Dalla sezione pipeline di istruzioni in Structured Computer Organization di Tanenbaum:
Pipelining allows a trade-off between latency (how long it takes to execute an instruction), and processor bandwidth (how many MIPS the CPU has).
With a cycle time of T nsec, and n stages in the pipeline, the latency is nT nsec because each instruction passes through n stages, each of which takes T nsec.
Since one instruction completes every clock cycle and there are 10^9/T clock cycles/second, the number of instructions executed per second is 10^9/T. To get the number of MIPS, we have to divide the instruction execution rate by 1 million to get (10^9/T)/10^6 = 1000/T MIPS.
Penso che la minore latenza sia, meglio è, e maggiore è la larghezza di banda, meglio è. Poiché i due sono correlati come reciproci, si muovono sempre simultaneamente in entrambe le direzioni, buone o cattive. Allora, come fanno le due forme di compromesso? Grazie.