On Tile-and-Energy Allocation in OFDMA Broadband Wireless Networks

1284 IEEE COMMUNICATIONS LETTERS, VOL. 15, NO. 12, DECEMBER 2011

On Tile-and-Energy Allocation in OFDMA Broadband Wireless Networks

Jia-Ming Liang, Student Member, IEEE, Jen-Jee Chen, Member, IEEE, Cheng-Wen Liu,

Yu-Chee Tseng, Senior Member, IEEE, and Bao-Shuh Paul Lin, Fellow, IEEE

Abstract—The Orthogonal Frequency Division Multiple Access (OFDMA) technique has been widely applied in broadband wireless networks, such as IEEE 802.16 (WiMAX) and 3GPP Long Term Evolution (LTE). Existing studies have targeted at improving network throughput by increasing the transmission rates of mobile stations (MSs). Nevertheless, this is at the cost of higher energy consumption of MSs. In the letter, we consider the tile-and-energy joint allocation problem for uplink transmissions in an OFDMA wireless network. We formulate this problem as a mixed integer programming, where the objective is to minimize MSs’ energy consumption subject to satisfying their traffic demands. We show this to be NP-hard and develop a heuristic taking advantage of the water-filling technique. Simulation results show that the performance of our heuristic is close to the optimum, especially when the network is under non-saturated condition.

Index Terms—3GPP LTE, energy conservation, IEEE 802.16, OFDMA, resource management, WiMAX.

I. INTRODUCTION

B

OTH IEEE 802.16 (WiMAX) and 3GPP Long Term Evo-lution (LTE) have employed OFDMA, which can support multiuser diversity and dynamic power adaptation, thus signif-icantly improving the spectral efficiency. Existing studies [1]– [4] for OFDMA networks consider only multiuser diversity and power allocation for higher throughput but neglect the energy conservation issue of MSs. The work [5] does try to reduce MSs’ energy consumption by modeling it as a multiple-choice knapsack problem, but it does not take the multiuser diversity and power constraint of MSs into account. In this letter, we consider the allocation of tiles and energy of MSs in the uplink direction of an OFDMA network. We model it as an optimization problem to minimize MSs’ energy consumption subject to satisfying their demands. We formulate this problem as a mixed integer programming problem, which has been shown to be NP-hard [6], and propose a low complexity, energy-efficient heuristic. The heuristic first allocates tiles to MSs in a greedy way to satisfy more demands. If there are remaining resources, a new water-filling technique is applied to adjust MSs’ transmission power to save their energy. The performance of the heuristic will be shown by simulations.

Manuscript received May 5, 2011. The associate editor coordinating the review of this letter and approving it for publication was F.-N. Pavlidou.

J.-M. Liang, C.-W. Liu, Y.-C. Tseng, and B.-S. P. Lin are with the Department of Computer Science, National Chiao-Tung University, Hsin-Chu 30010, Taiwan (e-mail:{jmliang, chengwen, yctseng}@cs.nctu.edu.tw,

bplin@mail.nctu.edu.tw).

J.-J. Chen is with the Department of Electrical Engineering, National Uni-versity of Tainan, Tainan 70005, Taiwan (e-mail: jjchen@mail.nutn.edu.tw).

Y.-C. Tseng’s research is co-sponsored by MoE ATU Plan, by NSC grants 97-3114-E-009-001, 97-2221-E-009-142-MY3, 2219-E-009-019, and 98-2219-E-009-005, 99-2218-E-009-005, by ITRI, Taiwan, by III, Taiwan, by D-Link, and by Intel.

Digital Object Identifier 10.1109/LCOMM.2011.101211.110938

II. SYSTEMMODEL

The resource in an OFDMA network is frames, where a frame is a two-dimensional (subchannel × time slot) array.

Each frame is composed of a downlink subframe and an uplink

subframe. The resource unit to be allocated to MSs is tiles,

where a tile is a time slot on a subchannel where an MS can transmit at a certain rate with a proper transmission power. This letter considers the resource allocation of uplink sub-frames. Each uplink subframe is modeled by 𝑁 subchannels

and 𝑀 time slots. There are 𝐾 MSs. Each MS𝑘, 𝑘 = 1..𝐾,

has a maximum transmission power of 𝑃𝑀𝐴𝑋

𝑘 mW at any

instant and an uplink traffic demand of 𝐷𝑘 bits per frame

granted by the BS. Let𝑥𝑖,𝑗_𝑘 be the binary indicator such that

𝑥𝑖,𝑗_𝑘 = 1 if the 𝑗th time slot of subchannel 𝑖 (or called tile (𝑖, 𝑗)) is allocated to MS𝑘. Since each tile is allocated to at

most one MS, we have ∑

𝑘=1..𝐾 𝑥𝑖,𝑗

𝑘 ≤ 1, ∀𝑖, 𝑗. (1)

Let𝑝𝑖,𝑗_𝑘 be the power that MS_𝑘 uses for tile(𝑖, 𝑗). It follows that

∑

𝑖=1..𝑁

𝑝𝑖,𝑗_𝑘 ≤ 𝑃𝑀𝐴𝑋

𝑘 , ∀𝑘, 𝑗. (2)

Assuming that the channel condition remains unchanged dur-ing a frame, we let 𝐺𝑖

𝑘 > 0 and 𝑁𝑘𝑖 > 0 be the channel

gain and the noise power of subchannel 𝑖 from MS𝑘 to the

BS, respectively. Here, we assume the background noise to be additive white Gaussian noise. Given a bandwidth of 𝐵 (in

Hz) per subchannel, according to the Shannon capacity, MS_𝑘 can transmit at rate𝑥𝑖,𝑗_𝑘 ⋅𝐵 log₂(1+𝑝𝑖,𝑗_𝑘 ⋅𝐺𝑖

𝑘/𝑁𝑘𝑖) at tile (𝑖, 𝑗).

Below, we let𝑓𝑖

𝑘= 𝐺𝑖𝑘/𝑁𝑘𝑖 be the channel gain to noise ratio.

To meet MS_𝑘’s demand, we impose ∑ 𝑖=1..𝑁 ∑ 𝑗=1..𝑀 𝑥𝑖,𝑗_𝑘 ⋅ 𝐵 log₂(1 + 𝑝𝑖,𝑗_𝑘 ⋅ 𝑓𝑖 𝑘) ≥ 𝐷𝑘, ∀𝑘. (3)

The energy consumption of MS𝑘, denoted by 𝐸𝑘, is the

summation of its energy costs in all tiles, i.e., 𝐸𝑘 =

∑

𝑖=1..𝑁

∑

𝑗=1..𝑀𝑥𝑖,𝑗𝑘 ⋅ 𝑝𝑖,𝑗𝑘 ⋅ 𝑇𝑠, where 𝑇𝑠 is the time slot

length (in second). Our goal is to minimize the total energy consumption of all MSs: min 𝑝𝑖,𝑗 𝑘 ,𝑥𝑖,𝑗𝑘 ,∀𝑖,𝑗,𝑘 ∑ 𝑘=1..𝐾 𝐸𝑘, (4)

by allocating tiles (i.e.,𝑥𝑖,𝑗_𝑘 ) and the corresponding power (i.e.,

𝑝𝑖,𝑗_𝑘 ) subject to Eqs. (1), (2), and (3). This problem is a mixed

integer programming (MIP) problem, which is known to be

NP-hard and thus intractable [6]. 1089-7798/11$25.00 c⃝ 2011 IEEE

LIANG et al.: ON TILE-AND-ENERGY ALLOCATION IN OFDMA BROADBAND WIRELESS NETWORKS 1285

III. ANENERGY-EFFICIENTHEURISTIC

The proposed heuristic has two phases. The first phase tries to satisfy MSs’ demands by selecting the best tiles and power for MSs such that the use of uplink frame space is minimized. If there are free tiles left from the first phase, the second phase will try to allocate them to MSs, which can help MSs lower down their transmission rates, and thus energy costs.

A. Phase 1: Meeting MSs’ Demands

To meet more MSs’ demands, phase 1 will try to utilize MSs’ largest power at all instances, i.e.,∑_𝑖=1..𝑁𝑥𝑖,𝑗_𝑘 ⋅ 𝑝𝑖,𝑗_𝑘 =

𝑃𝑀𝐴𝑋

𝑘 if ∑𝑖=1..𝑁𝑥𝑖,𝑗𝑘 ≥ 1 for any 𝑗. We define a reward function to calculate the benefit of assigning a tile to an MS

and apply a greedy approach for MSs to compete for tiles by this function iteratively. In this iterative process, we use a binary flagℐ𝑘 to reflect whether MS𝑘’s demand is satisfied or

not and𝑑𝑖,𝑗

𝑘 to reflect the amount of data that MS𝑘 transmits

in tile(𝑖, 𝑗).

1) Initially, set allℐ𝑘 = 0, all 𝑑𝑖,𝑗𝑘 = 0, and all 𝑥𝑖,𝑗𝑘 = 0.

2) Define the reward function for tile (𝑎, 𝑏) to be assigned to MS_𝑘 as:

𝑤((𝑎, 𝑏), 𝑘) = Δ𝑅((𝑎, 𝑏), 𝑘)_{𝐴(𝑏, 𝑘)} , (5) where𝐴(𝑏, 𝑘) is the maximal rate that MS𝑘can transmit

if all tiles in time slot𝑏 are allocated to MS𝑘, 𝐴(𝑏, 𝑘) = ∑ 𝑖=1..𝑁 𝐵 log₂(1 + ˜𝑝𝑖,𝑏_𝑘 ⋅ 𝑓𝑖 𝑘). Note that ˜𝑝𝑖,𝑏_𝑘 = (𝜆 − 1 𝑓𝑖 𝑘)

+ _{can be found using the}

rate-optimum water-filling technique [7], where 𝑧+ ₌

max{𝑧, 0} and 𝜆 is a water level constant subject to ∑

𝑖=1..𝑁˜𝑝𝑖,𝑏𝑘 = 𝑃𝑘𝑀𝐴𝑋. The numeratorΔ𝑅((𝑎, 𝑏), 𝑘) is

the additional rate that MS𝑘can transmit in time slot𝑏 if

we assign tile(𝑎, 𝑏) to MS𝑘. Since MS𝑘 has no tile

ini-tially, we haveΔ𝑅((𝑎, 𝑏), 𝑘) = 𝐵 log₂(1+𝑃𝑀𝐴𝑋 𝑘 ⋅𝑓𝑘𝑎).

Intuitively,𝑤((𝑎, 𝑏), 𝑘) is to compare the importance of

tile(𝑎, 𝑏) to MS𝑘. A higher ratio Δ𝑅((𝑎,𝑏),𝑘)_{𝐴(𝑏,𝑘)} means that

tile(𝑎, 𝑏) is more important to MS𝑘.

3) For each free tile (𝑎, 𝑏) and each unsatisfied MS𝑘,

compare their reward value 𝑤((𝑎, 𝑏), 𝑘). Pick the one,

say,𝑤((ˆ𝑎, ˆ𝑏), ˆ𝑘) with the maximum positive value and

assign tile (ˆ𝑎, ˆ𝑏) to MS_ˆ𝑘. Then we set 𝑥ˆ𝑎,ˆ𝑏_ˆ𝑘 = 1. Note that this assignment has implied that MS_ˆ𝑘 has to redistribute its total power𝑃𝑀𝐴𝑋

ˆ𝑘 to all tiles in time slot

ˆ𝑏 assigned to it. Again, this is obtained by the water-filling technique. This means that we need to update all

𝑝∗,ˆ𝑏_ˆ𝑘 and 𝑑∗,ˆ𝑏_ˆ𝑘 accordingly. Also, we need to update ℐ_ˆ𝑘

if MS_𝑘’s demand is already satisfied.

4) Since MS_ˆ𝑘 has won tile(ˆ𝑎, ˆ𝑏) in time slot ˆ𝑏, we need to update the numerator parts of all MSs’ reward functions properly.

a) For each MS_𝑘∕=MS_ˆ𝑘, setΔ𝑅((ˆ𝑎, ˆ𝑏), 𝑘) = 0. b) For MS_ˆ𝑘, since its power distribution in time

slot ˆ𝑏 has been changed, we need to recompute

Δ𝑅((𝑎, ˆ𝑏), ˆ𝑘), 𝑎 ∕= ˆ𝑎, for each free tile (𝑎, ˆ𝑏) in

advance by finding the additional rate that MS_ˆ𝑘 can transmit.

The above updates will change MSs’ reward functions accordingly. Note that we do not change the denomina-tor in Eq. (5) throughout the process.

5) If there is any remaining free tile and any unsatisfied MS, go to step 3; otherwise, terminate phase 1.

B. Phase 2: Reducing MSs’ Energy Costs

Phase 1 aims at reducing the frame usage. If there are remaining free tiles, we use “spread” the data transmissions of the allocated tiles to these free tiles. This normally imposes lower transmission rates on MSs, and thus lower energy consumption. Toward this goal, we define another energy

reward function for MSs to compete for free tiles. Phase 2

will repeat this process until all free tiles are allocated. First, we build our model of rewarding. Consider any free tile (𝑎, 𝑏) and any MS𝑘. Let 𝑥𝑖,𝑗𝑘 , 𝑝𝑖,𝑗𝑘 , and 𝑑𝑖,𝑗𝑘 be

the parameters reflecting the current allocation to MS_𝑘. The energy reward function of assigning tile (𝑎, 𝑏) to MS𝑘 is

defined as

𝑒((𝑎, 𝑏), 𝑘) = max{𝐸𝑠𝑡((𝑎, 𝑏), 𝑘), 𝐸𝑐ℎ((𝑎, 𝑏), 𝑘)}. 𝐸𝑠𝑡((𝑎, 𝑏), 𝑘) and 𝐸𝑐ℎ((𝑎, 𝑏), 𝑘) are the conserved energy

if we spread some of the data of MS_𝑘 to be transmitted in time slot 𝑏 and subchannel 𝑎 to tile (𝑎, 𝑏), respectively. In

particular, 𝐸𝑠𝑡((𝑎, 𝑏), 𝑘) and 𝐸𝑐ℎ((𝑎, 𝑏), 𝑘) can be presented

as the following equations.

𝐸𝑠𝑡((𝑎, 𝑏), 𝑘) = ( ∑ 𝑖=1..𝑁 𝑥𝑖,𝑏 𝑘 ⋅ 𝑝𝑖,𝑏𝑘 ) − (˜𝑝𝑎,𝑏𝑘 + ∑ 𝑖=1..𝑁 𝑥𝑖,𝑏 𝑘 ⋅ ˜𝑝𝑖,𝑏𝑘 ) 𝐸𝑐ℎ((𝑎, 𝑏), 𝑘) = ( ∑ 𝑗=1..𝑀 𝑥𝑎,𝑗 𝑘 ⋅ 𝑝𝑎,𝑗𝑘 ) − (˜𝑝𝑎,𝑏𝑘 + ∑ 𝑗=1..𝑀 𝑥𝑎,𝑗 𝑘 ⋅ ˜𝑝𝑎,𝑗𝑘 ),

where ˜𝑝𝑖,𝑏_𝑘 , 𝑖 = 1..𝑁, and ˜𝑝𝑎,𝑗_𝑘 , 𝑗 = 1..𝑀, are the allocated

power after assigning the free tile(𝑎, 𝑏) to MS𝑘 to spread the

data transmitted in time slot𝑏 and subchannel 𝑎, respectively.

We derive the allocating power ˜𝑝𝑖,𝑏_𝑘 and ˜𝑝𝑎,𝑗_𝑘 by the proposed

energy-optimum water-filling technique, where

˜𝑝𝑖,𝑏_𝑘 = { 𝑥𝑖,𝑏 𝑘 ⋅ (𝜆 −𝑓1𝑖 𝑘) +_{, ∀𝑖 = 1..𝑁, 𝑖 ∕= 𝑎} (𝜆 − 1 𝑓𝑖 𝑘) +_{, 𝑖 = 𝑎} , (6) subject to ∑_𝑖=1..𝑁𝑥𝑖,𝑏_𝑘 ⋅ 𝑑𝑖,𝑏_𝑘 = (𝐵𝑙𝑜𝑔2(1 + ˜𝑝𝑎,𝑏𝑘 ⋅ 𝑓𝑘𝑎) + ∑

𝑖=1..𝑁𝑥𝑖,𝑏𝑘 ⋅ 𝐵𝑙𝑜𝑔2(1 + ˜𝑝𝑖,𝑏𝑘 ⋅ 𝑓𝑘𝑖)). In the end of the section,

we will prove that the proposed energy-optimum water-filling technique can save the most energy when assigning an MS an additional free tile. Note that it is possible that the new allocated power is greater than the original one except for that of tile(𝑎, 𝑏), especially when a free tile is used to spread the data on a subchannel. To prevent this, we will recalculate

𝐸𝑐ℎ((𝑎, 𝑏), 𝑘) and ˜𝑝𝑎,𝑗_𝑘 by omitting this tile when this case

happens.

Below, we present the operations of phase 2 in detail. 1) Calculate𝑒((𝑎, 𝑏), 𝑘) for each free tile (𝑎, 𝑏) for MS𝑘.

2) From all𝑒((𝑎, 𝑏), 𝑘), pick the one, say, 𝑒((ˆ𝑎, ˆ𝑏), ˆ𝑘) with

the maximum positive value assigning tile(ˆ𝑎, ˆ𝑏) to MS_ˆ𝑘. a) Set𝑥ˆ𝑎,ˆ𝑏_ˆ𝑘 = 1. Update the allocated power (𝑝∗,ˆ𝑏_ˆ𝑘 and

1286 IEEE COMMUNICATIONS LETTERS, VOL. 15, NO. 12, DECEMBER 2011

𝑑ˆ𝑎,∗_ˆ𝑘 ) accordingly. Finally, set 𝑒((ˆ𝑎, ˆ𝑏), 𝑘) = 0 for

all MS_𝑘 and recompute𝑒((𝑎, ˆ𝑏), ˆ𝑘) and 𝑒((ˆ𝑎, 𝑏), ˆ𝑘)

for those free tiles on subchannelˆ𝑎 and time slot ˆ𝑏, respectively.

b) If there is any remaining free tile, go to step 2. Otherwise, terminate phase 2.

Below, we first show that the proposed energy-optimum water-filling technique can minimize the total power of the allocated tiles in a time slot for an MS by keeping to deliver the fixed amount of data. Then, we will prove that spreading the data transmissions of the allocated tiles to an additional free tile can further reduce an MS’s total energy consumption. Based on these two properties, we can minimize the total power when assigning an additional tile for an MS in a time slot.

Theorem 1: Consider the MS_𝑘has been allocated tiles with amount of data 𝐷 in time slot 𝑏, the power allocated by

the proposed water-filling technique can incur the minimum energy.

Proof: Without loss of generality, we assume that MS_𝑘

has been allocated 𝑛 tiles in time slot 𝑏 and the channel

gain to noise ratios of these tiles are 𝑓𝑖

𝑘, 𝑖 = 1..𝑛, and the

allocated power of each tile is𝑝𝑖,𝑏_𝑘 (we shorten𝑓𝑖

𝑘 as𝑓(𝑖) and 𝑝𝑖,𝑏_𝑘 as 𝑝(𝑖) _{for ease of presentation). The allocation problem}

to incur the minimum total energy on these 𝑛 tiles subject

to the amount of data 𝐷 in time slot 𝑏 can be formulated

as the following optimization: min_𝑝(𝑖)∑_𝑖=1..𝑛𝑝(𝑖) subject

to ∑_𝑖=1..𝑛𝐵 log₂(1 + 𝑝(𝑖) _{⋅ 𝑓}(𝑖)_{) = 𝐷. By the method of}

Lagrange multiplier, we can use𝑦 as the Lagrange multiplier

and rewrite the optimization as:min𝑝(𝑖)𝐿(𝑦, 𝑝(1), .., 𝑝(𝑛)) =

∑

𝑖=1..𝑛𝑝(𝑖)+ 𝑦 ⋅ (𝐷 −

∑

𝑖=1..𝑛𝐵 log2(1 + 𝑝(𝑖)⋅ 𝑓(𝑖))). By

differentiating 𝐿(𝑦, 𝑝(1)_{, .., 𝑝}(𝑛)_{) respect to 𝑦 and 𝑝}(𝑖)_{, 𝑖 =}

1..𝑛, and setting the results equal to zero yield 𝑝(𝑖) _{= (𝜆 −} 1

𝑓(𝑖))+, 𝑖 = 1..𝑛, where 𝜆 is the water level constant such that 𝐷 =∑_𝑖=1..𝑛𝐵 log₂(1 + 𝑝(𝑖)_{⋅ 𝑓}(𝑖)_).

Theorem 2: With the proposed water-filling technique,

un-der the same amount of allocated data in time slot𝑏, we can

save an MS’s energy by spreading the data transmissions of the allocated tiles to the additional free tile in time slot𝑏.

Proof: Without loss of generality, we assume that we

have allocated MS_𝑘 𝑛 tiles in time slot 𝑏 with the channel

gain to noise ratios of 𝑓𝑖

𝑘, 𝑖 = 1..𝑛 (we shorten 𝑓𝑘𝑖 as 𝑓(𝑖)

for ease of presentation). The allocating power of each tile (𝑖, 𝑏) is assumed to be (𝜆𝐴−_𝑓1(𝑖)) > 0, 𝑖 = 1..𝑛, where 𝜆𝐴

is the current level determined by the proposed water-filling technique. Now, given one more free tile, denoted as the (𝑛 + 1)th tile, from time slot 𝑏 with the channel gain to noise ratio of𝑓(𝑛+1)_{to the MS}

𝑘. Here, we assume the tile’s channel

gain to noise ratio is good enough, i.e.,(𝜆𝐴−_𝑓(𝑛+1)1 ) > 0;

otherwise, no power will be allocated to the free tile and the total energy allocation will be the same as the original. Let𝜆𝐵

be the new water level conducted by the proposed water-filling technique for those𝑛 + 1 tiles. The reduced energy Δ𝔼 by

spreading the data transmissions of the𝑛 tiles to the (𝑛 + 1)

tiles can be written as

Δ𝔼 = ∑ 𝑖=1..𝑛 (𝜆𝐴−_𝑓1_(𝑖)) − ∑ 𝑖=1..𝑛+1 (𝜆𝐵−_𝑓1_(𝑖)). (7)

Since we have to maintain the same amount of data

de-0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1 3 5 7 9 11 13 15 number of MSs ener gy cost ( m W x sec.) MP SA QD RE OPT ours

(a) energy consumption

0.6 0.7 0.8 0.9 1.0 21 23 25 27 29 31 33 35 number of MSs satisfaction ratio MP SA QD RE OPT ours (b) satisfaction ratio Fig. 1. The simulation results under different numbers of MSs.

livery, we have ∑_𝑖=1..𝑛𝐵 log₂(1 + (𝜆𝐴 − _𝑓1(𝑖)) ⋅ 𝑓(𝑖)) =

∑

𝑖=1..𝑛+1𝐵 log2(1 + (𝜆𝐵−_𝑓1(𝑖))⋅ 𝑓(𝑖)). This equation yields 𝜆𝐵 = 𝑛+1

√

(𝜆𝐴)𝑛

𝑓(𝑛+1). Thus, bring it to Eq. (7), we can rewrite

Eq. (7) as Δ𝔼 =_𝑓(𝑛+1)1 ⋅ (𝑛 ⋅ 𝜆𝐴⋅ 𝑓(𝑛+1)− (𝑛 + 1) ⋅ 𝑛+1 √ (𝜆𝐴)𝑛⋅ (𝑓(𝑛+1))𝑛+ 1). (8)

Let𝑥 = 𝜆𝐴⋅ 𝑓(𝑛+1) (and thus𝑥 > 0). Eq. (8) can be further

rewritten as Δ𝔼 = 1 𝑓(𝑛+1)(𝑛 ⋅ 𝑥 − (𝑛 + 1) ⋅ 𝑥 𝑛 𝑛+1 + 1) = 1 𝑓(𝑛+1)(𝑥 1 𝑛+1 − 1)(𝑛 ⋅ 𝑥𝑛+1𝑛 − 𝑥(𝑛−1)𝑛+1 − 𝑥𝑛−2𝑛+1 − ... − 1) = 1 𝑓(𝑛+1)(𝑥 1 𝑛+1 − 1)(𝑥𝑛−1𝑛+1 ⋅ (𝑥𝑛+11 − 1) + 𝑥𝑛−2𝑛+1 ⋅ (𝑥𝑛+12 − 1) + ... + (𝑥𝑛+11 − 1)(𝑥𝑛−1𝑛+1 + 𝑥𝑛−2𝑛+1 + ... + 1)) = 1 𝑓(𝑛+1)(𝑥 1 𝑛+1 − 1)2_(𝑥𝑛−1𝑛+1+𝑥𝑛−2𝑛+1⋅(𝑥𝑛+11 +1)+...+(𝑥𝑛−1𝑛+1+𝑥𝑛−2𝑛+1+...+1)) > 0. Thus, we can ensure that the MS_𝑘’s energy consumption in time slot𝑏 can be saved by spreading the data transmissions

of the allocated tiles to the additional free tile in time slot 𝑏.

Above two properties can be applied for the tiles on the same subchannel𝑎, since the channel gain to noise ratios of

the tiles on the same subchannel are the same, i.e., 𝑓(1) ₌

𝑓(2)_{= ... = 𝑓}(𝑛)_{= 𝑓}(𝑎)_{, which is a special case of Theorems}

1 and 2.

IV. PERFORMANCEEVALUATION ANDCONCLUSION In the simulation, we consider the scenario the same as [1], which is for the uplink of a single-cell OFDMA network. The cell radius is 1 km and the MSs are uniformly distributed over the cell. In particular, the uplink subframe duration is 2.5 ms with 𝑁 = 15 subchannels and 𝑀 = 15 time slots. The

subchannel bandwidth𝐵 is 180 KHz. The path loss follows

the modified Hata urban propagation model, including the frequency correlation and multipath fading. Each MS_𝑘 has a traffic demand up to 2.56 Kbits/frame (= 512 Kbits/s) [4] in average and a maximum power per time slot 𝑃𝑀𝐴𝑋

𝑘 = 50

mW.

We compare our heuristic against the maximal-rate pair

(MP) scheme [1], the sequential-allocation (SA) scheme

[2], the quota-determination (QD) scheme [3], the

resource-efficient (RE) scheme [4], and the optimal (OPT) scheme.

MP scheme iteratively finds the MS-subchannel pair with

the maximal rate to allocate. SA scheme sequentially allo-cates each subchannel to the unsatisfied MS which has the maximum incremental rate on the subchannel. QD scheme

LIANG et al.: ON TILE-AND-ENERGY ALLOCATION IN OFDMA BROADBAND WIRELESS NETWORKS 1287

assigns each MS a quota in terms of number of subchannels based on its demand. Then, QD allocates an MS subchannels when it has the best subchannel gains on them. RE scheme dynamically adjusts the number of subchannels allocated to the MSs over frames to reduce the amount of resources. OPT exploits Lingo software [8] to find the optimum results, but this incurs high computational cost. Note that we can not compare the scheme in [5] because it neglects the power constraint. Fig. 1(a) compares the energy consumption per frame of all schemes under different numbers of MSs. Our heuristic can approximate OPT scheme and save up to 70% of energy in average as compared to MP, SA, QD, and RE schemes, because our heuristic utilizes the free tiles to spread the allocated data to lower down MSs’ transmission power, which is neglected by above schemes. Fig. 1(b) investigates the number of MSs on satisfaction ratio, which is the ratio of the amount of satisfied demands to the total amount of demands per frame. Our heuristic has higher satisfaction ratio because it allocates the subchannels by consulting both the incremental rate and the maximum rate of each MS. This can utilize the deviation of subchannels for each MS and thus further exploit multiuser diversity to satisfy more MSs’ demands.

To conclude, this letter addressed the energy conservation issue in uplink transmissions of an OFDMA wireless network. This problem is formulated by an MIP problem and an energy-efficient heuristic is proposed to satisfy MSs’ demands while saving their energy. By employing multiuser diversity and efficient power allocation in additional free tiles, simulation results verified that, compared with the existing schemes, our heuristic can save more MSs’ energy and increase their satisfaction ratios.

REFERENCES

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