book volume-2 ACE ,memory management

Question

Consider a logical address space of eight pages of 1,024 words each, mapped on to a physical memory of

32 frames.

a)How many bits are in logical address?

b)How many bits are in physical address?

Comments

13,15 if memory is word addressable I guess..

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a-13 and b-15?

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how i'm not getting

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If memory is word addressable then to uniquely identify each word in LAS bits rqd should be:

Bits for(No. Of pages) +Bits for(Each page) =3+10=13 as LAS.

similarly for PAS :

Bits for(No. Of Frames) +Bits for(Each Page) =5+10=15 as PAS.

Answer 1

logical address space=total no. of pages *size of each page=$2^{10}*2^{3}=2^{13}$=13 bit to represent logical address.

physical address space=total no. of frame*size of each frame=$2^{10}*2^{5}=2^{15}$=15 bit to represent physical addresss.