Maximizing the Minimum Achievable Secrecy Rate

in a Two-User Gaussian Interference Channel

Meysam Mirzaee, Soroush Akhlaghi

Shahed University, Tehran, Iran

Emails: me.mirzaee,akhlaghi@shahed.ac.ir

Abstract—This paper studies a two-user Gaussian interferencechannel in which two single-antenna sources aim at sending theirconfidential messages to the legitimate destinations such that eachmessage should be kept confidential from non-intended receiver.Also, it is assumed that the direct channel gains are stronger thanthe interference channel gains and the noise variances at twodestinations are equal. In this regard, under Gaussian code bookassumption, the problem of secrecy rate balancing which aims atexploring the optimal power allocation policy at the sources inan attempt to maximize the minimum achievable secrecy rate isinvestigated, assuming each source is subject to a transmit powerconstraint. To this end, it is shown that at the optimal point,two secrecy rates are equal, hence, the problem is abstracted tomaximizing the secrecy rate associated with one of destinationswhile the other destination is restricted to have the same secrecyrate. Accordingly, the optimum secrecy rate associated with theinvestigated max-min problem is analytically derived leading tothe solution of secrecy rate balancing problem.

Index Terms—Achievable secrecy rate, Gaussian interferencechannel, Max-Min problem.

I. INTRODUCTION

Security is regarded as one of the main fundamental issues

in wireless networks due to the broadcast nature of wireless

medium. Using physical layer characteristics to secure the data

transmission was first proposed by Wyner in his landmark

paper in 1975 [1]. He investigated the discrete-memoryless

wiretap channel and showed that if the eavesdropper’s signal

is a degraded version of the destination signal, there exist

a positive rate in which transmission is accomplished with

perfect secrecy. This problem is further studied by Cheong

and Hellman for the Gaussian wiretap channel, where the

corresponding secrecy capacity is analytically derived [2].

This motivated researchers to seek for physical layer se-

curity for a variety of network topologies. For instance, the

secrecy rate of Interference channel is studied in [3] where an

inner and outer bound on the secrecy capacity region is de-

rived. Moreover, under Gaussian code book assumption, some

achievable secrecy rates for Gaussian interference channel are

derived.

Following the work done in [3], some researchers studied a

variation of the interference channel from the signal processing

viewpoint [4]–[6]. For instance, [4] explores the Multi-Input

Single-Output (MISO) two-user interference channel in which

the stronger transmitter attempts to maximize the difference

between its secrecy rate and the secrecy rate of the weaker

transmitter relying on transmit beamforming, while the weaker

transmitter attempts to minimize this discrepancy, all using a

game theoretic approach. Moreover, it is shown the studied

problem has a Nash Equilibrium point, where the optimal

transmit beamforming vectors associated with this point are

derived. In [5], a Single-Input Single-Output (SISO) symmet-

ric interference channel is considered where it is assumed

both links have the same direct and cross channel gains.

Also, each user sends a data signal as well as an artificial

noise, simultaneously. Accordingly, in an attempt to maximize

the minimum secrecy rate, i.e., the secrecy rate balancing,

a closed-form solution for the signal power as well as the

artificial noise power is derived. In [6], the secrecy rate

balancing problem is studied for MISO interference channel.

It is assumed the transmit beamforming vectors are linear

combinations of two Maximum Ratio Combining and Zero-

Forcing vectors, where the optimization problem is tackled

relying on search method.

In this paper, we investigate a single-antenna two-user

Gaussian interference channel where the channel coefficients

of two sources are not necessarily similar. We also suppose

that the direct channels have stronger gains than cross channels

and the variances of noise at two destinations are equal. It

is assumed that each user is subject to a maximum transmit

power constraint, where the objective is to find the best power

allocation strategy leading to the secrecy rate balancing. To

this end, having the idea in mind that at the optimal point one

link acts as bottleneck, the corresponding max-min problem

is divided into two separated maximization problems. First,

we assume the first link has the worst case secrecy rate, thus

attempting to maximize its corresponding secrecy rate when

the secrecy rate associated with the second link is restricted to

be higher than that of the first link and this problem is repeated

in a similar fashion for the second link. Accordingly, it is

proved that at the optimal point, two secrecy rates are equal,

so it is adequate to merely solve one of the above mentioned

problems.

In this paper, the notation |x| refers to absolute value of

complex variable x, and x ∼ CN (μ, σ2) denotes the variable

x is complex Gaussian random variable with mean μ and

variance σ2. Also, the function x+ is equivalent to max{0, x}.

II. SYSTEM MODEL

We consider two-user Gaussian interference channel with

single antenna nodes (Fig. 1). It is assumed, each transmitter

aims at sending confidential message to its legitimate des-

tination such that the information is kept private from the

2014 Iran Workshop on Communication and Information Theory (IWCIT)

978-1-4799-4877-2/14/$31.00 ©2014 IEEE

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Fig. 1. System Model

adversary. The received signals at the destinations can be

modeled as,

y1 = h11x1 + h21x2 + n1, (1)

y2 = h12x1 + h22x2 + n2, (2)

where xi ∼ CN (0, pi) is the transmitted signal of the ith

source, hij is the channel coefficient from source i to desti-

nation j and ni ∼ CN (0, σ2i ) is the received Additive White

Gaussian Noise (AWGN) at the ith destination. We assume

that each destination first decodes the intended message of

its respective source and then tries to decode the message of

the other source. According to [3], an achievable secrecy rate

region is the convex hull of the set of (R1, R2) satisfying,

0 ≤ R1 ≤ Rs1 and 0 ≤ R2 ≤ Rs2 (3)

where,

Rs1 = {I(X1;Y1)− I(X1;Y2|X2)}+

=

{log2

(1 +

p1|h11|2p2|h21|2 + σ2

1

)− log2

(1 +

p1|h12|2σ22

)}+

=

{log2

(p1|h11|2 + p2|h21|2 + σ2

1

p2|h21|2 + σ21

× σ22

p1|h12|2 + σ22

)}+

Rs2 = {I(X2;Y2)− I(X2;Y1|X1)}+

=

{log2

(1 +

p2|h22|2p1|h12|2 + σ2

2

)− log2

(1 +

p2|h21|2σ21

)}+

=

{log2

(p2|h22|2 + p1|h12|2 + σ2

2

p1|h12|2 + σ22

× σ21

p2|h21|2 + σ21

)}+

over all transmit powers p1 ∈ [0, P1] and p2 ∈ [0, P2]1.

III. ACHIEVABLE SECRECY RATE BALANCING

The objective here is to address the achievable secrecy

rate balancing problem. Mathematically speaking, referring to

(3), we are going to investigate the following optimization

problem,

maxp1,p2

min{Rs1, Rs2}s.t. 0 ≤ p1 ≤ P1 , 0 ≤ p2 ≤ P2, (4)

1 [3] drives more general achievable regions but we restrict to a specialinner bound on the achievable region.

where P1 and P2 are the maximum allowable transmit power

constraints associated with source1 and source2, respectively.

Noting at the optimal point of (4), either of inequalities

Rs2 ≥ Rs1 or Rs1 ≥ Rs2 should be satisfied, thus one of

the following optimization problems, gives the solution of (4),

maxp1,p2

Rs1

s.t. Rs2 ≥ Rs1, 0 ≤ p1 ≤ P1 , 0 ≤ p2 ≤ P2, (5)

and

maxp1,p2

Rs2

s.t. Rs1 ≥ Rs2, 0 ≤ p1 ≤ P1 , 0 ≤ p2 ≤ P2. (6)

In the sequel, we solve (5) and show that the first constraint

is satisfied with equality. By the same token, due to the

similarities in the mathematical forms, one can readily verify

that the first constraint of (6) is also satisfied with equality,

and therefore the optimal solutions of (5) and (6) are equal,

providing the solution of (4). To this end, using (3), (5) can

be rewritten, equivalently, as,

maxp1,p2

p1σ22 |h11|2 + p2σ

22 |h21|2 + σ2

1σ22

p1p2|h12|2|h21|2 + p1σ21 |h12|2 + p2σ2

2 |h21|2 + σ21σ

22

s.t. p1σ21 |h12|2 + p2σ

21 |h22|2 ≥ p1σ

22 |h11|2 + p2σ

22 |h21|2

and 0 ≤ p1 ≤ P1 , 0 ≤ p2 ≤ P2. (7)

Using the definitions a � σ22 |h11|2, b � σ2

1 |h22|2, c �

σ21 |h12|2, d � σ2

2 |h21|2, e � σ21σ

22 , f = |h12|2|h21|2, and

inverting the objective, (7) can be simplified to,

minp1,p2

fp1p2 + cp1 + dp2 + e

ap1 + dp2 + e

s.t. (a− c)p1 ≤ (b− d)p2

and 0 ≤ p1 ≤ P1 , 0 ≤ p2 ≤ P2. (8)

To address (8), we use the following lemma.

Lemma 1: Consider the following fractional programming

problem

minx

f(x)

g(x)

s.t. x ∈ X, (9)

where X is a polytope. Define the function,

F (x, λ) = f(x)− λg(x), (10)

and let’s define,

π(λ) = minx∈X

F (x, λ) and x(λ) = argminx∈X

F (x, λ). (11)

Then, it can be shown that π(λ) is a decreasing function of

λ and if λ∗ is the root of π(λ), then x(λ∗) is the optimum

solution of (9).

Proof : see [7].

Based on Lemma 1, and noting the objective of (8), we

define the function F (p1, p2, λ) as

F (p1, p2, λ) = fp1p2 + cp1 + dp2 + e− λ(ap1 + dp2 + e)

= fp1p2 + (c− λa)p1 + (1− λ)dp2 + (1− λ)e.(12)

To compute π(λ), we consider the following optimization

problem

minp1,p2

fp1p2 + (c− λa)p1 + (1− λ)dp2 + (1− λ)e

s.t. (a− c)p1 ≤ (b− d)p2,

and 0 ≤ p1 ≤ P1 , 0 ≤ p2 ≤ P2. (13)

In the sequel, we assume that the direct channels are stronger

than cross channels, i.e. |h11| ≥ |h12| and |h22| ≥ |h21|. This

assumption is reasonable for the cases in which Source1-to-

Destination1 is away from Source2-to-Destination2. Moreover,

the noise power at both destinations are assumed to be the

same, i.e. σ21 = σ2

2 = σ2. According to these assumptions, we

have a ≥ c and b ≥ d.

To solve (13), for fixed values of p2 and λ, one can readily

observe that the optimal value of p1, namely p∗1, can be

computed from the following optimization problem,

minp1

(fp2 + c− λa)p1 + (1− λ)dp2 + (1− λ)e

s.t. p1 ≤ b− d

a− cp2, 0 ≤ p1 ≤ P1. (14)

One can readily observe that for known parameters p2 and

λ, (14) is a single variable Linear Programming (LP) in

terms of p1 where depending on the coefficient of p1 and the

constraints, p∗1(p2, λ) will be either the start or the end point of

the interval [0 , min(P1,b−da−c

p2)]. Mathematically speaking,

we have,

p∗1(p2, λ) =

⎧⎪⎨⎪⎩0 p2 ≥ λa−c

f

P1 p2 < λa−cf

∧p2 ≥ a−c

b−dP1

b−da−c

p2 p2 < min(a−cb−d

P1,λa−c

f)

(15)

where substituting (15) in (12) results in,

F (p2, λ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(1− λ)dp2 + (1− λ)e for p2 ≥ λa−cf

(fP1 + (1− λ)d)p2 + (c− λa)P1 + (1− λ)e

for p2 < λa−cf

∧p2 ≥ a−c

b−dP1

f(b−d)a−c

p22 + (ad+bc−2cd−λ(ab−cd)a−c

)p2 + (1− λ)e

for p2 < min(a−cb−d

P1,λa−c

f)

(16)

One can readily verify that F (p2, λ) is a continuous function

of p2. Moreover, for 0 ≤ λ ≤ 1, the first and the second

terms of (16) are linear increasing functions with respect to

p2, while the third term is a convex quadratic function. Thus,

the minimum value of F (p2, λ) resides on the third term

and therefore the following optimization problem gives the

optimum value of p2, i.e., p∗2(λ),

minp2

f(b− d)

a− cp22 +

(ad+ bc− 2cd− λ(ab− cd)

a− c

)p2

+ (1− λ)e

s.t. 0 ≤ p2 ≤ min

(P2,c,

λa− c

f

), (17)

where,

P2,c = min

(P2,

a− c

b− dP1

). (18)

To find p∗2(λ), it should be noted that the objective function

of (17) has a minimum value at the point p̃2(λ), where,

p̃2(λ) =λa− c

2f− (1− λ)(a− c)d

2f(b− d). (19)

If p̃2(λ) satisfies the constraint of (17), it will be the optimum

solution of (17) as well; otherwise, either of the start or the

end point of interval[0,min

(P2,c,

λa−cf

)]yields the optimal

solution. As a result, noting p̃2(λ) in (19) is always lower thanλa−c

f, hence after some mathematics it turns out that p∗2(λ) is

given by,

p∗2(λ) =

⎧⎪⎨⎪⎩0 p̃2(λ) < 0λa−c2f − (1−λ)(a−c)d

2f(b−d) 0 ≤ p̃2(λ) ≤ P2,c

P2,c p̃2(λ) > P2,c,

(20)

or equivalently, using (19), the equation (20) can be rewritten

as,

p∗2(λ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 λ < ad+bc−2cdab−cd

λa−c2f − (1−λ)(a−c)d

2f(b−d)ad+bc−2cd

ab−cd≤ λ and

λ ≤ 2f(b−d)P2,c+ad+bc−2cdab−cd

P2,c λ >2f(b−d)P2,c+ad+bc−2cd

ab−cd.

(21)

For notational convenience, we define,

λ1 �ad+ bc− 2cd

ab− cd,

λ2 �2f(b− d)P2,c + ad+ bc− 2cd

ab− cd,

and therefore, (21) can be replaced by,

p∗2(λ) =

⎧⎪⎨⎪⎩0 λ < λ1

λa−c2f − (1−λ)(a−c)d

2f(b−d) λ1 ≤ λ ≤ λ2

P2,c λ > λ2,

(22)

According to (11), we should substitute (22) in (16) to

compute π(λ) as follows,

π(λ) = F(p∗2(λ), λ

)=

⎧⎪⎨⎪⎩(1− λ)e λ < λ1

αλ2 + βλ+ γ λ1 ≤ λ ≤ λ2

ωλ+ ν λ > λ2,

(23)

where the following definitions are being used,

α �− (ab− cd)2

4f(b− d)(a− c),

β �(ab− cd)(ad+ bc− 2cd)

2f(b− d)(a− c)− e,

γ �− (ad+ bc− 2cd)2

4f(b− d)(a− c)+ e,

ω �− (ab− cd

a− cP2,c + e),

ν �f(b− d)P 2

2,c + (ad+ bc− 2cd)P2,c

a− c+ e.

We know that the objective function in (8) has positive value

and can equal to one for p1 = p2 = 0, so the optimum value of

it lies in the interval [0, 1]. Moreover, according to Lemma 1,

the root of π(λ) is the optimum value of the objective function

in (8). Thus, the root of π(λ) will be placed in [0, 1]. Also, in

the Lemma 1, it is argued that π(λ) is a decreasing function

of λ and thus it has one root in the interval λ ∈ [0, 1]. It is

easy to show that λ1 < 1 and therefore either the second or

the third term of π(λ) contain the root. Thus, one can readily

compute π(λ2) and based on its sign, one of the second or the

third term of π(λ) is set to zero to obtain the root of π(λ),i.e. λ∗. Mathematically speaking, we have,

λ∗ =

{−β−

√β2−4αγ

2α π(λ2) ≤ 0−νω

otherwise.(24)

Finally, the optimum value of achievable secrecy rate is given

by,

R∗

s1 = R∗

s2 = log2(1/λ∗) (25)

IV. NUMERICAL RESULTS

This section aims at providing some numerical results to

explore the achievable secrecy rate as well as transmit power

of each source versus the maximum allowable transmit power.

As is mentioned in the preceding section, at the optimal point,

both achievable secrecy rates are equal. Thus, it is adequate to

merely investigate one of the two secrecy rates. Throughout

the simulations, we assume that σ21 = σ2

2 = 1 and that the

direct channel gains are assumed to be complex Gaussian

random variables, i.e. hii ∼ CN (0, 1) for i = 1, 2, while

the cross channel gains are related to direct gains through

equations h12 =√α1h11 and h21 =

√α2h22. Moreover,

for a maximum allowable transmit power PT , two cases of

P1 = P2 = PT and P1 = 2P2 = PT are considered.

Accordingly, Fig. 2 is provided to illustrate the achievable

secrecy rate versus PT for P1 = P2 = PT and three

different values of α1 and α2. It is shown that increasing the

maximum allowable transmit power, the achievable secrecy

rate approaches to a constant value. Moreover, increasing the

strength of the cross channel gains decreases the achievable

secrecy rate. Fig. 3 represents the transmit power of two

sources versus PT , showing the source with stronger cross

gain consumes more power in order to have the same secrecy

−5 0 5 10 15 20 25 30 35 400

0.2

0.4

0.6

0.8

1

1.2

1.4Achievable Secrecy Rate

PT (dB)

Rs1 =

Rs2 (

bits/tra

nsm

issio

n)

α1 = 0.2 , α

2 = 0.1

α1 = 0.2 , α

2 = 0.2

α1 = 0.2 , α

2 = 0.4

Fig. 2. Achievable secrecy rate versus PT for P1 = P2 = PT and differentvalues of α1 and α2

−5 0 5 10 15 20 25 30 35 40−10

−5

0

5

10

15

20Transmit Power

P(dB)

Tra

nsm

it P

ow

er

(dB

)

Source 1

Source 2

α1 = 0.2 , α

2 = 0.4

α1 = 0.2 , α

2 = 0.2

α1 = 0.2 , α

2 = 0.1

Fig. 3. Transmit power of sources versus PT for P1 = P2 = PT anddifferent values of α1 and α2

rate as the link with smaller cross gains. Moreover, it is evident

that as long as the cross gains are relatively low, sources make

use of more power, thus according to Fig. 2 the secrecy rate

is increased.

Figs. 4 and 5 are provided to demonstrate the impact of

transmit power imbalance for the case of P1 = 2P2 = PT ,

showing at high values of PT , both of secrecy rates as well

as transmit powers do not change that much as compared to

previous results with equal transmit powers, while for small

values of PT , the secrecy rate as well as transmit powers

are smaller than that of reported in previous results, since

the maximum allowable transmit power of source2 acts as

bottleneck.

V. CONCLUSION

This paper studies the achievable secrecy rate balancing

problem under transmit power constraint in a two-user Gaus-

−5 0 5 10 15 20 25 30 35 400

0.2

0.4

0.6

0.8

1

1.2

1.4Achievable Secrecy Rate

PT (dB)

Rs1 =

Rs2 (

bits/tra

nsm

issio

n)

α1 = 0.2 , α

2 = 0.1

α1 = 0.2 , α

2 = 0.2

α1 = 0.2 , α

2 = 0.4

Fig. 4. Achievable secrecy rate versus PT for P1 = 2P2 = PT and differentvalues of α1 and α2

−5 0 5 10 15 20 25 30 35 40−10

−5

0

5

10

15

20

PT (dB)

Tra

nsm

it P

ow

er

(dB

)

Source 1

Source 2

α1 = 0.2 , α

2 = 0.4

α1 = 0.2 , α

2 = 0.2

α1 = 0.2 , α

2 = 0.1

Fig. 5. Transmit power of sources versus PT for P1 = 2P2 = PT anddifferent values of α1 and α2

sian interference channel. It is shown that the corresponding

max-min problem can be simplified to a simplified maximiza-

tion problem, where a closed-form solution is derived.

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