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(96 中央資工) There is an unpipelined processor that has a 1 ns clock cycle and that uses 4 cycles, for ALU and branch operations and 5 cycles for memory operations. Assume that the relative frequencies of these operations are 40%, 20% and 40% respectively. Suppose that due to clock skew and setup, pipelining the processor adds 0.2 ns of overhead to the clock. Ignoring any latency impact how much speedup in the instruction rate will we gain from a pipeline implementation? 我知道沒有 pipeline的 avg time 是 4.4 ns 我想問的是 : 給了指令混合比例之後, 在pipeline中怎麼算avg time呢? 另外一個問題 (關於harzard) 為了要早些判定 branch 是否會發生 所以我們把branch的判定拉到ID stage 那如果說一個 lw 指令跟其後的 beg 存在 data harzard ex: lw $t1 0($t2) beg $t1 $t3 somewhere 這樣是否一定要 stall 兩個 clock 來處理 data harzard 而無法用forwarding的方式處理 ? 謝謝! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.114.32.62
christianSK:好像問了一個笨問題... 沒有stall的話 10/07 14:10
christianSK:每個指令都用一個clock cycle對吧?! 10/07 14:11