[理工] [記組]CPI

[discipile (孤獨的風)]

Consider three different processors P1,P2,P3 excuting the same instruction set
with the clock rates and CPIs given in the following table.

Processor      Clock rate         CPI
   P1              2 GHz          1.5
   P2            1.5 GHz          1.0
   P3              3 GHz          2.5
Q1.If the processors each excute a program in 10 seconds,find the number of
cycles and the number of instructions.
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我只知道
P1:1.5=C/I,但兩個未知數根本解不出來

Q2.We are trying to reduce the time by 30% but this leads to an increase of 20%
in the CPI.What clock rate should we have to get this time reduction?
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題目我就有點看不懂了reduce the time by 30%是指說假設我現在要花10秒,那我就必須變
成7秒,還是3秒?
另一個困擾,對我而言
0.7=(I/Program)*0.8*(seconds/cycles)
有未知數不知來自哪?

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推 lineageorc:Q1先求出IC 然後IC*CPI 就是# of Cycles 應該是這樣吧
推 lineageorc:減少30% 故是 exe. time =7s CPInew=1.2*CPIoriginal
→ discipile:但是IC不知道怎麼算,我知道的就是CPI=clock cycles/IC
推 lineageorc:exe. time = (IC*CPI)/clock rate
→ lineageorc:exe.time , CPI,Clock rate 都有了
推 maiklover:IC * CPI * cycle time = 10sec
→ maiklover:cycle time用clock rate倒數求 CPI給你了 IC就出來了
→ maiklover:第二題用同樣的式子 時間改7sec CPI變1.2倍
→ maiklover:IC第一題有算出來 所以未知數變成Clock rate