推 fzrmitsul:謝謝f大 05/18 07:32
推 dreamroyc:有情幫推 05/25 16:37
※ 引述《fzrmitsul (我的妹妹很可愛)》之銘言：
: 10.A program runs in 10 seconds on computer X, which has a 5 GHZ clock. You
: are trying by increasing the clock rate to build a computer Y, that will run
: the program in 6 seconds. However, the increase will cause the computer Y to
: require 1.2 times as many clock cycles as computer X. What clock rate should
: you design?
: (A) 10 GHZ (B) 9 GHZ (C) 8 GHZ (D) 7 GHZ
: 答案是A
: 請問該怎麼算呢??謝謝
cpu execution time = cpu clock cycles x cycle time
= cpu clock cycles / clock rate
令 Y 的clock rate為 C x 10^9
computer X: 10 = cpu clock cycles / 5 x 10^9
computer Y: 6 = cpu clock cycles / C x 10^9
又 Performance Y / Performance X = 1.2
=> ( 1 / cpu execution time Y ) / (1 / cpu execution time X) = 1.2
=> cpu execution time Y / cpu execution time X = 1.2
所以...
6 x C x 10^9 = 1.2 x (10 x 5 x 10^9)
=> 6 x C = 60
=> C = 10 (GHZ)
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