1
A companion technical report of “Joint Power
and Admission Control: Non-Convex Lq
Approximation and An Effective Polynomial
Time Deflation Approach”
Ya-Feng Liu, Yu-Hong Dai, and Shiqian Ma
I. P ROOF OF L EMMA 1 IN [1]
In this part, we prove Lemma 1 in [1], i.e., for any q ∈ [0, 1], problem
min kb − Axkqq + αp̄T x
x
s.t.
(1)
0≤x≤e
is equivalent to
min kb − Axkqq + αp̄T x
x
s.t.
Ax ≤ b,
(2)
0 ≤ x ≤ e.
The lemma is a consequence of the special structure of A and b.
Before showing the lemma, we first introduce some notations which will be used later. For any subset
I ⊆ K, we use AI to denote the matrix formed by the rows of A indexed by I . For a vector x, the
notation xI is similarly defined. Moreover, for any J ⊆ K, the notation AI,J will denote the submatrix
of A obtained by taking the rows and columns of A indexed by I and J respectively.
Notice that problem (2) is equivalent to
min kykqq + α p̄T x
x, y
s.t.
y = b − Ax,
0 ≤ x ≤ e, y ≥ 0.
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Thus, to show the equivalence of (1) and (2), it suffices to show that any optimal solution x̃ of (1) always
satisfies b − Ax̃ ≥ 0. For simplicity of notations, let z̃ = Ax̃ − b. Next, we show
z̃ = (z̃1 , z̃2 , . . . , z̃K )T ≤ 0.
Denote K+ = { k | z̃k > 0 }, K= = { k | z̃k = 0 }, and K− ={ k | z̃k < 0 }. We claim |K+ | = 0.
Assume the contrary so that |K+ | ≥ 1. We will derive a contradiction. Let I = K+ ∪ K= . Then
z̃I = AI x̃ − bI ≥ 0, which further implies
AI,I x̃I − (bI − AI,I c x̃I c ) ≥ 0.
Hence, x̃I is a feasible solution to the following linear subsystem in xI :
AI,I xI − (bI − AI,I c x̃I c ) ≥ 0.
(3)
Since AI,I c ≤ 0 (the off-diagonals of A are nonpositive), it follows that bI − AI,I c x̃I c ≥ bI > 0. It
can be checked that the other assumptions of Lemma 3.1 in [2] all hold for (3) so that there exists a
vector x̄I such that bI ≤ x̄I ≤ x̃I and AI,I x̄I − (bI − AI,I c x̃I c ) = 0. Define x̄I c = x̃I c . Then we
have
AI x̄ − bI = AI,I x̄I − (bI − AI,I c x̃I c ) = 0.
(4)
Moreover, we have 0 ≤ x̄ ≤ x̃ ≤ e, so x̄ is a feasible power allocation. With this new power allocation
x̄, there holds
AI c x̄ − bI c
= AI c ,I c x̃I c + AI c ,I x̄I − bI c
≥ AI c ,I c x̃I c + AI c ,I x̃I − bI c
(5)
= AI c x̃ − bI c ,
where the inequality follows from AI c ,I ≤ 0 and x̄ ≤ x̃. This further implies
(bI c − AI c x̄)+ ≤ (bI c − AI c x̃)+
where (·)+ denotes the projection to the nonnegative orthant. Since bI − AI x̄ = 0 (cf. (4)), it follows
that
(b − Ax̄)+ ≤ (b − Ax̃)+
(6)
with the inequality holds true strictly for entries indexed by K+ . Define the function
φ(x) = k(b − Ax)+ kqq + α p̄T x.
Clearly, we have
φ(x) ≤ kb − Axkqq + α p̄T x,
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where the equality holds whenever Ax − b ≤ 0. Since 0 ≤ x̄ ≤ x̃, we can use (6) to obtain
φ(x̄) = k(b − Ax̄)+ kqq + α p̄T x̄ < k(b − Ax̃)+ kqq + α p̄T x̃ = φ(x̃).
Moreover, it follows from |K+ | ≥ 1 that
kAx̄ − bk0 = K − |I| = K − |K= | − |K+ | < K − |K= | = kAx̃ − bk0 .
We claim that, without loss of generality, we can assume AI c x̄ − bI c ≤ 0 so that Ax̄ − b ≤ 0. This
is because otherwise we can repeat the above steps by replacing the vector x̃ with x̄ to find a new vector
0 ≤ x̂ ≤ x̄ such that
φ(x̂) < φ(x̄) < φ(x̃)
and
kAx̂ − bk0 < kAx̄ − bk0 < kAx̃ − bk0 .
Since the `0 -norm can be reduced at most finitely many times, it follows that by repeatedly applying the
above steps we will eventually obtain a power allocation (still denote by x̄) such that
0 ≤ x̄ ≤ x̃ ≤ e,
φ(x̄) < φ(x̃),
Ax̄ − b ≤ 0.
Notice that when Ax̄ − b ≤ 0 we have
kAx̄ − bkqq + α p̄T x̄ = φ(x̄) < φ(x̃) ≤ kAx̃ − bkqq + α p̄T x̃.
This contradicts the optimality of x̃. This completes the proof.
II. G LOBAL M INIMIZER OF P ROBLEM (12) IN [1]
In this part, we show that, for any q ∈ (0, 1), x∗ = (0.5, 0.5, 0)T is the unique global minimizer of
the non-convex optimization problem
min hq (x) = (0.5 − x1 + x3 )q + (0.5 − x2 + x3 )q + (0.5 + x1 + x2 − x3 )q + α
3
X
xi
i=1
s.t.
0.5 − x1 + x3 ≥ 0,
0.5 − x2 + x3 ≥ 0,
(7)
0.5 + x1 + x2 − x3 ≥ 0,
0 ≤ x1 , x2 , x3 ≤ 1,
where
0 < α ≤ ᾱq := min {1 + (0.5)q , 2q } − (1.5)q .
(8)
To show the desired result, let us first introduce the following useful facts.
Fact 1. Suppose f1 (x) and f2 (x) are concave functions, so is f (x) := min {f1 (x), f2 (x)} .
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Fact 2. Suppose f (x) is an univariate strictly concave function in x ∈ [a, b], then
f (x) > min {f (a), f (b)} , ∀ x ∈ (a, b).
Fact 3. For any q ∈ (0, 1), kAx + bkqq is concave in {x | Ax + b ≥ 0} . Furthermore, it is strictly
concave if A is column full rank.
Fact 4. For any q ∈ (0, 1), x1 ≥ 0, x2 ≥ 0, we have xq1 + xq2 ≥ (x1 + x2 )q , where the inequality holds
with “=” if and only if at least one of x1 and x2 is zero.
From Fact 3, we know hq (x) defined in (7) is strictly concave in its feasible region, and thus is also
strictly concave in each of its components. Fact 4 implies that ᾱq defined in (8) is positive.
We are now ready to show the desired result. It suffices to show that for any feasible set S ⊆ {1, 2, 3} ,
the optimal value of problem
min hq,S (x) =
x
s.t.
X
[b − Ax]qi + αeT x
i∈S
/
[b − Ax]i = 0, i ∈ S,
(9)
[b − Ax]i > 0, i ∈
/ S,
0 ≤ x ≤ e,
is strictly greater than hq (x∗ ) = (1.5)q + α except at the point x = x∗ .
It is simple to check that the sets {1, 3} , {2, 3} , and (thus) {1, 2, 3} are infeasible for problem (9). In
the sequential, due to the symmetry of x1 and x2 , it is sufficient to consider the following feasible sets
(including the empty set):
A: When S = {1, 2} , we have 0.5 − x1 + x3 = 0 and 0.5 − x2 + x3 = 0. Therefore,
hq,S (x) = (0.5 + x1 + x2 − x3 )q + α(x1 + x2 + x3 )
= (1.5 + x3 )q + α + 3αx3
≥ hq (x∗ ),
where “=” holds if and only if x = x∗ .
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B: When S = {1} , we have x3 = x1 − 0.5 ≥ 0 and x1 ≥ 0.5. Therefore,
hq,S (x) = (0.5 − x2 + x3 )q + (0.5 + x1 + x2 − x3 )q + α(x1 + x2 + x3 )
= (x1 − x2 )q + (1 + x2 )q + 2αx1 + αx2 − 0.5α
(10)
≥ min {(1 + x1 )q + 3αx1 − 0.5α, xq1 + 1 + 2αx1 − 0.5α}
(11)
≥ min {(1.5)q + α, (0.5)q + 1 + 0.5α}
(12)
= hq (x∗ ),
(13)
where (10) is due to x3 = x1 − 0.5, (11) comes from the fact that the function in the right hand of
(10) is strictly concave in x2 ∈ [0, x1 ] and Fact 2, (12) is due to the fact that the function in the
right hand of (11) is increasing in x1 ∈ [0.5, 1], and (13) is due to α ≤ ᾱq (cf. (8)). Concluding
the above analysis, we have hq,S (x) ≥ hq (x∗ ) and the equality holds true if and only if x = x∗ .
C: When S = {3} , we have x3 = x1 + x2 + 0.5. Then,
hq,S (x) ≥ (0.5 − x1 + x3 )q + (0.5 − x2 + x3 )q
= (1 + x2 )q + (1 + x1 )q
≥ 2 > hq (x∗ ),
where the last strict inequality is due to the fact α ≤ ᾱq ≤ 2q − (1.5)q < 2 − (1.5)q (cf. (8)).
D: When S = ∅, we further consider the following cases separately.
D1: When x1 = 0, x2 ∈ [0, 1], x3 ∈ [0, 0.5], we have
hq,∅ (x) = (0.5 + x3 )q + (0.5 − x2 + x3 )q + (0.5 + x2 − x3 )q + α(x2 + x3 )
≥ (0.5 + x3 )q + αx3 + min {(0.5 + x3 )q + (0.5 − x3 )q , 1 + α(0.5 + x3 )}
(14)
≥ min {(0.5)q + 1 + 0.5α, 3(0.5)q , 2 + 0.5α, 2 + 1.5α}
(15)
> hq (x∗ ),
(16)
where (14) is due to the fact that hq,∅ (x) is strictly concave in x2 ∈ [0, 0.5 + x3 ] and Fact
2, (15) is due to Fact 1 (the function in the right hand side of (14) is strictly concave in
x3 ∈ [0, 0.5]) and Fact 2, and the last strict inequality (16) is due to (8).
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D2: Similar as case D1, when x1 = 0, x2 ∈ [0, 1], x3 ∈ [0.5, 1], we have
hq,∅ (x) = (0.5 + x3 )q + (0.5 − x2 + x3 )q + (0.5 + x2 − x3 )q + α(x2 + x3 )
≥ (0.5 + x3 )q + αx3 + min {(1.5 − x3 )q + (x3 − 0.5)q + α, 1 + α(x3 − 0.5)} (17)
≥ min {(1.5)q + 2(0.5)q + 2α, (1.5)q + 1.5α + 1, 2 + 1.5α, 2 + 0.5α}
(18)
> hq (x∗ ),
(19)
where (17) is due to the fact that hq,∅ (x) is strictly concave in x2 ∈ [x3 − 0.5, 1] and Fact
2, (18) is due to Fact 1 (the function in the right hand side of (17) is strictly concave in
x3 ∈ [0.5, 1]) and Fact 2, and the last strictly inequality (19) is due to (8).
D3: When x1 ∈ (0, 1), x2 ∈ [0, 1], x3 = 0, we have
hq,∅ (x) = (0.5 − x1 )q + (0.5 − x2 )q + (0.5 + x1 + x2 )q + α(x1 + x2 )
≥ (1 − (x1 + x2 ))q + (0.5 + (x1 + x2 ))q + α(x1 + x2 )
(20)
≥ min {(1.5)q + α, 1 + (0.5)q }
(21)
= hq (x∗ ),
(22)
where (20) is due to Fact 4, (21) is due to Fact 2 and the fact that x1 ≤ 0.5 and x2 ≤ 0.5,
and (22) is due to (8). Hence, hq,∅ (x) ≥ hq (x∗ ) for any x1 ∈ (0, 1), x2 ∈ [0, 1], x3 = 0, and
the inequality holds with “=” if and only if x = x∗ .
D4: In a similar fashion as the case D3, when x1 ∈ (0, 1), x2 ∈ [0, 1], x3 = 1, we can show
hq,∅ (x) ≥ (1.5 − x1 )q + (1.5 − x2 )q + (x1 + x2 − 0.5)q
≥ (3 − x1 − x2 )q + (x1 + x2 − 0.5)q
≥ min {1 + (1.5)q , (2.5)q }
= (2.5)q > hq (x∗ ),
where the third inequality comes from the fact that (3 − (x1 + x2 ))q + (x1 + x2 − 0.5)q is
strictly concave in x1 + x2 ∈ [0.5, 2] and Fact 2.
D5: The remaining case is x1 ∈ (0, 1), x2 ∈ (0, 1), x3 ∈ (0, 1). Recall that hq (x) in problem (7) is
strictly concave in x. Combining this, the fact S = ∅, and Fact 2, we know that the minimum
of problem (9) will not lie in this region.
From the above analysis, we conclude that x∗ is the unique global minimizer of problem (7).
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R EFERENCES
[1] Y.-F. Liu, Y.-H. Dai, and S. Ma, “Joint power and admission control: Non-convex `q approximation and an effective
polynomial time deflation approach,” accepted for publication in IEEE Trans. Signal Process.
[2] Y.-F. Liu, Y.-H. Dai, and Z.-Q. Luo, “Joint power and admission control via linear programming deflation,” IEEE Trans.
Signal Process., vol. 61, no. 6, pp. 1327–1338, Mar. 2013.
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