Updated: 29-September-2017

CBSE NET JAN 2017 PAPER III

OPERATING SYSTEM QUESTIONS

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

Ans:- 3

Explanation:

Page size = frame size for minimizing the internal fragmentation.

Page size = frame size for minimizing the internal fragmentation.

LOGICAL ADDRESS CALCULATION:

Number of bits for logical address = Number of bits to represent pages + Number of bits to represent bytes per page size.

Here, number of pages = 64, and 512 bytes page size.

64 = 26 . So, number of bits to represent pages = 6.

512=29. So, number of bits to represent bytes per page size = 9.

So, the number of bits for logical address = 6 + 9 = 15

PHYSICAL ADDRESS CALCULATION:

Number of bits for physical address = Number of bits to represent frames+Number of bits to represent bytes per frame size.

Here, number of page frames = 32, and frame size is the same as page size which is 512 bytes.

32 = 25. So, number of bits to represent frames = 5.

512=29. So, number of bits to represent bytes per frame size = 9.

So, the number of bits for physical address = 5 + 9 = 14.

Therefore, the number of bits required in logical and physical address are 15 and 14 respectively.

So, the correct answer is 3.

32 = 25. So, number of bits to represent frames = 5.

512=29. So, number of bits to represent bytes per frame size = 9.

So, the number of bits for physical address = 5 + 9 = 14.

Therefore, the number of bits required in logical and physical address are 15 and 14 respectively.

So, the correct answer is 3.