25. fast algorithms for resource allocation in wireless cellular networks

7/31/2019 25. Fast Algorithms for Resource Allocation in Wireless Cellular Networks

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Fast Algorithms for Resource Allocation in

Wireless Cellular NetworksRitesh Madan, Stephen P. Boyd, and Sanjay Lall

Abstract We consider a scheduled orthogonal frequency di-vision multiplexed (OFDM) wireless cellular network where thechannels from the base station to the n mobile users undergoflat fading. Spectral resources are to be divided among theusers in order to maximize total user utility. We show that thisproblem can be cast as a nonlinear convex optimization problem,and describe an O(n) algorithm to solve it. Computationalexperiments show that the algorithm typically converges inaround 25 iterations, where each iteration has a cost that is O(n),with a modest constant. When the algorithm starts from an initialresource allocation that is close to optimal, convergence typicallytakes even fewer iterations. Thus, the algorithm can efficiently

track the optimal resource allocation as the channel conditionschange due to fading. We also show how our techniques canbe extended to solve resource allocation problems that arise inwideband networks with frequency selective fading and when theutility of a user is also a function of the resource allocations inthe past.

I. INTRODUCTION

Resource allocation in wireless networks is fundamentally

different than that in wireline networks due to the time varying

nature of the wireless channel [1]. There has been much prior

work on scheduling policies in wireless networks to allocate

resources among different flows based on the channels they

see and the flow state [1], [2]. The flow state can consistof the average rate seen by the flow in the past [3], [4],

the delay of the head-of-line packet [5], or the length of the

queue [1]. Much prior work in this area can be divided into

two categories:

1) Scheduling for Best Effort flows: The end-user experi-

ence for a best effort (elastic) flow is modeled by a con-

cave increasing utility function of the rate experienced

by the flow [6]. The proportional fair algorithm (see,

for example, [7]) where all the resources are allocated

to the flow with the maximum ratio of instantaneous

spectral efficiency (which depends on the channel gain)

to the average rate has been analyzed in [8], [9],

R. Madan is at Qualcomm-Flarion Technologies, NJ, USA. This work wasperformed when he was at the Department of Electrical Engineering, StanfordUniversity. [email protected]

S. P. Boyd is at the Department of Electrical Engineering, StanfordUniversity. [email protected]

S. Lall is at the Department of Aeronautics and Astronautics, StanfordUniversity. [email protected]

This work was funded in part by the MARCO Focus Center for Cir-cuit & System Solutions (C2S2, www.c2s2.org), under contract 2003-CT-888, by AFOSR grant AF F49620-01-1-0365, by NSF grant ECS-0423905,by NSF grant 0529426, by DARPA/MIT grant 5710001848, by AFOSRgrant FA9550-06-1-0514, DARPA/Lockheed contract N66001-06-C-2021, byAFOSR/Vanderbilt grant FA9550-06-1-0312, the Stanford URI Architecturefor Secure and Robust Distributed Infrastructures (AFOSR DoD award number49620-01-1-0365), and the Sequoia Capital Stanford Graduate Fellowship.

[3]. Under general assumptions on channel fading, the

above allocation was shown to converge weakly to an

allocation which maximized the sum of log utilities

of the average rates of the flows when the averaging

was done using an exponential filter with a large time

constant [3]. The analysis also extends to a wider class

of concave increasing utility functions [3]. A more

general scheduling rule where potentially multiple users

can be scheduled simultaneously has been considered

in [10]. The analysis in the above papers assumes that

the congestion control algorithm, i.e., TCP, has a largeenough time-out such that when a flow is not scheduled

because of bad channel states, TCP does not time-out.

Joint design of scheduling and congestion control has

been considered in, for example, [11], [12], [13], [4].

2) Scheduling for QoS Flows: Quality-of-Service (QoS)

flows are typically modeled by a pre-determined arrival

process and a delay deadline for each packet. For such

flows, we can roughly define the stability region as

follows: The stability region for a set of queues is

defined as the set of arrival rates at the queues for which

there exists a scheduling policy such that the length of

any queue does not grow without bound over time (see,

for example, [14]). A stabilizing policy is one whichensures that the queue lengths do not grow without

bound. Stabilizing policies for a vector of arrival rates

within the stability region for different wireless network

models have been characterized in, for example, [15],

[16], [17], [1], [5], [14]. The scheduling policy in [5]

minimizes the percentage of packets lost because of

deadline expiry, while the delay performance of the

exponential rule (introduced in [1]) was empirically

studied in [18].

We note that policies to schedule a mixture of best effort

and QoS flows have been considered in [18]. Distributed

algorithms for interference management to maximize the sumutilities of user signal-to-noise ratios (SNR) in cellular net-

works have been studied in [19], [20]. Also, related cross-

layer optimization problems for resource allocation in wireless

networks with different objectives have been analyzed in,

for example, [21], [22], [23]. The above summary is only

a representative sample of the work in the general area of

resource allocation in wireless networks. For a more complete

description of prior work, we refer the reader to [1], [2], and

the references therein.

In this paper, we focus on best effort flows. We study the

problem of resource allocation in wideband OFDM wireless


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cellular networks like Ultra Mobile Broadband (UMB) [24]

and Long Term Evolution path for 3GPP [25]. In particular,

we study the assignment of power and spectral resources to

maximize the sum-utility of the achieved data rates the

user utility can be a function of instantaneous rate or average

rate over time. For both these cases, the solution in general

can result in the distribution of resources to multiple flows

at the same time. We show that the problem is a convex

optimization problem. Hence, it can be solved in O(n3) timefor n users using a general purpose barrier method (see, forexample, [26]). However, the time-varying nature of wireless

channels necessitates re-computation of an optimal resource

allocation in an online manner. This requires the design of

faster computational algorithms to track the optimal resource

allocation. We exploit the underlying structure of the problem

to derive a specialized barrier method that has a complexity of

O(n). We also illustrate the generality of our computationaltechniques through extensions to frequency selective fading,

where we exploit frequency diversity.

We note that our work focusses on computational algorithms

and is complementary to that in [8], [9], [3], [10]. The focus ofthose papers is on the asymptotic analysis when the user utility

is a function of the rate averaged over a very long time. With

the exception of [10], all the above works focus on scheduling

a single user at any given time, i.e., all the resources are

allocated to a single user. Even in [10], at each scheduling

instant, it is sufficient to consider only the gradient of the

utility function at the vector of average rates of the users and

compute a resource allocation that maximizes the incremental

utility assuming a first order Taylor-series expansion such a

scheme reduces to one where a single user is scheduled at any

given time for the system model we consider in this paper. We

note that schemes which allocate all the resources to a single

user at any given time do not scale to large bandwidths anda large number of flows we postpone a detailed discussion

of the similarities and the differences between our approach

of solving a utility maximization problem and previous work

to Sec. V, where we also show that the solution computed by

our approach reduces to that in [8], [9], [3], [10] in a special

case.

A. Organization

The rest of the paper is organized as follows. We firstconsider the utility for each flow to be a function of the

instantaneous rate. We describe the mathematical model and

problem formulation, and prove the existence of a unique

positive solution in Sec. II. We exploit the structure of the

underlying optimization problem to obtain an O(n) algorithmand illustrate its typical behavior through computational results

in Sec. III. In Sections IV and V, we consider frequency

selective fading and the case where the utility of a user

is a function of its average rate, respectively. In Sec. VI,

we compare our algorithm with other standard computational

approaches.

I I . PROBLEM FORMULATION

A. System Model

We model an OFDM wireless cellular network where spec-

trum and power need to be divided between communication

flows (users) on n links in a cell. We formulate an optimizationproblem which is applicable to the downlink as we show

later, extensions to the uplink can be similarly obtained. We

assume an M-Quadrature Amplitude Modulation (MQAM)scheme for transmission and a total system bandwidth, B.Then, the maximum rate (in nats/sec) at which a user, i, cantransmit is given by

Ri = Bi log

1 +

PiGiKN0Bi

,

where Pi is the transmit power, Gi is the channel gain over thelink to user i, Bi is the bandwidth allocated to user i, N0 isthe noise power spectral density, and K = 1.5/ log(5BER),where BER is the desired (constant) bit error rate [27].

We denote the effective flow rate in nats/sec/Hz for user iby ri = Ri/Bi 0, and the fraction of bandwidth allocatedto it by bi 0. We denote the associated vectors of rates andbandwidth-fractions as r Rn and b Rn, respectively. Thepower consumption to support flow ri > 0 can be modeled as

pi(ri, bi) = aibi

eri/bi 1

, ai = N0B/(GiK).

When ri = 0, the power required is 0. Hence, the powerconsumption of user i as a function of ri and bi is aif(ri, bi)where the function f: S R is defined as follows:

f(x, y) =

y(ex/y 1) if y > 0,

0 otherwise.

The set S R2

is given byS = {0} { (x, y) R2 | x 0, y > 0 }.

We assume that each cell has a (weighted) total power

constraint of the form

P(r, b) =ni=1

wiaifi(ri, bi) Pmax,

where P(r, b) is the (weighted) total power, Pmax > 0 isthe given maximum (weighted) total power, and wi > 0 arethe weights. This constraint can be used to model a sum-

power constraint, with wi = 1, for the downlink in a cell.For the uplink, it can also be used to model the requirement

that the total interference at a neighboring interfering base

station should be kept below some threshold; the weights then

represent the power gains to the neighboring base station1. We

will normalize the power constraint by defining the normalized

power p : Sn R by p(r, b) =n

i=1 cif(ri, bi) whereci = wiai/Pmax. The power constraint is then p(r, b) 1.

1In general we can have a total interference budget constraint at more thanone base-station our analysis extends to this case as well. Also, a totalinterference budget constraint is a reasonable way to keep interference low atneighboring base stations when the frequency tones in neighboring cells hoprandomly and independently of each other [7]. In this case, for the uplink,N0 now represents the noise plus average interference power spectral density.


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We first observe that pi is a convex function ofri and bi. Thefunction g(x, y) = yex/y, defined for y > 0, is the perspectiveof the exponential function, and so is convex in x and y (see,e.g., [26, Sec. 3.2.6]). The function pi is obtained from g byan affine composition, and the addition of a linear term, and

so is convex. The total power P is therefore also a convexfunction of r and so, the total power constraint is a convexconstraint for r, b > 0.

B. User Utility Functions

The utility for user i is a function of its instantaneous rate,given by Ui(ri), so the total utility is

U(r) =ni=1

Ui(ri).

We assume that the utility functions Ui : (0, ) R arethrice continuously differentiable with

Ui(x) > 0, Ui (x) < 0,

for all x > 0 and

limx0+

Ui(x) = .

Thus, Ui (and therefore also U) is strictly increasing andstrictly concave, and the marginal utility increases without

bound as the rate converges to zero. Examples of common

utility functions satisfying these conditions include log x andxa, for 0 < a < 1.

Note that the above utility function does not take into

account past allocations to users. We consider this extension

in Secion V and show that we can use our computational

techniques to efficiently compute a scheduling policy that is a

generalization of the proportional fair scheduling policy.

C. Maximum Utility Resource Allocation

Our goal is to choose r and b to maximize the total utility,subject to the power constraint, and the bandwidth-fraction

constraint:maximize U(r),subject to 1Tb = 1,

r > 0, b > 0,p(r, b) 1,

(1)

where 1 denotes the vector with all entries one. The opti-

mization variables are ri and bi; the problem data are ci and

the functions Ui. The vector inequalities are componentwise;r 0 means ri 0, i = 1, . . . , n. For convenience welldefine the feasible set D by

D =

(r, b) R2n | 1Tb = 1, p(r, b) 1, r > 0, b > 0

.

We now have the equivalent problem

maximize U(r),subject to (r, b) D.

(2)

In the following section we will show that there is a unique

optimal allocation (r, b) which is achieved at a point withr > 0 and b > 0. Hence relaxing these strict inequalities

to nonstrict inequalities, and appropriately interpreting p andU, does not change the optimal solution.

The resource allocation problem (2) is a convex optimiza-

tion problem, with 2n variables and 2n + 2 constraints.Roughly speaking, this means that its global solution can be

efficiently computed, for example by a general interior-point

method. These methods typically converge in a few tens of

iterations; each iteration in a general-purpose implementation

requires O(n3) arithmetic operations (see, e.g., [26, Ch. 11]or [28]). The algorithm we describe in the next section solves

the resource allocation problem much faster by exploiting its

special structure. The resulting interior point method converges

in about 30 iterations, where each iteration requires O(n)operations with a modest constant.

D. Existence and Uniqueness of a Positive Solution

In this section, we show that the resource allocation prob-

lem (1) has a unique solution (r, b), with r > 0 andb > 0. We will do this by constructing a sequence of pointsconverging to the maximum, which must therefore lie in the

closure of the feasible set. We first show the following the proofs of the next three lemmas have been moved to the

Appendix.

Lemma 1: The closure of D satisfies D Sn.The interpretation of this result is that allocating zero

bandwidth-fraction and positive rate to a user requires infinite

power. Hence for every point (r, b) in the feasible set, we musthave bi > 0 whenever ri > 0, and in fact this holds for theclosure of the feasible set also.

The next result shows that a point (r, b) with (ri, bi) =(0, 0) for some i cannot be optimal. The idea here is that sinceUi has infinite slope at 0, slightly increasing ri and bi will givean increase in utility Ui which outweighs the decrease in theother rates necessary to maintain the power constraint.

Lemma 2: Suppose (rk, bk) is a sequence in Sn with limit

limk

(rk, bk) = (r, b)

and (r, b) Sn, with 1Tb = 1 and p(r, b) 1. Suppose alsothat for all i = 1, . . . , n either ri > 0 or (ri, bi) = (0, 0).If there is some i such that (ri, bi) = (0, 0) then there exists(x, y) D such that

limk

U(rk) < U(x).

The final lemma needed shows that a point (r, b) with ri = 0

for some i must also have bi = 0. If this were not the case, wecould decrease bi to zero, spreading this bandwidth-fractionamong the other users, who can use the extra bandwidth-

fraction to increase their rates without increasing their powers,

thus giving a feasible point with larger total utility. Then using

Lemma 2, we can rule out the possibility that a maximizing

sequence converges to (r, b) = 0.Lemma 3: Suppose (rk, bk) is a sequence in Sn with limit

limk

(rk, bk) = (r, b)

and (r, b) Sn, with 1Tb = 1 and p(r, b) 1. If there issome i such that ri = 0, bi > 0, then there exists (x, y) D


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such that

limk

U(rk) < U(x).

We now have the following theorem showing the existence

and uniqueness of the solution.

Theorem 1: There exists a unique (r, b) D withr, b > 0 such that

U(r

, b

) = sup

U(r) | (r, b) D

.Proof: First notice that problem (2) is feasible. That is,

the set D is nonempty, since for small enough > 0 the choiceb = (1/n)1, r = 1 satisfies (r, b) D. Let

U = sup

U(r) | (r, b) D

.

Then U is finite, since D is bounded and U is concave.We must show that this optimal value is actually achieved.

Suppose (rk, bk) is a maximizing sequence in D, so thatU(rk, bk) U. By extracting a subsequence, we can assumethat (rk, bk) converges to a point (r, b) D. Lemma 1 impliesthis point lies in Sn and since it is optimal on D Lemma 3implies that r > 0 and b > 0. Hence the optimal value isachieved in D. Uniqueness now follows from strict concavityof U.

III . FAS T ONLINE RESOURCE ALLOCATION ALGORITHM

In this section, we describe the barrier method to compute

an optimal resource allocation. Such a method, in general, has

complexity O(n3). However, we exploit the structure of theproblem to reduce the complexity to O(n).

A. Barrier Method

We use the barrier method to solve the optimization problem

in (2) [26]. The central point (r

(t), b

(t)) for a given valueof t are given by the solution of the following problem

minimize tU(r) n

i=1 (log ri + log bi) log (1 p(r, b)) ,

subject to 1Tb = 1.(3)

As t increases, (r(t), b(t)) becomes a more accurate approx-imation to the solution to the problem in (2). Note that the

objective function above is convex, and the above problem is

a convex optimization problem. Moreover, the solution to the

above problem is unique. This follows, in particular, from the

positive-definiteness of the Hessian of the objective function,

as argued in Sec. III-C.

We collect the variables into one vector x R2n, x =(r1, b1, . . . , rn, bn). Note that we have interleaved the rateand bandwidth-fraction variables here, so that the variables

associated with a given user are adjacent. Also, we denote the

barrier function as

(x) = ni=1

(log ri + log bi) log (1 p(r, b)) ,

and

t(x) = tU(r) (x).

The barrier method is then as follows:

Given strictly feasible starting point x, t := t(0), > 1, tolerance .Repeat

1) Centering Step. Minimize t(x) subject to1Tb 1 = 0, starting at x.

2) Update. x := x(t).3) Stopping Criterion. quit if (2n + 1)/t < .

4) Increase t. t := t.

B. Newton Method

We now describe the Newton method to compute the central

point x(t), i.e., solve the problem in (3) for a given value oft. The Newton step x at x, and the associated dual variableare given by following equations

2t(x) ddT 0

x

= tU(r) t(x)0 ,

(4)

where d = [0 1 . . . 0 1]T. For the Newton method, we use abacktracking line search to ensure an adequate decrease in (see, e.g., [26, Ch.11] or [29]). The method is then as follows:

Given starting point x such that 1Tb = 1, tolerance, (0, 1/2), (0, 1).Repeat

1) Compute x and 2 := t(x)x.2) Stopping Criterion. quit if 2/2 3) Backtracking line search on t(x). s := 1.

while t(x + sx) > t(x) s2,

s := s.4) Update. x := x + sx.

C. Fast Computation of Newton Step

We now describe how we can exploit the structure of the

problem to compute the Newton step in O(n) time rather thanusing matrix inversion in (4) which has a cost of O(n3). Thegradient of the barrier function is given by

(x)

ri=

1

ri+

cieri/bi

1 p(r, b),

(x)bi= 1bi

+ cieri/bi

(1 1/bi) ci1 p(r, b)

.

The Hessian of the barrier function is given by

2(x) =

1/r211/b21

. . .

1/r2n1/b2n

+1

(1 p(r, b))2p(r, b)p(r, b)T +

1

1 p(r, b)2p(r, b).


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Hence, it follows that

2t(x) = t2U(r) + 2(x)

=1

(1 p(r, b))2p(r, b)p(r, b)T

+

H1H2

. ..

Hn

,

where the blocks not shown are all zero, and

Hi =

tUi (ri) + 1/r

2i 0

0 1/b2i

+1

1 p(r, b)

eri/bici/bi eri/biciri/b

2i

eri/biciri/b2i e

ri/bicir2i /b

3i

.

The gradient, p(r, b), of p(r, b) is given by

p(r, b)

ri= cie

ri/bi

p(r, b)bi

= cieri/bi(1 1/bi) ci.

Let us denote

g =1

(1 p(r, b))p(r, b),

h = tU(r) t(x).

Then we have

2t(x) =

H1H2

. . .

Hn

+ ggT.

It is easy to show that Hi > 0. Since ggT 0, it follows

that 2t(x) > 0. Since d is a nonzero vector, it follows thatthe KKT matrix on the left in equation (4) is invertible. Also,

the KKT matrix on the left in (4) is the sum of a block-arrow

matrix and a rank-one matrix. We exploit this structure to

compute the Newton step in O(n) time. Let us denote H =diag(H1, . . . , H n). In particular, we have (see, for example,[26, App. C])

x

= u

[gT 0]u

1 + [gT 0]vv,

where H ddT 0

u =

h0

, (5)

and H ddT 0

v =

g0

.

We now obtain analytical formulas for u and v, which can becomputed in O(n) time. We consider the computation of u indetail; the computation for v is identical. It follows from (5)that

u2i1u2i

= H1i

h2i1

h2i u2n+1

.

Substituting these back in (5), it follows that

u2n+1 =1n

i=1 H1i2,2

ni=1

(H1i2,1h2i1 + H1i2,2

h2i).

To compute u, we first obtain u2n+1, and then obtain the otheruis. Both these operations cost O(n).

D. Convergence AnalysisWe now prove the convergence of the Newton method for

this problem for a given t. The convergence of the barriermethod then follows. Consider the minimization of t(x).Define the set of iterates for the Newton method by L =L(x(0)), where the initial point (x(0)) is chosen to be strictlyfeasible. For the initial value of t, such a point is easy to findby allocating equal bandwidth fractions, and powers to users

such that the total power is less than 1, i.e., p(r(0), b(0)) < 1;for other iterations of the barrier method, the solution for the

previous value of t is guaranteed to be strictly feasible. TheNewton method is a descent method, i.e., t(x

(k)) t(x(0)),

for any iteration k.

We first consider the following two lemmas the proofshave been moved to the Appendix.

Lemma 4: For all iterations k of the Newton method, x(k)

is strictly feasible.

Now, it can be shown that the iterates belong to a closed

and bounded set.

Lemma 5: The set L L, where for any (r, b) L, ri, bisare bounded above and bounded away from zero.

Since the KKT matrix on the left in equation (4) is in-

vertible, and is a continuous function of (r, b), it followsthat its inverse is bounded on the closed set L. Also, 2tis a continuously differentiable function of (r, b) and hence,2t is Lipschitz continuous on L, and

2t is boundedabove on L. The convergence of the Newton method thenfollows (see, for example, [26, Ch. 10]).

A formal complexity analysis (i.e., a bound on the number

of Newton steps required to attain an accurate solution) can be

carried out, but this seems irrelevant to us, given the extremely

fast convergence of the algorithm in practice. A typical number

of steps required is 25, and often less.

E. Warm Start

The Newton method can be initialized with b = (1/n)1,and r = 1, where > 0 such that (r, b) is strictly feasible,

i.e., p(r, b) < 1. It can also be initialized with an approximatesolution, such as the solution of a resource allocation problem

that is close. Consider, for example, the situation where we

have computed the optimal resource allocation, and then the

problem changes, but not drastically; for example, the utility

functions change, or the channel parameters ai change, orthe maximum available power Pmax changes. Running thebarrier method starting from the previously computed optimal

point and a larger value of t typically cuts the number ofiterations required to 10 to 15. This can be repeated, in order toefficiently track the optimal resource allocation as the physical

parameters or requirements change.


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5 10 15 20 25 30 3510

3

102

101

100

101

102

103

104

total number of Newton iterations

dualitygap

0 5 10 15 20 25 3010

4

103

102

101

10

0

101

102

103

total number of Newton iterations

U U

Fig. 1. Typical convergence of the barrier method. Top. Norm of residualversus iteration for two different instances. Bottom. Convergence ofU Uversus iteration.

F. Numerical Results

In this section, we show the typical behavior of the algo-

rithm described in this paper. We consider a system ofn = 200users in a cell. The utility function for user i is taken to beUi(ri) = ki log ri, where ki are generated as independentuniform random variables on [1, 10]. We take wi = 1, i.e.,we model the sum-power constraint for the downlink.

We first study the convergence of our algorithm for ran-

domly generated cis. In particular, we consider each ci to berandomly distributed over [0.1, 5], i.e., the received signal tonoise ratio (SNR) at the mobile can vary over the large range

of -9.6 dB to 20dB. Figure 1 (top) shows the convergence of

the norm of the residual, versus cumulative Newton iteration,for two different instances of the problem. The bottom plot

shows the convergence of the utility to its optimal value; note

that all intermediate iterates are feasible. This plot shows that

the resource allocation obtained is close to optimal, from a

practical point of view, within 20 or so Newton iterations.Highly accurate solutions can be obtained in about 30 iter-

ations or so. Both plots are quite typical; similar results are

obtained as n and other problem parameters are varied.To illustrate warm-start methods, we simulated a wireless

network with time varying fading channels. The resulting

scheduling policy obtained by solving (1) has the following

0 100 200 300 400 50040

35

30

25

20

15

10

5

0

5

10

time (ms)

SNR

(dB)

Fig. 2. Typical realization of a Rayleigh fading channel with mean SNR of0 dB and Doppler frequency 5 Hz.

properties. Users with a higher average channel gain get more

resources on an average. Users get allocated more resources

when their instantaneous channel gain is relatively high than

when their instantaneous channel gain is low. In our simu-lation, each users channel undergoes mutually independent

Rayleigh fading with a Doppler frequency of 5Hz and mean

SNR of 0dB. We re-computed the optimal resource allocation

at every time step of 1ms. Thus, the channel completely de-

correlates after 200 time-steps or so. Also, the variation in

channel gains over time is very high. A typical realization of

such a channel is shown in Fig. 2 the channel can easily

swing over a range of 30 dB.

Figure 3 shows the number of Newton steps required to

re-converge to a very accurate optimal resource allocation,

starting from the previously computed one. The first compu-

tation (from a generic initial resource allocation) requires 29

cumulative Newton steps. For the rest of time-steps we useda larger value of t(0) such that only 2 centering steps wererequired for a guaranteed duality gap of less than 103. About80% of the time, the number of Newton iterations required forre-convergence is less than 15. A larger number of Newton

iterations is occasionally required at times when the rate of

change of channel is high; for example during deep fades.

IV. FREQUENCY SELECTIVE FADING

In this section, we describe an extension to the case where

there are m frequency bands such that over a given frequencyband, each users channel undergoes flat fading. For example,

it is sufficient to choose the bandwidth of each band to be

less than the minimum coherence bandwidth of the users [30].Denote by Gji , the channel gain on the jth frequency band foruser i. Similarly, denote the rate and bandwidth for user i onthe jth frequency band by rji and b

ji , respectively. Then the

total rate allocated to user i is

ri =

mj=1

rji .

Also, the total (weighted) power consumption is given by

p(r, b) =ni=1

mj=1

cjif(rji , b

ji ),


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0 20 40 60 80 1000

5

10

15

20

25

30

35

time

numb

erofNewtoniterations

10 15 20 25 30 350

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

number of Newton iterations

cdf

Fig. 3. Number of Newton iterations needed for re-convergence withRayleigh fading channels. Top. Number of Newton iterations for re-convergence during the first 100 time steps. Bottom. CDF of number ofNewton iterations for re-convergence over 500 time-steps.

where cji = wiN0B/(GjiK).

We again would like to compute a resource allocation to

maximize the total utility, i.e., solve the following optimization

problem.

maximizen

i=1 Ui

mj=1 r

ji

,

subject to 1Tbj = 1, j = 1, . . . , m ,

mj=1 r

ji > 0, i = 1, . . . , n ,

(rji , b

ji ) S, i = 1, . . . , n, j = 1, . . . , m ,

p(r, b) 1,(6)

where rj and bj are in Rn and denote the vectors of the ratesand the bandwidth-fractions given to the n users in frequencyband j, respectively.

Analysis to show the existence of a solution and con-

vergence of the barrier method is similar to that before.

We now illustrate an efficient method to compute the New-

ton step during each Newton iteration. Again, we inter-

leave all the variables into one vector x R2nm, x =(r11, b

11, . . . , r

m1 , b

m1 , . . . , r

1n, b

1n, . . . , r

mn , b

mn ).

The barrier function is given by

(x) = ni=1

mj=1

(log rji + log bji ) log(1 p(r, b)).

Also, denote

t(x) = tU(r) (x),

where now U(r) =n

i=1 Uim

j=1 r

j

i

.Then, at each iteration of the barrier method, we solve the

following problem using Newtons method.

minimize t(x),subject to 1Tbj = 1, j = 1, . . . , m .

(7)

The Newton step for this problem can be computed through

the solution of the linear equation in (4), where now, d is a2mn m matrix give by

d =

duser

...

duser

,

where duser is a 2m m matrix whose (2i, i) entry is one fori = 1, . . . , n, and all other entries are zero. Now,

2t(x) = t2U(r) + 2(x)

=1

(1 p(r, b))2p(r, b)p(r, b)T

+

K1K2

. . .

Kn

,

where the blocks not shown are all zero, and Kis are 2m2m

matrices given by the following.

Ki = tUi

m

j=1

rji

1 0 1 0 . . . 1 00 0 0 0 . . . 0 0

...

1 0 1 0 . . . 1 00 0 0 0 . . . 0 0

+

H1i. . .

Hmi

where

Hji =

1/(rj

i )2

00 1/(bji )

2

+

1

1 p(r, b)

er

j

i/bj

i cji/bji e

rji/bj

i cji rji /(b

ji )2

erji/bj

i cji rji /(b

ji )2 er

ji/bj

i cji (rji )2/(bji )

3

Thus, Ki is the sum of a block diagonal matrix (where theblocks are 2 2) and a rank one matrix. Hence, Ki can beinverted in O(m) time. Now, the Hessian of t(x) is the sumof a rank one matrix and a block diagonal matrix with blocks

given by the Kis, each of which can be inverted in O(m)time. Using the elimination of variables as before, it can be

shown that each Newton iteration can be performed in O(nm)


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8

time compare this with a general purpose method which

costs O(n3m3). Thus, the reduction in complexity is huge,especially, because in many systems the number of users, n,is much larger than the number of frequency bands, m [25],[24].

V. SCHEDULING ALGORITHMS WITH MEMORY

We now show how our computational techniques can be

used in a straightforward manner to track an optimal resource

allocation when the user utility is a function of the average

rate allocated to the user. The average is computed in an online

manner using an exponential filter. This can be used to model

the behavior that the end-user experience is a function of

the scheduled rates over multiple consecutive time-slots rather

than a single scheduling decision.

A. Utility Functions

The utility for user i is a function of its average rate.We consider an exponential averaging filter; in particular the

average rate, yi(), for user i is computed at time as follows:

yi() = ri() + (1 )yi( 1), (8)

where ri() is the rate allocated to user i at time , and 0 < < 1. Also, we assume all users are initialized with (possiblyvery small) non-zero average rates yi(0) > 0. Then the utilityof user i at time is given by Ui(yi()), so the total utility is

ni=1

Ui(yi())).

The assumptions on Ui are the same as those in previoussections. However, note that now Ui(ri()+(1)yi(1))is well defined for ri() = 0 because yi(0) > 0 (and hence,yi() > 0 for all finite ).

B. Resource Allocation

The total (weighted) normalized power consumption when

each user i is allocated rate ri() and bandwidth-fraction bi()is

p(r(), b()) =

ni=1

ci()f(ri(), bi()),

where ci() = wiN0B/(Gi()KPmax) and Gi() is thechannel gain for the ith user at time .

Our goal is to choose r() and b() at each time tomaximize the total utility, subject to the power constraint and

the total bandwidth constraint. Thus, at each time , we solvethe following resource allocation problem:

maximizen

i=1 Ui(ri() + (1 )yi( 1)),subject to 1Tb() = 1,

(r(), b()) S,p(r(), b()) 1.

(9)

The optimization variables are ri() and bi(); the problemdata are ci(), yi( 1), and the functions Ui.

For convenience well re-define the feasible set D by

D =

(r(), b()) R2n | 1Tb() = 1, p(r(), b()) 1,

(r(), b()) S

.

We now have the equivalent problem

maximizen

i=1 Ui(ri() + (1 )yi( 1)),subject to (r(), b()) D.

(10)

C. Relation to Proportional Fair Scheduling

Note that when we take to be small enough, a solution tothe optimization problem (9) is one where all the bandwidth

and power is allocated to a user i for which

log

1 +

1

ci()

Ui(yi( 1))

log

1 +

1

cj()

Uj(yj( 1)), j = 1, . . . , n .

(11)

This scheduling scheme has been widely studied in the liter-

ature. It has been shown that under appropriate assumptions

on the channel gain processes Gi()s, the above schedulingscheme converges (as ) weakly to one which maxi-mizes the total utility of average rates, U(y()) [3].

The above scheduling scheme is a good one for narrowband

systems and when there are few users in the system it

exploits multi-user diversity well and users get scheduled afterrelatively short intervals of time. However, with the advent of

fourth generation wideband systems (e.g. LTE, WiMax, and

UMB) we need to consider values of which will distributethe resources among multiple users simultaneously due to the

following reasons:

1) Wideband systems can have a total bandwidth of 20

MHz, and if all the bandwidth is allocated to one user

(cell-phone), the user (cell-phone) may not even have

enough processing power to decode the huge burst of

data. In fact, the UMB spec specifies an upper bound

on the amount of data that can be transmitted to a user

in a single time slot [24].

2) Fourth generation systems will have thousands of flowsand hybrid ARQ mechanisms. Consider the case where

there are 5000 flows and each time slot is 1ms. More-

over, assume that it takes 3 hybrid ARQ re-transmissions

to transmit a packet. Then if all the flows experience

i.i.d. channels, on an average each flow will get sched-

uled roughly every 15 seconds this is clearly not

acceptable for many types of traffic even when the

individual packets do not have strict delay requirements.

In many applications (e.g., web browsing), a users

utility, i.e., the end-user experience is a function of the

average rate it sees over a short time horizon in the past

rather than over a very long time horizon. Also, in many

practical systems, this will lead to TCP time-outs andhence, the long inter-scheduling time will be interpreted

as congestion thereby deprecating performance.

We note that the problem formulation in (10) is for a general

value of (0, 1).

D. Existence and Uniqueness of Solution

In this sub-section, we simplify notation we drop the

dependence of the variables on . Also,we denote

U(r) =ni=1

Ui((1 )yi( 1) + ri). (12)


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9

We show that the resource allocation problem (1) has a

unique solution (r, b). The proof of the following lemmacan be found in the Appendix.

Lemma 6: The set D is closed.

We now have the following theorem showing the existence

and uniqueness of the solution.

Theorem 2: There exists a unique (r, b) D such that

U(r) = sup

U(r, b) | (r, b) D

.Proof: First notice that problem (10) is feasible. That is,

the set D is nonempty, since for small enough > 0 the choiceb = (1/n)1, r = 1 satisfies (r, b) D. The boundedness ofD is easy to see. Since, D is closed, the supremum is achieved.Uniqueness follows from strict concavity of U.

E. Fast Barrier Method

The barrier method to solve problem (10) is identical to that

in Sec. III except that the utility function is now given by that

in (12). Hence, using our approach we can solve problem (10)

in O(n) time.

F. Numerical Results

We considered a time varying channel model similar to that

in Sec. III-F. In particular, we consider 300 users with i.i.d.

Rayleigh fading channels with 25 Hz Doppler and mean gain

of 0dB. A typical sample path for this channel is shown in

Fig. 4. We again set Ui(yi()) = ki log(yi()), where ki weregenerated as independent uniform random variables on [1, 10].Also, we set 1/ = 100ms. Thus, if a user, i, does not getscheduled for 100 ms, its average rate, yi(), decays by about33%. The problem in (9) was again re-solved every one ms.

In Fig. 4, we plot the utility function as a function of

time (after initial transients) for the following three resource

allocation schemes.

1) Utility maximization: This scheme corresponds to allo-

cating resources according to the solution of (9) which

is updated every ms.

2) Proportional Fair: All the resources are allocated to a

single user according to the scheduling policy in (11).

3) Equal Resource: In this scheme, power and spectrum

are equally distributed among all users at all times.

Since, we use log utilities for our computations, the difference

in utilities is a reasonable metric for comparison (vs. ratios

of utilities which can change a lot depending on the unitsof ris). Also, note that the large negative values for the totalutility are because we consider normalized rates ri()s, and sori() 1 always. We see that the proportional-fair algorithmtracks the optimal utility function poorly over time this

is to be expected because the proportional-fair algorithm is

designed to be optimal for very large time constants, i.e.,

small values of . The main reason for the poor performanceof the proportional fair algorithm is the large delay between

successive times when a flow is scheduled. In fact, the equal

resource allocation algorithm outperforms the proportional-fair

algorithm.

0 200 400 600 800 100050

40

30

20

10

0

10

time (ms)

S

NR

(dB)

400 500 600 700 8001.15

1.1

1.05

1

0.95x 10

4

time (ms)

utility

utility maximization

proportional fair

equal resource

Fig. 4. Scheduling with memory and log utilities. Top. Typical sample pathof channel gain. Bottom. Evolution of utility functions with time for twodifferent scheduling policies.

We show the evolution of the average rate of a single user in

Fig. 5. Specifically, we plot the average rate for the above three

schemes, and for a proportional-fair scheme with a smaller

= 1/500. At any time , the increase in average rateis due to resources allocated to that user, while the decay

is due to the exponential averaging when no resources are

allocated. We can see that the utility maximization scheme

dominates the equal resource scheme this is because the

equal resource scheme does not take advantage of multi-user

diversity by allocating more resources to users which have

strong channels at any given time. Also, for most of the time,the utility maximization scheme has a higher average rate

than that for the proportional fair scheme with = 1/100.This is because the proportional fair scheme does not allocate

any resources to the user for 100s of ms, and then allocates

all the power and spectrum to the user. We also show the

evolution of average rate for a proportional-fair scheme with

= 1/500. Since, the exponential filter is different in thiscase, we cannot directly compare its average rate to those

of the other schemes. However, we can see that there are

larger intervals in this case when the user is not allocated

any resources.


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10

0 500 1000 1500 20000

0.002

0.004

0.006

0.008

0.01

time (ms)

av

eragerate

utility maximization

equal resource

0 500 1000 1500 20000

0.002

0.004

0.006

0.008

0.01

0.012

0.014

time (ms)

averagerate

proportional fair: =1/100

proportional fair: =1/500

Fig. 5. Evolution of single users average rate. Top. Utility maximizationand equal resource schemes. Bottom. Proportional-fair scheme for differentvalues of.

VI . DISCUSSION: COMPARISON WITH OTHER

COMPUTATIONAL METHODS

Many resource allocation problems in wireless networks are

either convex or can be approximated by convex problems

(e.g., [21], [22], [31]). While a general interior point method

can be used to solve these problems, in many cases it is

possible to exploit the structure of the optimization problem

to obtain fast and/or distributed algorithms. Next, we compare

our approach with two other such approaches.

A. Dual Subgradient MethodThe subgradient method (applied to the dual) can also

be used to solve the optimization problem (1) (see [19] for

such a method for CDMA systems). Such a method has an

economic interpretation where the dual variables act as prices

for violating constraints [6]. However, the rate of convergence

of this method is highly dependent on the various condition

numbers in the problem, and it will typically converge much

more slowly than the algorithm presented here. Moreover, each

iteration of the subgradient method also has O(n) complex-ity, which is the same as that for our method. Unlike the

subgradient approach, the fast convergence of our method

enables it to be used for fading channels, as the number of

iterations required for re-convergence after a warm start is

small. However, we note that the subgradient method can

be used to derive (typically slow) distributed algorithms for

resource allocation problems in an adhoc wireless network

(e.g., [23]), or the internet [6]; for such problems exploiting

the structure in the computation of the Newton step is typically

not possible. Dual decomposition, primal decomposition, or

joint primal-dual decomposition can be used (e.g., [12]).

B. Waterfilling

For the special case of log-utility functions, a waterfilling

algorithm can be obtained to solve the problem (1), where dur-

ing each iteration, we adjust a dual variable and recompute riand bi. This is similar to the waterfilling algorithm to computethe capacity of a wireless channel see for example, [30,

Ch. 4]. While this might appear to be a better algorithm, the

complexity of this method is quite similar to the complexity of

the barrier method described in this paper. In both algorithms,

(i) each iteration has a cost that isO(n)

, (ii) around10

25or so steps are needed to solve the problem, and (iii) a good

initial condition gives convergence within fewer steps. We also

note that the waterfilling approach can be used to solve the

problem in [19].

APPENDIX

Proof: [Lemma 1] Suppose (xk, yk) is a sequence ofpoints in D converging to (r, b) D. Now suppose i is suchthat bi = 0, and ri > 0. Then we have limk f(x

ki , y

ki ) =

and hence p(xk, yk) also tends to infinity, contradicting theassumption that (xk, yk) D.

Proof: [Lemma 2] If limk U(rk) = then we aredone. Suppose not, and let T = { i | ri = 0 }. For > 0 definey() by

yi() =

if i T,

bi |T|

n |T|otherwise.

Then 1Ty() = 1 for all > 0. Also define x() by

xi() =

if i T,

ri otherwise,

where > 0 and > 0. For > 0 sufficiently large we have

for all i Tdf

xi(), yi()

d

=0

< 0.

Pick such a . Hence

dp

x(), y()

d= |T|(e 1) +iT

d

df

xi(), yi()

and therefore for > 0 sufficiently small

dp

x(), y()

d

=0

< 0


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11

and hence for > 0 sufficiently small we have p

x(), y()

0 sufficiently small

limk

U(rk) < U

x()

as desired.

Proof: [Lemma 3] If limk U(rk) = then we are

done. Suppose not, and let T = { i | ri = 0 and bi > 0 }.Define y Rn by

yi =

0 if i T

bi +

jT bj

n |T|otherwise.

Then 1Ty = 1 and y 0. For any x > 0 we have

f(x, z1) > f(x, z2) if 0 < z1 < z2.

If r = 0 then for some i T we have ri > 0 and hencep(r, y) < p(r, b) 1. Also clearly if r = 0 then p(r, y) < 1.Now for > 0 define x() by

xi() = ri + if ri > 0 and bi > 0ri otherwise.

Since p is continuous, there exists > 0 sufficiently small sothat p

x(), y

< 1. Pick such an . Then since Ui is increasingwe have

U

x()

> limk

U(rk).

Now either x > 0 and y > 0, in which case the proof iscomplete, or there is some i such that (xi(), yi) = (0, 0). Inthis case the conditions of Lemma 2 hold, and this then gives

the desired result.

Proof: [Lemma 4] x(0) is strictly feasible by assumption.Now we use induction to prove the lemma.

Consider iteration k +1, and assume that x(k) = (r(k), b(k))is strictly feasible. Denote the Newton step by (r(k), b(k)).Now, let l be the minimum value of l such that for some i,we have r

(k)i + lr

(k)i = 0 or b

(k)i + lb

(k)i = 0, or p(r

(k) +lr(k), b(k) + lb(k)) = 1. Thus, l is the minimum value of lfor which (r(k)+ lr(k), b(k)+ lb(k)) is not strictly feasible.We claim that as l l, f(r(k) + lr(k), b(k) + lb(k)) ,i.e., the step length returned by the line search algorithm is

less than l, which implies that the (k + 1)th iterate is strictlyfeasible.

Note that r(k)i + lr

(k)i and b

(k)i + lb

(k)i are finite for all

i. Now assume that l < l. Then U(r(k) + lr(k)) is upper

bounded. Similarly log(r(k)i + lr

(k)i ) and log(b

(k)i + lb

(k)i )

are upper bounded for all i. Also, (1 p(r(k) + lr(k), b(k) +lb(k))) is upper bounded by 1. Hence, it follows from thedefinition of f(r, b) that that as l l, f(r(k) + lr(k), b(k) +lb(k)) , as claimed above.

Proof: [Lemma 5] For all (r, b) L, 1Tb = 1. By the

above lemma, all iterates are strictly feasible. Since b > 0 forall (r, b) L, the bis are bounded above by 1, which impliesthatn

i=1 log bi is bounded above. Also, 0 < p(r, b) < 1 forall (r, b) L, i.e., log(1 p(r, b)) is bounded above by zero.Since p(r, b) is an increasing function of the ris and decreasingfunction of the bis, and bi 1 for all (r, b) L, it followsthat ris are bounded above by a constant for all (r, b) L.This also implies that U(r) is bounded above by some U for(r, b) L.

Now, we show that ris and bis are bounded away from zerofor all (r, b) L. To see this, first note that U(r),

ni=1 log bi,

ni=1 log ri, and log(1 p(r, b)) are all bounded above for all

(r, b) L. Thus, it follows that t(r, b) as ri 0 or

bi 0 for any i. Then, the claim follows since the Newtonmethod is a descent method, i.e., t(r

(k), b(k)) t(r(0), b(0))

for any iteration k.

Proof: [Lemma 6] We show that the complement of D,i.e., DC is open. Note that DC is the union of the followingsets:

O1 = {(x, y) R2n | 1Ty = 1},

O2 = {(x, y) R2n | x < 0},

O3 = {(x, y) R2n | x > 0, y 0},

O4 = {(x, y) R2n | x = 0, y < 0},

O5 = {(x, y) R2n | x 0, p(x, y) > 1, y > 0}.

It is easy to see that O1 and O2 are open. Since, the union ofopen sets is open, it is sufficient to show that O3 O4 O5

is open. To do this, consider a point (x, y) O3 O4 O5.Hence, either (x, y) O3 or (x, y) O4 or (x, y) O5 ineach of these cases there exists an ball around (x, y) whichis contained in O3 O4 O5.

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PLACEPHOTOHERE

Ritesh Madan Ritesh Madan is at Qualcomm-Flarion Technologies, Bridgewater, NJ. He receiveda Ph.D. in 2006 and a M.S. in 2003, from StanfordUniversity, and a B.Tech. from IIT Bombay in 2001,all in Electrical Engineering. He was a recipient ofthe Sequoia Capital Stanford Graduate Fellowship.He has held visiting research positions at MitsubishiElectric Research Labs (MERL), Cambridge, MA,and at the Tata Institute of Fundamental Research(TIFR), Mumbai. His research interests include

wireless networks, convex optimization, networkedcontrol, and dynamic programming.

PLACEPHOTOHERE

Stephen Boyd Stephen P. Boyd is the SamsungProfessor of Engineering, in the Information Sys-tems Laboratory, Electrical Engineering Department,Stanford University. His current interests include

convex programming applications in control, signalprocessing, and circuit design.

He received an AB degree in Mathematics, summacum laude, from Harvard University in 1980, and aPhD in EECS from U. C. Berkeley in 1985. He is theauthor of Linear Controller Design: Limits of Per-formance (with Craig Barratt, 1991), Linear Matrix

Inequalities in System and Control Theory (with L. El Ghaoui, E. Feron, andV. Balakrishnan, 1994), and Convex Optimization (with Lieven Vandenberghe,2004). He received an ONR Young Investigator Award, a Presidential YoungInvestigator Award, and the 1992 AACC Donald P. Eckman Award. He hasreceived the Perrin Award for Outstanding Undergraduate Teaching in theSchool of Engineering, and an ASSU Graduate Teaching Award. In 2003, hereceived the AACC Ragazzini Education award. He is a fellow of the IEEE,a Distinguished Lecturer of the IEEE Control Systems Society, and holds anhonorary PhD from Royal Institute of Technology (KTH), Stockholm.

PLACEPHOTOHERE

Sanjay Lall Sanjay Lall is Associate Professor ofAeronautics and Astronautics and Vance D. andArlene C. Coffman Faculty Scholar at Stanford Uni-versity. Until 2000 he was a Research Fellow at theCalifornia Institute of Technology in the Departmentof Control and Dynamical Systems, and prior to thathe was NATO Research Fellow at Massachusetts

Institute of Technology, in the Laboratory for Infor-mation and Decision Systems. He received the Ph.D.in Engineering from the University of Cambridge,England. Professor Lalls research focuses on the

development of advanced engineering methodologies for the design of controlsystems which occur in a wide variety of aerospace, mechanical, electrical andchemical systems. Professor Lall is a senior member of the IEEE. In 2007 hereceived the NSF Career award, and in 2005 he received the Graduate ServiceRecognition Award from Stanford university.

25. fast algorithms for resource allocation in wireless cellular networks

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