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We consider a distributed IoT network, where each node wants to minimize its age of information and there is a cost to make any transmission. A collision model is considered, where any transmission is successful from a node to a common monitor if no other node transmits in the same slot. There is no explicit communication/coordination between any t...
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Context 1
... that P oA ≥ 1, and a value close to 1 indicates that the algorithm is close to optimal. Figure 6 plots the P oA of learning algorithm (3) for different values of N . Initially when N increases, P oA increases as well. ...
Context 2
... minimum (i.e., close to 0). Since U sys (P) is maximum at P = P OP T (in (15), we assumed P OP T to be the optimal transmission probability vector which maximizes U sys (·)), therefore for P LA (i.e., P N E ) to be close to P OP T (and P oA ≈ 1 according to (15)), (30) must be close to 0 for each node j. However, when N is small (less than 4 in Fig. 6), then with addition of every new node in the system, number of positive terms in the summation on RHS of (30) increases, thereby taking the value of (30) far from 0 for each node j. Therefore, P oA increases. But because (1−p N E k ) < 1 (∵ ∀k, p N E k ∈ (0, 1)), therefore if N increases, then for each ℓ, (p N E ℓ ) 2 e −ρ ℓ2 k =ℓ,j ...
Context 3
... that P oA ≥ 1, and a value close to 1 indicates that the algorithm is close to optimal. Figure 6 plots the P oA of learning algorithm (3) for different values of N . Initially when N increases, P oA increases as well. ...
Context 4
... minimum (i.e., close to 0). Since U sys (P) is maximum at P = P OP T (in (15), we assumed P OP T to be the optimal transmission probability vector which maximizes U sys (·)), therefore for P LA (i.e., P N E ) to be close to P OP T (and P oA ≈ 1 according to (15)), (30) must be close to 0 for each node j. However, when N is small (less than 4 in Fig. 6), then with addition of every new node in the system, number of positive terms in the summation on RHS of (30) increases, thereby taking the value of (30) far from 0 for each node j. Therefore, P oA increases. But because (1−p N E k ) < 1 (∵ ∀k, p N E k ∈ (0, 1)), therefore if N increases, then for each ℓ, (p N E ℓ ) 2 e −ρ ℓ2 k =ℓ,j ...
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