[理工] 清大104 計算機系統

[evor (托托)]

想請問一下關於104清大計算機系統題目
2.

Consider a variant of the RR scheduling algorithm where the entries in the
ready queue are pointers to the PCBs.

a. What would be the effect of putting two pointers to the same process in
the ready queue?
b. What would be the major advantages and disadvantages of this scheme?
c. How would you modify the basic RR algorithm to achieve the same effect
without the duplicate pointers?

Answer:

a. In effect, that process will have increased its priority since by getting
time more often it is receiving preferential treatment.

b. The advantage is that more important jobs could be given more time, in
other words, higher priority in treatment. The consequence, of course, is
that shorter jobs will suffer.

c. Allot a longer amount of time to processes deserving higher priority.  In
other words, have two or more quantums possible in the Round-Robin scheme.

有找到英文解答，但還是不懂他的意思。

請各位高手解惑!! 謝謝

也預祝大家大家考試順利!

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→ evor: 另外想對第9題答案 a. x:3EA00000 y:BDC00000  b.BCF00000
→ evor: 第十題 不會 請人幫忙回答 謝謝!
推 amge1524: 第10題應該是(c) 正 - 負 變 負 , 負 - 正 變 正
→ lemonsheep: 我覺得應該是 RR根據ready queue中的entries
→ lemonsheep: 來決定接下來輪到哪process如果很多個entries都只向同
→ lemonsheep: 一process
→ lemonsheep: 那該process分配到CPU的時間就比其他process多
→ lemonsheep: 僻如其他process都只有被1entry指到
→ lemonsheep: 那被兩個entry指到的process就會在同一輪中得到CPU兩
→ lemonsheep: 所以優點就是比較重要的可以讓多個entry指向他
→ lemonsheep: 讓重要的process在同一輪中執行比較多次
→ evor: 謝謝回答~~
推 positMIT: push
→ evor: 想再問一下 第十題 ab選項 是都不會發生overflow嗎?