Valor del suelo

In general, a non-convex optimization problem is difficult to solve. In this section, we aim at designing the optimal solution to the non-convex problem (4.93). To this end, we first show that strong duality holds in spite of non-convexity and thus the KKT conditions are necessary optimality conditions. Using this property, the KKT solutions are derived. The optimality of each KKT solution is then characterized.

Concerning problem (4.93) and ε ∈ [0, 1], it is easy to verify that the solution for

ε = 1 is P₂?(t) = 0. Therefore, the remaining task is to solve (4.93) for 0 ≤ ε < 1.

In Appendix A.8, we apply the theorems in [127, 138] and show that strong duality holds between (4.93) and its dual problem. This motivates us to solve the problem in the Lagrangian dual domain.

Since the optimization problem (4.93) has differentiable objective and constraint functions for which strong duality holds, the optimal solution can be attained by solving the KKT conditions [111]. Assuming the non-negative parameter v? is the

inverse of an optimal Lagrangian multiplier λ? and applying the KKT conditions for functional optimization [85, 87] to (4.93), we derive the KKT solutions as follows. Proposition 4.3.1. Solving the KKT conditions of (4.93) yields

P₂?(t) = PS, if t ≤T1(v?) (4.95) P₂?(t) = 0, if t≥T2(v?) (4.96) P₂?(t) = x1(t, v?) ∨ P2?(t) =x2(t, v?), if t ∈ Ft (4.97) where x1(t, v) = 1 −2v+√1−4v+4avt 2av (4.98) x2(t, v) = 1−2v−√1−4v+4avt 2av (4.99) T1(v) = (1+aPS)2 a v−PS (4.100) T2(v) = v a. (4.101)

The feasible regions Ft to constrain P₂?(t) = x1(t, v?) ∈ (0, PS) and P2?(t) = x2(t, v?) ∈

(0, PS) are analyzed in Appendix A.10. If 0≤ ε≤1/(1+aPS), the solution is P₂?(t) = PS, cf. (4.91). Thus the calculation of v? is not required. If 1/(1+aPS) < ε< 1, v? needs to be

selected to let the constraint Qc(P₂?(t)) ≥ε be satisfied with equality.

Proof. See Appendix A.9.

Therefore, we only need to focus on deriving the non-trivial optimal solution for 1/(1+aPS) < ε < 1. Note that KKT conditions are necessary but not sufficient op-

timality conditions under such circumstances, i.e., the primal and dual optimum are among the KKT solutions. We thus characterize the optimality of the KKT solutions by checking sufficient conditions [14, 58].

Lemma 4.3.1. Given a non-negative v?, P₂?(t) = x1(t, v?) yields a local minimum and P₂?(t) = x2(t, v?) yields a local maximum in the corresponding feasible region of t given in Appendix A.10.

Proof. See Appendix A.11.

Proposition 4.3.1 combined with Lemma 4.3.1 implies that the optimal solution to (4.93) is made of three functions: two functions at the boundary points, i.e., P₂?(t) = PS and P₂?(t) = 0, and the function at the interior points P₂?(t) = x2(t, v?). The regions of t for these functions are given in (4.95), (4.96), and Appendix A.10, respectively. We assume that the optimal solution is a piecewise continuous function which is a common assumption in optimal control theory, e.g., [74]. In addition, there is no re- movable discontinuity since this kind of discontinuity can be removed to make the function continuous at the discontinuity. Therefore, the finite number of discontinu- ities are located in the corresponding feasible region of t for each function. Figure 4.7

0 2 4 6 8 10 0 5 10 15 P2(t) =PS P2(t) =0 P2(t) =x2(t, v) T1(v) T2(v) t Case 1 − 2v? ≥ 2av?_P S 0 2 4 6 8 10 0 5 10 15 20 P2(t) =PS P2(t) =0 T2(v) T1(v) t Case 1 − 2v? ≤ 0 10.5 11 11.5 12 12.5 0 2 4 6 8 10 P₂(t) =P_S P2(t) =0 P2(t) =x2(t, v) T3(v) T2(v)_T1(v) t Case 0 < 1 − 2v? ≤ av?_P S 0 2 4 6 8 10 0 10 20 P2(t) =PS P₂(t) =0 P2(t) =x2(t, v) T3(v) T1(v) T2(v) t Case av?PS< 1 − 2v? < 2av?PS

Figure 4.7: Exemplified candidate solutions for different cases of v?. The functions T1(v), T2(v), and T3(v) are given in (4.100), (4.101), and (A.51), respec- tively.

exemplifies the candidate solutions for the four cases of v?. The feasible regions of t for the three functions are not mutually exclusive under the following circumstances:

• The regions of t for P₂?(t) = PS and P2?(t) =0 overlap if 1−2v? ≤av?PS.

• The regions of t for P₂?(t) = PS and P₂?(t) = x2(t, v?) overlap if 0 < 1−2v? < 2av?PS.

Due to the fact that the discontinuities can be arbitrarily located in the corresponding feasible region of t for each function and t is continuous, there is an infinite number of candidate solutions. In order to further reduce the set of candidate optimal solutions, we provide an additional characteristic of the optimal solution to (4.93).

Proof. See Appendix A.12.

Lemma A.12 indicates that the optimal power allocation strategy is to allocate a large amount of power when the CSI of the ST-SR link to interference plus noise ratio is high and allocate a small amount of power when this ratio is low. This property is reasonable since the ST needs to allocate more power during favorable channel states in order to maximize the achievable rate of the secondary link while keeping the outage probability of the PU under a certain limit.

Remark: Lemma 4.3.1 can be partly verified by Lemma 4.3.2. More particularly, a solution containing P₂?(t) = x1(t, v?) is not the optimal solution since x1(t, v?) is an increasing function of t. However, Lemma 4.3.1 additionally shows that a solution containing P₂?(t) = x2(t, v?) might be a optimal solution. This part cannot be vali- dated by Lemma 4.3.2 since it only provides a necessary optimality condition tailored to the problem (4.93).

Given the above-mentioned characteristics of the optimal solution, we present the optimal solution of (4.93) in the following theorem:

Theorem 4.3.1. The optimal solution of (4.93) is given as 1) P₂?(t) = PS, if 0 ≤ε≤1/(1+aPS).

2) P₂?(t) = 0, if ε=1.

3) If 1/(1+aPS) < ε<1, the optimal solution P₂?(t)is divided into three cases:

• If 1−2v? ≥2av?PS, P₂?(t) =      PS, if 0≤t≤T1(v?) x2(t, v?), if T1(v?) <t <T2(v?) 0, otherwise. (4.102)

The value of v? is chosen to let the equality Qc(P2?(t)) =ε hold. • If 1−2v? ≤0,

P₂?(t) =

(

PS, if 0≤t≤T2(v?)

0, otherwise. (4.103)

The value of v? is chosen to let the equality Qc(P₂?(t)) =ε hold.

• If 0<1−2v? <2av?PS, the solution P₂?(t) is a function of v? where v? =arg max v {R(P2(t))} (4.104) with P2(t) =      PS, if 0≤t≤γ(v) x2(t, v), if γ(v) < t< T2(v) 0, otherwise. (4.105)

where the non-negative γ(v)is in[T3(v), T1(v)]and it is chosen to satisfy Qc(P₂?(t)) =

ε. The functions T1(v), T2(v), and T3(v) are given in (4.100), (4.101), and (A.51), respectively.

Proof. It is easy to verify that the optimal solution is P₂?(t) = PS and P2?(t) = 0 for 0 ≤ ε ≤ 1/(1+aPS) and ε = 1, respectively. Therefore, we only focus on the case 1/(1+aPS) <ε <1. In the following, the optimal solution is provided as a function

of v?. Incorporating them into the outage probability constraint (4.91) yields the left side as the function of v? only. Then v?can be obtained by solving (4.91) with equality.

1. 1−2v? ≥2av?PS:

The case is exemplified in the upper-left subfigure in Figure 4.7. We can ver- ify that T1(v?) < T2(v?) holds. The solutions P2?(t) = 0, P2?(t) = PS, and P₂?(t) = x2(t, v) have non-overlapping feasible regions of t. Therefore, the op- timal solution is given in (4.102) as the unique piecewise continuous function consisting of P₂?(t) =0, P₂?(t) = PS, and P2?(t) = x2(t, v).

2. 1−2v? ≤0:

One example under this case is given in the upper-right subplot in Figure 4.7. For this case T2(v?) < T1(v?) holds. The optimal solution is a piecewise contin- uous function combined with P₂?(t) = PS and P₂?(t) =0 with the feasible region of t as t ≤ T1(v?) and t ≥ T2(v?), respectively. The discontinuity can thus be chosen in [T2(v?), T1(v?)]. However, if the optimal power allocation strategy is such a binary power switching scheme, there exists a unique discontinuity to let the outage probability constraint be satisfied with equality, i.e., the value of the discontinuity is fixed. Without loss of generality, we give the optimal solution in (4.103) where the discontinuity is located at T2(v?).

3. 0<1−2v? <2av?PS:

Two different examples are given in the lower-left and lower-right subplots in Figure 4.7 in which T2(v?) ≤ T1(v?) and T1(v?) < T2(v?) holds, respectively. The feasible regions of t for P₂?(t) = PS, P2?(t) = 0, and P2?(t) = x2(t, v) are t ≤T1(v?), t ≥T2(v?), and T3(v?) < t<T2(v?), respectively, and they overlap. Therefore, the optimal solution can be a piecewise continuous monotonically decreasing function composed of the three functions in which the discontinu- ities are located at arbitrary points in the overlapping region. The influence on the objective function of the variation of the discontinuities is difficult to eval- uate, resulting in difficulty in further reducing the search space of the optimal solution. We give the solution as a function of v and claim that the optimal solu- tion can be obtained by solving an optimization problem over a single bounded variable v.

• T1(v) ≥ T2(v): The lower-left subplot in Figure 4.7 indicates one example. a) For a discontinuity at γ(v) when switching from P₂?(t) = PS to P2?(t) = x2(t, v) for an increase of t, it is located at an arbitrary point in the region [T3(v), T2(v)]. Note that there is no discontinuity switching

from P₂?(t) = x2(t, v) to P₂?(t) = PS for an increase of t since it violates the optimality condition in Lemma 4.3.2. In addition, the candidate solution switches from P₂?(t) = x2(t, v) to P2?(t) = 0 at T2(v). Due to Lemma 4.3.2, only P₂?(t) = 0 holds for t ≥ T2(v). The solution is expressed in (4.105) with γ(v) ∈ [T3(v), T2(v)] and γ(v) is chosen to satisfy the constraint Qc(P2(t)) = ε.

b) For a discontinuity at γ(v) when switching from P₂?(t) = PS to P2?(t) = 0 for an increase of t, then γ(v) is located at an arbitrary point in the region[T2(v), T1(v)]. Similar to the last case, only P2?(t) = 0 holds for t≥γ(v) due to the optimality condition in Lemma 4.3.2. The solution

is also expressed in the form of (4.105) with γ(v) ∈ [T2(v), T1(v)]and

γ(v) is chosen to let the constraint Qc(P2(t)) =ε be satisfied.

Combined with the above two cases, the power allocation strategy is given in (4.105) with γ(v) ∈ [T3(v), T1(v)]. The value γ(v) should be chosen to satisfy Qc(P2(t)) =ε.

• T1(v?) < T2(v?): As shown in the exemplified in the lower-right subplot in Figure 4.7, a discontinuity at γ(v) when switching from P₂?(t) = PS to P₂?(t) = x2(t, v) for an increase of t can be located at an arbitrary point in the region [T3(v), T1(v)]. According to Lemma 4.3.2, there exists no discontinuity when switching from P₂?(t) = x2(t, v) to P₂?(t) = PS. Further- more, the candidate solution switches from P₂?(t) = x2(t, v)to P2?(t) =0 at T2(v). For t ≥ T2(v), the solution is P₂?(t) = 0. The solution also matches the form in (4.105). The value γ(v) ∈ [T3(v), T1(v)]is chosen to satisfy the constraint Qc(P2(t)) =ε.

The proof is concluded by combining the results discussed in each case.

For the third case 1/(1+aPS) < ε<1 in Theorem 4.3.1, the value of v?is critical to

determine the optimal solution. For example, P₂?(t)in (4.102) and (4.103) is a function of v? since T1(v?), T2(v?), T3(v?), and x2(t, v?) are all functions of v?, incorporating them into Qc(P₂?(t)) results in Qc(·) being dependent on v?. We can then obtain v? by solving Qc(P₂?(t)) =ε. Similarly, P2(t)in (4.105) is also provided as a function of v. Taking it into Qc(P2(t)) and solving Qc(P2(t)) =ε yields a feasible v. Incorporating v into (4.105), we obtain a feasible solution P2(t). Finally, searching over all feasible v, we select v? corresponding to P₂?(t)that results in the largest secondary achievable rate in (4.104). Briefly speaking, (4.104) is a nonlinear maximization problem over a single bounded variable v for 0 < 1−2v < 2avPS. Some optimization toolboxes can be applied for its efficient implementation, e.g., fminbnd in Matlab©.