Exact(1)
Under the linear deterministic model it is straightforward to see that by using interference avoidance or TDMA, it takes three time slots to reconstruct the induced subgraphs of Fig. 4b d, which results in a normalized sum rate of (alpha =frac {1}{3}).
Similar(59)
If the per layer conflict graphs are not fully connected and the network does not fall in category (a), then a normalized sum rate of α=1/2 is easily achievable.
Given the assumption of 1-local view, in order to achieve a normalized sum rate of α, each source should transmit at a rate greater than or equal to α n−τ (in linear deterministic model) for some constant τ.
This is due to the fact that from each source's point of view, it is possible that the other S-D pairs have capacity 0. Therefore, in order to achieve a normalized sum rate of α, it should transmit at a rate of at least α n−τ.
Fig. 18 A path exists from source S 1 to destination D 2; all solid edges have capacity n (for linear deterministic model) or h (for the Gaussian model) and the rest have capacity 0. In order to guarantee a normalized sum rate of α with 1-local view, each source has to transmit at a rate greater than or equal to α n−τ (for the linear deterministic model) or α log(1+|h|2)−τ (for the Gaussian model).
For a K-user two-layer network (phantom {dot {i}!}mathcal {G} = (mathcal {V},mathcal {E},{ w_{ij} }_{ i,j) in mathcal {E}})) with 1-local view if there exists a valid assignment of transmit and receive matrices, then all induced subgraphs can be reconstructed in T time slots and a normalized sum rate of (alpha = frac {1}{T}) is achievable.
In Section 3, we showed that a normalized sum rate of α=1/2 is achievable.
For networks in category (b), a simple TDMA achieves a normalized sum rate of 1/3.
Hence, a normalized sum rate of (alpha = left (frac {1}{2} right)^{2}) is achievable for the network in Fig. 16.
This would immediately imply that a normalized sum rate of (frac {1}{2}) is achievable1.
Hence, a normalized sum rate of (alpha =frac {1}{2}) is achievable.
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