Zero-load latency has four terms: injection, serialization, router pipeline over H hops, and link traversal over H hops. Queueing adds a fifth term under load.
Serialization is packet flits divided by flits per cycle. A 5-flit packet on a 1-flit-per-cycle link costs 5 cycles before pipelining helps the next packet.
Packet 5 flits, link 1 flit per cycle, path H = 6 hops, router 3 cycles, link 1 cycle per hop
serialization = 5 flits / 1 flit per cycle = 5 cycles
router pipeline = 6 hops * 3 cycles = 18 cycles
link traversal = 6 hops * 1 cycle = 6 cycles
zero-load latency = 5 + 18 + 6 = 29 cyclesBisection bandwidth bounds cross-partition traffic before any link saturates. Offered load above the cap queues; it never gets through.
8-link bisection, 4 bytes per cycle per link, 1 GHz clock
bisection throughput = 8 * 4 B * 1e9 = 32 GB/s per direction
offered 40 GB/s across the cut -> achieved at most 32 GB/s; the extra 8 GB/s queuesTail latency is the KPI that exposes the cap. Averages stay flat while the queue builds; p99 climbs first, which is why the triage lesson watches p99 and not the mean.
Little's law turns a round-trip time into a buffer depth. A credit pool must cover bandwidth times round-trip delay or the link idles waiting for returns.
Little's law: occupancy = arrival rate * residence time
link 1 flit per cycle, credit round trip 6 cycles -> depth at least 1*6 = 6 flits
link 1 flit per cycle, credit round trip 12 cycles -> depth at least 12 flitsDepth below the product throttles throughput to depth over round-trip. The credit lesson proves this with a running link: depth 2 against a 6-cycle round trip delivers 20 flits in 60 cycles.
Router pipeline and link traversal double with H. Serialization and injection stay flat because they depend on packet size, not distance.
12 cycles: 24 divided by 2. If the credit round trip is longer than 12 cycles, the pool is too shallow.