※ 引述《mqazz1 (無法顯示)》之銘言：
assume that the breakdown of dynamic instructions into various instruction
categories is as follows:
R-type 50%, beq 15%, jump 10%, lw 15%, sw 10%
branch predictor accuracies: always-taken 40%, always not-taken 60%, 2-bit 80%
some branch instructions are much more predictable tan others
if we know that 80% of all executed branch instructions are easy-to-predict
loop-back branches that are always predicted correctly, what is the accuracy
of the 2-bit predictor on the remaining 20% of the branch instructions?
let the total number of branch instructions executed in the program be B
correctly predicted: B * 0.80
correctly predicted non-loop-back: B*0.00
請問黃字是為什麼?
題目的意思是
branch中有80%是簡單一定會猜對的
那請問在2-bit情況下
剩下的20%猜中的機率是多少??
所以依照題目列出等式
2-bit 猜中機率 = 80%*100% + 20% * X
=> 80% = 80% +0.2X
=> X = 0
如果有理解有錯誤,請跟我說一聲,謝謝
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