JANUARY
2017 – PAPER III
OPERATING
SYSTEM QUESTIONS
49. 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:
Always page size = frame size for minimizing the
internal fragmentation.
LOGICAL
ADDRESS CALCULATION
In the above question, 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
In the above question, 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.
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