Table 2 An iterative algorithm to derive b ^(p) and v ^(p)

Compute Λ ₁,Λ _θ using (7) and (49)

Initialize \( \mathbf {b}^{(p)} = \frac {p_{1}}{M}\mathbf {I}_{M} \) satisfying (60)

Repeat

1)

Find v ^(p) (i.e., the solution to problem (57)) for the fixed b ^(p) using

\( v_{l}^{(p)} = \left [\sqrt {\left (\frac {\lambda _{1,l}}{2 \gamma \lambda _{\theta,l}}b_{l}^{(p)}\right)^{2} + \frac {\lambda _{1,l}}{\gamma \lambda _{\theta,l}} b_{l}^{(p)} \mu _{v}} - \frac {\lambda _{1,l}}{2 \gamma \lambda _{\theta,l}}b_{l}^{(p)} - \frac {1}{\gamma \lambda _{\theta,l}} \right ]^{+}, \)

where μ _v >0 satisfies

\( \sum _{l=1}^{L} {\left [\sqrt {\left (\frac {\lambda _{1,l}}{2 \gamma \lambda _{\theta,l}}b_{l}^{(p)}\right)^{2} + \frac {\lambda _{1,l}}{\gamma \lambda _{\theta,l}} b_{l}^{(p)} \mu _{v}} - \frac {\lambda _{1,l}}{2 \gamma \lambda _{\theta,l}}b_{l}^{(p)} - \frac {1}{\gamma \lambda _{\theta,l}} \right ]^{+}} = p_{2}. \)

Compute \( \mathcal {\dot {I}}_{erg}(\mathbf {b}^{(p)},\mathbf {v}^{(p)})^{(old)}. \)

2)

Find b ^(p) (i.e., the solution to problem (59)) with the obtained v ^(p) using

\( b_{l}^{(p)} = \left [\sqrt {\left (\frac {\gamma \lambda _{\theta,l}}{2 \lambda _{1,l}}v_{l}^{(p)}\right)^{2} + \frac {\gamma \lambda _{\theta,l}} { \lambda _{1,l}} v_{l}^{(p)} \mu _{b}} - \frac {\gamma \lambda _{\theta,l}}{2 \lambda _{1,l}}v_{l}^{(p)} - \frac {1}{\lambda _{1,l}} \right ]^{+}, \)

where μ _b >0 meets

\( \sum _{l=1}^{L} {\left [\sqrt {\left (\frac {\gamma \lambda _{\theta,l}}{2 \lambda _{1,l}}v_{l}^{(p)}\right)^{2} + \frac {\gamma \lambda _{\theta,l}} { \lambda _{1,l}} v_{l}^{(p)} \mu _{b}} - \frac {\gamma \lambda _{\theta,l}}{2 \lambda _{1,l}}v_{l}^{(p)} - \frac {1}{\lambda _{1,l}} \right ]^{+}} = p_{1}. \)

Compute \( \mathcal {\dot {I}}_{erg}(\mathbf {b}^{(p)},\mathbf {v}^{(p)})^{(new)}. \)

Until \( \mathcal {\dot {I}}_{erg}(\mathbf {b}^{(p)},\mathbf {v}^{(p)})^{(new)} - \mathcal {\dot {I}}_{erg}(\mathbf {b}^{(p)},\mathbf {v}^{(p)})^{(old)} \leq \epsilon. \) Here, ε>0 denotes

a desired accuracy.