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Table 2 Low-complexity suboptimal two-step algorithm
| Two-step algorithm | |
|---|---|
| 1. | Initialization: S = ∅, K = ∅, L = ∅ |
| 2. | Clustering Procedure: |
| 3. | For each resource pattern with only one RB rb ∈ {1,2,…,K rb } |
| 4. | Find the CUE sub-cluster and DUE sub-cluster \( \left\{\overline{k},\overline{l}\right\}= \arg \underset{\left\{k,l\right\}}{ \max }{t}_{k,l,rb} \) |
| 5. | \( K=K{\displaystyle \cup}\kern0.5em \overline{k} \) |
| 6. | \( L=L\;{\displaystyle \cup}\kern0.5em \overline{l} \) |
| 7. | End |
| 8. | Searching Space Reduction Procedure: |
| 9. | For each \( \widehat{k}\in K;\widehat{l}\in L \) sets |
| 10. | \( {s}_0=\left\{i\Big|\left(\widehat{k}-1\right)*{N}_{all,d}{N}_{all,rb}+\left(\widehat{l}-1\right)*{N}_{all,rb}+1\le i\le \left(\widehat{k}-1\right)*{N}_{all,d}{N}_{all,rb}+\widehat{l}*{N}_{all,rb}\right\} \) |
| 12. | S = S ∪ s 0 |
| 13. | End |
| 14. | Joint Solution Procedure: |
| 15. | Obtain the reduced-dimension optimization problem |
| 16. | \( {x}_s=\underset{{\mathbf{x}}_S}{ \min}\left\{-{\mathbf{t}}_{\boldsymbol{S}}^{\boldsymbol{T}}{\mathbf{x}}_S\right\} \), s.t. \( {\mathbf{R}}_{:,S}{\mathbf{x}}_S={\mathbf{1}}_{\left({N}_c{N}_d{N}_{rb}+{K}_c+{K}_d\right)\times 1} \) |
| 17. | Find its solution based on the BBS algorithm |
| 18. | Return \( {x}_s \) |
| 19. | End |
