research article optimal resource allocation for energy...

Research ArticleOptimal Resource Allocation for Energy-EfficientOFDMA Networks

Fan Wu Yuming Mao Xiaoyan Huang and Supeng Leng

Key Laboratory of Optical Fiber Sensing and Communications School of Communication and Information EngineeringUniversity of Electronic Science and Technology of China Chengdu 611731 China

Correspondence should be addressed to Supeng Leng splenguestceducn

Received 22 September 2014 Revised 27 February 2015 Accepted 1 March 2015

Academic Editor Alex Elıas-Zuniga

Copyright copy 2015 Fan Wu et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper focuses on radio resource allocation in OFDMA networks for maximizing the energy efficiency subject to the data raterequirements of users We propose the energy-efficient water-filling structure to obtain the closed-form optimal energy-efficientpower allocation for a given subcarrier assignment Moreover we establish a new sufficient condition for the optimal energy-efficient subcarrier assignment Based on the theoretical analysis we develop a joint energy-efficient resource allocation (JERA)algorithm to maximize the energy efficiency Simulation results show that the JERA algorithm can yield optimal solution withsignificantly low computational complexity

1 Introduction

As the high data rate applications are going to dominate themobile services energy efficiency (EE) is becoming morecrucial inwireless communication networks Energy-efficientradio resource allocation is one of the effective ways toimprove the EE of the OFDMA (orthogonal frequency divi-sion multiple access) networks [1] Although radio resourceallocation in OFDMAnetworks has been extensively studiedthemajor focus is on improving spectral efficiency (SE)whichmay not always coincide with EE [2]

Different from SE-based resource allocation schemesin which the total transmitting power is fixed the EE-based schemes adjust the power level adaptively based onthe channel conditions [3] Accordingly the classic water-filling power allocation method cannot be applied directlydue to the unknown total transmitting power In order todetermine the proper transmitting power the perturbationfunctions of EE have been studied in [4 5] It has been shownthat EE is strictly quasi-concave in SE [4] and in the totaltransmitting power [5] However the perturbation functionsof EE cannot be expressed in an analytic expression only

the approximation algorithms have been proposed to find thenear-optimal solutions [4ndash6]

In this paper we focus on joint subcarrier assignmentand power allocation in OFDMA networks for maximiz-ing the energy efficiency problem subject to the data raterequirements of users The main contributions of our workare summarized as follows

(i) We prove that EE is strictly pseudo-concave withrespect to power vector for a given subcarrier allo-cation which guarantees that the solution satisfyingthe KKT conditions is also the global optimal Usingthis property we show that the optimal solution hasa special EE water-filling structure that is determinedby only one variable Based on this observation wefurther provide the first closed-form expression forthe optimal energy-efficient power allocation

(ii) According to the analysis we propose an opti-mal energy-efficient power allocation algorithm bysequentially searching within a finite number ofwater-level intervals The computational complexityof the proposed algorithm is much lower than that of

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 594024 10 pageshttpdxdoiorg1011552015594024

2 Mathematical Problems in Engineering

the algorithms approaching the optimal solution withiterative searching method

(iii) We provide a sufficient condition for the optimal en-ergy-efficient subcarrier assignment This conditionis the basis of the quick search method because itallows us to easily determine whether a subcarrierassignment is optimal As a result by combining thesufficient condition and the EE water-filling solutionwe design a joint energy-efficient resource allocation(JERA) algorithm which not only achieves the opti-mal EE but also outperforms the existing algorithmsin terms of convergence rate

The state of the art to solve the EEmaximization problemis JIOO (joint inner- and outer-layer optimization) algorithm[4 5] Relying on the strict quasi-concavity of the pertur-bation function of EE with respect to the total transmittingpower the JIOO algorithm solves the problem via two-layeriterations where the inner layer aims to find the maximumEE and its derivative for a given total transmitting powerand the outer layer targets to search for the total transmittingpower which results in the maximum EE by bisection searchThe computational complexity is mainly determined by theouter layer iterations Unfortunately the number of the outerlayer iterations becomes infinite when tolerance approacheszero Different from this approach the complexity of ouralgorithm depends on the finite number of the feasiblesubcarrier assignments Moreover the search space can begreatly reduced by using the sufficient condition provided bythis paper In addition the EE water-filling structure ensuresthat the results obtained by our algorithm are the exactoptimal solutions other than approximation ones

The rest of the paper is organized as follows In Section 2we describe the system model and formulate the EE maxi-mization problem for the downlink OFDMA networks Theoptimal energy-efficient power allocation for a given subcar-rier assignment and the sufficient condition for the optimalenergy-efficient subcarrier assignment are studied in Sections3 and 4 respectively An optimal energy-efficient powerallocation algorithm and a low-complexity joint resourceallocation algorithm are elaborated in Section 5 Section 6provides the performance evaluation of the proposed algo-rithms via simulations Finally we conclude this paper inSection 7

2 System Model

Consider a downlink OFDMA network with one base stationand 119870 users Let K = 1 119870 and N = 1 119873

be the set of users and subcarriers respectively Define thesubcarrier assignment matrix 120588 = (120588119896119899)119870times119873 where 120588119896119899 = 1

means that subcarrier 119899 is allocated to user 119896 and otherwise120588119896119899 = 0 The transmitting power allocation matrix isdefined as p = (119901119896119899)119870times119873 where 119901119896119899 ge 0 represents thetransmitting power allocated to user 119896 on subcarrier 119899 Thenthe maximum achievable data rate of user 119896 on subcarrier 119899

is given by

119903119896119899 = 119861log2

(1 + 119901119896119899120574119896119899) (1)

where 119861 is the bandwidth of subcarrier and 120574119896119899 denotes thenormalized channel power gain of user 119896 on subcarrier 119899Accordingly the overall system data rate is

119877 =

119870

sum

119896=1

119873

sum

119899=1

119903119896119899 sdot 120588119896119899 (2)

and the total transmitting power is

119875119879 =

119870

sum

119896=1

119873

sum

119899=1

119901119896119899 sdot 120588119896119899 (3)

In addition to transmitting power the energy consump-tion also includes circuit power which is consumed by deviceelectronics The circuit power is modeled as a constant119875119862 which is independent of data transmission rate [7]Accordingly we define EE as the amount of bits transmittedper Joule of energy that is 120578EE ≜ 119877(120577119875119879+ 119875119862) where 120577 is thereciprocal of drain efficiency of power amplifier

In our work we consider maximizing EE under theminimum data rate requirements 119877119896 and the total transmit-ting power constraint 119875max Accordingly this optimizationproblem can be formulated as

P1 maxp120588

sum119870

119896=1sum119873

119899=1119903119896119899 sdot 120588119896119899

120577119875119879 + 119875119862

(4a)

st119873

sum

119899=1

119903119896119899 sdot 120588119896119899 ge 119877119896 forall119896 isin K (4b)

119875119879 ≜

119870

sum

119896=1

119873

sum

119899=1

119901119896119899 sdot 120588119896119899 le 119875max (4c)

119870

sum

119896=1

120588119896119899 = 1 forall119899 isin N (4d)

120588119896119899 isin 0 1 119901119896119899 ge 0 forall119896 isin K forall119899 isin N

(4e)

where (4b) indicates the minimum data rate requirement ofeach user (4c) is the total transmitting power constraint and(4d) is the constraint on subcarrier assignment to ensure thateach subcarrier is only assigned to one user

Similar to the traditional spectral-efficient resourcemodel P1 is a mixed integer nonlinear programming prob-lem and it is not trivial to obtain the global optimal solutionto this problem To solve the problem we first decompose P1into two subproblems which include (1) the energy-efficientpower allocation for a fixed subcarrier assignment 120588 and (2)the energy-efficient subcarrier assignment for a given totaltransmitting power 119875119879 Then based on the properties of thesubproblems we develop an algorithm to find the solutionof joint energy-efficient power allocation and subcarrierassignment to maximize EE

3 Optimal Energy-Efficient Power Allocation

In this section we analyze the optimal energy-efficient powerallocation (EPA) based on the EE water-filling structure All

Mathematical Problems in Engineering 3

the major results are given by some theorems In particularTheorems 2 and 5 demonstrate that the global optimalsolution to the energy-efficient power optimization problemis with the EE water-filling structureTheorem 3 provides thecorresponding closed-form water-level whose optimality isproved byTheorem 4

31 EE Water-Filling Structure Given the subcarrier assign-ment matrix 120588 the set of subcarriers assigned to user 119896 canbe denoted by N120588(119896) = 119899 | 120588119896119899 = 1 and the power vectorp = (1199011 sdot sdot sdot 119901119899)

119879 Then P1 is reduced to the following EPAproblem

P2 120578(120588)

EE ≜ maxpge0

120578(120588)

EE (p)

st 119903119896 (p) ge 119877119896 forall119896 isin K

119901119879 (p) le 119875max

(5)

where 120578(120588)

EE (p) ≜ sum119873

119899=1(119861log2(1 + 120574119899119901119899)(120577 sum

119873

119899=1119901119899 + 119875119862))

with 120574119899 = 120574119896(119899)119899 and 119896(119899) denotes the index of the userassigned on subcarrier 119899 119903119896(p) ≜ sum

119899isinN120588(119896) 119861log2(1 + 120574119899119901119899)and 119901119879(p) ≜ sum

119873

119899=1119901119899 Since the numerator of 120578

(120588)

EE (p) isdifferentiable positive and strictly concave function in pand the denominator is positive and affine in p 120578

(120588)

EE (p) isa strictly pseudo-concave function with respect to p [8]Besides 119903119896(p) is differentiable and concave for all 119896 and119901119879(p) is positive and affine Therefore according to the KKTsufficient optimality theorem [9] any feasible solution psatisfying the KKT conditions is also globally optimal for P2

When the feasible solution set of P2 is nonempty theminimum power vector p to guarantee the minimum data-rate requirement of each user must be a feasible solution toP2 which can be obtained by solving the following marginadaptive (MA) problem [10]

P3 119875min ≜ min119901119899ge0

119873

sum

119899=1

119901119899

st 119903119896 (p) = 119877119896 forall119896 isin K

(6)

Resorting to the Lagrange dual theory the optimal solutionto P3 is given by p = ((119896(119899) minus1120574119899)

+)119873times1 where 119896 is the root

of the equation

sum

119899isinN120588(119896)

119861log2

[1 + 120574119899 (119896 minus1

120574119899

)

+

] = 119877119896 forall119896 isin K (7)

We call 119896 the lowest power water-level of user 119896More importantly a series of feasible solutions to P2 can

be constructed based on p by raising the water-levels of someusers and maintaining that of the others To be specific bysorting all the users in ascending order of their lowest powerwater-levels such that 1 le sdot sdot sdot le 119870+1 with 119870+1 = +infinthe region of the promotable water-level can be divided into119870 intervals that is (1 2] (119870 119870+1] named as water-level rise interval For the 119897th (0 lt 119897 le 119870) interval given 119909 isin

(119897 119897+1] called EE water-level if we raise the water-levels of

the first 119897 users to 119909 while maintaining that of other 119870 minus 119897

users a new feasible power solution p(119909) can be obtained bythe following EE water-filling structure

119899 (119909) =

(119909 minus1

120574119899

)

+

119896 isin K1

(119896(119899) minus1

120574119899

)

+

119896 isin K2

(8)

whereK1 = 1 sdot sdot sdot 119897 andK2 = K minus K1 The correspondingdata rate of each user satisfies

119903119896 (p (119909))

gt 119877119896 119896 (119899) isin K1

= 119877119896 119896 (119899) isin K2(9)

It can be found that the data rate of each user with the samewater-level is greater than the minimum requirement whilethat of the other users is equal to the minimum requirement

It is noteworthy that since 119896 in (8) can be obtainedby solving (7) and hence p(119909) can be expressed solely asa function of the water-level 119909 accordingly P2 can betransformed into a single variable problem Furthermore forany given total transmitting power 119875119879 P2 is equivalent toP4 shown in the following If the optimal solution to P4 hasthe EE water-filling structure we can deduce that P2 mustbe maximized at a power vector with the EE water-fillingstructure It can be proved byTheorem 1

P4 max119901119899ge0

119877(120588)

(p)


119901119879 (p) = 119875119879

(10)

where 119877(120588)

(p) ≜ sum119873

119899=1119861log2(1 + 120574119899119901119899)

Theorem 1 Given 119875119879 ge 119875min the optimal solution plowast to P4has the EE water-filling structure (8)

Proof Since P4 is a convex programming the optimal solu-tion plowast must satisfy the KKT conditions that is there existscalars 119906119896 ge 0 (119896 = 1 sdot sdot sdot 119870) and 120582 such that

nabla119877(120588)

(plowast) +

119870

sum

119896=1

119906119896nabla119903119896 (plowast) + 120582nabla119901119879 (plowast) = 0 (11)

119906119896 sdot (119903119896 (plowast) minus 119877119896) = 0 119896 = 1 sdot sdot sdot 119870 (12)

According to (11) we have

119901lowast

119899= (119861 sdot

(1 + 119906119896(119899))

(ln 2 sdot 120582)minus

1

120574119899

)

+

forall119899 (13)

Besides based on the complementary slackness conditions(12) we can get that 119906119896 = 0 if 119903119896(plowast) gt 119877119896 Let 119909 ≜ 119861(ln 2 sdot 120582)

and 119909119896 ≜ (1 + 119906119896(119899))119909 Therefore plowast can be further expressedas follows

119899 (119909) =

(119909 minus1

120574119899

)

+

119896 (119899) isin K10158401

(119909119896(119899) minus1

120574119899

)

+

119896 (119899) isin K10158402

(14)


where K10158401

≜ 119896 | 119903119896(plowast) gt 119877119896 and K10158402

≜ 119896 | 119903119896(plowast) = 119877119896To maximize the overall data rate the power allocated to theusers in K1015840

2should be minimized And then 119909119896 = 119896 forall119896 isin

K10158402 Since 119906119896 ge 0 we get 119896 = 119909119896 ge 119909 On the other hand 119909 gt

119896 (forall119896 isin K10158401) in order to satisfy 119903119896(plowast) gt 119877119896 Consequently

K10158401

= K1 and K10158402

= K2 We can now conclude that plowast hasthe EE water-filling structure as (8)

According to the water-levels of p(119909) the subcarrier setcan be further divided into three subsets N0 = 119899 | 119901119899 =

0 119899 isin N N1 = 119899 | 119901119899 gt 0 119896(119899) isin K1 and N2 = 119899 |

119901119899 gt 0 119896(119899) isin K2 The partial derivatives of 120578(120588)

EE (p) havethe following properties which will be used in the proof ofthe following theorems

Property 1 If p(119909) is a feasible solution to P2 then

(a)

120597120578(120588)

EE (p (119909))

120597119901119899

=

119861 (ln 2 sdot 119909) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N1

119861 (ln 2 sdot 119896(119899)) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N2

119861120574119899 ln 2 minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N0

(15)

(b)

1198891 (p (119909)) = sdot sdot sdot = 119889119897 (p (119909)) ge 119889119897+1 (p (119909)) ge sdot sdot sdot ge 119889119870 (p (119909))

(16)

if 119897 lt 119909 le 119897+1 where 119889119896(p(119909)) ≜ maxforall119899isinN(119896)120597120578(120588)

EE (p(119909))120597119901119899

32 EE Water-Level Interval [ 119909] The minimum powervector p = p() is also with the EE water-filling structurewhose water-level is 1 and the total transmitting power is119875min On the other hand suppose p is the optimal solutionto P4 when 119875119879 = 119875max according to Theorem 1 we have p =

p(119909) where119909 is the EEwater-level Since the feasible region ofP1 is nonempty the total transmitting power 119875119879 must satisfy119875min le 119875119879 le 119875max Hence the corresponding power vectorwith the EE water-filling structure in the feasible region of P1should be subject to p(119909) isin P = p(119909) 119909 isin [ 119909] Based onthe strict pseudo-concavity of 120578

(120588)

EE (p) we have the followingtheorem

Theorem 2 Assume 119889119896 = 119889119896(p) and 119889119896 = 119889119896(p) p is theoptimal solution to P2 if and only if 1198891 le 0 and p is the optimalsolution to P2 if and only if 1198891 ge 0

However when 1198891 gt 0 and 1198891 lt 0 whether there exists aEE water-level to make p(119909) optimal is still not answered Weshould study the relation between 120578

(120588)

EE (p) and the EE water-level 119909

33 Analytical Expression of 120578(120588)

EE (119909) According to (8) p(119909) isa piecewise function and 1120574119899 is its discontinuity point If wesort all discontinuity points in an ascending order such that11205741 le sdot sdot sdot le 1120574119873 the interval (11205741 infin) can be dividedinto 119873 subintervals that is (11205741 11205742] (1120574119873 1120574119873+1)

with 1120574119873+1 = +infin Moreover p(119909) is continuous when119909 isin (1120574119888 1120574119888+1] forall119888 isin [1 119873] ((1120574119888 1120574119888+1] is named as thecontinuous power interval hereafter) To simplify our analysisand get the closed-form 120578

(120588)

EE (p(119909)) we further assume thatthe water-level rise interval is (119897 119897+1] and the continuouspower interval is (1120574119888 1120574119888+1] sube (119897 119897+1] Then accordingto the definition of the subcarrier subset we haveN1 = 119899 |

119909 gt 1120574119899 119896(119899) isin K1 and let 119872 = |N1| In this case120578(120588)

EE (p(119909)) can be transformed into a continuous function ofthe water-level 119909 that is

120578(120588)

EE (119909) ≜(119861119872log

2119909 + 1198611199030)

(120577119872119909 + 1199010) (17)

where

1199010 ≜ 119875119862 minus 120577

119872

sum

119899=1

1

120574119899

+ 120577 sum

119899isinN2

119899

1199030 ≜

119872

sum

119899=1

log2120574119899 + sum

119896isinK2

119877119896

119861

(18)

Thedomain of 120578(120588)EE (119909) isD119897 = 119909 isin (1120574119888 1120574119888+1] sube (119897 119897+1]It is noteworthy that 1199010 and 1199030 are constant as long as thewater-level rise interval and continuous power interval aredetermined

Similar to 120578(120588)

EE (p) 120578(120588)

EE (119909) is also strictly pseudo-concaveand the first-order derivative is

119889120578(120588)

EE (119909)

119889119909=

119872

(120577119872119909 + 1199010)sdot 119891 (119909) (19)

where

119891 (119909) ≜119861

(ln 2 sdot 119909)minus 120577120578(120588)

EE (119909) (20)

According to the first-order optimality condition a stationarypoint 1199090 of 120578

(120588)

EE (119909) is the root of the equation 119891(119909) = 0 Theclosed-form expression of 1199090 is given byTheorem 3

Theorem 3 If there exists a stationary point 1199090 in the domainof 120578(120588)

EE (119909) its closed-form expression is given by

1199090 =1199010120577119872

119882 ((1199010120577119872) sdot 21199030119872119890minus1) (21)

where 119882(sdot) represents the Lambert-119882 function

The proof of the theorem can be found in [11]

34 EE-OptimalWater-Level According to the strict pseudo-concavity 120578

(120588)

EE (119909) is maximized at the stationary point 1199090However whether the corresponding p(1199090) is the globaloptimal solution to P2 still needs to be verified


Theorem 4 If 1199090 isin D119897 is the stationary point of 120578(120588)

EE (119909) givenby (21) then p(1199090) is the global optimal solution to P2

Proof Since 1199090 isin D119897 it can be verified that p(1199090) is afeasible solution to P2 Besides because 1199090 is a stationarypoint of 120578

(120588)

EE (119909) we have 120597120578(120588)

EE (p(1199090))120597119901119899 = 0 (119899 isin N1)According to Property 1 we can show that 120597120578

(120588)

EE (p(1199090))120597119901119899 le

0 (119899 isin N0 or N2) Then it can be verified that p(1199090) satisfiesthe KKT conditions Hence p(1199090) is the optimal solution toP2

In addition the existence of 1199090 is proved byTheorem 5

Theorem 5 If neither p nor p is the optimal solution to P2there must exist 1199090 isin ( 119909) such that p(1199090) is the optimalsolution to P2

Proof According to the intermediate value theorem to proveTheorem 5 we should show that there must exist a continu-ous power interval (119886 119887] such that 119891(119886) gt 0 119891(119887) le 0

In fact if neither p nor p is the optimal solution to P2 wecan verify that119891() gt 0 and119891(119909) lt 0 according to Property 1Assume ( 119909) is divided into 119871 water-level rise intervals Itcan be proved that there must exist a water-level rise interval(119888 119889] such that 119891(119888) gt 0 119891(119889) le 0 If there does not existsuch an interval it can be deduced that119891()sdot119891(119909) gt 0 whichyields a contradiction

Furthermore assume that there are 1198731 discontinuitypoints in (119888 119889] such that 119888 le 1120574119899 le sdot sdot sdot le 1120574119899+119873

1

le 119889Similarly there must exist an interval (119886 119887] such that 119891(119886) gt

0 119891(119887) le 0 among1198731+1 continuous power intervals Hencethere must be a 1199090 isin ( 119909) satisfying 119891(1199090) = 0 Based onTheorem 4 p(1199090) is the optimal solution to P2

4 Optimal Energy-EfficientSubcarrier Assignment

In this section we will provide a sufficient condition forthe optimal energy-efficient subcarrier assignment (ESA)based on the relation between EE and the total transmittingpower119875119879 By utilizing this sufficient condition a quick searchmethod can be devised to obtain the optimal ESA which willbe described in the next section

According to (4a)ndash(4e) a feasible ESA can be obtainedby solving a rate adaptive (RA) problem for a given totaltransmitting power 119875119879 Moreover the maximum EE canonly be achieved at one of three different total transmittingpowers including two boundary points (119875min and 119875max) anda stationary point 119875119879 of the perturbation function of P1 [5]To obtain the optimal ESA it should first determine 119875119879which is an unknown value Unfortunately 119875119879 is difficult tobe determined and only an approximation can be found bythe iterative algorithms [5] Therefore only the suboptimalESA can be obtained according to the approximate 119875119879

On the other hand based on the EEwater-filling structurediscussed in the previous section the optimal ESA can beobtained by calculating the exact optimal EE for every feasiblesubcarrier assignment 120588 and then selecting the one with the

maximum value This exhaustive search is prohibitive forlarge 119870 and 119873 in a practical system However combiningthe EE water-filling framework and the property of theperturbation function of P1 a sufficient condition for theoptimal ESA can be established to greatly simplify the search

Define 120578EE(119875119879) ≜ 119877max(119875119879)(120577119875119879+119875119862) as the perturbationfunction of P1 where 119877max(119875119879) represents the maximumoverall data rate of the rate adaptive (RA) problem [10] withuser data requirements for a given total transmitting power119875119879 Then we have the following

Theorem 6 For a feasible subcarrier assignment 120588 theoptimal-EE water-level 1199090 and the maximum EE 120578

(120588)

EE (1199090) aregiven by (21) and (17) respectively Correspondingly the totaltransmitting power 119875119879(1199090) = 1119879 sdot p(1199090) and the overall datarate 119877

(120588)(119875119879(1199090)) = 120578

(120588)

EE sdot (120577119875119879(1199090) + 119875119862) If 119875min lt 119875119879 lt 119875maxand 119877max(119875119879(1199090)) = 119877

(120588)(119875119879(1199090)) 120588 is the optimal ESA of P1

Proof To prove Theorem 6 we should show 119875119879(1199090) derivedfrom the optimal EPA for the fixed subcarrier assignment120588 is a stationary point of 120578EE(119875119879) In this case consider thederivative of 120578EE(119875119879)

119889120578EE (119875119879)

119889119875119879

=119889119877max (119875119879) 119889119875119879 minus 120577120578EE (119875119879)

120577119875119879 + 119875119862

(22)

If 119877max(119875119879(1199090)) = 119877(120588)

(119875119879(1199090)) we can get 120578EE(119875119879(1199090)) =

119877(120588)

(119875119879(1199090))(120577119875119879(1199090) + 119875119862) = 120578(120588)

EE (1199090) Since 1199090 is thestationary point of 120578

(120588)

EE (119909) it has that 120578(120588)

EE (1199090) = 119861(120577sdot ln 2sdot1199090)

based on (20) On the other hand 119889119877max(119875119879(1199090))119889119875119879 =

119861(ln 2 sdot 1199090) [11] Then 119889120578EE(119875119879(1199090))119889119875119879 = 0 that is 119875119879(1199090)is a stationary point of 120578EE(119875119879) When 119875min lt 119875119879 lt 119875max EEis maximized at the stationary point of 120578EE(119875119879) Therefore 120588is the optimal ESA

Based on Theorem 6 the following proposition can beeasily verified

Proposition 7 If 119877max(119875119879(1199090)) gt 119877(120588)

(119875119879(1199090)) then 120578(120588119890)

EE gt

120578(120588)

EE where 120588119890 is the subcarrier assignment obtained by solvingthe RA problem with 119875119879 = 119875119879(1199090)

5 Joint Energy-Efficient ResourceAllocation Algorithm

Based on the analysis in the previous sections we develop anoptimal energy-efficient resource allocation algorithm withlow complexity to solve P1 named as joint energy-efficientresource allocation (JERA) algorithm Different from theexisting algorithms proposed in [4ndash6] the JERA algorithmconsists of two layers to iteratively perform subcarrier assign-ment and power allocation so as to achieve the optimalsolution The aim of the outer layer is to find a feasiblesubcarrier assignment for a given total transmitting powerand the inner layer is in charge of energy-efficient powerallocation based on the obtained subcarrier assignmentBased on the EE water-filling framework the optimal EPA


Input 119875max and 119877119896 forall119896 isin K

Output 120578EE120588

and p

(1) Obtain p and by solving the MA problem and then get 119889120578EE(119875119879)119889119875119879|119875119879=119875min

by (22) where 119875min = 1119879 sdot p(2) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875min

le 0 then(3) return p

= p120588

= and 120578EE = 120578EE(119875min)

(4) else(5) Obtain p and by solving the RA problem with the power constraint 119875max and get 119889120578EE(119875119879)119889119875119879|119875

119879=119875max

by (22)(6) end if(7) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875max

ge 0 then(8) return p

= p120588

= and 120578EE = 120578EE(119875max)

(9) else(10) 120588

119890larr

(11) repeat(12) 120588larr 120588

119890

(13) Obtain 119875119879 by performing the OEPA algorithm (see detail in Algorithm 2) with input 120588 and calculate 120588119890by solving

the RA problem with 119875119879(14) until 119877max(119875119879(1199090)) = 119877

(120588)(119875119879(1199090))

(15) return p= p120588

= 120588 and 120578

EE = 120578EE(119875

119879)

(16) end if

Algorithm 1 Joint energy-efficient resource (JERA) allocation

can be easily obtained in the inner layerMeanwhile the outerlayer finds a series of subcarrier assignments such that theoptimal EEs increase monotonously

The JERA algorithm is shown in Algorithm 1 FromLine (1) to Line (8) the algorithm first verifies whether theboundary point (119875min or119875max) is optimal by utilizing the strictquasi-concavity of 120578EE(119875119879) shown in [5] If neither is optimalthe algorithm will search the optimal ESA and calculate thecorresponding EPA as shown from Line (10) to Line (15)Specifically starting from a feasible subcarrier assignment 120588JERA calculates the optimal EPA and hence a new 119875119879 whichis further used to find a new subcarrier assignment 120588

119890by

solving a RA problem [10] This procedure repeats until thesufficient condition inTheorem 6 is satisfied

It is noteworthy that the sequence of EE generated bythe JERA algorithm is monotonously increasing according toProportion 1 Besides the optimal EE is upper bounded sothat JERA algorithm must converge to the optimal solution

The OEPA algorithm in Line (13) is devised to get theexactly optimal power allocation based on the EE water-filling framework which is summarized as Algorithm 2According toTheorems 2 and 5 the OEPA algorithm consistsof three phases In the first phase the minimum water-levels119896rsquos are obtained If the power vector pderived from 119896rsquos is notoptimal the algorithm proceeds to the second phase wherethe maximumwater-level 119909 under the total transmitting 119875maxis found If the maximum power vector p with 119909 is optimalthe algorithm returns Otherwise it searches sequentiallyfor the optimal water-level 1199090 in the feasible region ( 119909)

in the last phase Since the number of the searched water-level intervals only depends on that of the users and thesubcarriers the average computational complexity of theOEPA algorithm is O(119873 + 119870) in the worst case while that ofthe BPA (bisection-based power adaptation) algorithm [5] isproportional to 1120575

2 where 120575 is the convergence tolerance

More importantly the value of EE obtained by the OEPAalgorithm is the exact optimum whereas the BPA algorithmcan only provide an approximation result

6 Performance Evaluation

In this section simulation results are given to verify thetheoretical analysis and the performance of the proposedalgorithms In our simulation the number of data subcarriersis set to be 72 and the bandwidth of each subcarrier is 15 kHz[5] The block Rayleigh fading channel model is consideredand the Okumura-Hata path loss model is followed that is119875119871(119889) = 13774 + 522 log(119889) in decibels where 119889 is thedistance between transmitter and receiver in kilometers Thestandard deviation of shadowing is 7 dB and the thermalnoise spectral density is minus174 dBmHz [4]The circuit poweris 20W and the maximum transmitting power is 40W forthe base station [12] The drain efficiency of power amplifieris assumed to be 38 [4] Each user in the simulation has thesame minimum rate requirement of 100 kbps

First we compare the performance of the OEPA algo-rithm with the other two algorithms the BPA algorithm [5]and the MWF (multilevel water filling) algorithm [10] for afixed subcarrier allocation Although the MWF algorithm isa classical SE-based scheme rather than a EE-based schemeit is used as a benchmark to measure the difference in theenergy efficiency between the two classes of scheme In thissimulation example the number of users is set to 30 Theusers are uniformly distributed in a circle centered at the BSwith a variable radius The results in Figures 1(a) and 1(b)show that the average EE and the system throughput of allthe three algorithms decrease with the channel power gainThis is due to the fact that the average channel-gain-to-noiseratio (CNR) of each user decreases with the increase of thedistance between the user and the BS such that more power


Input 120588 and 119877119896 forall119896 isin K

Output p and 119875119879 = 1119879 sdot p(1) Based on 120588 calculateN120588(119896) = 119899|120588119896119899 = 1 119901119899 ge 0 forall119896 isin K(2) for each user 119896 isin K do(3) sort user 119896rsquos subcarriers in ascending order such that 11205741 le sdot sdot sdot le 1120574119873

119896

where 119873119896 = |N120588(119896)| Let 1120574119873119896+1 = infin forall119896

(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894

(6) if 1120574119894 lt 119896 le 1120574119894+1 then(7) break(8) end if(9) end for(10) end for(11) if 1198891 le 0 then(12) return p = p and 119875119879 = 1119879 sdot p(13) end if(14) sort 1120574119899 119899 = 1 sdot sdot sdot 119873 and 119896 119896 = 1 sdot sdot sdot 119870 jointly in ascending order to divide the interval (11205741 infin) into 119870 water-level

rise intervals (1 2] (119870 119870+1] and each (119897 119897+1] into 119873119897 + 1 continuous power intervals(119886119897119894

119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1

119894 = 0 sdot sdot sdot 119873119897 minus 1

119897+1

119894 = 119873119897

(15) 119891119897119886119892 larr 0(16) for each water-level rise interval (119897 119897+1] do(17) for each continuous power interval (119886119897119894 119887119897119894] do(18) if flag = 0 then(19) Calculate 119909 = (119875max + 1198751 minus 1198752)119872119897 where 1198751 = sum

119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|

(20) if 119909 isin (119886119897119894 119887119897119894] then(21) 119891119897119886119892 larr 1(22) if 1198891 ge 0 then(23) return p = p and 119875119879 = 1119879 sdot p(24) end if(25) end if(26) else(27) calculate 1199090 by (21)(28) if 1199090 isin (119886119897119894 119887119897119894] then(29) return p = p(1199090) and 119875119879 = 1119879 sdot p(1199090)(30) end if(31) end if(32) end for(33) end for

Algorithm 2 Optimal energy-efficient power allocation (OEPA) algorithm

is required to combat with the severe channel fading to satisfythe data rate constraint of each user Figure 1(c) illustratesthat the SE-based scheme that is theMWF algorithm alwaysdepletes all the power to maximize the system throughputOn the contrary the power consumption of the EE-basedscheme that is the OEPA algorithm and the BPA algorithmis adaptive to the channel condition It is worth noting thatthere exists a gap between the EE obtained by the OEPAalgorithm and the BPA algorithmThe difference is especially1015 bitjoule on average when 120575 = 10

minus4 This stems fromthe fact that the OEPA algorithm always finds the exactoptimal solution while the BPA algorithm only approaches

the optimal value by iterative method More importantly thecomputational complexity of the OEPA algorithm is muchlower than that of the BPA algorithm which is demonstratedin the following simulation example

In order to compare the computational complexityassume that there are 10 users evenly distributed in thenetwork and 10000 independent experiments (with differentuser location) are conducted in the MATLAB environmentFigure 2 shows the CDF of the required CPU time forconvergence in different algorithms where 120575 represents theerror tolerance of BPA algorithm From Figure 2 it is obviousthat the convergence speed of theOEPA algorithm is superior


01 02 03 04 05 06 070

100

200

300

400

500

600

Distance to the BS (km)

Aver

age E

E (k

bitJ

)

(a) Evaluation and comparison of average EE

2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)

01 02 03 04 05 06 07Distance to the BS (km)

(b) Evaluation and comparison of average throughput

0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF


(c) Evaluation and comparison of average transmitting power

Figure 1 Performance comparison of different algorithms

to that of the BPA algorithm In addition the complexityof the BPA algorithm increases evidently when the errortolerance becomes tighter This is due to the fact that theOEPA algorithm can obtain the exact optimal solution bychecking at most 119870 + 119873 continuous power intervals in theworst case based on the closed-form expression of EPA Onthe contrary the BPA algorithm is to search bidirectionallyfor the optimal transmitting power which results in a highercomputational complexity On average the CPU time forconvergence of the OEPA algorithm is about 1564 and1303 of that of the BPA algorithm with 120575 = 01 and 120575 =

0001 respectivelyMoreover in order to verify the optimality of the pro-

posed JERA algorithm we compare the EE obtained by JERAwith the global optimum obtained by exhaustive search Inthis case we consider a system with 9 subcarriers and 3 usersto reduce the complexity of exhaustive search As shown inFigure 3 the achieved optimal EE in both algorithmsdecreasewith the distance between the BS and users Moreover

the two curves match with each other very well It demon-strates the proposed ESA searchmethod based onTheorem 6is effective and the JERA algorithm can obtain the globaloptimal ESA and EPA simultaneously

In addition we compare the performance of the JERAalgorithm with that of the JIOO algorithm [5] with differentnumber of users and subcarriers The result is shown inTable 1 it can be observed that the EE of the JERA algorithmis superior to that of the JIOO algorithmdue to the optimalityof the solution obtained by the JERA algorithmMore impor-tantly the convergence rate of the JERA algorithm is signifi-cantly faster than that of the JIOO algorithm Specifically thenumber of iterations for convergence of the JERA algorithmis less than 5 in average while the JIOO algorithm requiresat least 29 iterations in average to approximate the optimumIt is worth noting that each iteration in both algorithmsneeds to solve a RA problem with rate requirements and totaltransmitting power constraintDespite of the nonconvexity ofthis type RA problem it has been proven that it can be solved


0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA

The CPU time (120583s)

BPA 120575 = 01

BPA 120575 = 0001

Figure 2 Comparison of convergence performance of differentalgorithms

02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450

Average distance to BS (km)

Aver

age E

E (k

bits

J)

Global optimumJERA

Figure 3 Evaluation and comparison of average EE

efficiently by the Lagrange dual decomposition method withzero duality gap [13]

7 Conclusion

In this paper we investigated the EE maximization problemunder both the user rate requirements and the transmittingpower constraint Utilizing the EE water-filling structure weobtain the closed-form of the optimal EPA The sufficientcondition for optimal ESA is also derived based on therelation between EE and power Furthermore we proposea low-complexity algorithm with joint ESA and EPA toaddress the energy-efficient resource allocation in downlink

Table 1 Performance comparison of the two algorithms

(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317

OFDMA-based networks Simulation results show that theproposed algorithm achieves the optimal energy-efficientresource allocation with significantly reduced computationalcomplexity compared with the iterative methods

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported by the National S amp T MajorProject of China under Grant no 2014ZX03004-003 theFundamental Research Funds for the Central Universities(no ZYGX2013J009) China and EU FP7 Project CLIMBER(PIRSESGA-2012-318939)

References

[1] D Feng C Jiang G Lim L J Cimini Jr G Feng and G Y LildquoA survey of energy-efficient wireless communicationsrdquo IEEECommunications Surveys and Tutorials vol 15 no 1 pp 167ndash178 2013

[2] G Y Li Z Xu C Xiong et al ldquoEnergy-efficient wireless com-munications tutorial survey and open issuesrdquo IEEE WirelessCommunications vol 18 no 6 pp 28ndash35 2011

[3] G Miao N Himayat and G Y Li ldquoEnergy-efficient linkadaptation in frequency-selective channelsrdquo IEEE Transactionson Communications vol 58 no 2 pp 545ndash554 2010

[4] C Xiong G Y Li S Zhang Y Chen and S Xu ldquoEnergy- andspectral-efficiency tradeoff in downlink OFDMA networksrdquoIEEE Transactions on Wireless Communications vol 10 no 11pp 3874ndash3886 2011

[5] C Xiong G Y Li S Zhang Y Chen and S Xu ldquoEnergy-efficient resource allocation in OFDMA networksrdquo IEEE Trans-actions on Communications vol 60 no 12 pp 3767ndash3778 2012

[6] C Xiong G Y Li Y Liu Y Chen and S Xu ldquoEnergy-efficientdesign for downlink OFDMAwith delay-sensitive trafficrdquo IEEETransactions on Wireless Communications vol 12 no 6 pp3085ndash3095 2013

[7] S Cui A J Goldsmith and A Bahai ldquoEnergy-efficiency ofMIMO and cooperativeMIMO techniques in sensor networksrdquoIEEE Journal on Selected Areas in Communications vol 22 no6 pp 1089ndash1098 2004


[8] A Cambini and L Martein Generalized Convexity and Opti-mization Theory and Applications Springer Berlin Germany2009

[9] M S Bazaraa H D Sherali and C M Shetty NonlinearProgramming Theory and Algorithms Wiley-Interscience 3rdedition 2006

[10] S Sadr A Anpalagan and K Raahemifar ldquoRadio resourceallocation algorithms for the downlink of multiuser OFDMcommunication systemsrdquo IEEE Communications Surveys andTutorials vol 11 no 3 pp 92ndash106 2009

[11] F Wu Y Mao X Huang and S Leng ldquoLow-complexity opti-mal energy-efficient resource allocation in downlink OFDMAnetworksrdquo in Proceedings of the International Conference onComputational Problem-Solving (ICCP rsquo12) pp 46ndash51 ChengduChina October 2012

[12] Technical Specification Group Radio Access Network ldquoFurtheradvancements for E-UTRA physical layer aspects (Release 9)rdquo3GPP Standard TR 36814 V900 2010

[13] K Seong M Mohseni and J M Cioffi ldquoOptimal resourceallocation for OFDMA downlink systemsrdquo in Proceedings of theIEEE International Symposiumon InformationTheory (ISIT rsquo06)pp 1394ndash1398 IEEE Seattle Wash USA July 2006

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of


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Differential EquationsInternational Journal of

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Applied MathematicsJournal of


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International Journal of


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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


the algorithms approaching the optimal solution withiterative searching method

(iii) We provide a sufficient condition for the optimal en-ergy-efficient subcarrier assignment This conditionis the basis of the quick search method because itallows us to easily determine whether a subcarrierassignment is optimal As a result by combining thesufficient condition and the EE water-filling solutionwe design a joint energy-efficient resource allocation(JERA) algorithm which not only achieves the opti-mal EE but also outperforms the existing algorithmsin terms of convergence rate

The state of the art to solve the EEmaximization problemis JIOO (joint inner- and outer-layer optimization) algorithm[4 5] Relying on the strict quasi-concavity of the pertur-bation function of EE with respect to the total transmittingpower the JIOO algorithm solves the problem via two-layeriterations where the inner layer aims to find the maximumEE and its derivative for a given total transmitting powerand the outer layer targets to search for the total transmittingpower which results in the maximum EE by bisection searchThe computational complexity is mainly determined by theouter layer iterations Unfortunately the number of the outerlayer iterations becomes infinite when tolerance approacheszero Different from this approach the complexity of ouralgorithm depends on the finite number of the feasiblesubcarrier assignments Moreover the search space can begreatly reduced by using the sufficient condition provided bythis paper In addition the EE water-filling structure ensuresthat the results obtained by our algorithm are the exactoptimal solutions other than approximation ones

The rest of the paper is organized as follows In Section 2we describe the system model and formulate the EE maxi-mization problem for the downlink OFDMA networks Theoptimal energy-efficient power allocation for a given subcar-rier assignment and the sufficient condition for the optimalenergy-efficient subcarrier assignment are studied in Sections3 and 4 respectively An optimal energy-efficient powerallocation algorithm and a low-complexity joint resourceallocation algorithm are elaborated in Section 5 Section 6provides the performance evaluation of the proposed algo-rithms via simulations Finally we conclude this paper inSection 7

2 System Model

Consider a downlink OFDMA network with one base stationand 119870 users Let K = 1 119870 and N = 1 119873

be the set of users and subcarriers respectively Define thesubcarrier assignment matrix 120588 = (120588119896119899)119870times119873 where 120588119896119899 = 1

means that subcarrier 119899 is allocated to user 119896 and otherwise120588119896119899 = 0 The transmitting power allocation matrix isdefined as p = (119901119896119899)119870times119873 where 119901119896119899 ge 0 represents thetransmitting power allocated to user 119896 on subcarrier 119899 Thenthe maximum achievable data rate of user 119896 on subcarrier 119899

is given by

119903119896119899 = 119861log2

(1 + 119901119896119899120574119896119899) (1)

where 119861 is the bandwidth of subcarrier and 120574119896119899 denotes thenormalized channel power gain of user 119896 on subcarrier 119899Accordingly the overall system data rate is

119877 =

119870

sum

119896=1

119873

sum

119899=1

119903119896119899 sdot 120588119896119899 (2)

and the total transmitting power is

119875119879 =

119870

sum

119896=1

119873

sum

119899=1

119901119896119899 sdot 120588119896119899 (3)

In addition to transmitting power the energy consump-tion also includes circuit power which is consumed by deviceelectronics The circuit power is modeled as a constant119875119862 which is independent of data transmission rate [7]Accordingly we define EE as the amount of bits transmittedper Joule of energy that is 120578EE ≜ 119877(120577119875119879+ 119875119862) where 120577 is thereciprocal of drain efficiency of power amplifier

In our work we consider maximizing EE under theminimum data rate requirements 119877119896 and the total transmit-ting power constraint 119875max Accordingly this optimizationproblem can be formulated as

P1 maxp120588

sum119870

119896=1sum119873

119899=1119903119896119899 sdot 120588119896119899

120577119875119879 + 119875119862

(4a)

st119873

sum

119899=1

119903119896119899 sdot 120588119896119899 ge 119877119896 forall119896 isin K (4b)

119875119879 ≜

119870

sum

119896=1

119873

sum

119899=1

119901119896119899 sdot 120588119896119899 le 119875max (4c)

119870

sum

119896=1

120588119896119899 = 1 forall119899 isin N (4d)

120588119896119899 isin 0 1 119901119896119899 ge 0 forall119896 isin K forall119899 isin N

(4e)

where (4b) indicates the minimum data rate requirement ofeach user (4c) is the total transmitting power constraint and(4d) is the constraint on subcarrier assignment to ensure thateach subcarrier is only assigned to one user

Similar to the traditional spectral-efficient resourcemodel P1 is a mixed integer nonlinear programming prob-lem and it is not trivial to obtain the global optimal solutionto this problem To solve the problem we first decompose P1into two subproblems which include (1) the energy-efficientpower allocation for a fixed subcarrier assignment 120588 and (2)the energy-efficient subcarrier assignment for a given totaltransmitting power 119875119879 Then based on the properties of thesubproblems we develop an algorithm to find the solutionof joint energy-efficient power allocation and subcarrierassignment to maximize EE

3 Optimal Energy-Efficient Power Allocation

In this section we analyze the optimal energy-efficient powerallocation (EPA) based on the EE water-filling structure All





P2 120578(120588)

EE ≜ maxpge0

120578(120588)

EE (p)


119901119879 (p) le 119875max

(5)

where 120578(120588)

EE (p) ≜ sum119873

119899=1(119861log2(1 + 120574119899119901119899)(120577 sum

119873

119899=1119901119899 + 119875119862))



119873


(120588)


(120588)



P3 119875min ≜ min119901119899ge0

119873

sum

119899=1

119901119899

st 119903119896 (p) = 119877119896 forall119896 isin K

(6)



of the equation

sum

119899isinN120588(119896)

119861log2

[1 + 120574119899 (119896 minus1

120574119899

)

+

] = 119877119896 forall119896 isin K (7)






119899 (119909) =

(119909 minus1

120574119899

)

+

119896 isin K1

(119896(119899) minus1

120574119899

)

+

119896 isin K2

(8)


119903119896 (p (119909))

gt 119877119896 119896 (119899) isin K1

= 119877119896 119896 (119899) isin K2(9)



P4 max119901119899ge0

119877(120588)

(p)


119901119879 (p) = 119875119879

(10)

where 119877(120588)

(p) ≜ sum119873

119899=1119861log2(1 + 120574119899119901119899)



nabla119877(120588)

(plowast) +

119870

sum

119896=1




119901lowast

119899= (119861 sdot

(1 + 119906119896(119899))


1

120574119899

)

+

forall119899 (13)



119899 (119909) =

(119909 minus1

120574119899

)

+

119896 (119899) isin K10158401

(119909119896(119899) minus1

120574119899

)

+

119896 (119899) isin K10158402

(14)


where K10158401

≜ 119896 | 119903119896(plowast) gt 119877119896 and K10158402





K10158401

= K1 and K10158402







(a)

120597120578(120588)

EE (p (119909))

120597119901119899

=

119861 (ln 2 sdot 119909) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N1

119861 (ln 2 sdot 119896(119899)) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N2

119861120574119899 ln 2 minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N0

(15)

(b)


(16)


EE (p(119909))120597119901119899



(120588)




(120588)





(120588)




120578(120588)

EE (119909) ≜(119861119872log

2119909 + 1198611199030)

(120577119872119909 + 1199010) (17)

where

1199010 ≜ 119875119862 minus 120577

119872

sum

119899=1

1

120574119899

+ 120577 sum

119899isinN2

119899

1199030 ≜

119872

sum

119899=1

log2120574119899 + sum

119896isinK2

119877119896

119861

(18)


Similar to 120578(120588)

EE (p) 120578(120588)


119889120578(120588)

EE (119909)

119889119909=

119872

(120577119872119909 + 1199010)sdot 119891 (119909) (19)

where

119891 (119909) ≜119861

(ln 2 sdot 119909)minus 120577120578(120588)

EE (119909) (20)


(120588)




1199090 =1199010120577119872

119882 ((1199010120577119872) sdot 21199030119872119890minus1) (21)




(120588)






(120588)

EE (119909) we have 120597120578(120588)


(120588)

EE (p(1199090))120597119901119899 le







1










(120588)


(120588)(119875119879(1199090)) = 120578

(120588)




119889120578EE (119875119879)

119889119875119879

=119889119877max (119875119879) 119889119875119879 minus 120577120578EE (119875119879)

120577119875119879 + 119875119862

(22)

If 119877max(119875119879(1199090)) = 119877(120588)

(119875119879(1199090)) we can get 120578EE(119875119879(1199090)) =

119877(120588)

(119875119879(1199090))(120577119875119879(1199090) + 119875119862) = 120578(120588)


(120588)

EE (119909) it has that 120578(120588)

EE (1199090) = 119861(120577sdot ln 2sdot1199090)





(119875119879(1199090)) then 120578(120588119890)

EE gt

120578(120588)






Output 120578EE120588

and p



119879=119875min


= p120588

= and 120578EE = 120578EE(119875min)


119879=119875max

by (22)(6) end if(7) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875max


= p120588

= and 120578EE = 120578EE(119875max)

(9) else(10) 120588

119890larr

(11) repeat(12) 120588larr 120588

119890



(120588)(119875119879(1199090))

(15) return p= p120588

= 120588 and 120578

EE = 120578EE(119875

119879)

(16) end if




119890by













119896


(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894



119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1


119897+1

119894 = 119873119897


119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|








01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













P2 120578(120588)

EE ≜ maxpge0

120578(120588)

EE (p)


119901119879 (p) le 119875max

(5)

where 120578(120588)

EE (p) ≜ sum119873

119899=1(119861log2(1 + 120574119899119901119899)(120577 sum

119873

119899=1119901119899 + 119875119862))



119873


(120588)


(120588)



P3 119875min ≜ min119901119899ge0

119873

sum

119899=1

119901119899

st 119903119896 (p) = 119877119896 forall119896 isin K

(6)



of the equation

sum

119899isinN120588(119896)

119861log2

[1 + 120574119899 (119896 minus1

120574119899

)

+

] = 119877119896 forall119896 isin K (7)






119899 (119909) =

(119909 minus1

120574119899

)

+

119896 isin K1

(119896(119899) minus1

120574119899

)

+

119896 isin K2

(8)


119903119896 (p (119909))

gt 119877119896 119896 (119899) isin K1

= 119877119896 119896 (119899) isin K2(9)



P4 max119901119899ge0

119877(120588)

(p)


119901119879 (p) = 119875119879

(10)

where 119877(120588)

(p) ≜ sum119873

119899=1119861log2(1 + 120574119899119901119899)



nabla119877(120588)

(plowast) +

119870

sum

119896=1




119901lowast

119899= (119861 sdot

(1 + 119906119896(119899))


1

120574119899

)

+

forall119899 (13)



119899 (119909) =

(119909 minus1

120574119899

)

+

119896 (119899) isin K10158401

(119909119896(119899) minus1

120574119899

)

+

119896 (119899) isin K10158402

(14)


where K10158401

≜ 119896 | 119903119896(plowast) gt 119877119896 and K10158402





K10158401

= K1 and K10158402







(a)

120597120578(120588)

EE (p (119909))

120597119901119899

=

119861 (ln 2 sdot 119909) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N1

119861 (ln 2 sdot 119896(119899)) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N2

119861120574119899 ln 2 minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N0

(15)

(b)


(16)


EE (p(119909))120597119901119899



(120588)




(120588)





(120588)




120578(120588)

EE (119909) ≜(119861119872log

2119909 + 1198611199030)

(120577119872119909 + 1199010) (17)

where

1199010 ≜ 119875119862 minus 120577

119872

sum

119899=1

1

120574119899

+ 120577 sum

119899isinN2

119899

1199030 ≜

119872

sum

119899=1

log2120574119899 + sum

119896isinK2

119877119896

119861

(18)


Similar to 120578(120588)

EE (p) 120578(120588)


119889120578(120588)

EE (119909)

119889119909=

119872

(120577119872119909 + 1199010)sdot 119891 (119909) (19)

where

119891 (119909) ≜119861

(ln 2 sdot 119909)minus 120577120578(120588)

EE (119909) (20)


(120588)




1199090 =1199010120577119872

119882 ((1199010120577119872) sdot 21199030119872119890minus1) (21)




(120588)






(120588)

EE (119909) we have 120597120578(120588)


(120588)

EE (p(1199090))120597119901119899 le







1










(120588)


(120588)(119875119879(1199090)) = 120578

(120588)




119889120578EE (119875119879)

119889119875119879

=119889119877max (119875119879) 119889119875119879 minus 120577120578EE (119875119879)

120577119875119879 + 119875119862

(22)

If 119877max(119875119879(1199090)) = 119877(120588)

(119875119879(1199090)) we can get 120578EE(119875119879(1199090)) =

119877(120588)

(119875119879(1199090))(120577119875119879(1199090) + 119875119862) = 120578(120588)


(120588)

EE (119909) it has that 120578(120588)

EE (1199090) = 119861(120577sdot ln 2sdot1199090)





(119875119879(1199090)) then 120578(120588119890)

EE gt

120578(120588)






Output 120578EE120588

and p



119879=119875min


= p120588

= and 120578EE = 120578EE(119875min)


119879=119875max

by (22)(6) end if(7) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875max


= p120588

= and 120578EE = 120578EE(119875max)

(9) else(10) 120588

119890larr

(11) repeat(12) 120588larr 120588

119890



(120588)(119875119879(1199090))

(15) return p= p120588

= 120588 and 120578

EE = 120578EE(119875

119879)

(16) end if




119890by













119896


(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894



119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1


119897+1

119894 = 119873119897


119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|








01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where K10158401

≜ 119896 | 119903119896(plowast) gt 119877119896 and K10158402





K10158401

= K1 and K10158402







(a)

120597120578(120588)

EE (p (119909))

120597119901119899

=

119861 (ln 2 sdot 119909) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N1

119861 (ln 2 sdot 119896(119899)) minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N2

119861120574119899 ln 2 minus 120577120578(120588)

EE (p (119909))

120577119875119879 + 119875119862

119899 isin N0

(15)

(b)


(16)


EE (p(119909))120597119901119899



(120588)




(120588)





(120588)




120578(120588)

EE (119909) ≜(119861119872log

2119909 + 1198611199030)

(120577119872119909 + 1199010) (17)

where

1199010 ≜ 119875119862 minus 120577

119872

sum

119899=1

1

120574119899

+ 120577 sum

119899isinN2

119899

1199030 ≜

119872

sum

119899=1

log2120574119899 + sum

119896isinK2

119877119896

119861

(18)


Similar to 120578(120588)

EE (p) 120578(120588)


119889120578(120588)

EE (119909)

119889119909=

119872

(120577119872119909 + 1199010)sdot 119891 (119909) (19)

where

119891 (119909) ≜119861

(ln 2 sdot 119909)minus 120577120578(120588)

EE (119909) (20)


(120588)




1199090 =1199010120577119872

119882 ((1199010120577119872) sdot 21199030119872119890minus1) (21)




(120588)






(120588)

EE (119909) we have 120597120578(120588)


(120588)

EE (p(1199090))120597119901119899 le







1










(120588)


(120588)(119875119879(1199090)) = 120578

(120588)




119889120578EE (119875119879)

119889119875119879

=119889119877max (119875119879) 119889119875119879 minus 120577120578EE (119875119879)

120577119875119879 + 119875119862

(22)

If 119877max(119875119879(1199090)) = 119877(120588)

(119875119879(1199090)) we can get 120578EE(119875119879(1199090)) =

119877(120588)

(119875119879(1199090))(120577119875119879(1199090) + 119875119862) = 120578(120588)


(120588)

EE (119909) it has that 120578(120588)

EE (1199090) = 119861(120577sdot ln 2sdot1199090)





(119875119879(1199090)) then 120578(120588119890)

EE gt

120578(120588)






Output 120578EE120588

and p



119879=119875min


= p120588

= and 120578EE = 120578EE(119875min)


119879=119875max

by (22)(6) end if(7) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875max


= p120588

= and 120578EE = 120578EE(119875max)

(9) else(10) 120588

119890larr

(11) repeat(12) 120588larr 120588

119890



(120588)(119875119879(1199090))

(15) return p= p120588

= 120588 and 120578

EE = 120578EE(119875

119879)

(16) end if




119890by













119896


(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894



119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1


119897+1

119894 = 119873119897


119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|








01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













(120588)

EE (119909) we have 120597120578(120588)


(120588)

EE (p(1199090))120597119901119899 le







1










(120588)


(120588)(119875119879(1199090)) = 120578

(120588)




119889120578EE (119875119879)

119889119875119879

=119889119877max (119875119879) 119889119875119879 minus 120577120578EE (119875119879)

120577119875119879 + 119875119862

(22)

If 119877max(119875119879(1199090)) = 119877(120588)

(119875119879(1199090)) we can get 120578EE(119875119879(1199090)) =

119877(120588)

(119875119879(1199090))(120577119875119879(1199090) + 119875119862) = 120578(120588)


(120588)

EE (119909) it has that 120578(120588)

EE (1199090) = 119861(120577sdot ln 2sdot1199090)





(119875119879(1199090)) then 120578(120588119890)

EE gt

120578(120588)






Output 120578EE120588

and p



119879=119875min


= p120588

= and 120578EE = 120578EE(119875min)


119879=119875max

by (22)(6) end if(7) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875max


= p120588

= and 120578EE = 120578EE(119875max)

(9) else(10) 120588

119890larr

(11) repeat(12) 120588larr 120588

119890



(120588)(119875119879(1199090))

(15) return p= p120588

= 120588 and 120578

EE = 120578EE(119875

119879)

(16) end if




119890by













119896


(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894



119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1


119897+1

119894 = 119873119897


119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|








01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Output 120578EE120588

and p



119879=119875min


= p120588

= and 120578EE = 120578EE(119875min)


119879=119875max

by (22)(6) end if(7) if 119889120578EE(119875119879)119889119875119879|119875

119879=119875max


= p120588

= and 120578EE = 120578EE(119875max)

(9) else(10) 120588

119890larr

(11) repeat(12) 120588larr 120588

119890



(120588)(119875119879(1199090))

(15) return p= p120588

= 120588 and 120578

EE = 120578EE(119875

119879)

(16) end if




119890by













119896


(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894



119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1


119897+1

119894 = 119873119897


119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|








01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119896


(4) for 119894 = 119873119896 1 do

(5) 119896 =2119877119896(119861sdot119894)

(prod119894

119895=1120574119895)1119894



119887119897119894

] (119886119897119873119897+1

119887119897119873119897+1

] where

119886119897119894

=

119897

119894 = 0

1

120574119899119897+119894

119894 = 1 sdot sdot sdot 119873119897

119887119897119894

=

1

120574119899119897+119894+1


119897+1

119894 = 119873119897


119872119897

119899=11120574119899 1198752 = sum

119870

119896=119897+1sum119899isinN120588(119896)

119899 and119872119897 = |119899|1120574119899 lt 119886119897119894 119899 isin N1|








01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










01 02 03 04 05 06 070

100

200

300

400

500

600


Aver

age E

E (k

bitJ

)


2

4

6

8

10

12

14

16

Aver

age t

hrou

ghpu

t (M

bps)



0

5

10

15

20

25

30

35

40

Aver

age t

rans

mitt

ing

pow

er (W

)

OEPABPAMWF










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 25 50 75 100 125 150 175 200 225 2500

01

02

03

04

05

06

07

08

09

10

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

OEPA


BPA 120575 = 01

BPA 120575 = 0001


02 03 04 05 06 07 08 09 10 1150

100

150

200

250

300

350

400

450


Aver

age E

E (k

bits

J)

Global optimumJERA



7 Conclusion



(Userssubcarriers)

JERA JIOOAve EE

(kbitsJoule)Ave

iterationsAve EE

(kbitsJoule)Ave

iterations(5 16) 819 26 818 296(10 32) 1263 28 1262 298(15 64) 1766 32 1765 301(20 64) 1792 33 1791 301(25 72) 1795 38 1574 304(30 72) 1598 43 1576 317




Acknowledgments


References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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