No. of bits in physical and logical address space

Question

Is it possible to have different number of pages and frames or they need to be equal in number?

Comments

Don't we need offset bits (page size) to determine the number of address bits?

Is it possible to have different number of pages and frames or they need to be equal in number?

They can be different. Different pages can be mapped to same frames. 

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It is given that each page contains 1024 words.

Answer 1

Always page size=frame size for minimizing the internal fragmentation.

Here 8 pages are there. Number of entries in page table = 8.  
CPU generates logical address which will be divided into 2 halves - to identify page and to identify how much deeper I have to go in that page. So 3 (8=2^3) bits for uniquely identifying page and 10 bit for uniquely(1024=2^10) identifying every word. Similarly the offset that is page size will be equal to frame size so here again 10 bits for uniquely identifying each word while here 32 frames are available which can be uniquely identified by 5 bits hence answer is 13 and 15. Here they are saying logical memory is less than physical memory. Generally main memory is less in size then physical memory.

Comments

Yes, thats restricted by main memory address lines. But that is not affecting this question rt?

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how to find physical address and logical address when size of ram and physical address space is given?

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How to find physical and logical address size, No. of frames and Frame size when physical address space and size of ram and page size is given??

Answer 2

logical address=13 bit(3 bit for page 10 bit for offset)

physical address=15 bit( 5 bit for frame 10 bit for offset)

Answer 3

Here the number of pages are $8 = 2^3$, so you need 3 bits to represent page no field, $1024 = 2^(10)$ words per page hence you need  10 bits to represent offset. Now the number of offset bits in both logical and physical address will be the same.

Physical memory is divided into $32 = 2^5$ frames hence we need 5 bits to represent frame no.

So the logical address would be 3 + 10 = 13 bits long

and the physical address would be 5 + 10 = 15 bits long.