[理工] [OS ] 96台大電機 15題 (round robin)
台灣大學96年的計算機結構考古題
Process arrival time Burst time
P0 0 7ms
P1 1 3ms
P2 3 3ms
P3 5 5ms
先問FCFS下average response time
答案是21/4
再來問Round robin的 average waiting time (quantum=4)
有五個答案選擇
48/4
30/4
39/4
29/4
21/4
可是我自己算出來是25/4耶..我哪裡計算錯了Orz..
我的算法和我弄的gantt chart
0 4 7 10 14 17 18
-------------------------------------
| P0 | P1 | P2 | P3 | P0 | P3 |
-------------------------------------
(結束時間-進入時間)-burst time
P0=17-0-7=10
P1=7-1-3=3
P2=10-3-3=4
P3=18-5-5=8
total waiting time:10+3+4+8=25
我哪裡算錯了呢
thanks!!
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