看板 Grad-ProbAsk 關於我們 聯絡資訊
assume that main memory accesses take 70ns and that memory accesses are 30% of all instructions. the following table shows data for L1 caches attached to each of two processors, P1 and P2 L1 size L1 miss rate L1 hit time P1 1 KB 11.4% 0.62ns P2 2 KB 8.0% 0.66ns (2) what is the AMAT for P1 0.62 + (11.4% * 70) = 8.6ns 然後還有另一個數據 13.87 cycles 請問這個是怎麼來的? (3) 這題是P2比較快 為什麼? 謝謝 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.228.27.152
compulsory:(2)8.6ns/0.62ns 07/15 21:42
compulsory:(3)P1 13.87*1.3=18.03 07/15 21:42
compulsory: 0.66+(8%*70)=6.26 07/15 21:46
compulsory: P2 6.26ns/0.66ns=9.48 9.48*1.3=12.32 07/15 21:49
compulsory:P1 CPI=18.03 P2 CPI=12.32 在乘clk cycle time 比較 07/15 21:53
lexa:請問樓上1.3哪來的? 07/17 02:04
compulsory:指令100%+30%記憶體指令 07/17 22:26