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Consider a logical address space of $8$ pages of $1024$ words each, mapped onto a physical memory of $32$ frames. How many bits are there in the physical address and logical address respectively?

  1. $5, 3$
  2. $10, 10$
  3. $15, 13$
  4. $15, 15$
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Why aren't we converting words to bits?
Just because nothing is mentioned, we are not assuming anything?
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2 Answers

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Best answer

Answer 15, 13

Physical Memory 32 frames = 25

Size of word 1024 = 210

Total Physical Address Space =15 bits

Logical Memory 8 pages = 23

Size of word = 210

Total Logical Address Space= 10 +3 = 13 bits

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3 Comments

When questions says 8 pages of 1024 words each the why it considered as size of word in solution ? I am confused in the language only that when we use 1024 words it means if each page has 1024 words but how do we understand that it is the size of word?
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Why aren't we converting words to bits?
Just because nothing is mentioned, we are not assuming anything?
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0
we can't expect same clarity as GATE in ISRO.  10% ~ 20% questions in every ISRO paper would be ambiguous....
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0 votes
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Given, Logical address space= 8 pages → 2^3

words → offset → 1024 -->2^10

Frames = 32 -->2^5

Logical address = 10 bits + 3 bits = 13 bits

Physical address = 10-bits + 5-bits = 15 bits

Answer:

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