Exponential backoff timeline for τ with τ_min = 100 ms, τ_max = 1 hour:
Starting from τ = 100 ms, the interval doubles each round when the network is consistent:
\[\tau_0 = 100 \text{ ms}, \quad \tau_1 = 200 \text{ ms}, \quad \tau_2 = 400 \text{ ms}, \quad \tau_3 = 800 \text{ ms}, \quad \ldots\]
How many rounds until τ reaches τ_max = 3,600,000 ms?
\[\tau_n = \tau_{\min} \times 2^n = 100 \times 2^n\]
Solving for n when \(\tau_n = 3{,}600{,}000\):
\[2^n = \frac{3{,}600{,}000}{100} = 36{,}000 \quad \Rightarrow \quad n = \log_2(36{,}000) = 15.14\]
So after 16 doubling steps, τ reaches the 1-hour cap. Each step takes one full interval duration before proceeding to the next – so the total elapsed time to reach τ_max is the sum of all interval durations: 100 ms + 200 ms + 400 ms + … + 2^15 × 100 ms ≈ 2 × 3,276,800 ms ≈ 109 minutes to complete all 16 steps. From that point onward, nodes broadcast at most once per hour. In a stable 1,000-node network, this means only ~1,000 transmissions per hour network-wide (1 per node) = 0.28 transmissions per second – nearly silent compared to the initial update phase with hundreds of transmissions per second.