Register

UGC NET CSE | January 2017 | Part 3 | Question: 49

• reopened by
7,823 views
1 votes
1 votes

A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively:

  1. 14 and 15
  2. 14 and 29
  3. 15 and 14
  4. 16 and 32
• reopened by
User Avatar

6 Answers

0 votes
0 votes
Option C is correct.

Explanation :

page offset = 9.

Number of pages = 64, so bits required = 6.

Virtual address = 9+6= 15.

Number of frame pages  = 32, so bits reserved for frame = 5.

Physical address = page offset + 5 = 9 + 5 = 14.
User Avatar
0 votes
0 votes
64 pages and 512 B that means that total VAS is 64*512 B that is 32KB

and PAS is 16KB which gives the no of bits as 15 and 14
User Avatar
Page:
Voting & Accepting an answer helps improve the community
Answer:
← PreviousNext →
← Previous in categoryNext in category →

Related questions

4 votes
4 votes
5 answers
5
go_editor asked Mar 24, 2020
1,780 views
UGC NET CSE | January 2017 | Part 3 | Question: 2
Which of the following is incorrect for virtual memory? Large programs can be written More I/O is required More addressable memory available Faster and easy swapping of process
Which of the following is incorrect for virtual memory?Large programs can be writtenMore I/O is requiredMore addressable memory availableFaster and easy swapping of proce...
User Avatar
0 votes
0 votes
4 answers
7
go_editor asked Mar 24, 2020
1,566 views
UGC NET CSE | January 2017 | Part 3 | Question: 51
Match the following for Unix file system : ... $\text{a-iv, b-iii, c-ii, d-i}$ $\text{a-iv, b-iii, c-i, d-ii}$
Match the following for Unix file system : $\begin{array}{clcl} & \textbf{List-I} & {} & \textbf{List-II} \\ \text{a.} & \text{Boot block} & \text{i.} & \text{Informat...
User Avatar
Link Copied!