When we have a shared transport and 5 channels on top of it, then every channel will run the discovery protocol. If it is the first node in a cluster, this will take <5 * Discovery.timeout> ms.

Now, if the 5 channels didn't just share the transport, but also the discovery protocol, then only the first channel to start would have to wait for Discovery.timeout ms. It would then cache the results of that discovery and, when view changes are received, replace the contents of the cache with view information.

The remaining 4 channels would then not even need to run the discovery phase, but the discovery protocol would simply use the current view to return the coordinator. This means that instead of 5 * timeout, we have 1 * timeout !