System Software and Operating System - Deadlocks

2. A state is safe if the system can allocate resources to each process (up to its maximum) in some order and still avoid deadlock. Then

  • Option : D
  • Explanation : A system is in a safe state only if there exists a safe sequence of processes P1, P2, P3, Pn where:
    For each Pi, the resources that Pi can still request can be satisfied by the currently available resources plus the resources help by all Pj, j<i.
    If a system is in safe state, there is no deadlock.
    If the system is deadlocked, it is in an unsafe state.
    If a system is in unsafe state, there is a possibility for a deadlock.
    Avoidance: making sure the system will not enter an unsafe state.
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3. A computer system has 6 tape drives, with 'n' processes competing for them. Each process may need 3 tape drives. The maximum value of 'n' for which the system is guaranteed to be deadlock free is

  • Option : C
  • Explanation : Two processes can never lead to deadlock as the peak time demand of 6 (3 + 3) tape drives can be satisfied. But 3 processes can lead to a deadlock if each hold 2 drives and then demand one more.
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4. 'm' processes share 'n' resources of the same type. The maximum need of each process doesn't exceed 'n' and the sum all the their maximum needs is always less than m + n. In this set up

  • Option : A
  • Explanation : Using Banker's algorithm, one can show that one process has to acquire all its needed resources. This process, after completing its task, will release all its resources, thereby avoiding any possible deadlock.
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5. Consider a system having 'm' resources of the same type. These resources are shared by 3 processes A, B, C, which have peak time demands of 3, 4, 6 respectively. The minimum value of 'm' that ensures that deadlock will never occur is

  • Option : A
  • Explanation : Having 11 resources ensures that atleast 1 process will have no pending request. This process after using will release the resources and so deadlock can never occur.
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