推 mqazz1:謝謝 08/07 14:23
※ 引述《mqazz1 (無法顯示)》之銘言：
: consider two different implementations, P1 and P2, of the same instruction set
: There are five classes of instructions(A,B,C,D,E) in the instruction set.
: P1 has a clock rate of 4GHz. P2 has a clock rate of 6GHz. The average number
: of cycles for each instruction class for P1 and P2 is as follows:
: Class CPI on P1 CPI on P2
: A 1 2
: B 2 2
: C 3 2
: D 4 4
: E 3 4
: (1)assume that peak performance is defined as the fastest rate that a computer
: can execute any instruction sequence. What are the peak performances of P1
: and P2 expressed in instructions per second?
: 張凡上冊247頁 請問這題要怎麼解?
: 謝謝
看起來是要算MIPS這類的東西
題目說道peak performance
所以要這樣算
IC IC f
MIPS= ───── = ────── = ──────
TIME*10^6 IC*CPI CPI*10^6
─── *10^6
f
至於CPI，分別選P1跟P2 CPI最少的指令
P1選CPI=1的A；P2選CPI=2
4G
peak per. of P1 = ────── = 4000 MIPS
1*10^6
6G
peak per. of P1 = ────── = 3000 MIPS
2*10^6
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.171.208.239