Algorithm 3: USBA-MC Algorithm

1:Input:$B_0,B_0^D,B_0^U$ 2:$(t_{n,C}^U,t_{n,C}^D,t_{dn,C},E_{n,C}^{\mathrm{M}})\leftarrow \mathbf{MC}(s^{\mathrm{L}},s^{\mathrm{G}})$ 3:${\mathcal{S}}^0\leftarrow$GetS($B_0^U,B_0^D,B_0$) 4:for$\tau \in \{1,\dots ,T\}$do 5:   $(B_{\tau }^U,B_{\tau }^D,B_{\tau })\leftarrow$GetB(${\mathcal{S}}^{\tau -1}$) 6:   $(t_{n,C}^U,t_{n,C}^D,t_{dn,C},E_{n,C}^{com})\leftarrow \mathbf{MC}(s^{\mathrm{L}},s^{\mathrm{G}})$ 7:   ${\mathcal{S}}^{\tau }\leftarrow$GetS($B_{\tau }^U,B_{\tau }^D,B_{\tau }$) 8:   if ${\mathcal{S}}^{\tau }=={\mathcal{S}}^{\tau -1}$ and $(B_{\tau }^U,B_{\tau }^D,B_{\tau })==(B_{\tau -1}^U,B_{\tau -1}^D,B_{\tau -1})$ then 9:     break 10:   end if 11:end for