Q:

How to solve calculations about  logical address space and physical address space?

How to calculate number of bits in logical address  and physical address  when 
 logical address space of 8 pages of 1024 word each, mapped to physical memory of 32 frames?

A:

15 is the correct answer
i think this is correct way
size of logical address space is No. of pages * Page size = 8 * 1024 = 2^3 * 2 ^10 = 2^13 No. of bits for logical address is 13
Size of Physical address space is 2^5 * 2^10 = 2^15 No. of bits for physical address is 15

A:

There are 8 pages in logical address space so, 2^3 = 8 then page size of 3-bits
We have 1024 words(1 word = 2-bytes) then, 1024 * 2 = 2048 bytes
which we can say that 2^11 = 2048 then so there are 11 + 3 = 14-bits are the total number of bits in a logical address.
Now coming towards the Physical address:
we have 32 frames so 2^5 = 32 we have 5-bits for frame + 11 bits = 16-bits
then we have 16-bits for our physical address.

A:

Consider the following room/floor analogy:  Each floor in a hotel contains 10 rooms. The door in each room is labeled 01, 02, 03, ..., 10. Then you get off the elevator, there is a plaque with the floor number.  There are 3 floors in this hotel: floors 1, 2, and 3.  Therefore, you can say that, to eliminate the ambiguity in room numbers, you concatenate the floor number to the room in the following format: floor:room.  So, 1:01 is different than 2:01, or 3:01.
Viewing this graphically:
1 | 01 | 02 | 03 | 04 | 05 | 06 | 07 | 08 | 09 | 10 |
2 | 01 | 02 | 03 | 04 | 05 | 06 | 07 | 08 | 09 | 10 |
3 | 01 | 02 | 03 | 04 | 05 | 06 | 07 | 08 | 09 | 10 |
The floor number can be expressed with one digit.  The room number can be expressed with two digits.  To express the unique location of the room (floor:room concatenation), you need three digits.  Replace floor with frame, and room with page.

