To help you solve this question, we need to get our details right:
- 16 bit of virtual address space = 2^16 = 65,536 address space
- 16 bit of physical address space = 2^16 = 65,536 address space
- 4096 Byte page size determines the offset, which is Log(4096) / Log (2) = 12 bit. This means, 2^12 for Page size
- As per @Akash Mahapatra, the offset from virtual address is directly mapped to the offset onto physical address
As such, we now have:
- 2^16 (16bit) for virtual address - 2^12 (12bit) for offset = 4-bit for pages, or rather total number of pages available.
- I won't repeat the calculation for physical since it's the same numbers.
- 2^4 (4bit) for pages = 16, which correlates to the number of table entries above!
We're getting there... be patient! :)
Memory Address 0xE12C in hex notation is also known to be holding 16-bit of address. (Since it's stated in the question.)
Let's butcher the address now...
We first remove '0x' from the info.
We can convert E12C to binary notation like @Tony Tannous, but I am going to apply a little short-cut.
I simply use a ratio. Well, the address is notated in 4 characters above, and since 16/4 = 4, I can define the first letter as virtual address, while the other 3 are offset address.
With the information, 'E' in hexadecimal format, I need to convert to Decimal = 14. Then I look at your table provided, and I found page frame '3'. Page frame 3 is noted in decimal format, which then need to be converted back to Hexadecimal format... Duh!... which is 3!
So, the Physical address mapping of the virtual memory location of 0xE12C can be found at 0x312C on the physical memory.
You will then go back to the table, and refer to the reference bit column and put a '1' to the row 14.
Apply the same concept for these -
0x3A9D → 0xAA9D
0xA9D9 → 0x59D9
0x7001 → 0xF001
0xACA1 → 0x5CA1
If you notice, the last 3 digits are the same (which determines the offset).
And the 1st of the 4-digits are mapped according to the table:
table entry 3 -> page frame 10 -> hex notation A
table entry A (10) -> page frame 5 -> hex notation 5
table entry 7 -> page frame 15 -> hex notation F
table entry A (10) -> page frame 5 -> hex notation 5
Hope this explanation helps you and others like me! :)