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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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2 Answers

Best answer
13 votes
13 votes

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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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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