A computer system has 7 tape drives. There are ‘n’ processes competing for…

Concept

For a single type of resource with R identical units, suppose each of n processes may claim at most M units. The system is guaranteed deadlock-free as long as R ≥ n(M − 1) + 1. Reason: the tightest situation to break is when every process is simultaneously blocked holding (M − 1) units — one short of its maximum. Together they then hold n(M − 1) units; if the total R exceeds this, at least one unit is still free, so some process can obtain its last unit, finish, and release — and no deadlock can occur.

Application

1. Here R = 7 tape drives, and each process needs at most M = 2 drives, so M − 1 = 1.

2. Substitute into the safe condition: 7 ≥ n·(1) + 1, i.e. 7 ≥ n + 1.

3. Solve for n: n ≤ 6. The largest process count that keeps the system guaranteed deadlock-free is n = 6.

Cross-check

Test the boundary. With n = 6, the worst case gives each process one drive (6 in use) and leaves 1 drive free; that spare lets some process take its second drive, finish, and release both — the rest then unblock, so no deadlock. With n = 7, all seven drives could be held one-per-process with none free and every process waiting for a second drive — a genuine deadlock. Hence 6 is exactly the safe boundary, and the maximum value of n is 6.