推 future1234:1.2 times as many clock cycles as computer X 05/18 18:21
→ future1234:我把它看成效能來做計算了 , 那就算clock cycle就好哩 05/18 18:22
※ 引述《future1234 (Low)》之銘言：
: ※ 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)
中間那段過程似乎怪怪的
computer X: 10 = cpu clock cycles / 5 GHz
computer Y: 6 = 1.2 * cpu clock cycles / C GHz
=> C = 10
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.45.54.125