In a system, there are three types of resources: E, F and G. Four processes…

2018

In a system, there are three types of resources: E, F and G. Four processes P0, P1, P2and P3 execute concurrently. At the outset, the processes have declared their maximum resource requirements using a matrix named Max as given below. For example, Max[P2,F] is the maximum number of instances of F that P2 would require. The number of instances of the resources allocated to the various processes at any given state is given by a matrix named Allocation.

Consider a state of the system with the Allocation matrix as shown below, and in which 3 instances of E and 3 instances of F are the only resources available.

From the perspective of deadlock avoidance, which one of the following is true?

Answer: A. The system is in safe state. — Compute Need = Max - Allocation. P0 needs (E=3, F=3, G=0). P1 needs (E=1, F=0, G=2). P2 needs (E=0, F=3, G=0). P3 needs (E=3, F=4, G=1). Initial available…

  1. A.

    The system is in safe state.

  2. B.

    The system is not in safe state, but would be safe if one more instance of E were available.

  3. C.

    The system is not in safe state, but would be safe if one more instance of F were available.

  4. D.

    The system is not in safe state, but would be safe if one more instance of G were available.

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Correct answer: A

Compute Need = Max - Allocation.

  • P0 needs (E=3, F=3, G=0).

  • P1 needs (E=1, F=0, G=2).

  • P2 needs (E=0, F=3, G=0).

  • P3 needs (E=3, F=4, G=1).

Initial available resources: (E=3, F=3, G=0).

Find a process whose need is <= available, let it finish, add its allocation to available, and repeat to build a safe sequence.

  1. Choose P0 since its need (3,3,0) <= available (3,3,0). After P0 finishes it releases its allocation (1,0,1). New available = (4,3,1).

  2. Choose P2 since its need (0,3,0) <= available (4,3,1). After P2 finishes it releases (1,0,3). New available = (5,3,4).

  3. Choose P1 since its need (1,0,2) <= available (5,3,4). After P1 finishes it releases (1,1,2). New available = (6,4,6).

  4. Finally P3's need (3,4,1) <= available (6,4,6), so P3 can finish.

All processes can finish in this order, therefore the system is in a safe state. One safe sequence is: P0 → P2 → P1 → P3.

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