Question

I have a homework problem here which im a bit confused how to start. I have to translate the logical addresses 0,4,9,15,30 into their physical addresses. Ive looked online but havent been able to find similar examples. I have found questions but with much different given information (such as giving me the page index and offset for each logical address). Below is the question, and i did do part 1 which gives me m=32 and n-3. A help in the right direction would be great!

Answer 1

It’s almost impossible to figure the above page table structure, it doesn’t provide the essential details. Though I’ll try to help you.
However, I’ll assume that those values are the physical addresses (probably, they are).

We must consider the following details:
Address space: 32 bytes
Total pages: 4
Page size: 8 bytes
Addressing: 1 byte

Page table entries:
6
8
0
1

Strictly speaking, we would have to index the page table entry, and consequently get the offset into it.
So we need only 2 bits to index the page table entry (4 entries), whereas 6 bits to the offset in the page (8 bytes – 64 bits).
Besides, we should assume one-byte addressing.

Virtual address:    (8 bits) - Total
0 0                 (2 bits) - Page table entry.
    0 0 0 0 0 0     (6 bits) - Offset into page. 

The next question is to translate logical into physical:
0, 4, 9, 15, 30 (It’s always better to convert such numbers into binary format)

One example:
     PTE      OFFSET
0 = (0 0) (0 0 0 0 0 0)

It indexes the first page table entry which physical address is 6,
and the offset is 0.
So 6 + 0 = 6 (Physical address)

Another example:
     PTE      OFFSET
4 = (0 0) (0 0 0 1 0 0)

It indexes the first page table entry which physical address is 6, 
and the offset is 4.
So 6 + 4 = 10 (Physical address)

I hope it will help you.
Regards, Raphael S.Carvalho