I'm reading through an example of page tables and just found this:
Consider a system with a 32-bit logical address space. If the page size in such a system is 4 KB (2^12), then a page table may consist of up to 1 million entries (2^32/2^12). Assuming that each entry consists of 4 bytes, each process may need up to 4 MB of physical address space for the page table alone.
I don't really understand what this 4MB result represents. Does it represent the space the actual page table takes up?
Since we have a virtual address space of 2^32 and each page size is 2^12, we can store (2^32/2^12) = 2^20 pages. Since each entry into this page table has an address of size 4 bytes, then we have 2^20*4 = 4MB. So the page table takes up 4MB in memory.
My explanation uses elementary building blocks that helped me to understand. Note I am leveraging #Deepak Goyal's answer above since he provided clarity:
We were given a logical 32-bit address space (i.e. We have a 32 bit computer)
Consider a system with a 32-bit logical address space
This means that every memory address can be 32 bits long.
"A 32-bit entry can point to one of 2^32 physical page frames"[2], stated differently,
"A 32-bit register can store 2^32 different values"
We were also told that
each page size is 4 KB
1 KB (kilobyte) = 1 x 1024 bytes = 2^10 bytes
4 x 1024 bytes = 2^2 x 2^10 bytes => 4 KB (i.e. 2^12 bytes)
The size of each page is thus 4 KB (Kilobytes NOT kilobits).
As Depaak said, we calculate the number of pages in the page table with this formula:
Num_Pages_in_PgTable = Total_Possible_Logical_Address_Entries / page size
Num_Pages_in_PgTable = 2^32 / 2^12
Num_Pages_in_PgTable = 2^20 (i.e. 1 million)
The authors go on to give the case where each entry in the page table takes 4 bytes. That means that the total size of the page table in physical memory will be 4MB:
Memory_Required_Per_Page = Size_of_Page_Entry_in_bytes x Num_Pages_in_PgTable
Memory_Required_Per_Page = 4 x 2^20
Memory_Required_Per_Page = 4 MB (Megabytes)
So yes, each process would require at least 4MB of memory to run, in increments of 4MB.
Example
Now if a professor wanted to make the question a bit more challenging than the explanation from the book, they might ask about a 64-bit computer. Let's say they want memory in bits. To solve the question, we'd follow the same process, only being sure to convert MB to Mbits.
Let's step through this example.
Givens:
Logical address space: 64-bit
Page Size: 4KB
Entry_Size_Per_Page: 4 bytes
Recall: A 64-bit entry can point to one of 2^64 physical page frames
- Since Page size is 4 KB, then we still have 2^12 byte page sizes
1 KB (kilobyte) = 1 x 1024 bytes = 2^10 bytes
Size of each page = 4 x 1024 bytes = 2^2 x 2^10 bytes = 2^12 bytes
How Many pages In Page Table?
`Num_Pages_in_PgTable = Total_Possible_Logical_Address_Entries / page size
Num_Pages_in_PgTable = 2^64 / 2^12
Num_Pages_in_PgTable = 2^52
Num_Pages_in_PgTable = 2^2 x 2^50
Num_Pages_in_PgTable = 4 x 2^50 `
How Much Memory in BITS Per Page?
Memory_Required_Per_Page = Size_of_Page_Entry_in_bytes x Num_Pages_in_PgTable
Memory_Required_Per_Page = 4 bytes x 8 bits/byte x 2^52
Memory_Required_Per_Page = 32 bits x 2^2 x 2^50
Memory_Required_Per_Page = 32 bits x 4 x 2^50
Memory_Required_Per_Page = 128 Petabits
[2]: Operating System Concepts (9th Ed) - Gagne, Silberschatz, and Galvin
In 32 bit virtual address system we can have 2^32 unique address, since the page size given is 4KB = 2^12, we will need (2^32/2^12 = 2^20) entries in the page table, if each entry is 4Bytes then total size of the page table = 4 * 2^20 Bytes = 4MB
Suppose logical address space is 32 bit so total possible logical entries will be 2^32 and other hand suppose each page size is 4 kilobyte then size of one page is 2^22^10=2^12...
now we know that no. of pages in page table is
pages=total possible logical address entries/page size
so pages=2^32/2^12 =2^20
Now suppose that each entry in page table takes 4 bytes then total size of page table in physical memory will be=2^22^20=2^22=4mb**
Since the Logical Address space is 32-bit long that means program size is 2^32 bytes i.e. 4GB.
Now we have the page size of 4KB i.e.2^12 bytes.Thus the number of pages in program are 2^20.(no. of pages in program = program size/page size).Now the size of page table entry is 4 byte hence the size of page table is 2^20*4 = 4MB(size of page table = no. of pages in program * page table entry size). Hence 4MB space is required in Memory to store the page table.
yes it represents the space the actual page table takes for one process.
if each page is 4KB -> 12 bits for offset (how?)
1 kb is 2^10 bytes => 4 kb is 4*2^10 bytes which is 2^12 => hence 12 bits for offset and the remaining 20 bits for VPN => 2^20 translations which means 2^20 pages are there which means 2^20 entries in the page table.
Hence, size of page table = number of entries in the page table * size of one entry
=> size of page table = 2^20 * 4 KB = 2^22 kB and 1 MB is 2^20 KB => 4 MB
Related
I am studying for my computer architecture exam that is due tomorrow and am stuck on a practice exercise regarding tag size and the total number of cache bits. Here is the question:
Question 8:
This question deals with main and cache memory only.
Address size: 32 bits
Block size: 128 items
Item size: 8 bits
Cache Layout: 6 way set associative
Cache Size: 192 KB (data only)
Write Policy: Write Back
Answer: The tag size is 17 bits. The total number of cache bits is 1602048.
I know that this a failry straight-forward exercise, but I seem to be lacking the proper formulas. I also know that the structure of a N set way associative is |TAG 25 bits|SET 2 bits|OFFSET 5 bits|. And that Tag size = AddrSize - Set - Offset (- item size if any) thus giving the answer of 17 bits tag size.
However, how do I calculate the total number of cache bits please?
cache size in bytes: 192*1024 = 196608
number of blocks: 196608 / 128 = 1536
number of sets: 1536 / 6 = 256
set number bits: log2(256) = 8
offset number bits: log2(128) = 7
tag size: 32-(8+7) = 17
metadata: valid+dirty = 2 bits
total tag + metadata: (17+2)*1536 = 29184 bits
total data: 1536*128*8 = 1572864 bits
total size: 29184 + 1572864 = 1,602,048
There could also be bits used for the replacement policy, but we can assume it's random to make the answer work.
suppose there is a computer with 18 bits address space and the cell size is 8 bits, then
What is the smallest and highest address?
What will be the possible largest memory size of this computer in bytes, kilobytes and megabytes?
What is the smallest and highest address?
smallest Address = 0x0000 (HEX)
highest address = 2^18 / 8
= 32768 = 0x8000 (HEX)
What will be the possible largest memory size of this computer in
bytes, kilobytes and megabytes?
Memory Size = 2^18 X 8 bits
Memory Size = 262,144B (Bytes)
Memory Size = 262,144/1024 KB = 256 KB
Memory Size = 256/1024 MB = 0.25 MB
Note::
For more convenient representation the units KiB and MiB are used:
in that Case The memory size is 2048KiB OR 2MiB
i am really confused on the topic Direct Mapped Cache i've been looking around for an example with a good explanation and it's making me more confused then ever.
For example: I have
2048 byte memory
64 byte big cache
8 byte cache lines
with direct mapped cache how do i determine the 'LINE' 'TAG' and "Byte offset'?
i believe that the total number of addressing bits is 11 bits because 2048 = 2^11
2048/64 = 2^5 = 32 blocks (0 to 31) (5bits needed) (tag)
64/8 = 8 = 2^3 = 3 bits for the index
8 byte cache lines = 2^3 which means i need 3 bits for the byte offset
so the addres would be like this: 5 for the tag, 3 for the index and 3 for the byte offset
Do i have this figured out correctly?
Do i figured out correctly? YES
Explanation
1) Main memmory size is 2048 bytes = 211. So you need 11 bits to address a byte (If your word size is 1 byte) [word = smallest individual unit that will be accessed with the address]
2) You can calculating tag bits in direct mapping by doing (main memmory size / cash size). But i will explain a little more about tag bits.
Here the size of a cashe line( which is always same as size of a main memmory block) is 8 bytes. which is 23 bytes. So you need 3 bits to represent a byte within a cashe line. Now you have 8 bits (11 - 3) are remaining in the address.
Now the total number of lines present in the cache is (cashe size / line size) = 26 / 23 = 23
So, you have 3 bits to represent the line in which the your required byte is present.
The number of remaining bits now are 5 (8 - 3).
These 5 bits can be used to represent a tag. :)
3) 3 bit for index. If you were trying to label the number of bits needed to represent a line as index. Yes you are right.
4) 3 bits will be used to access a byte withing a cache line. (8 = 23)
So,
11 bits total address length = 5 tag bits + 3 bits to represent a line + 3 bits to represent a byte(word) withing a line
Hope there is no confusion now.
This is a question from a Computer Architecture exam and I don't understand how to get to the correct answer.
Here is the question:
This question deals with main and cache memory only.
Address size: 32 bits
Block size: 128 items
Item size: 8 bits
Cache Layout: 6 way set associative
Cache Size: 192 KB (data only)
Write policy: Write Back
What is the total number of cache bits?
In order to get the number of tag bits, I find that 7 bits of the address are used for byte offset (0-127) and 8 bits are used for the block number (0-250) (250 = 192000/128/6), therefore 17 bits of the address are left for the tag.
To find the total number of bits in the cache, I would take (valid bit + tag size + bits per block) * number of blocks per set * number of sets = (1 + 17 + 1024) * 250 * 6 = 1,536,000. This is not the correct answer though.
The correct answer is 1,602,048 total bits in the cache and part of the answer is that there are 17 tag bits. After trying to reverse engineer the answer, I found that 1,602,048 = 1043 * 256 * 6 but I don't know if that is relevant to the solution because I don't know why those numbers would be used.
I'd like if someone could explain what I did wrong in my calculation to get a different answer.
I get why we have the number 1024 instead of 1000 to use the suffix "kilo" in computing (computer uses base 2, so 2 ^ 10, blah blah blah). So I get the kilo part, but why is it called a kilo - "byte"? To make a kilo - "byte", we need to use bits with 10 digits from 0000000000 to 1111111111. That is not 8 digits, shouldn't it be called something else.
I.e. a kilobyte is not 1024 groupings of 8 bit binary digits, it is 1024 groups of 10 bit binary digits and a megabyte has even more than 10 binary digits - not 8. If asked how many bits are in 1 kilobytes, people calculate it as 1*1024*8. But that's wrong! It should be 1*1024*10.
I.e. a kilobyte is not 1024 groupings of 8 bit binary digits, it is
1024 groups of 10 bit binary digits
You are confusing the size of a byte with the size of the value needed to address those bytes.
On most systems a byte is 8 bits, which means 1000 bytes is exactly 1000*8 bits, and 2000 bytes is exactly 2000*8 bits (i.e. exactly the double, which makes sense).
To address or index those bytes you need 10 bits in the first example (2^10) and 11 bits in the second (2^11 up to 2048 bytes). It wouldn't make a lot of sense if the size of a byte was changing when there are more bytes in a data structure.
As for the 1000 (kilobyte) vs 1024 (kibibyte):
1 kB (kilobyte) = 10^3 = 1000
1 KiB (kibibyte) = 2^10 = 1024
A kilobyte used to be generally accepted as being 1024 bytes. However at some point hard disk manufacturers started to count 1 kB as 1000 bytes (kilo being 1000 which is actually correct):
1 GB = 1000^3 = 1000000000
1 GiB = 1024^3 = 1073741824
Windows still used 1 kB = 1024 bytes to show the hard disk size, i.e. it showed 954MB for 1GB of hard disk space. I remember a lot of customers complaining about that when checking, for example, the size of their 250GB drive which only showed 233GB in Windows.