[理工] [OS] - deadlock free

[nypgand1 (祈附‧征前御祭)]

Consider a computer system consisting of 4 resources of the type A
and 4 resources of the type B that are shared by 3 processes.
Type A and type B resources are boundled in the sense that
a process always gets an A if it gets a B, and vice versa.
Each of these processes requires no more than 2 resources of type A
and 2 resources of type B.

Is this system deadlock-free?
Show proof for your answer.


這類題目我都是用鴿籠原理去想
如果是問答題好像比較難下筆

解答寫說:
ΣMax[i] = 6
ΣMax[i] = ΣNeed[i] + ΣAllocation < 4 + 3

想問最後的不等式怎麼得到?

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推 christianSK:ΣMax[i] = ΣNeed[i] + ΣAllocation
→ christianSK:是假設所有資源被配置出去的情況下
→ christianSK:此時 ΣAllocation = m => ΣNeed[i] < n
→ christianSK:由鴿籠可以知道有人已經不需要資源 可順利完工
→ nypgand1:原來是這樣 我是直接想 "每人都不足1個時資源還有剩"
→ nypgand1:q是每個process最多要求   所以 n(q-1) < m
→ nypgand1:也就是說解答給的是 nq < m + n 囉?
推 christianSK:可以這樣想吧