A system contains three programs and each requires three tape units for its…
2014
A system contains three programs and each requires three tape units for its operation. The minimum number of tape units which the system must have such that deadlocks never arise is _________.
Attempted by 144 students.
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Correct answer: 7
Key idea: If there are n programs and each may need up to m tape units, deadlock can be avoided only if the total number of tape units R satisfies R ≥ n·(m−1)+1.
Reasoning for this problem:
If there were only 6 tape units, each of the three programs could hold 2 units (since each needs 3) and then wait for one more unit. That would be 3 × 2 = 6 units held and every program waiting — a deadlock.
With 7 tape units, at least one program can obtain all 3 units it needs, finish, and release its units. Those released units can then be used by others, so a circular wait (deadlock) cannot persist.
Applying the formula with n = 3 and m = 3 gives 3·(3−1)+1 = 3·2+1 = 7.
Answer: 7 tape units.
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