Algorithm 2: Computing Resource Allocation based on Fairness Perception

Input: $\mathcal{C}$, $\mathcal{X}$, tolerance $\xi$

Output: $ℱ$, $\tau$, ${\tau }_{ave}$

1: for $k\in ℳ$ do

2: for $n\in {\mathcal{N}}_k^{exe}$ do

3:    ${\tau }_{\min }=Q_n$;

4:    ${\tau }_{\max }={\displaystyle \sum _{n\in {\mathcal{N}}_k^{exe}}\left(\frac{{\lambda }_n}{f_k}+Q_n\right)}$;

5:  while $\left|{\tau }_{\max }-{\tau }_{\min }\right|\ge \xi$

6:    ${\tau }_{mid}=\frac{{\tau }_{\max }-{\tau }_{\min }}{2}$;

7:    if ${\displaystyle \sum _{n\in {\mathcal{N}}_k^{exe}}\frac{{\lambda }_n}{{\tau }_{mid}-Q_n}}\ge f_k$

8:       ${\tau }_{\min }={\tau }_{mid}$;

9:    else

10:     ${\tau }_{\max }={\tau }_{mid}$; 11:     end if

12:  end while

13:  ${\tau }_n={\tau }_{\min }$;

14: end for

15:  $ℱ\leftarrow f_{kn}\leftarrow {\tau }_n$, according to Equation (14);

16: end for

17:  ${\tau }_{\max }=\underset{n\in {\mathcal{N}}_{off}}{\max }\left\{{\tau }_n\right\}$;

18:  ${\tau }_{ave}=\frac{{\displaystyle \sum _{n\in {\mathcal{N}}_{off}}{\tau }_n}}{\left|{\mathcal{N}}_{off}\right|}$;