈ + π2π+
πΈ
2 ln 2
(1β πΌ)πΈ
(1β πΌ)πΈ + π2π(60)
and(π
(πΌπΈ/π
(1β πΌ)πΈ/π + π2
)π
)β²β²
=πΈ2
2π ln 2
(1
(π2π/(1β πΌ) + πΈ)2β 1
(π2π + πΈ)2
)< 0
where the last inequality follows as πΈ > 0. Hence,π(
πΌπΈ/π(1βπΌ)πΈ/π+π2
)π is a strictly concave function of π . Note
that each term in the right hand side of (60) goes to 0 asπ β β. Thus,
limπββ
(π
(πΌπΈ/π
(1β πΌ)πΈ/π + π2
)π
)β²= 0.
Combining with the strict concavity, we conclude that the firstderivative is positive when π < β, and the monotonicityfollows. β
We then prove the optimality of the algorithm. In order toprove that the algorithm is optimal, we need to prove that π1
is optimal. Once we prove the optimality of π1, the optimalityof π2, π3, . . . follows. Since the solution obtained usingour algorithm always has the optimal structure described inLemma 4, the optimality of the power allocation also impliesthe optimality of the rate selection, thus, the optimality of the
algorithm follows. Therefore, in the following, we prove thatπ1 is optimal.
First, we note that π1 is the minimal slope up to π . Weneed to prove that π1 is also the minimal slope up to the finaltransmission completion time, π . Let us define π β² as follows
π β² =βοΏ½ΜοΏ½1
π=0πΈπ
π1(61)
Assume that with π1, we allocate πΌπ1 to the first user, andfinish (π΅1, π΅2) using constant rates. Then, we allocate πΌπ1
to the first user, and the rest to the second user. Based onLemma 6, we have
π(πΌπ1)πβ² β₯ π(πΌπ1)π = π΅1 (62)
π
(πΌπ1
(1 β πΌ)π1 + π2
)π β² β₯ π
(πΌπ1
(1β πΌ)π1 + π2
)π = π΅2
(63)
Therefore, π β² is an upper bound for the optimal transmissioncompletion time. Since π1 is the minimal slope up to π β²,we conclude that π1 is optimal throughout the transmission.Following similar arguments, we can prove the optimality ofthe rest of the power allocations. This completes the proof ofoptimality.
In order to prove that the allocation is feasible, we need toshow that the power allocation for the first user is alwaysfeasible in each step. Therefore, in the following, we firstprove that π1 is feasible when we assume that ππ = π1. Thefeasibility of π1 also implies the feasibility of the rest of thepower allocation. With the assumption that ππ = π1, the finaltransmission time for the first user is
π1 =π΅1
π(π1)β€ π΅1
π(πΌπ1)(64)
Based on (62) and (64), we know that π1 < π β². Sinceπ1 is feasible up to π β², therefore, π1 is feasible when weassume that ππ = π1. The feasibility of the rest of thepower allocations follows in a similar way. This completesthe feasibility part of the proof.
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Jing Yang received the B.S. degree in electronicengineering and information science from the Uni-versity of Science and Technology of China, Hefei,China in 2004, and the M.S. and Ph.D. degreesin electrical and computer engineering from theUniversity of Maryland, College Park in 2010. SinceOctober 2010, she has been a research associate inthe department of electrical and computer engineer-ing at the University of Wisconsin-Madison. Herresearch interests are in learning, inference, opti-mization, and scheduling in complex networks, with
particular emphasis on energy management in energy harvesting networks.
Omur Ozel received the B.Sc. and the M.S. degreeswith honors in electrical and electronics engineeringfrom the Middle East Technical University (METU),Ankara, Turkey, in June 2007 and July 2009, respec-tively. Since August 2009, he has been a graduateresearch assistant at the University of Maryland Col-lege Park, working towards Ph.D. degree in electri-cal and computer engineering. His research focuseson information and network theoretical aspects ofenergy harvesting communication systems. He is astudent member of IEEE and a member of IEEE
Information Theory Society.
Sennur Ulukus is a Professor of Electrical andComputer Engineering at the University of Mary-land at College Park, where she also holds a jointappointment with the Institute for Systems Research(ISR). Prior to joining UMD, she was a SeniorTechnical Staff Member at AT&T Labs-Research.She received her Ph.D. degree in Electrical andComputer Engineering from Wireless InformationNetwork Laboratory (WINLAB), Rutgers Univer-sity, and B.S. and M.S. degrees in Electrical andElectronics Engineering from Bilkent University.
Her research interests are in wireless communication theory and networking,network information theory for wireless communications, signal processingfor wireless communications, information-theoretic physical-layer security,and energy-harvesting communications.
Dr. Ulukus received the 2003 IEEE Marconi Prize Paper Award in WirelessCommunications, the 2005 NSF CAREER Award, and the 2010-2011 ISROutstanding Systems Engineering Faculty Award. She served as an AssociateEditor for the IEEE TRANSACTIONS ON INFORMATION THEORY between2007-2010, as an Associate Editor for the IEEE TRANSACTIONS ON COM-MUNICATIONS between 2003-2007, as a Guest Editor for the Journal ofCommunications and Networks for the special issue on energy harvestingin wireless networks, as a Guest Editor for the IEEE TRANSACTIONS ON
INFORMATION THEORY for the special issue on interference networks, asa Guest Editor for the IEEE JOURNAL ON SELECTED AREAS IN COM-MUNICATIONS for the special issue on multiuser detection for advancedcommunication systems and networks. She served as the TPC co-chair of theCommunication Theory Symposium at the 2007 IEEE Global Telecommuni-cations Conference, the Medium Access Control (MAC) Track at the 2008IEEE Wireless Communications and Networking Conference, the WirelessCommunications Symposium at the 2010 IEEE International Conference onCommunications, the 2011 Communication Theory Workshop, the Physical-Layer Security Workshop at the 2011 IEEE International Conference onCommunications, the Physical-Layer Security Workshop at the 2011 IEEEGlobal Telecommunications Conference. She was the Secretary of the IEEECommunication Theory Technical Committee (CTTC) in 2007-2009.