Discussion
An operating system contains 3 user processes each requiring 2 unit of resources 'R'. The minimum number of units of 'r' such that no deadlocks will ever arise is :
Answer: C
The minimum is 4 units.
With n processes each needing at most k units, deadlock is impossible if there are at least \(n(k - 1) + 1\) units. In the worst case each process holds k - 1 units and waits for one more; one extra unit lets one process finish and release its resources.
\[ 3(2 - 1) + 1 = 4 \]
With 3 units, each process can hold one and wait forever for a second, which is a deadlock. Five or six units also avoid deadlock but are not the minimum.
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