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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
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3 Comments

It should be option C
1
1
15 and 14
1
1
  1. 15 and 14 repectively
1
1

6 Answers

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Best answer

we know, Number of page =Virtual adress space / Page size 

so virtual adrress space=Number of page*page size

                                  =26*29=215 

so we need 15 bit for VA

and Physical dress=Page size*frames =29*25=214

so 14 bit required for PA

C is correct answer Here

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3 votes
3 votes

Ans will be (3) 15 and 14

Total pages =64 and bytes pe page =512 so total logical memory = 64x512=215  hence 15 bits are needed for logical memory

Total frames =32 and bytes per frame =512  so total physical memory =32x512=214  hence 14 bits are needed for physical memory

2 votes
2 votes

answer will be (c) because 

LAS: page no.(no.of bits required to represent pages) + page offset                                                           

page no.=log2(no.of pages) bits

              =6 bits

page offset=no. of bits required to represent page size

               =log2(512) bits

              =9 bits

 so for LAS 15 bits required

 PAS=  no. of bits required for represents frames+ page offset (page size=frame size)

              = log2(32) bits + 9 bits

               =14 bits

         

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0 votes

logical address : #pages (64->6 bit) + page size(512 ->9 bit)= 15 bit

physical address: #frame (32->5bit )+ page size(9bit)=14bit

option c is correct!

Answer:

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