| SU1 | SU2 | DSU1(t) | DSU2(t) | OC1(t) | OC2(t) | OC3(t) |
---|
t=1 | C1 | C1 | 1 | 1 | 2 | 0 | 0 |
t=2 | C2 | C3 | 0 | 0 | 0 | 0 | 0 |
t=3 | C2 | C2 | 1 | 1 | 0 | 2 | 0 |
t=4 | C3 | C2 | 0 | 0 | 0 | 0 | 0 |
t=5 | C3 | C3 | 1 | 1 | 0 | 0 | 2 |
t=6 | C1 | C1 | 1 | 1 | 2 | 0 | 0 |
- In this case, the total number of collisions for the two users is \(D(n) = \sum \limits _{k=1}^{U} D_{k}(n) = D_{SU1}(n) + D_{SU2}(n)\). The number of collisions in all channels produced by the users is \(O_{C}(n) = \sum \limits _{i=1}^{C} O_{i}(n)= O_{C1}(n)+ O_{C2}(n)+O_{C3}(n)\), while the number of collisions in the best channels, i.e., C1 and C2, is \(O_{U}(n)= \sum \limits _{k=1}^{U} O_{k}(n) = O_{C1}(n)+O_{C2}(n)\)