Simple Example

• Component performs an operation in 100ns

• Simple bandwidth: 10 Mops

• Internally pipeline depth 10 => bandwidth 100 Mops

  • Rate determined by slowest stage of pipeline, not overall latency

• Delivered bandwidth on application depends on initiation frequency

• Suppose application performs 100 M operations. What is cost?

  • op count * op latency gives 10 sec (upper bound)
  • op count / peak op rate gives 1 sec (lower bound)
    • assumes full overlap of latency with useful work, so just issue cost
  • if application can do 50 ns of useful work before depending on result of op, cost to application is the other 50ns of latency

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