Algorithm 1 Joint optimization of τ and K

(1)initialize n = 0, K(n) = K0 where K0 is any integer within interval $(0,\min (\frac{T}{\epsilon },N))$, τ(n) = 0 and the estimation error $\delta$; (2) with given K(n), select the K(n) largest SNRs as ${\left\{{\gamma }_i\right\}}_{i=1}^{K^{(n)}}$ and obtain ${\tau }^{*}(K^{(n)})$ through the following steps: a) initialize ${\tau }_{\min }=0$ and ${\tau }_{\max }=T- K^{\left(n\right)}\epsilon$; b) set $\tau =({\tau }_{\min }+{\tau }_{\max })/2$; c) if $\nabla P_{Acc}(\tau )\text{==}\nabla P_{Acc}({\tau }_{\min })$: let $\tau ={\tau }_{\min }$; d) else if $\nabla P_{Acc}(\tau )\text{==}\nabla P_{Acc}({\tau }_{\max })$: let $\tau ={\tau }_{\max }$; e) repeat Steps (b) to (d) until $\left|{\tau }_{\max }-{\tau }_{\min }\right|<\delta$; f) set ${\tau }_M=({\tau }_{\min }+{\tau }_{\max })/2$; g) calculate ${\tau }^{*}(K^{(n)})=\max \left(\pi (K),{\tau }_M,{\tau }_l|K=K^{(n)}\right)$; (3)set τ(n+1) = ${\tau }^{*}(K^{(n)})$; (4)with given τ(n+1), obtain $K^{*}({\tau }^{(n+1)})$ from Equation (37); (5)set K(n+1) = $K^{*}({\tau }^{(n+1)})$ and n = n + 1; (6)repeat Steps(2)~(5) until $\left|{\tau }^{(n)}-{\tau }^{(n-1)}\right|\le \delta$ and $\left|K^{(n)}-K^{(n-1)}\right|\le \delta$; (7)output ${\tau }^{*}={\tau }^{(n)}$ and $K^{*}=K^{(n)}$.