ADDRESS TRANSLATION

Question

• 16 bit virtual address space
• 20 bit physical address space

Page Table

0x1F

0x3F

0x23

0x17

What are the physical address for virtual address 0xF0F0 and 0x001F?

Comments

But in ans they broke VA in 2 and 14 bits, plz chk it

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looks like i was correct

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@Anu Sir, the page table is supposed contain all entries for each page present in the process and trap is generated when the valid bit is zero. You were right in choosing 2 bits for the Pg.no since there are only 4 entries in the page table. And moreover simply hex with 2 digits doesn't mean that the frame number is supposed to contain 8 bits. If you observe the frame numbers given, i.e:
 

0x1F	0001 1111
0x3F	0011 1111
0x23	0010 0011
0x17	0001 0111

The 2 MSB bits are always 0 here. So, saying that 2-hex digits denote 8 bits isn't right. So please make it an answer.

Answer 1

Here Page table entry is described by 8 bits.

Hence block offset = 20-8 = 12 bits.

Now in virtual address => total bits - block offset = bits for pages = > 16-12 = 4 bits (Hence pages must be 16) but given only 4 in PT.

Still for virtual address 0xF0F0 = 1111 0000 1111 0000 => last 12 bits = block offset AND first 4 bits = page Number. Page is not present  Hence Physical address =NA

for virtual address 0x001F = 0000 0000 0001 1111 => last 12 bits = block offset AND first 4 bits = page Number. I have taken it as first page(since page number =0000). Hence Physical address = Frame No + offset = 0x1F01F

Comments

Page table entry is described by 8 bits.how? Clarification is needed.Becsuse we can't solve question like this.There is a chance of committing mistake.Logic and calculation both should be clear.

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0x1F this hex code can be written as 0001 1111

hence this each entry has 8 bits