推 bennylu:你是對的 10/19 15:26
→ yesa315:3Q 10/19 15:47
推 opcan:請問為何不是1.5*0.25+0.1+0.52+1.25*0.11+2*0.02＝1.1725？ 10/19 22:57
25% loads, 10% stores,11% branches, 2% jumps, and 52% ALU instructions
For the pipelined design, load take 1 clock cycle when there is no load-use
dependence and 2 when there is.
Branches take 1 when predicted correctly and 2 when not.
Jumps always pay 1 full clock cycle of delay, so their average time is
2 clock cycles. Other instructions take 1 clock cycle.
For pipelined execution, assume that half of the load instructions are
immediately followed by an instruction that uses the result and that
one-quarter of the branches are mispredicted.
Ignore any other hazards.
答案給
Pipelined: Average CPI=1+0.5*1+0.25*1+0.02*1=1.77
好像不太對...
我認為是 CPI=1 + 0.25*1/2*1 + 0.11*1/4*1 + 0.02*1 = 1.11
(load use) (branch) (jump)
有高手可以幫我驗證嗎?
謝謝
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※ 編輯: yesa315 來自: 140.127.208.96 (10/19 11:59)
※ 編輯: yesa315 來自: 140.127.208.96 (10/19 12:00)