3.3 Problem Statement and Proposed Solution
3.3.2 Data Sensor Resource Allocation
Available channels detected by the spectrum sensors are allocated to the B CR transceivers that are mounted on the sink. If the number of available channels is less than B, then all of the available channels are allocated. Alternatively, the available channels are sorted with respect to their sojourn time, and the channels with the largest sojourn time values are allocated to transceivers. Let K¯ be the number of allocated channels, and note thatK¯ ≤B. Because all of the channels have the same
1The Frobenius norm is defined as the square root of the sum of the absolute squares of the elements of the matrix. For example, if
A= a11a12
a21a22
, then
||A||Fr =
|a11|2+ |a12|2+ |a21|2+ |a22|2.
3.3 Problem Statement and Proposed Solution 35 bandwidth and average power gain, a long average sojourn time implies a large capacity.
Recall thatαkis thek-th channel’s available time. However, scheduling the data sensors to transmit for the entireαkincreases the chance of collision between the data sensor and the returning PU. Letα¯kbe the maximum access time of thek-th channel, where α¯k< αk. It is important to design α¯k such that a low collision probability pkcoll(α¯k)is maintained on thek-th channel. The probability of collision pcollk (α¯k)is determined in the following Theorem:
Theorem 3.1 Given that the PU behavior on each channel is a stationary exponen- tial ON–OFF random process, the probability of collision pkcoll(α¯k)can be expressed by
pkcoll(α¯k)=PO F Fk ã(1−e−μkα¯k), (3.16) where PO F Fk is the probability that PU is not present on the k-th channel at the beginning of the data transmission phase, and(1−e−μkα¯k)captures the probability that PU returns in[0,α¯k].
Proof Let TO F Fk be the sojourn time of a OFF/Inactive period with the probabil- ity density function (p.d.f) fTO F Fk (α). Given the exponentially distributed ON/OFF period, the p.d.f of the Inactive period is equal to [14]
fTO F Fk (α)=μke−μkα.
The probability that the OFF/Inactive period is less thanα¯k, i.e., the PU on channel kreturns in [0,α¯k], can be derived to be
Pr(TO F Fk <α¯k)= α¯k
0
fTO F Fk (α)dα
=1−e−μkα¯k.
Since channelkis available with probabilityPO F Fk , we can have Eq.3.16.
To maintain a target collision probabilitypcollk , the channel access time should not exceed,
¯
αk≤ −ln(1−pkcoll/PO F Fk ) μk
. (3.17)
Furthermore,α¯kis bounded by the duration of the data transmission phaseT −τs. Thus,
¯
αk=min
−ln(1−pkcoll/PO F Fk ) μk
,T −τs
. (3.18)
LetTandPwith elementstn,kandpn,kdenote the transmission time and power allocation matrices of sizeN× ¯K. Lettn,kandpn,kdenote the transmission time and
power of then-th data sensor over thek-th channel, respectively. The total energy consumption of the data sensors is determined by
N n=1
¯
K k=1
tn,kpn,k. (3.19)
The transmission time of all of the data sensors over thek-th channel is limited by the channel access timeα¯k,
N n=1
tn,k≤ ¯αk,∀k. (3.20)
Furthermore, the transmission time of then-th data sensor is bounded by the duration of the data transmission phase, namely,
¯
K k=1
tn,k ≤T −τs,∀n. (3.21) The data amount that is required from the n-th data sensor is denoted by Dn. During the data transmission phase, then-th data sensor transmits sensed data over thek-th channel to the sink at a transmission power ofpn,kand duration oftn,k. The data transmission rate is given by
Rn,k=Wlog2
1+δn,kpn,k
, (3.22)
whereδn,krepresents then-th sensor channel gain over thek-th channel at the sink.
The allocated rate should be sufficiently large to support the generated data. This relationship is captured by
¯
K k=1
tn,kRn,k≥Dn. (3.23) The transmission time tn,k and power pn,k are nonnegative. Additionally, pn,k is constrained by the maximum transmission power pmax. Thus, we have
tn,k≥0,∀k, ∀nand (3.24)
0≤ pn,k≤ pmax. (3.25)
We allocate the transmission time Tand powerP to minimize the energy con- sumption of all of the data senors, which can be formulated as
3.3 Problem Statement and Proposed Solution 37
(DSRA) min
T,P
N n=1
¯
K k=1
tn,kpn,k
s.t.
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ N
n=1tn,k≤ ¯αk,∀k, K¯
k=1tn,k≤T −τs,∀n, K¯
k=1tn,kWlog2(1+δn,kpn,k)≥Dn,∀n, tn,k≥0,∀k,n,
0≤ pn,k≤ pmax,∀k,n.
The amount of data to transmit is determined by the product of the transmission timetn,kand logarithm of the powerpn,k. These structures lead to the non-convexity of the problem DSRA with potentially multiple local optima and generally imply difficulty in determining the global optimal solution [19]. However, by showing that DSRA is biconvex in Theorem3.2, we gain access to algorithms that efficiently solve biconvex problems [20].
Theorem 3.2 If we fix one set of variables inTor P, then DSRA is convex with respect to the other set of variables. Thus, DSRA is biconvex.
Proof We first determine a feasibleP, and then, DSRA becomes a problem of deter- miningTto satisfy
(DSRA-1) min
T
N n=1
¯
K k=1
tn,kpn,k
s.t. (3.20)(3.21)(3.23)(3.24),
which is linear and convex due to the linear objective function and linear feasible set. DSRA-1 can be solved using the simplex method [21]. Additionally, by fixing T, DSRA becomes a problem of determiningPto satisfy
(DSRA-2) min
P
N n=1
¯
K k=1
tn,kpn,k
s.t. (3.23)(3.25).
DSRA-2 can be solved by the interior point method. Both DSRA-1 and DSRA-2 are convex and can be solved efficiently. Therefore, the objective functionN
n=1K¯ k=1
tn,kpn,kis biconvex on the feasible set which makes DSRA a biconvex problem.
3.3.2.1 Joint Time and Power Allocation (JTPA) Algorithm
Because DSRA is biconvex, the variable space is divided into two disjoint subspaces.
Therefore, the problem is divided into two convex subproblems that can be solved efficiently: time allocation (DSRA-1) and power allocation (DSRA-2).
In the following, we adopt the alternate convex search in [20] to solve the DSRA problem. In every step of the proposed algorithm, one of the variables is fixed, and the other is optimized, and vice versa in the subsequent step. The proposed algorithm solves the two problems iteratively and converges to a partially optimal solution [20].
The detailed procedure of the proposed algorithm is given as follows:
Algorithm 1:Joint Time and Power Allocation (JTPA)
Data:Network parameters, stopping criterionεand maximum number of iterationsimax Result:The optimal (T∗,P∗).
1 Choose an arbitrary starting point (T0,P0) and set the iteration index asi=0, and the initial solution asz0=0;
2 whilezi+1−zi−1< εor i≥imaxdo
3 FixPiand determine the optimalTi+1by solving DSRA-1 via the Simplex method [21];
4 FixTi+1, determine the optimalPi+1and objective function valueziby solving DSRA-2 via the Interior Point method [22];
5 i=i+1;
6 Return(Ti+1,Pi+1)
The convergence of the proposed algorithm to the global optimum is not guaran- teed since DSRA is biconvex and could have several local optima. However, because the objective function is differentiable and biconvex over a biconvex set, convergence to a stationary point that is partially optimal is guaranteed [20]. Data sensors transmit their data to the sink using the transmission time and power that is determined by the proposed JTPA algorithm.