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GO Classes DPP | GATE CS | Operating System | Priority Preemption Scheduling

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5 5 votes

A single-CPU system contains three processes with priorities:

$$X > Y > Z$$

The scheduler always runs the highest-priority runnable process and performs preemption when necessary.

At a particular point:

  • $X$ is BLOCKED waiting for disk I/O.
     
  • $Y$ is RUNNING.
     
  • $Z$ is READY.

The disk I/O requested by $X$ now completes.

What are the states immediately after the scheduler reacts?

  1. $X$: READY, $Y$: RUNNING, $Z$: READY
     
  2. $X$: RUNNING, $Y$: READY, $Z$: READY
     
  3. $X$: RUNNING, $Y$: BLOCKED, $Z$: READY
     
  4. $X$: READY, $Y$: READY, $Z$: RUNNING

1 Answer

4 4 votes

Initially:

  • $X$ is BLOCKED because it is waiting for disk I/O.
     
  • $Y$ is RUNNING because $X$ is unavailable.
     
  • $Z$ is READY.

Now, the disk I/O of $X$ completes.

This causes the transition:

$X:$ BLOCKED $\rightarrow$ READY

At this moment, both $X$ and $Z$ are ready, while $Y$ is currently running.

Given the priority order:

$$X > Y > Z$$

$X$ has a higher priority than the currently running process $Y$. Since the scheduler is preemptive, $X$ immediately preempts $Y$.

Therefore:

  • $X$: READY $\rightarrow$ RUNNING
     
  • $Y$: RUNNING $\rightarrow$ READY
     
  • $Z$: remains READY

So the final states are:

$X:$ RUNNING,$\quad Y:$ READY, $\quad Z:$ READY


Answer : B

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