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## Responses

There are N processes and they share M resources.

Sum of the requirement of all processes will be " Sum of all S

_{i}" ie \(\sum\limits_{i=1}^n S_i\) . Now If you dont want the deadlock to occur,then you should share M resources such that each process will get 1 less than their maximum requirement and then add 1 resource to any one of the processes. This way you can avoid deadlock. So the expression should be like this : \(\sum\limits_{i=1}^n S_i\)- N+ 1= M meaning we are summing all requirements, then subtracting N,that is 1 resource from each of the processes' requirement, so that all Processes will be waiting. then adding 1, that is allocating 1 resource to any 1 of the processes so that it can complete its execution and leave its resources so that other process can use. This should be equal to M.

Now if you move N to RHS then expression will be like \(\sum\limits_{i=1}^n S_i\)-1 = M+N.

If clearly, now if we ignore 1, then the expression wil be reduced to \(\sum\limits_{i=1}^n S_i\) < M+N. Right.

I hope I made my point and doubts are clear.

every process Pi require Si units

so in worst case we can give Si-1 to a process Pi

so total we have n processes and m resources

so every process we r giving Si-1 i.e (S1-1)+(S2-1)+.....+(Sn-1)<m (in order to avoid deadlock)

(S1+S2+........+Sn)-(1+1+......+n times)<m

sigma S1-n<m

sigma Si<m+n

What's wrong with option D)..? It seems true to me.