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Bankers algorithm

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Consider the following system snapshot

Which of the following is a safe sequence?

  • P1, P3, P0, P4, P2
  • P3, P1, P4, P0, P2
  • P3, P4, P1, P0, P2
  • All of these
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1 Answer

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Answer is option (a). A safe sequence will not start with P3 here because, P3 needs 1 resource of C, but for C, none is available at present. If P1 completes and releases all its currently held resources, then P3 can proceed. So here P1, P3, P0, P4, P2 is a safe sequence.

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