Interview Q&A / Operating Systems

1. 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
2. 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
3. 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
4. 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
5. 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
6. 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
7. 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
8. 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