2610 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 4, NO. 5, SEPTEMBER 2005

Comments and CorrectionsComments on “Optimal Resource Allocation

in Multiservice CDMA Networks”

Sun-Mog Hong

Abstract—In this note, counterexamples to Proposition 2 in the above-mentioned paper are presented, and a correct proof of Proposition 3 isgiven.

Index Terms—Code-division multiple-access (CDMA) networks, radioresource allocation.

In the above-mentioned paper [1], a joint transmission power andspreading gain allocation for non-real-time (NRT) users was obtained,which maximizes the aggregate NRT throughput, subject to peaktransmission power and total received-power constraints. Specifically,the jointly optimal transmission power and spreading gain allocationwas presented in Proposition 2 in the above-mentioned paper. Werestate the proposition.

Proposition 2: The optimal transmission power p∗ and spreadinggain N∗ allocations of the ith NRT user are given by

p∗i =

{pmax, if i < min{iT , Mc}pmax1{iT <Mc} + pc1{iT ≥Mc}, if i = min{iT , Mc}0, if i > min{iT , Mc}

(15)

N∗i =

γ∗(I + α

∑min{iT ,Mc}j=1,j �=i

pmaxgj

)pmaxgi

(16)

i = 1, 2, . . . M . If there exists j such that

j−1∑i=1

pmaxgi + pcgj = PmaxR for some 0 ≤ pc ≤ pmax (17)

then Mc = j; otherwise, Mc = M . iT is the largest j ∈ {1, 2, . . . M}satisfying ∆j > 0, and γ∗ is given by (6).

Notice that ∆j in the above proposition denotes the net change inthe aggregate throughput when allocating the maximum transmissionpower pmax to the additional user with the next highest channel gain,under the assumption that pmax has been allocated to the j userswith the highest channel gains. To be consistent with the recursiveprocedure for determining iT using (18), the description of iT in theabove proposition should be modified. The modification is as follows:iT is, if it exists, the smallest j ∈ {1, 2, . . . M − 1} satisfying ∆j ≤ 0;otherwise, iT = M . In this note, counterexamples to Proposition 2with and without this modification are given. These counterexamplesshow that the transmission-power allocations of Proposition 2 are notnecessarily optimal.

Manuscript received September 29, 2003; revised May 6, 2004; acceptedAugust 31, 2004. The editor coordinating the review of this paper and approvingit for publication is N. Mandayam. This work was supported by the KoreaScience and Engineering Foundation (KOSEF) under Grant R05-2003-000-11118.

The author is with the School of Electronic and Electrical Engineering,Kyungpook National University, Taegu 702-701 Republic of Korea (e-mail:smhong@ee.knu.ac.kr).

Digital Object Identifier 10.1109/TWC.2005.853906

Counterexample 1: Suppose that there are four users with chan-nel gains g1 = g, g2 = (1 − ε)g, g3 = µg, and g4 = (1 − ε)µg, andsuppose that Pmax

R = ∞ and 1 < η <√

3, where η = I/αpmaxg. If(3 − η2)/2η < µ < (3 − η2)/(1 + η), it holds for sufficiently smallε > 0 that ∆1 > 0, ∆2 < 0, and ∆2 + ∆3 > 0. This implies thatiT should be 4, which is a contradiction to Proposition 2 (with themodification), which gives iT = 2.

As an example, consider a case with α = 1 and η = 1.5. If 0.25 <µ < 0.3, it holds for sufficiently small ε > 0 that ∆1 > 0, ∆2 < 0,and ∆2 + ∆3 > 0. For instance, if µ = 0.275 and ε = 10−3, then∆1 ≈ 1.33 × 10−1, ∆2 ≈ −6.76 × 10−4, and ∆2 + ∆3 ≈ 1.48 ×10−3. Here, the net change ∆j is normalized with respect to theconstant Wβf(γ∗)/γ∗; see (9) in the above-mentioned paper.Counterexample 2: If µ = 0.24 and ε = 10−3 in the above ex-

ample, then ∆1 ≈ 1.33 × 10−1, ∆2 ≈ −1.47 × 10−3, ∆3 ≈ 1.00 ×10−3, but ∆2 + ∆3 ≈ −4.72 × 10−4. This is an example where iTshould be 2, which is a contradiction to Proposition 2 (without themodification), which gives iT = 3.

As a consequence, in proving Proposition 3 in the above-mentionedpaper, we need to show that

∑k

j=1∆j < 0 holds for all k =

1, 2, . . . , M − 1. Let us define ∆k,1 =∑k−1

j=1∆j , k = 2, 3, . . . , M .

∆k,1 is the net change in the aggregate throughput when allocatingpmax to the additional k − 1 users with the next highest channel gains,under the assumption that pmax has been allocated to the user with thehighest channel gain (user with channel gain g1).

Proof of Proposition 3: Assume that pmax has been allocatedto the user with channel gain g1, and consider the net change ∆k,1,normalized with respect to the constant Wβf(γ∗)/γ∗, given as

∆k,1 =

k∑i=1

gi

Ipmax

+ α∑k

j=1,j �=igj

− g1

Ipmax

=

k∑i=2

gi

Ipmax

+ α∑k

j=1,j �=igj

−g1α

∑k

j=2gj

Ipmax

(I

pmax+ α

∑k

j=2gj

) .

Define g0 = I/pmax, and Tk = g0 + α∑k

i=1gi. Since g1 > g2 >

· · · > gM , it holds that

∆k,1 ≤∑k

i=2gi

Tk − αg2

−g1α

∑k

j=2gj

g0(Tk − αg1)(1)

for all k = 2, 3, . . . , M . A sufficient condition for ∆k,1 < 0 is that theright-hand side of (1) is less than zero. This leads to

pmaxg1

I>

1

α· Tk − αg1

Tk − αg2

.

It is easy to verify the following inequality

Tk − αg1

Tk − αg2

≤ TM − αg1

TM − αg2

, k = 2, 3, . . . , M.

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IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 4, NO. 5, SEPTEMBER 2005 2611

From the last two inequalities, we can obtain a sufficient condition for∆k,1 < 0, such that

pmaxg1

I>

1

α· TM − αg1

TM − αg2

. (2)

This inequality holds independently of k. Thus, it follows that∆k,1 < 0 for all k = 2, 3, . . . , M , which implies iT = 1. If

pmaxg1/I > 1/α, the inequality (2) also holds. This completesthe proof.

REFERENCES

[1] S.-J. Oh, D. Zhang, and K. M. Wasserman, “Optimal resource allocationin multiservice CDMA networks,” IEEE Trans. Wireless Commun., vol. 2,no. 4, pp. 811–821, Jul. 2003.