The BER expressions at SU-R for MQAM (21) or MPSK modulation (22) [22] over the AWGN channel are written as

$$\begin{array}{*{20}l} {}\text{BER}_{l,\text{MQAM}}^{n} &=\frac{4}{b}\left({1 - \frac{1}{{\sqrt M }}} \right)Q\left({\sqrt {\frac{{3b({\text{SINR}_{l,eq}^{n}}/b)}}{{M - 1}}}} \right) \\ &=\frac{4}{b}\left({1 - \frac{1}{{\sqrt M }}} \right)Q\left({\sqrt {\frac{{3}}{{M - 1}}} {{\left({\text{SINR}_{l,eq}^{n}} \right)}^{2}}} \right) \end{array} $$

((21))

$$\begin{array}{*{20}l} {\text{BER}_{l,\text{MPSK}}^{n} = \frac{2}{b}Q\left({\sqrt {2b ({\text{SINR}_{l,eq}^{n}}/b){{\sin }^{2}}\left({\frac{\pi }{M}} \right)}} \right)} \end{array} $$

((22))

where \(Q\left (\bar {x} \right) = \frac {1}{{\sqrt {2\pi } }}\int _{\bar x}^{\infty } {{e^{- \frac {{{w^{2}}}}{2}}}} dw\) is a Gaussian Q-function, *b*=log2*M*, and *M* is the number of bits of the modulation symbols.

In this paper, a worst-CSI PC algorithm is presented to limit the BER of the SUs while keeping the interference leakage to the PUs below the IT level, the maximum transmit power of the SU and relay below certain thresholds. Here, we introduce the SINRs at SU-R and relay in order to guarantee the requirements for each hop. Thus, the optimization problem is formulated as **OP1**

$$ {}\begin{aligned} \textbf{OP1}\quad &\mathop {\min }\limits_{P_{l,1}^{n},P_{l,2}^{n}} \quad \mathop {\max }\limits_{\forall l} \quad \text{BER}_{l}^{n} \\ s.t.\quad \;\; &C1: 0 \le \sum\limits_{n = 1}^{N} {P_{l,1}^{n}} \le P_{l,1}^{\text{max} },\quad \forall l\\ \quad \quad \quad &C2:0 \le \sum\limits_{n = 1}^{N} {P_{l,2}^{n}} \le P_{l,2}^{\text{max} },\quad \forall l \\ \quad \quad \quad &C3: \text{SINR}_{l,1}^{n} \ge \text{SINR}_{l,1,\text{th}}^{n},\quad \forall l,\forall n \\ \quad \quad \quad &C4: \text{SINR}_{l,2}^{n} \ge \text{SINR}_{l,2,\text{th}}^{n},\quad \forall l,\forall n \\ \quad \quad \quad &C5: \sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({O_{p}^{n}})P_{md}^{n}P_{l,1}^{n}{{|h_{l,p,1}^{n}|^{2}}}}} \le {I_{p,\text{th}}},\quad \forall p \\ \quad \quad \quad &C6: \sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({O_{p}^{n}})}} P_{md}^{n}P_{l,2}^{n}{{|h_{l,p,2}^{n}|^{2}}} \le {I_{p,\text{th}}},\quad \forall p \end{aligned} $$

((23))

where \(P_{l,1}^{\text {max} }\) and \(P_{l,2}^{\text {max} }\) are the maximum power budgets of SU-T and relay, respectively. \(\text {SINR}_{l,1,\text {th}}^{n}\) and \(\text {SINR}_{l,2,\text {th}}^{n}\) are the SINR thresholds at the relay and SU-R, respectively. *I*
_{
p,th} is the interference threshold prescribed by the *p*th PU receiver. *C*1 and *C*2 represent the transmit power constraints at the transmitters of the source node and relay node, respectively. *C*3 and *C*4 are the SINR constraints to keep basic communication requirements at the *l*th relay and SU-R. *C*5 and *C*6 denote the interference power constraints at tge source and relay nodes, respectively. Since the objection of **OP1** is a monotonic function about the equivalent SINR \(\text {SINR}_{l,eq}^{n}\), **OP1** can be converted into

$$\begin{array}{*{20}l} \textbf{OP2}\quad &\mathop{\max}\limits_{P_{l,1}^{n},P_{l,2}^{n}} \quad \mathop{\min}\limits_{\forall l}~\text{SINR}_{l,eq}^{n} \\ \quad \quad \quad s.t.\quad \quad &C1 \sim C6 \end{array} $$

((24))

Therefore, the original optimization problem **OP1** becomes a worst-CSI SINR maximization problem **OP2**. The criterion about the selection of the worst-CSI user is

$$\begin{array}{*{20}l} {\left| {h_{l,1}^{n}} \right|^{2}}{\left| {h_{l,2}^{n}} \right|^{2}} \le {\left| {h_{j,1}^{n}} \right|^{2}}{\left| {h_{j,2}^{n}} \right|^{2}} \end{array} $$

((25))

If the channel gain of two hops can satisfy (25), we regard the *l*th SU as the worst case. **OP2** is not convex due to the constraints *C*3 and *C*4. In order to simplify our analysis, we take *C*3 and *C*4 on reciprocal as

$$\begin{array}{*{20}l} C3:\frac{1}{{\text{SINR}_{l,1}^{n}}} \le \frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}} \end{array} $$

((26))

$$\begin{array}{*{20}l} C4:\frac{1}{{\text{SINR}_{l,2}^{n}}} \le \frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}} \end{array} $$

((27))

i.e.,

$$\begin{array}{*{20}l} {\frac{{\frac{{N_{l,1}^{n}}}{{{ {{{|h_{l,1}^{n}|^{2}} }}}}} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}\frac{{{{ {{|g_{p,l,1}^{n}|^{2}}} }}}}{{{{ {{|h_{l,1}^{n}|^{2}}} }}}}} }}{{P_{l,1}^{n}}} \le \frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}}} \end{array} $$

((28))

$$\begin{array}{*{20}l} {\frac{{\frac{{N_{l,2}^{n}}}{{{ {{{|h_{l,2}^{n}|^{2}} }}}}} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}\frac{{{{ {{|g_{p,l,2}^{n}|^{2}}} }}}}{{{{ {{|h_{l,2}^{n}|^{2}}} }}}}} }}{{P_{l,2}^{n}}} \le \frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}}} \end{array} $$

((29))

Define

$$ \left\{ \begin{array}{l} F_{l,1}^{n} = {\frac{{N_{l,1}^{n}}}{{{ {|h_{l,1}^{n}|^{2}} }}}}\\ F_{l,2}^{n} = {\frac{{N_{l,2}^{n}}}{{{ {|h_{l,2}^{n}|^{2}} }}}} \\ \end{array} \right. $$

((30))

$$ {\left\{ \begin{array}{l} G_{p,l,1}^{n} = {\frac{{{ {|g_{p,l,1}^{n}|^{2}} }}}{{{ {|h_{l,1}^{n}|^{2}} }}}}\\ G_{p,l,2}^{n} ={\frac{{{ {|g_{p,l,2}^{n}|^{2}} }}}{{{ {|h_{l,2}^{n}|^{2}} }}}} \\ \end{array} \right.} $$

((31))

then *C*3 and *C*4 can be written as

$$ {\frac{{F_{l,1}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,1}^{n}} }}{{P_{l,1}^{n}}} \le \frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}}} $$

((32))

$$ {\frac{{F_{l,2}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,2}^{n}} }}{{P_{l,2}^{n}}} \le \frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}}} $$

((33))

Then, the equivalent SINR is

$$ {{\begin{aligned} {}{\text{SINR}_{l,eq}^{n} \,=\, \frac{{\frac{{P_{l,1}^{n}}}{{F_{l,1}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,1}^{n}} }}\! \times\! \frac{{P_{l,2}^{n}}}{{F_{l,2}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,2}^{n}} }}}}{{\frac{{P_{m,1}^{n}}}{{F_{l,1}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,1}^{n}} }} + \frac{{P_{l,2}^{n}}}{{F_{l,2}^{n} + \sum\limits_{p = 1}^{P} {P_{l,2}^{n}G_{p,l,2}^{n}} }} + 1}} \,=\, \frac{{{a_{l}^{n}}P_{l,1}^{n}{b_{l}^{n}}P_{l,2}^{n}}}{{{a_{l}^{n}}P_{l,1}^{n} \,+\, {b_{l}^{n}}P_{l,2}^{n} \,+\, 1}}} \end{aligned}}} $$

((34))

where \({a_{l}^{n}}\) and \({b_{l}^{n}}\) are given by

$$ {\left\{ {\begin{array}{l} {{a_{l}^{n}} = \frac{1}{{F_{l,1}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,1}^{n}} }}} \\ {{b_{l}^{n}} = \frac{1}{{F_{l,2}^{n} + \sum\limits_{p = 1}^{P} {{P_{p}^{n}}G_{p,l,2}^{n}} }}} \\ \end{array}} \right.} $$

((35))

Furthermore, to make the equivalent SINR tractable, we adopt the following approximation [23]

$$ {\text{SINR}_{l,eq}^{n} \approx \frac{{{a_{l}^{n}}P_{l,1}^{n}{b_{l}^{n}}P_{l,2}^{n}}}{{{a_{l}^{n}}P_{l,1}^{n} + {b_{l}^{n}}P_{l,2}^{n}}}} $$

((36))

Define \(P_{l,1}^{n} = {x_{1}}\), \(P_{l,2}^{n} = {x_{2}}\), \(t = \frac {1}{{\text {SINR}_{l,eq}^{n}}}\), then

$$ {t = \frac{1}{{\text{SINR}_{l,eq}^{n}}} = \frac{1}{{{b_{l}^{n}}}}\frac{1}{{{x_{2}}}} + \frac{1}{{{a_{l}^{n}}}}\frac{1}{{{x_{1}}}}} $$

((37))

Therefore, **OP2** can be rewritten as

$$\begin{array}{*{20}l} \textbf{OP3}\quad &\mathop {\min }\limits_{{x_{1}},{x_{2}}} \quad \mathop {\max }\limits_{\forall l} \quad t \\ s.t.\quad &C1: 0 \le \sum\limits_{n = 1}^{N} {{x_{1}}} \le P_{l,1}^{\text{max} },\quad \forall l \notag\\ \quad \quad \;&C2: 0 \le \sum\limits_{n = 1}^{N} {{x_{2}}} \le P_{l,2}^{\text{max} },\quad \forall l \\ \quad \quad \;&C3:\frac{1}{{{a_{l}^{n}}}}\frac{1}{{{x_{1}}}} \le \frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}},\quad \forall l,\forall n \\ \quad \quad \;&C4: \frac{1}{{{{b_{l}^{n}}}}}\frac{1}{{{x_{2}}}} \le \frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}},\quad \forall l,\forall n \\ \quad \quad \;&C5: \sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({O_{p}^{n}})P_{md}^{n}{x_{1}}{{|h_{l,p,1}^{n}|^{2}}}} \le {I_{p,\text{th}}},\quad \forall p }\\ \quad \quad \;&C6: \sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({O_{p}^{n}})}} P_{md}^{n}{x_{2}}{{|h_{l,p,2}^{n}|^{2} }} \le {I_{p,\text{th}}},\quad \forall p \end{array} $$

((38))

Now **OP3** is a convex problem which can be solved by the dual decomposition method [24]. First, we give a Lagrange function with Lagrange multipliers \(\lambda _{l,1},\lambda _{l,2},\lambda _{l,3}^{n},\lambda _{l,4}^{n},\lambda _{p,5},\lambda _{p,6} \ge 0\) as follows

$$ {{\begin{aligned} &{}L\left({t,\{ {\lambda_{l,1}} \},\{ {\lambda_{l,2}} \},\{ {\lambda_{l,3}^{n}} \},\{ {\lambda_{l,4}^{n}} \},\{ {\lambda_{p,5}} \},\{ {\lambda_{p,6}} \}}\right) \\ &{}= t + \sum\limits_{l = 1}^{L} {\left({ {\lambda_{l,1}\left({\sum\limits_{n = 1}^{N}{x_{1}} - P_{l,1}^{\text{max} }} \right)}} \right)} + \sum\limits_{l = 1}^{L} {\left({ {\lambda_{l,2}\left({\sum\limits_{n = 1}^{N}{x_{2}} - P_{l,2}^{\text{max} }}\right)}} \right)} \\ & \;\;\;\;{}+ \sum\limits_{l = 1}^{L}\! {\left(\!{\sum\limits_{n = 1}^{N}\! {\lambda_{l,3}^{n}\!\left(\!{\frac{1}{{{a_{l}^{n}}}}\frac{1}{{{x_{1}}}} - \frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}}}\!\right)}}\!\right)} \,+\,\! \sum\limits_{l = 1}^{L}\! {\left(\!{\sum\limits_{n = 1}^{N}\! {\lambda_{l,4}^{n}\left(\! {\frac{1}{{{b_{l}^{n}}}}\!\frac{1}{{{x_{2}}}} \,-\, \frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}}}\!\right)}}\!\right)} \\ & \;\;\;\;{}+ \sum\limits_{p = 1}^{P} {\left({{\lambda_{p,5}\left({\sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({{O_{p}^{n}}})P_{md}^{n}{x_{1}}{|h_{l,p,1}^{n}|^{2}} }} - {I_{p,\text{th}}}} \right)}}\right)} \\ & \;\;\;\;{}+ \sum\limits_{p = 1}^{P} {\left({{\lambda_{p,6}\left({\sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({{O_{p}^{n}}})P_{md}^{n}{x_{2}}{|h_{l,p,2}^{n}|^{2}}}} - {I_{p,\text{th}}}}\right)}}\right)}\\ \end{aligned}}} $$

((39))

The dual problem of the Lagrange function (39) is

$$ {\begin{array}{l} D\left(t, {\{ \lambda_{l,1}\},\{ \lambda_{l,2}\},\{ \lambda_{l,3}^{n}\},\{ \lambda_{l,4}^{n}\},\{ \lambda_{p,5}\},\{ \lambda_{p,6}\}}\right) \\ = \sum\limits_{l = 1}^{L} {\left({\sum\limits_{n = 1}^{N} {\mathop {\min }\limits_{{x_{1}},{x_{2}}} \;{L_{l}^{n}}\left(t, {\lambda_{l,1},\lambda_{l,2},\lambda_{l,3}^{n},\lambda_{l,4}^{n},\{ \lambda_{p,5}\},\{ \lambda_{p,6}\}} \!\right)}} \right)} \\ \;\;\;\;- \sum\limits_{l = 1}^{L} {\left({{\lambda_{l,1}P_{l,1}^{\text{max} }}}\right)} - \sum\limits_{l = 1}^{L} {\left({ {\lambda_{l,2}P_{l,2}^{\text{max}}}}\right)} \\ \;\;\;\;- \sum\limits_{l = 1}^{L} {\left({\sum\limits_{n = 1}^{N} {\lambda_{l,3}^{n}\frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}}}}\right)} - \sum\limits_{l = 1}^{L} {\left({\sum\limits_{n = 1}^{N} {\lambda_{l,4}^{n}\frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}}}}\right)} \\ \;\;\;\;- \sum\limits_{p = 1}^{P} {\left({ { {{\lambda_{p,5}{I_{p,\text{th}}}}} }}\right)} - \sum\limits_{p = 1}^{P} {\left({{ {{\lambda_{p,6}{I_{p,\text{th}}}}} }}\right)} \\ \end{array}} $$

((40))

Define \({L_{l}^{n}}\) as a function of *x*
_{1} and *x*
_{2}

$$ {{\begin{aligned} &{}{L_{l}^{n}}\left(t, {\lambda_{l,1},\lambda_{l,2},\lambda_{l,3}^{n},\lambda_{l,4}^{n},\{ \lambda_{p,5}\},\{ \lambda_{p,6}\}}\right) \\ &{}=\! t \,+\, \lambda_{l,1}{x_{1}} \,+\, \lambda_{l,2}{x_{2}} \,+\, \lambda_{l,3}^{n}\frac{1}{{{a_{l}^{n}}}}\frac{1}{{{x_{1}}}} \,+\, \lambda_{l,4}^{n}\frac{1}{{{b_{l}^{n}}}}\frac{1}{{{x_{2}}}} \,+\, {x_{1}}\!\sum\limits_{p = 1}^{P} \!{\lambda_{p,5}{Pr}\!\left(\!{{O_{p}^{n}}}\right)\!P_{md}^{n}{{|h_{l,p,1}^{n}|^{2}}}}\\ &{}\;\;\;\;+ {x_{2}}\sum\limits_{p = 1}^{P} {\lambda_{p,6}{Pr}\left({{O_{p}^{n}}}\right)P_{md}^{n}{{|h_{l,p,2}^{n}|^{2}}}} \end{aligned}}} $$

((41))

Since the primal problem in (38) is convex, strong duality holds, and the dual problem can be solved by an iterative manner using the gradient projection method [24]. By the Karush-Kuhn-Tucker (KKT) condition, the optimal transmit power \(P_{l,1}^{n}\) and \(P_{l,2}^{n}\) at SU-T and relay can be calculated by

$$ {{\begin{aligned} {}{\frac{{\partial {L_{l}^{n}}}}{{\partial {x_{1}}}} \!= - \frac{1}{{{a_{l}^{n}}}}\frac{1}{{{x_{1}^{2}}}} \!+ \!\lambda_{l,1} - \lambda_{l,3}^{n}\frac{1}{{{a_{l}^{n}}}}\frac{1}{{{x_{1}^{2}}}} \,+\, \sum\limits_{p = 1}^{P} {\lambda_{p,5}Pr\left(\!{{{O_{p}^{n}}}} \right)P_{md}^{n}{{|h_{l,p,1}^{n}}|^{2}}} = 0} \end{aligned}}} $$

((42))

$$ {{\begin{aligned} {}{\frac{{\partial {L_{l}^{n}}}}{{\partial {x_{2}}}} \!= \!- \frac{1}{{{b_{l}^{n}}}}\frac{1}{{{x_{2}^{2}}}} \,+\, \lambda_{l,2} \,-\, \lambda_{l,4}^{n}\frac{1}{{{b_{l}^{n}}}}\frac{1}{{{x_{2}^{2}}}} + \sum\limits_{p = 1}^{P} {\lambda_{p,6}Pr\left({{{O_{p}^{n}}}} \right)P_{md}^{n}{{|h_{l,p,2}^{n}|^{2}}}} = 0} \end{aligned}}} $$

((43))

The optimal solutions are

$$ {\fontsize{8.5}{6}\begin{aligned} P_{l,1}^{n}{~}^{*} = x_{1}^{*}= \sqrt {\frac{{\frac{1}{{{a_{l}^{n}}}}\left({1 + \lambda_{l,3}^{n}} \right)}}{{\lambda_{l,1} + \sum\limits_{p = 1}^{P} {\lambda_{p,5}Pr\left({{{O_{p}^{n}}}} \right)P_{md}^{n}{{|h_{l,p,1}^{n}|^{2}}}} }}} \end{aligned}} $$

((44))

$$ {\fontsize{8.5}{6}\begin{aligned} P_{l,2}^{n}{~}^{*} = x_{2}^{*}= \sqrt {\frac{{\frac{1}{{{b_{l}^{n}}}}\left({1 + \lambda_{l,4}^{n}} \right)}}{{\lambda_{l,2} + \sum\limits_{p = 1}^{P} {\lambda_{p,6}Pr\left({{{O_{p}^{n}}}}\right)P_{md}^{n}{{|h_{l,p,2}^{n}|^{2}}}} }}} \end{aligned}} $$

((45))

The lagrange multipliers *λ*
_{
l,1}, *λ*
_{
l,2}, \(\lambda _{l,3}^{n}\), \(\lambda _{l,4}^{n}\), *λ*
_{
p,5}, and *λ*
_{
p,6} must be carefully chosen to ensure a fast convergence rate. A simple but effective way to decide these multipliers is to employ the subgradient method as follows

$$ {}{{\lambda_{l,1}}({d + 1}) = {\left[ {{\lambda_{l,1}}(d) + {\beta_{1}}\left({\sum\limits_{n = 1}^{N} {P_{l,1}^{n}} - P_{l,1}^{\textit{max} }}\right)}\right]^ +}} $$

((46))

$$ {}{{\lambda_{l,2}}({d + 1}) = {\left[ {{\lambda_{l,2}}(d) + {\beta_{2}}\left({\sum\limits_{n = 1}^{N} {P_{l,2}^{n}} - P_{l,2}^{\textit{max} }}\right)}\right]^ + }} $$

((47))

$$ {}\lambda_{l,3}^{n}({d +1})= \left[{\lambda_{l,3}^{n}(d)+{\beta_{3}}\left({\frac{1}{{{a_{l}^{n}}}}\frac{1}{{P_{l,1}^{n}}} - \frac{1}{{\text{SINR}_{l,1,\text{th}}^{n}}}}\right)}\right]^+ $$

((48))

$$ {}\lambda_{l,4}^{n}({d +}1)= \left[{\lambda_{l,4}^{n}(d)+{\beta_{4}}\left({\frac{1}{{{b_{l}^{n}}}}\frac{1}{{P_{l,2}^{n}}} - \frac{1}{{\text{SINR}_{l,2,\text{th}}^{n}}}}\right)} \right]^{+} $$

((49))

$$ {{\begin{aligned} {}{{\lambda_{p,5}}({d + 1}) = {\left[{{\lambda_{p,5}}(d) \,+\, {\beta_{5}}\left({\sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({{O_{p}^{n}}})P_{md}^{n}{{|h_{l,p,1}^{n}|^{2}}} - {I_{p,\text{th}}}}} }\right)} \right]^ + }} \end{aligned}}} $$

((50))

$$ {{\begin{aligned} {}{{\lambda_{p,6}}({d + 1}) = {\left[ {{\lambda_{p,6}}(d) \,+\, {\beta_{6}}\left({\sum\limits_{l = 1}^{L} {\sum\limits_{n = 1}^{N} {{Pr}({{O_{p}^{n}}})P_{md}^{n}{{|h_{l,p,2}^{n}|^{2}}} - {I_{p,\text{th}}}}} } \right)} \right]^ + }} \end{aligned}}} $$

((51))

where [·]^{+}= max(0,·). *d* denotes the iteration number. *β*
_{1}∼*β*
_{6} are the small step sizes which satisfy *β*
_{
q
}>0, *q*={1,2,3,4,5,6}. Apparently, *λ*
_{
l,1}(*d*+1), *λ*
_{
l,2}(*d*+1), \(\lambda _{l,3}^{n}({d + 1})\), and \(\lambda _{l,4}^{n}({d + 1})\) are locally updated, whereas *λ*
_{
p,5}(*d*+1) and *λ*
_{
p,6}(*d*+1) are updated through cooperation. In addition, the Lagrange multipliers *λ*
_{
p,5}(*d*+1) and *λ*
_{
p,6}(*d*+1) in (50) and (51) can only be updated by obtaining the interference channels information (i.e., \(h_{l,p,1}^{n}\) and \(h_{l,p,2}^{n}\)) about other SUs and relays, respectively.

Finally, taking the optimal solutions \({P_{l,1}^{n}}{~}^{*}\) and \({P_{l,2}^{n}}{~}^{*}\) into (21) and (22), respectively, the optimal BER can be calculated.

Based on the above development, we get our algorithm reaching the optimum control power at SU-T and relay for the optimization problem. And the specific power allocation algorithm can be given in Algorithm 1.