[理工] 計組 99清大資工

看板Grad-ProbAsk作者 (古月小楓)時間14年前 (2011/11/03 12:04), 編輯推噓0(009)
留言9則, 3人參與, 最新討論串1/3 (看更多)
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

11/03 14:37, , 1F
題目寫balanced stages,然後是4-stage pipelined
11/03 14:37, 1F

11/03 14:38, , 2F
所以一個stage的時間是(10 / 4)
11/03 14:38, 2F

11/03 14:39, , 3F
第二段寫addtional 1ns delay to each stage,所以 + 1
11/03 14:39, 3F

11/03 16:53, , 4F
可是切4個stage 為什麼不是+4?
11/03 16:53, 4F

11/03 16:53, , 5F
他不是說each 嗎?
11/03 16:53, 5F

11/03 22:40, , 6F
我的想法是,若以instruction數為n,而n很大的話
11/03 22:40, 6F

11/03 22:41, , 7F
single-cycle平均一個instruction的時間為(n*10)/n=10
11/03 22:41, 7F

11/03 22:47, , 8F
pipeline平均instruction的時間為(n+3)*(2.5+1)/n=2.5+1
11/03 22:47, 8F

11/29 16:00, , 9F
就算每個stage+1ns 算pipeline還是只會+1阿@@
11/29 16:00, 9F
文章代碼(AID): #1EiXBE7x (Grad-ProbAsk)
文章代碼(AID): #1EiXBE7x (Grad-ProbAsk)