I am currently studing exam questions but stuck on this one, I hope someone can help me out to understand.
Question: Assume that we have a paged virtual memory with a page size of 4Ki byte.
Assume that each process has four segments (for example: code, data, stack,
extra) and that these can be of arbitrary but given size. How much will the
operating system loose in internal fragmentation?
The answer is: Each segment will in average give rise to 2Ki byte of fragmentation.
This will in average mean 8 Ki byte per process.
If we for example have 100 processes this is a total loss of 800 Ki byte.
My question:
How the answer get the 2Ki byte of fragmentation for each segement, how is that possible we can calculate the size, am I missing something here?
If we have 8Ki byte per process, that would not even fit in a 4Ki byte page isn't that actually a external fragmentation?
This is academic BS designed to make things confusing.
They are saying probability wise, the last page in the sections in the executable file will only use 1/2 the page size on average. You can't count that size, they are just doing simple combinatorics. That presumes behavior of the linker.
This question already has an answer here:
Minimum associativity for a PIPT L1 cache to also be VIPT, accessing a set without translating the index to physical
(1 answer)
Closed last year.
Homework question, so please just nudge me in the right direction.
Consider a system with physically-addressed caches, and assume that 40-bit virtual addresses and 32-bit physical addresses are used, and the memory is byte-addressable. Further assume that the cache is 4-way set-associative, the cache line size is 64 Bytes and the total size of the cache is 64 KBytes.
What should be the minimum page size in this system to allow for the overlap of the TLB access and the cache access?
I've been stuck on this question and have no idea how to even begin. Can someone give me a hint towards finding the solution?
I think the most important piece of information in the question is
overlap of the TLB access and the cache access
This means, we access the Cache at the same time we access the TLB. In practice, what we really do is, we index the cache with the index bits from the virtual address and by the time we have located the entry in the cache, we will have the data (physical address) from the TLB. Then we can do the tag comparison with physical address. In other words cache acts as a Virtually indexed, Physically tagged (VIPT) cache.
Even though the scheme sounds efficient, the thing to lookout is, number of bits used to index the cache, cannot be higher than the number of bits needed to represent the page size. Simply, size of a page can put an upper limit on the number of cache entries.
Now coming back to your question,
its a 64KBytes cache with 4 way set assoc. and cacheline of 64Bytes.
Number of cachelines = (64KBytes/4)/64Bytes = 2^8 cachelines
That means if a page is 256Bytes or bigger, we can use this mechanism. If a page is smaller than 256 Bytes, then we cannot assume the index bits of the virtual address and the physical address are going to be the same.
What should be the minimum page size in this system to allow for the
overlap of the TLB access and the cache access?
256Bytes
Calculating offset of branch instruction:
Hey guys,
My professor sent us a study guide with answers. He never actually went over how he got the answers though. I've searched online but I did not have any luck finding an explanation so at this point I'm a bit desperate.
Does anyone know how the author arrived at those answers?
0xfffb is the 16-bit signed two's complement representation of -5. So in this machine, the offset is scaled by the (presumably fixed) instruction length to get a byte address. (It could have byte sized instructions, but that is not possible since the offset itself is 16-bits.) The architecture is such that by the time the branch is executed the PC is already incremented so a 0 offset is a NOP, a -1 offset branches to the branch itself, -2 branches to the instruction before the branch, etc. Count backwards until you get to loop:. (Either there is some more info attached to the question, or known context, giving the details of the architecture I have used in making the answer, or it is a fairly badly written question.)
For the cache question, you mostly just have to know the names used to describe cache architectures (or "cache geometry", "cache shape" etc.). "2-way set associative" means there are two places in the cache any given address can be placed. There are 128 sets, each of which can hold two blocks because it is 2-way associative, and each block is 32-bytes. (I usually call the 32-byte structure a "cache line", though it appears here the word "block" refers specifically to the data that is stored there and "line" also includes the valid bit and tag, etc.) You then break down the address starting from the least significant bit going outwards in the cache geometry.
It looks like this is an instruction cache so we're going to insist the bottom two bits are 0 and organize the cache in 32-bit items. The block is 32-bytes or 5-bits. 2 are the "byte offset" which probably should just be 0, and then 3 bits to complete the 5-bit part which gets called a "block offset" (really offset within the block). (This subdivison of the 5 low bits doesn't really change anything here.) 128 entries in the set gives a 7-bit "index." The rest of the address, 20 bits, has to be used to tag the block to make sure it holds the address being looked up. (I.e. to determine cache hit or miss.) Plus we need one more bit to say whether there's actually data in the block.
Then we just add it all up -- 32-bytes or 256 bits for the data, plus 20 bits of tag and 1 valid bit, multiply by 128 sets and 2 ways.
I have a question that I am trying to answer that gives the following situation:
16K Pages
32-bit Virtual Addresses
512MB hard disk, sector-addressable with 16K sectors
8 processes currently running
I am asked:
i) How many process page tables are required?
I think this is a trick question? Surely the answer is just 1.
ii) If a process address register PAR can be up to 32 bits, what is the maxmimum amount of physical memory that can be supported on this machine?
iii) How wide in bits should each entry in a process table be if 64MB physical memory is installed?
Please could anyone give me help/hint with the last two parts as I'm really stuck on them? Thanks!
In case you look on here before the exam later today, it is because it doesn't mean Process address register, it means Page address register!
Try looking at http://cseweb.ucsd.edu/classes/fa03/cse120/Lec08.pdf for some more information including help about segmentation and paging combined
Also, the book in the IC library called Operating Systems concepts with code 005.43SIL says that each process has it's own process page table and can even be segmented itself!
i) I said 8
ii) Well, 32 bits of virtual memory addressing with 14 bits of offset in the page table (2^14 = 16K page length) means there are 18 bits left for the page number. In 32 bits of PAR, this means 14 bits for the page location. If you multiple the amount of page locations by the page size, you get 2^14 * 2^14 = 2^18 which is 256MB of RAM
iii) I got 30 bits. 64MB is 2^26 divided by the page size is 2^26/2^14 = 2^12 which means 12 bits for the page location. From (ii) I calculated that 18 bits are left in the virtual memory address for the page number meaning that it should be 30 bits wide. I also put a comment that since it should be byte-aliged maybe the extra 2 bits can be used so that we know whether it has been written to and whether it is currently being stored on the disk.
Hope this helps!
Just asked by my 5 year old kid: what is the biggest number in the computer?
We are not talking about max number for a specific data types, but the biggest number that a computer can represent.
Infinity is not allowed.
UPDATE my kid always wants to print as
well, so lets say the computer needs
to print this number and the kid to
know that its a big number. Of course,
in practice we won't print because
theres not enough trees.
This question is actually a very interesting one which mathematicians have devoted a fair bit of thought to. You can read about it in this article, which is a fascinating and accessible read.
Briefly, a guy named Tibor Rado set out to find some really big, but still well-defined, numbers by defining a sequence called the Busy Beaver numbers. He defined BB(n) to be the largest number of steps any Turing Machine could take before halting, given an input of n symbols. Note that this sequence is by its very nature not computable, so the numbers themselves, while well-defined, are very difficult to pin down. Here are the first few:
BB(1) = 1
BB(2) = 6
BB(3) = 21
BB(4) = 107
... wait for it ...
BB(5) >= 8,690,333,381,690,951
No one is sure how big exactly BB(5) is, but it is finite. And no one has any idea how big BB(6) and above are. But at least these numbers are completely well-defined mathematically, unlike "the largest number any human has ever thought of, plus one." ;)
So how about this:
The biggest number a computer can represent is the most instructions a program small enough to fit in its available memory can perform before halting.
Squared.
No, wait, cubed. No, raised to the power of itself!
Dammit!
Bits are not numbers. You, as a programmer, give them the meaning you want, possibly numbers.
Now, I decide that 1 represents "the biggest number ever thought by a human plus one".
Errr this is a five year old?
How about something along the lines of: "I'd love to tell you but the number is so big and would take so long to say, I'd die before I finished telling you".
// wait to see
for(;;)
{
printf("9");
}
roughly 2^AVAILABLE_MEMORY_IN_BITS
EDIT: The above is for actually storing a number and treats all media (RAM, HD, cloud etc.) as memory. Subtracting the OS footprint (measured in KB) doesn't make "roughly" less accurate...
If you want to "represent" a number in a meaningful way, then you probably want to go with what the CPU provides: unsigned 32 bit integers (roughly 4 Gigs) or unsigned 64 bit integers for most computers your kid will come into contact with.
NOTE for talking to 5-year-olds: Often, they just want a factoid. Give him a really big and very accurate number (lots of digits), like 4'294'967'295. Then, once the glazing leaves his eyes, try to see how far you can get with explaining how computers represent numbers.
EDIT #2: I once read this article: Who Can Name the Bigger Number that should provide a whole lot of interesting information for your kid. Obviously he's not your normal five-year-old. So this might get you started in a cool direction about numbers and computation.
The answer to life (and this kids question): 42
That depends on the datatype you use to represent it. The computer only stores bits (0/1). We, as developers, give the bits meaning. (65 can be a number or the letter A).
For example, I can define my datatype as 1^N where N is unsigned and represented by an array of bits of arbitrary size. The next person can come up with 10^N which would be ten times larger than my biggest number.
Sure, there would be gaps but if you don't need them, that doesn't matter.
Therefore, the question is meaningless since it doesn't have context.
Well I had the same question earlier this day, so thought why not to make a little c++ codes to see where the computer gonna stop ...
But my laptop wasn't with me in class so I used another, well the number was to big but it never ends, i'll run it again for a night then i'll share the number
you can try the code is stupid
#include <stdlib.h>
#include <stdio.h>
int main() {
int i = 0;
for (i = 0; i <= i; i++) {
printf("%i\n", i);
i++;
}
}
And let it run till it stops ^^
The size will obviously be limited by the total size of hard drives you manage to put into your PC. After all, you can store a number in a text file occupying all disk space.
You can have 4x2Tb drives even in a simple box so around 8Tb available. if you store as binary, then the biggest number is 2 pow 64000000000000.
If your hard drive is 1 TB (8'000'000'000'000 bits), and you would print the number that fits on it on paper as hex digits (nobody would do that, but let's assume), that's 2,000,000,000,000 hex digits.
Each page would contain 4000 hex digits (40 x 100 digits). That's 500,000,000 pages.
Now stack the pages on top of each other (let's say each page is 0.004 inches / 0.1 mm thick), then the stack would be as 5 km (about 3 miles) tall.
I'll try to give a practical answer.
Common Lisp number crunching is particularly powerful. It has something called "bignums" which are integers that can be arbitrarily large, limited by the amount of available.
See: http://en.wikibooks.org/wiki/Common_Lisp/Advanced_topics/Numbers#Fixnums_and_Bignums
Don't know much about theory, but I far as I understood from your question, is: what is the largest number that the computer can represent (and I add: in a reasonable time, and not printing "9" until the Earth will "be eaten by the Sun"). And I put my PC to make one simple calculation (in PHP or whatever language): echo pow(2,1023) - resulting: 8.9884656743116E+307. So I guess this is the largest number that my PC can calculate. On the other side, I think the respresentation of the largest negative number can be: -0,(0)1
LE: That computed value was obataind through PHP, but I tried to figure out what's the largest number that my windows calculator can compute, and it is pow(2, 33219) = 8.2304951207588748764521361245002E+9999. Now I guess this is the largest number my PC can handle.
I think you should be very proud that your 5 year old is already asking questions like this.
And you should continue to promote that! This is truly amazing! With that said, I would say that saying Infinity does not
count is thinking incorrectly about what numbers mean in computer memory.
I feel like this way of thinking is a handicap.
Mathematicians will never be able to write out ALL the digits of pi or eulers number, BUT we FULLY understand it.
Pi, as an example, is perfectly represented by infinite this series: (Pi / 4) = 1 - 1/3 + 1/5 - 1/7 + 1/9 - …
Just because you literally can’t go to inf. or print every single digit in a console means nothing.
You could have printed the symbol representing pi and therefore capturing the inf. series.
Computer Algebra Systems (CAS) represent numbers symbolically all the time. Pi, for instance,
may be a Symbolic object in memory (the binary in memory did not DIRECTLY represent the number. It represents an "mathematical algorithm" for producing the answer to arbitrary precision).
Then you do some math with it, transforming from one expression to the next.
At no point in time did we not represent the number COMPLETELY.
At the end, you can do 2 things with this:
A) Evaluate the expression, turning it into a number of some kind (or Matrix or whatever).
BUT this number could very well be an approximation (say like 20 digits of pi).
B) Keep it in its symbolic form for reference. Obviously we don’t like staring at symbols because we
need to eventually turn the nobs on the apparatii.
NOTE: sometimes you can get a finite (non-irrational) number perfectly represented in memory (like number 1)
by taking limits or going to inf. Not literally having an inf. number in memory, but symbolically representing it.
Just throw this in Wolfram alpha: Lim[Exp[-x], x --> Inf]; It gives you the number 0. Which is EXACT.
In short:
It was the HUMANS need to have some binary in memory that DIRECTLY represented the number that caused
the number to degrade. Symbolically it was perfectly represented. You could design some algorithm that
just continues to calculate the next digits of pi or eulers number giving you an arbitrary amount of precision (Now, this is obviously not practical of course).
I hope this was at least somewhat useful or interesting to you, even if you disagree =)
Depends on how much the computer can handle. Although there are some times when the computer can handle numbers greater than (2^(bits-1)-1)... For example:
My computer is 64 bit (9223372036854775807), however the calculator that comes with the computer itself can handle numbers of up to 10^9999.
Many other supercomputers can exceed these limits, and the one with the most memory (bits) might as well be the one with the record (current largest number that can be held by computers).
Or, if it comes to visually seeing it on computers, you can just make a program that, on monitor, repeats writing 9 and not skips that line to form an ever-growing bunch of 9. :P
go on chrome then go on three dots above and click them then go on tools and then go on developer tool click on console and type Number.MAX_VALUE