Translate virtual address to physical address

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Discussion Overview

The discussion revolves around translating virtual addresses to physical addresses in a virtual memory system. Participants explore the mechanics of virtual memory, including page sizes, virtual page numbers (VPN), and physical page frame numbers (PFN). The conversation includes attempts to solve a homework problem involving specific virtual addresses and their corresponding physical addresses.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • Participants discuss the need to determine the virtual page number from a virtual address by dividing the address by the page size (512 bytes).
  • Some participants suggest using the modulus operation to find the offset within the page.
  • There is a debate about whether converting addresses to binary is necessary for finding physical addresses.
  • One participant emphasizes the importance of the page table for translating virtual page numbers to physical page numbers.
  • Participants explore the formula for calculating the physical address as a combination of the physical page number and the offset.
  • There is a clarification on the number of bits required for representing the physical address based on the offset and physical page number.

Areas of Agreement / Disagreement

Participants generally agree on the method for calculating the physical address but express uncertainty about the necessity of converting to binary. The discussion remains unresolved regarding the best approach to represent the physical address.

Contextual Notes

Participants mention the page size and the mapping of virtual pages to physical frames, but there are unresolved questions about the representation of addresses in different bases and the implications of using binary versus decimal formats.

Who May Find This Useful

This discussion may be useful for students studying computer architecture, particularly those interested in virtual memory systems and address translation techniques.

naspek
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Homework Statement



A virtual memory system has a page size of 512 bytes, seven virtual pages, and four
physical page frames. The page table is as follows:
VPN...PFN
0.....2
1.....0
2.....-
3.....1
4.....3
5.....-
6.....-
Note: VPN -> Virtual Page Number, PFN -> Page Frame Number


What physical address, if any, would each of the following virtual addresses
correspond to?
i. 0
ii. 2010
iii. 2046
iv. 3030
v. 1026

The Attempt at a Solution



i. 0============ 0000 0000 0000
ii. 2010========= 0111 1101 1010
iii. 2046========= 0111 1111 1110
iv. 3030======== 1011 1101 0110
v. 1026======== 0100 0000 0010

i've translate it to binary number already..
but.. i don't how to proceed..
can someone please guide me..
 
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naspek said:

Homework Statement



A virtual memory system has a page size of 512 bytes, seven virtual pages, and four
physical page frames. The page table is as follows:
VPN...PFN
0.....2
1.....0
2.....-
3.....1
4.....3
5.....-
6.....-
Note: VPN -> Virtual Page Number, PFN -> Page Frame Number


What physical address, if any, would each of the following virtual addresses
correspond to?
i. 0
ii. 2010
iii. 2046
iv. 3030
v. 1026

The Attempt at a Solution



i. 0============ 0000 0000 0000
ii. 2010========= 0111 1101 1010
iii. 2046========= 0111 1111 1110
iv. 3030======== 1011 1101 0110
v. 1026======== 0100 0000 0010

i've translate it to binary number already..
but.. i don't how to proceed..
can someone please guide me..

I don't think converting the virtual address to binary is much help. I'm not certain, but I think what you need to do is to find the virtual page number for each virtual address, and then see which physical page that VPN is mapped to.

To do this, divide the virtual address by 512 using integer division. For example, your second virtual address is 2010. 2010/512 == 3, so the virtual page number is 3. From the information you provided, virtual page 3 maps to physical page 1.
 
Mark44 said:
I don't think converting the virtual address to binary is much help. I'm not certain, but I think what you need to do is to find the virtual page number for each virtual address, and then see which physical page that VPN is mapped to.

To do this, divide the virtual address by 512 using integer division. For example, your second virtual address is 2010. 2010/512 == 3, so the virtual page number is 3. From the information you provided, virtual page 3 maps to physical page 1.

thanks for the idea..

now.. i proceed...

ii. 2010===== 2010/512
==========3.93 = 3 , virtual page number 3 maps to physical page 1..
but.. how do i get the physical address then?
 
The remainder (or modulus) gives you the offset in the memory page. In C notation, this is 2010 % 512 == 474. In mathematical notation, this is 2010 (mod 512) = 474.
 
Mark44 said:
The remainder (or modulus) gives you the offset in the memory page. In C notation, this is 2010 % 512 == 474. In mathematical notation, this is 2010 (mod 512) = 474.
get it.. then i need to change 474 to binary number to get my physical address?
 
I don't see why you would need to do that. It's just an address. It could be 474 (decimal) or 1DA (hex) or 111011010 (binary). These all represent the same number, but in different bases.
 
Mark44 said:
I don't see why you would need to do that. It's just an address. It could be 474 (decimal) or 1DA (hex) or 111011010 (binary). These all represent the same number, but in different bases.
thanks for the clarification.. but..
why do i need the table if, i can just do the calculation(divide by 512 and get the answer already)..?
 
There's an important part you're missing. Rather than me telling you what it is, I will let you figure it out.

If the virtual address is 2010, what is the physical address?
 
Mark44 said:
There's an important part you're missing. Rather than me telling you what it is, I will let you figure it out.

If the virtual address is 2010, what is the physical address?

the physical address is 0001 1101 1010.. right?
 
  • #11
Mark44 said:
How did you get that?
2010 mod 512 = 474

i change 474 into 12 bits binary number.. so.. i get 0001 1101 1010.. right?
 
  • #12
The 474 is just the offset within the page, but which page are we talking about? And by that, I mean, which physical page?
 
  • #13
Mark44 said:
The 474 is just the offset within the page, but which page are we talking about? And by that, I mean, which physical page?

ok... so..

VPN...PFN
3.....1

it's on physical page 1.. right?

is it 1 is my physical address? can't be.. too odd...
 
  • #14
Yes, physical page 1. That's where the table comes in, so that you can translate a virtual page number to a physical page number.

OK, you have physical page 1, and an offset of 447. What's the physical address?
 
  • #15
Mark44 said:
Yes, physical page 1. That's where the table comes in, so that you can translate a virtual page number to a physical page number.

OK, you have physical page 1, and an offset of 447. What's the physical address?

ok... physical page number constitutes the upper portion of the physical address, while the page offset, constitutes the lower portion

so.. my physical address is...

1 | 0001 1101 1010

correct?
 
  • #16
No, that's too big. It would be helpful if you didn't change the numbers into binary. Your physical address is calculated using this formula:
physical address = (physical page number) * 512 + offset
 
  • #17
Mark44 said:
No, that's too big. It would be helpful if you didn't change the numbers into binary. Your physical address is calculated using this formula:
physical address = (physical page number) + offset

ok.. thanks for the formula..

*got question here.. is it [(physical page number)*512] + offset or..
....... (physical page number) * (512 + offset)
the first one or the second formula is correct?

physical address = (1) * 512 + 474
......= 986 @ 0011 1101 1010
now.. this is the correct one?
 
  • #18
It's this one: (physical page number)*512 + offset

And 986 is the correct address.

To check: 986 / 512 = 1
and 986 (mod 512) = 474
 
  • #19
geez! thanks mark44!
but.. if I'm going to answer it in binary number format..
how should i know the number of bits of the physical address?
 
  • #20
The offset is 0 - 511, or 0 - 0x1ff, or 1 1111 1111, so that takes 9 bits.

The physical page number is a multiple of 512 (0, 1, 2, or 3 times 512), so the largest physical address will be 3*512 + 511 = 2047 = 0x7ff, which fits in 11 bits.
 

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