Page 159 - Proceedings of the 2017 ITU Kaleidoscope

Challenges for a data-driven society

problem of (12). To this end, we design a two-tier power al- After some simple manipulations, we can obtain the follow-
location iterative algorithm based on the following Theorem ing inequality:
1.
⌣ ∗
⌣ ∗
log (1 + γ ) − log (1 + γ )
2
c
2
e
Theorem 1 The optimization problem (12) and (13) are e- ⌣ ∗ ⌣ ∗
c
z
quivalent if and only if f(η ∗ ) = 0. P + P + P b
SEE ∗ (17)
= η SEE
log (1 + γ c ) − log (1 + γ e )
cP c 2 2
f(η SEE ) = max log (1 + ) ≥ ,
2
P c ,P z dP z + σ 2 P c + P z + P b
c
eP c ⌣ ∗ ⌣ ∗
− log (1 + ) − η SEE (P c + P z + P b )
2 2 from which one can observe that (P , P ) is equal to
fP z + σ c z
e ⌢ ∗ ⌢ ∗
(13a) (P , P ) if and only if f(η ∗ ) = 0.
c z SEE
cP c eP c Letting
s.t. log (1+ )−log (1+ ) ≥ 0 (13b)
2
2
dP z + σ 2 fP z + σ 2
c e
cP c ) + log (fP z + σ )
2
f 1(P c , P z , η SEE ) =log (1 +
max
0 ≤ P c + P z ≤ P CBS . (13c) 2 σ c 2 2 e (18)
− η SEE (P c + P z + P b ),
Proof 1 By comparing the problems (12) and (13), we can
note that they have the same constraints, which means the set and
of feasible solutions to (12) is also applied to (13). Firstly, we f 2 (P c , P z ) = log (eP c + fP z + σ ), (19)
2
⌢ ∗ ⌢ ∗ 2 e
denote (P , P ) as the optimal solution of problem (12). The
c z we can further rewrite (13) as
maximum SEE η ∗ can be obtained by using the following
SEE
formula.
max f 1(P c , P z , η SEE ) − f 2 (P c , P z ) (20a)
P c ,P z
log (1 + γ c ) − log (1 + γ e )
∗
2
2
η SEE = max 2 2
c
P c ,P z P c + P z + P b s.t. P z ≥ eσ − cσ e
⌢ ∗ ⌢ ∗ cf
log (1 + γ ) − log (1 + γ )
2
c
e
2
= ⌢ ∗ ⌢ ∗ (14) (20b)
0 ≤ P c + P z ≤ P max . (20c)
P + P + P b
c z CBS
log (1 + γ c ) − log (1 + γ e )
≥ 2 2 , However, the problem (20) is still non-convex, due to the fact
P c + P z + P b that the logarithmic function (19) is concave.
Next, we apply the D.C. approximation method [17] to ap-
⌢ ∗
⌢ ∗
⌢ ∗ cP ⌢ ∗ eP
¯ ¯
where γ = c and γ = c . Since P c + P z + proximate f 2 (P c , P z ) into a linear one. Assuming (P c , P z )
c ⌢ ∗ e ⌢ ∗
dP +σ c 2 fP +σ 2 e
z
z
P b > 0, we can further transform (14) into the following is a feasible solution of f 2 (P c , P z ), the first-order Taylor se-
ries expansion of f 2 (P c , P z ) is given by
form, i.e.,
¯
¯
¯ ¯
⌢ ∗ ⌢ ∗ ∗ ⌢ ∗ ⌢ ∗ f 2 (P c , P z ) ≈ f 2 (P c , P z ) + e(P c − P c ) + f(P z − P z )
log (1 + γ ) − log (1 + γ ) − η SEE (P + P + P b ) (eP c + fP z + σ ) ln 2 .
¯
¯
2
z
c
2
c
e
2
e
= 0 (21)
≥ log (1 + γ c ) − log (1 + γ e ) − η ∗ SEE (P c + P z + P b ), By substituting (21) into the objective function (20a) , we
2
2
(15) can obtain the optimal solution to problem (20) through the
following iterative procedure, i.e.
which shows that the maximum value of (13) is equal to zero n
⌢ ∗ ⌢ ∗ ⌣ ∗ ⌣ ∗ i+1 i+1 i i
at the optimal solution (P , P ). Then, we assume (P , P ) (P c , P z ) = max f 1 (P c , P z , η SEE ) − f 2 (P , P )
c
z
c z c z P c ,P z
to be the optimal solution of problem (13) and f (η ∗ ) = )
i
i
SEE e(P c − P ) + f(P z − P )
⌣ ∗
⌣ ∗
⌣ ∗ cP ⌣ ∗ eP − c z
0. By defining γ = c and γ = c , one can i i
c ⌣ ∗ e ⌣ ∗ 2
dP +σ 2 fP +σ 2 (eP + fP + σ ) ln 2
z
c
e
z c z e
obtain (22a)
f(η ∗ SEE ) = max log (1 + γ c ) − log (1 + γ e ) eσ − cσ 2
2
2
2
s.t. P z ≥ (22b)
P c ,P z c e
− η ∗ SEE (P c + P z + P b ) cf
max
⌣ ∗ ⌣ ∗ ∗ ⌣ ∗ ⌣ ∗ 0 ≤ P c + P z ≤ P CBS , (22c)
= log (1 + γ ) − log (1 + γ ) − η SEE (P + P + P b )
2
2
c
z
c
e
i
i
≥ log (1 + γ c ) − log (1 + γ e ) − η ∗ SEE (P c + P z + P b ). where (P i+1 , P i+1 ) and (P , P ) are the optimal solutions
c
2
c
z
z
2
(16) in (22) at iterations i and i + 1, respectively. Then, we can
– 143 –