Re: [理工] 計算機組織與系統

[BF3716 (JS)]

We want to compare the maximum bandwidth for a synchronous and asynchronous bus.
The synchronous bus has a clock cycle time of 50ns,and each bus transmission
takes 1 clock.The asynchronous bus requires 40 ns per handshake.The data portion
of both buses is 32 bits wide.Find the bandwidth for each bus when performing
one-word reads from a 220-ns memory.

[同步]

   1. 傳送位址給記憶體: 50ns
   2. 記憶體存取: 220ns
   3. 傳送資料: 50ns

  傳送one-word 合計需要320ns

               one-word     4bytes
  匯流排頻寬: ---------- = --------- = 12.5 MB/Sec
                320ns        320ns


[非同步]  交握協定具有7個step

   1. step1 : 40ns
   2. step234 : max(3*40ns,220ns) = 220ns
   3. step5 : 40ns
   4. step6 : 40ns
   5. step7 : 40ns

   傳送one-word 合計需要380ns

                 one-word     4bytes
   匯流排頻寬: ---------- = --------- = 10.52 MB/Sec
                  380ns        380ns





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推 sak1346:原PO計組神 快拜
推 newfantasy:推神手~幫我看一下18068^^
推 beautyanzi:step234 : max(3*40ns,220ns) = 220ns ，有寫錯嗎?
→ beautyanzi:3*40ns=?
→ Westind:感謝 又懂一題了
→ BF3716:Step2+Step3+Step4 跟記憶體存取會並行，要取比較久的時間
→ BF3716:可以參考交握協定的時序圖