A virtual memory system has an address space of 8 k words, a memory space of 4 k words, and page and block sizes of 1 k words. The number of page faults using LRU policy, for following page references is

1 0 2 4 6 2 1 5 7 0 0

This question was previously asked in
ESE Electronics 2012 Paper 2: Official Paper
View all UPSC IES Papers >
  1. 5
  2. 7
  3. 9
  4. 10

Answer (Detailed Solution Below)

Option 3 : 9
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
6.3 K Users
20 Questions 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

Given:

Memory space = 4 K words

Page size = Frame size = 1 K words

∴ frames \(= \frac{{4\;K\;words}}{{1K\;words}} = 4\)

F1 Tapesh 24.11.20 Pallavi D 3

Hence, the Total Number of page faults is 9.

Latest UPSC IES Updates

Last updated on Jun 23, 2025

-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.

->  UPSC ESE admit card 2025 for the prelims exam has been released. 

-> The UPSC IES Prelims 2025 will be held on 8th June 2025.

-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.

-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.

More Non Contiguous Questions

More Memory Management Questions

Get Free Access Now
Hot Links: teen patti live teen patti baaz teen patti bodhi teen patti game online teen patti - 3patti cards game downloadable content