Consider four processes P, Q, R, S scheduled with Round Robin (time quantum = 4 units), all arriving

Q: What is GATE-CS?

GATE-CS is a competitive examination conducted in India. This question is from the 2022 GATE-CS Operating Systems paper.

Q: How can I prepare for GATE-CS Operating Systems?

To prepare for GATE-CS Operating Systems, practice previous year questions (PYQs), understand concepts thoroughly, and solve mock tests regularly. This 2022 question covers CPU Scheduling topic.

Q: What is the difficulty level of this question?

This GATE-CS 2022 Operating Systems question has a hard difficulty level and is marked as an important question.

Q: Where can I find more GATE-CS previous year questions?

You can find more GATE-CS previous year questions on AspirantMitraa. Browse by subject, topic, year, and difficulty level to practice effectively.

Question

Consider four processes P, Q, R, S scheduled with Round Robin (time quantum = 4 units), all arriving at t=0. There is exactly one context switch S-->Q, one R-->Q, two Q-->R, no S-->P. Which is NOT possible as CPU burst times?

Accepted Answer

Correct Answer: (D) P=3, Q=7, R=7, S=3 is NOT possibleConcept -- Round Robin Scheduling and Context Switch Analysis (from Silberschatz, Galvin, Gagne, Operating System Concepts):Round Robin with quantum=4, all arrive at t=0, order P,Q,R,S. Constraints: exactly one S-->Q, one R-->Q, two Q-->R, no S-->P.For option (D): P=3, Q=7, R=7, S=3. Execution: P runs 3 (finishes at t=3, switch to Q). Q runs 4 (t=3 to 7, switch to R). R runs 4 (t=7 to 11, switch to S). S runs 3 (finishes at t=14, switch to Q). Q runs 3 remaining (finishes at t=17... wait Q has 7 total, used 4, remaining 3. Switch Q-->R? No. Q runs 3 and finishes. Then R runs remaining 3. Context switches: P finishes-->Q (1), Q quantum-->R (1), R quantum-->S (1), S finishes-->Q (1 S-->Q), Q finishes-->R (1 Q-->R)... let's count: Q-->R appears twice requires Q to be preempted twice by its quantum. Q=7 means Q uses 4+3 (one full quantum and one partial), so Q-->R only once. But we need two Q-->R switches. With Q=7, Q runs two slots: 4 units first (Q-->R switch 1) and Q needs to be scheduled again with 3 remaining. After S runs, Q gets 3 units and finishes (no second Q-->R). Only one Q-->R switch, but we need two. This scenario is impossible for these burst times, making (D) not achievable.

Consider four processes P, Q, R, S scheduled with Round Robin (time quantum = 4 units), all arriving at t=0. There is exactly one context switch S-->Q, one R-->Q, two Q-->R, no S-->P. Which is NOT possible as CPU burst times?

Master GATE-CS with Our Test Series

More Questions from CPU Scheduling

Frequently Asked Questions

Consider four processes P, Q, R, S scheduled with Round Robin (time quantum = 4 units), all arriving at t=0. There is exactly one context switch S-->Q, one R-->Q, two Q-->R, no S-->P. Which is NOT possible as CPU burst times?

Master GATE-CS with Our Test Series

More Questions from CPU Scheduling

Frequently Asked Questions