524 B. Wang et al. / Computer Networks 129 (2017) 522–535
Fig. 1. An example green heterogeneous wireless network architecture. (For interpretation of the references to color in this figure legend, the reader is referred to the web
version of this article.)
the practical traffic distribution, so as to minimize the total energy
consumption cost.
3. System model
3.1. Network model
In this paper, we consider a heterogeneous wireless network
consisting of both macro BSs and pico BSs. Each macro BS covers a
larger area, and each pico BS within a macro cell covers a smaller
area. Mobile users are assumed to be evenly distributed in the net-
work. For energy supply, all the BSs in our model are powered by
both on-grid energy and renewable energy sources. We consider
the use of solar panels as the source of green energy. Fig. 1 illus-
trates an example green heterogeneous network with hybrid en-
ergy supplies.
Let N
1
, N
2
, and M denote the set of macro BSs, pico BSs and
mobile users, respectively. |N
1
| = N
1
, |N
2
| = N
2
, and |M| = M. We
use N = { 1 , 2 , ··· , N} to denote the set of all BSs in the network,
i.e., N = N
1
∪ N
2
, and N = N
1
+ N
2
. We use the subscript i ∈ N to
denote the i th BS ,
1
and j ∈ M index the j th user. The operational
time of our algorithm is divided into K =
|
K
|
time slots, the length
of each slot is τ seconds and k ∈ K denotes the k th slot.
3.2. Traffic model
The mobile traffic shows both temporal and spatial diversi-
ties [35] . In the temporal domain, individual BS exhibits high traffic
dynamics over time. We find that the peak hour spans from 10 AM
to 6 PM, and off peak hours are from 1 AM to 5 AM. However, the
traffic volume has near-term stability. It is almost constant over a
short term like several minutes of the same time in consecutive
days. Thus, we can predict the average traffic load across several
time slots based on the historical mobile traffic statistics.
In the temporal domain, we use a peak and off-peak tempo-
ral traffic model for mobile users. The mean number of users in
the peak period is much larger than that in the off-peak period. In
each period, the number of users is uniformly distributed around
the mean value. In the spatial domain, we assume that mobile
users are randomly distributed in the area.
1
Without specifically stated, a BS can be either a macro BS, or a pico BS.
3.3. Data transmission model
In this paper, we focus on the downlink data transmission as
the main energy consumption of all BSs. Let X = { X
1
, X
2
, . . . , X
K
}
denote the user-BS association matrix. We use X
k
to denote the
user-BS association relationship at the k th slot. The element X
k
( i, j )
stands for the connection relationship between user j and BS i at
the k th slot, i.e.,
X
k
(
i, j
)
=
1 , user j is served by BS i,
0 , otherwise .
(1)
Note that a user in the system can be associated with
only one BS, either a macro BS or a pico BS. That is,
i ∈N
X
k
(
i, j
)
= 1 , ∀ j ∈ M , ∀ k ∈ K .
For simplicity, we ignore the time slot index k in this subsec-
tion below. During the connection period, according to the Shan-
non Theorem, we can obtain the downlink transmission data rate
of user j :
R
j
= W
i, j
log
2
(1 +
g
i, j
P
i, j
N
0
W
i, j
) , (2)
where P
i,j
is the transmission power of BS i for user j data trans-
mission, and g
i, j
is the channel gain between user j and BS i ,
which in general includes path loss, shadowing and antenna gain.
N
0
denotes the noise power level, and W
i, j
is the bandwidth of
user j allocated by its associated BS i . To reduce the computa-
tional complexity, we adopt a simple equal share strategy to allo-
cate the available bandwidth of each BS to its associated users. We
use L
i
=
j∈M
X
k
(
i, j
)
to denote the number of users served by the
BS i . So the bandwidth allocated for a user j associated with BS i
is computed by W
i, j
=
W
i
L
i
, where W
i
is the available bandwidth of
BS i .
3.4. Energy consumption model
In this paper, we assume that each user has the same data rate
requirement R
0
when admitted to the network. But different users
may have different service times due to their different traffic de-
mands. By letting R
j
= R
0
, we can obtain the transmission power
for data transmission of user j from its associated BS i as:
P
i, j
=
N
0
W
i, j
( 2
R
0
/ W
i, j
− 1)
g
i, j
. (3)