Safe State - aryanjoshi0823/5143-Operating-System GitHub Wiki

In operating systems, the concept of safe and unsafe states is used to manage resource allocation and avoid deadlocks.

Figure - Safe, unsafe, and deadlocked state spaces.

Safe State

A state is considered safe if:

  • The system can allocate all resources requested by every process (up to their maximum needs) without entering a deadlock.
  • A safe sequence of processes exists, defined as: { P0, P1, P2, ... PN}

    In this sequence:

    • Each process 𝑃𝑖 can complete using the resources currently allocated to it and those released by processes 𝑃𝑗 , where 𝑗 < 𝑖.
    • Once 𝑃𝑖 completes, it releases its resources, making them available for subsequent processes in the sequence.

Unsafe State

  • A state is unsafe if no safe sequence exists.
  • Important:
    • An unsafe state may lead to deadlock, but it does not guarantee that a deadlock will occur.
    • Conversely, all safe states are guaranteed to be deadlock-free.

Example: Resource Allocation for Tape Drives

Consider a system with 12 tape drives and the following allocations:

Process Maximum Needs Current Allocation
P0 10 5
P1 4 2
P2 9 2

To determine if the system is in a safe state, follow these steps:

  1. Check the available resources:
  • Total resources = 12
  • Current allocation = 5+2+2=9
  • Available resources = 12−9=3.
  1. Evaluate possible sequences:
  • Process P1: Needs 4−2=2, which can be satisfied with the available 3. After P1 finishes, it releases 2 more, making the total available = 3+2=5.
  • Process P0: Needs 10−5=5, which can now be satisfied. After P0 finishes, it releases 5, making the total available = 5+5=10.
  • Process P2: Needs 9−2=7, which can now be satisfied with the available 10.
  • Safe Sequence: P1,P0,P2.

Effect of Granting P2 One More Tape Drive

Process Maximum Needs Current Allocation
P0 10 5
P1 4 2
P2 9 3

Updated Calculation:

  1. Available resources = 12 − ( 5 + 2 + 3 ) = 2

  2. Process P1: Needs 4−2=2, which can be satisfied. After P1 finishes, it releases 2, making the total available = 2+2=4.

  3. Process P0: Needs 10−5=5, which cannot be satisfied with the available 4.

No safe sequence exists; the system is now in an unsafe state.

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