※ 引述《SiriusCloud (古月小楓)》之銘言：
: Assume that a single cycle datapath with the critical path
: of 10 ns can be partitioned into arbitary number of balanced
: stages for pipelining , and there is no dependency between
: instructions .
: If the pipeling will introduce an addtional 1 ns delay to
: each stage. What is the speed up for the 4-stage pipelined
: datapath when compared with the single-cycle one?
: -----------------------------------------------
: answer :
: The instruction time for a single-cycle machine = 10 ns
: The instruction time for pipeline = (10 / 4) + 1 = 3.5 ns
: ^^^^^^^^^^^^why?
: speedup = 10 / 3.5 = 2.86
這部分我也有點困惑 看原文推文還是有所不解
single-cycle的instruction time 是10ns沒錯 (一條指令而言)
pipeline 的instruction time就不大懂是3.5ns了 3.5不是才一個stage的cycle time嗎
若是不想這麼多 單純代SpeedUp公式 = S * N * T / [(S-1)+N ] * T
= 10 ns * N / [ 3 + N ] * 3.5ns
我假設N=1(一條指令)
那SpeedUp也是 10/ "4" * 3.5 不是嗎?
一時迷失在這裡 請各位指教~
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