Interview Q&A / Operating Systems

1. 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
2. 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
3. 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
4. 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
5. A system has 32-bit logical addresses and 4 KB pages. How many bits are the page number and offset, and how large is the page table?

   4 KB = 2^12, so the offset is 12 bits and the page number is 32 - 12 = 20 bits. That gives 2^20 entries; at 4 bytes each, the page table is 4 MB per process.

   IntermediateQuick answer1 min read
6. Burst times 24, 3 and 3, all arriving at time 0, with a time quantum of 4 ms. What is the average waiting time under Round Robin?

   P1 runs 0-4, P2 4-7, P3 7-10, then P1 alone finishes at 30. Waiting times are P1 6, P2 4, P3 7, so the average is 17 / 3, about 5.67 ms.

   IntermediateQuick answer1 min read
7. Five processes arrive together with bursts 10, 1, 2, 1, 5 and priorities 3, 1, 4, 5, 2 (lower number is higher priority). What is the average waiting time under non-preemptive priority scheduling?

   The order is P2, P5, P1, P3, P4. Waiting times are 0, 1, 6, 16 and 18, a total of 41, so the average is 41 / 5 = 8.2 ms.

   IntermediateQuick answer1 min read
8. Four processes arrive at 0, 1, 2 and 3 with bursts 8, 4, 9 and 5. What is the average waiting time under preemptive SJF (SRTF)?

   Run order by remaining time: P1 0-1, P2 1-5, P4 5-10, P1 10-17, P3 17-26. Waiting times are P1 9, P2 0, P3 15, P4 2, so the average is 26 / 4 = 6.5 ms.

   IntermediateQuick answer1 min read
9. Given a segment table, how do you translate the logical address (1, 500)?

   Look up segment 1, compare the offset with its limit, and add the offset to its base. If the offset is not below the limit the hardware traps. With limit 400 here, (1, 500) is invalid; (0, 500) maps to 1900.

   IntermediateQuick answer1 min read
10. How does the time quantum affect Round Robin performance?

    A very large quantum makes Round Robin behave like FCFS. A very small one wastes time on context switches. A good quantum is longer than most CPU bursts, often cited as 80% of bursts finishing within one quantum.

    IntermediateQuick answer1 min read
11. If a TLB lookup takes 10 ns, memory access 100 ns and the hit ratio is 90%, what is the effective access time?

    EAT = 0.9 x (10 + 100) + 0.1 x (10 + 100 + 100) = 99 + 21 = 120 ns. A TLB hit needs one memory access; a miss needs two, one for the page table and one for the data.

    IntermediateQuick answer1 min read
12. What is the convoy effect in FCFS scheduling?

    The convoy effect is when many short processes queue behind one long CPU-bound process, so the CPU and I/O devices sit idle in turns and average waiting time rises sharply.

    IntermediateQuick answer1 min read
13. How does the Banker's algorithm decide whether a resource request is safe?

    It pretends to grant the request, then checks whether some order exists in which every process can still finish with the resources left. If such a safe sequence exists, the request is granted; otherwise it waits.

    AdvancedQuick answer2 min read