Interview Q&A / Operating Systems
Operating Systems interview questions
Processes, threads, scheduling, memory and deadlocks, in plain words.
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- A system has 4 KB pages. Translate logical address 20000 if page 4 is stored in frame 9.
20000 = 4 x 4096 + 3616, so page number 4 and offset 3616. Frame 9 starts at 9 x 4096 = 36864, so the physical address is 36864 + 3616 = 40480.
BeginnerQuick answer1 min read - For burst times 24, 3 and 3 (all arriving at time 0), what is the average waiting time under non-preemptive SJF?
SJF runs the shortest burst first: P2, P3, then P1. Waiting times are P2 0, P3 3, P1 6, so the average is (0 + 3 + 6) / 3 = 3 ms, which is the minimum possible.
BeginnerQuick answer1 min read - Three processes with burst times 24, 3 and 3 arrive together. What is the average waiting time under FCFS?
Run in arrival order P1, P2, P3. Waiting times are 0, 24 and 27, so the average is (0 + 24 + 27) / 3 = 17 ms. Waiting time is the time a process spends in the ready queue before it gets the CPU.
BeginnerQuick answer1 min read - What is the difference between turnaround time, waiting time and response time?
Turnaround time is completion minus arrival. Waiting time is turnaround minus burst, the total time spent in the ready queue. Response time is the time from arrival to the first time the process gets the CPU.
BeginnerQuick answer1 min read