Everyone keeps saying that
1'000 KB = 1 MB (decimal) -> Megabyte (MB) 10^6 Byte = 1 000 000 Byte
1'024 KB = 1 MiB (binary) -> Mebibyte (MiB) 2^20 Byte = 1 048 576 Byte
But when you look at windows properties of a folder, you get:
Which clearly uses 1KB = 1024bytes (and it's not this "MiB" but still uses 1024). So, what's the veredict?
Is 1KB = 1.024 bytes OR 1.000 ?
Yes, in causal discourse, it is either one - context sensitive.
Memory size tends to use 1024.
File size tends to use 1000.
Exceptions are common.
Else see Kilobyte.
Pedantic concerns like this include using K, when k should be used as in 1kB.
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.
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.
I am trying to calculate the max volume and file size for fat 32 and hdfs. for fat32 i have 4096 bytes sector size and 2^32 possible sectors . so 2^32 * 4096 = 1.759218604×10¹³ bytes or 17.6TB for the Volume size. But this should be 16TB according to the texts.
The same for HDFS: i have a block size (sector size) of 64mb and 63 bits to index the sectors and I do the same calculation : 2^63 * 64 = 590.29 YB . But this should equal 512 YB according to the texts . YB = Yota bytes = 10^12TB
I'm not sure where you got your "according to the texts" from. Max volume size for a FAT32 file system is 32 GB for Windows 2000 and 127.53 GB for Windows 98 (Reference)
You also need to be careful with your byte calculations. Make sure you know if the texts you are referring to are using 2^n or 10^n for their reporting. 1 Terabyte (TB) = 10^12 bytes and 1 Tibibyte (TiB) = 2^40 bytes. So, 2^32 * 2^12 (4096) = 2^44 = 16 * 2^40 = 16 TiB.
Similarly, 2^63 * 2^6 (64) = 2^69 = 2^9 * 2^60 = 512 EiB
Your calculation of 590.29 YB is not correct... it works out to EB not YB. 1 EB = 10^18 bytes and 1 YB = 10^24 bytes
It's also worth noting that a lot times TB is used as short hand for TiB.
How to measure memory size of an image in direct coding 24-bit RGB color model & in 24-bit 256-entry loop-up table representation. For example: Given an image of resolution 800*600. how much spaces are required to save the image using direct coding and look-up table.
For a regular 24-bit RGB representation most probably you just have to multiply the number of pixel on number of bytes per pixel. 24 bits = 3 bytes, so the size is 800 * 600 * 3 bytes = 1440000 bytes ≈ 1.37 MiB. In some cases you may have rows of an image aligned on some boundary in memory, usually 4 or 8 or 32 bytes. But since 800 is divisible by 32, this will not change anything, still 1.37 MiB.
Now, for a look-up table, you have 1 byte per pixel, since you have only to address one entry in the table. This yields 800 * 600 * 1 = 480000 bytes ≈ 0.46 MiB. Plus the table itself: 256 colors, 24 bits (3 bytes) each - 256 * 3 = 768 bytes. Negligible comparing to the size of the image.
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