Page 95 - ITU Kaleidoscope 2016
P. 95
ICTs for a Sustainable World
¯ c
c
owing are considered. As an example, the channel gain be- 0 ≤ P ≤ P , ∀l ∈ C, (4k)
l
l
tween CU k and the receiver of CU l (i.e., the base station to d ¯ d
0 ≤ P m ≤ P , ∀m ∈ D. (4l)
m
which CU l is communicating) on channel n is
When a CU does not use an uplink channel, its required SINR
g cc = Kβ k,n,l ζ k,n,l L −α , (3) and transmit power on that channel are zero. Constraints
k,n,l
k,n,l
(4d) and (4e) guarantee the required SINR for each CU and
where K is a constant that depends on system parameters,
for the admissible D2D pairs, respectively. Constraint (4f)
β k,n,l is the fast fading gain with exponential distribution, d
makes ρ m,n a binary variable. Constraint (4g) ensures that
and ζ k,n,l is the slow fading gain with log-normal distribution
each channel is reused by at most one D2D pair. Constraint
between CU k and the receiver of CU l on channel n. Also, ¯ d
(4h) guarantees at least one and at most I m channels are as-
α is the path loss exponent and L k,n,l is the distance between
signed to each admissible D2D pair. Constraints (4i) and (4j)
CU k and the receiver of CU l on channel n.
guarantee that transmit power levels do not exceed their up-
We assume additive white Gaussian noise in each channel. per bounds. Finally, constraints (4k) and (4l) guarantee that
Noise power at the receiver of CU l in channel n is σ c , and
l,n the aggregate transmit power of each user does not exceed its
at the receiver of D2D pair m in channel n is σ d . upper bound.
m,n
This problem is a mixed integer linear programming (MILP)
problem, which is difficult to solve directly. In what follows,
3. RESOURCE ALLOCATION PROBLEM
we present our approach for solving it.
When the required SINRs of a D2D pair and CUs using the
same channel can be satisfied with bounded transmit power 4. OPTIMAL RESOURCE ALLOCATION
levels, the D2D pair is admissible and the channel is a candi-
0
0
date for reuse. Let D (D ⊆ D) be the set of all admissible We divide the optimization problem into two sub-problems,
D2D pairs. Each channel is reused by at most one D2D pair. solve each sub-problem separately, and combine the results
This constraint eliminates mutual interference between dif- via our proposed algorithm.
ferent D2D pairs, and simplifies the problem as well. If D2D In the first sub-problem, we determine if D2D pair m is ad-
pair m reuses channel n, then ρ d m,n is 1, otherwise it is 0. We missible to reuse channel n. We also determine the transmit
wish to maximize the number of admissible D2D pairs and power levels of the transmitter of D2D pair m and those CUs
the number of reused channels while minimizing the total that use the same channel n with a view to minimize the to-
transmit power for all users (P sum ). The problem is tal transmit power of all transmitters that use channel n (the
D2D pair and CUs).
d
ρ m,n , ∀m ∈ D, ∀n ∈ N,
d
Determine P m,n , ∀m ∈ D, ∀n ∈ N, (4a) In the second sub-problem, we identify the optimal reuse
P , ∀l ∈ C, ∀n ∈ N,
c channels for all admissible D2D pairs among all candidate
l,n reuse channels so that the total transmit power of all users is
X X d
To Maximize ρ , (4b) minimized.
m,n
m∈D n∈N
X X X c d d
To Minimize (P + ρ P ), (4c) 4.1. D2D Admissibility and Optimal Power Control
l,n m,n m,n
l∈C m∈D n∈N
The D2D pair m can reuse channel n if (4d), (4e), (4i), (4j),
Subject to:
(4k), and (4l) are satisfied. In this case we have
c
cc
P
g l,n,l l,n
c
ξ l,n = c P cc c P dc ( d d
σ + g P + ρ d g P d ρ m,n = 1, P m,n 6= 0, (5a)
l,n k,n,l k,n m,n m,n,l m,n
k∈C m∈D d d
k6=l ρ i,n = 0, P i,n = 0, ∀i ∈ D, i 6= m. (5b)
≥ ξ ˆ c l,n , ∀l ∈ C, ∀n ∈ N, (4d) Since P ¯ c and P ¯ d are obtained by (1) and (2), respectively,
m,n
l,n
g dd P d when (4i) and (4j) are satisfied, (4k) and (4l) are satisfied as
m,n,m m,n
d
ξ m,n = P well. Hence, (4d), (4e), (4i), (4j), (4k), and (4l) become
σ d + g cd P c
m,n k,n,m k,n
k∈C g cc P c
c l,n,l l,n
ξ
0
≥ ξ ˆ d , ∀m ∈ D , ∀n ∈ N, (4e) l,n = c P cc c dc d
P
P
m,n σ l,n + g k,n,l k,n + g m,n,l m,n
k∈C
ρ d m,n ∈ {0, 1} , ∀m ∈ D, ∀n ∈ N, (4f) k6=l
X d
ρ ≤ 1, ∀n ∈ N, (4g) ≥ ξ ˆ c , ∀l ∈ C, (6a)
m,n l,n
m∈D dd d
P
d g m,n,m m,n
X d 0 ξ = ≥ ξ ˆ d , (6b)
¯ d
1 ≤ ρ ≤ I , ∀m ∈ D , (4h) m,n d P cd c m,n
m,n m σ m,n + g P
k,n,m k,n
n∈N k∈C
c
c
0 ≤ P l,n ≤ P ¯ c , ∀l ∈ C, ∀n ∈ N, (4i) 0 ≤ P l,n ≤ P ¯ c , ∀l ∈ C, (6c)
l,n
l,n
d
d
0 ≤ P m,n ≤ P ¯ d , ∀m ∈ D, ∀n ∈ N, (4j) 0 ≤ P m,n ≤ P ¯ d . (6d)
m,n
m,n
– 77 –