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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 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 219.223.252.31
vendor47:題目寫balanced stages,然後是4-stage pipelined 11/03 14:37
vendor47:所以一個stage的時間是(10 / 4) 11/03 14:38
vendor47:第二段寫addtional 1ns delay to each stage,所以 + 1 11/03 14:39
SiriusCloud:可是切4個stage 為什麼不是+4? 11/03 16:53
SiriusCloud:他不是說each 嗎? 11/03 16:53
vendor47:我的想法是,若以instruction數為n,而n很大的話 11/03 22:40
vendor47:single-cycle平均一個instruction的時間為(n*10)/n=10 11/03 22:41
vendor47:pipeline平均instruction的時間為(n+3)*(2.5+1)/n=2.5+1 11/03 22:47
robert527152:就算每個stage+1ns 算pipeline還是只會+1阿@@ 11/29 16:00