I have been working on a virtual stock market game using PHP. The formula that I have been using for deciding the price of a stock is
$price += $ran*0.001*$price + $ratio*0.005*$price
where
$ran = rand(-1*$intensity, 2*$intensity)
$intensity is a number between -5 to 5 depending upon whether the news is good or bad for the company and
$ratio = (1.0*($buy-$sell))/($buy + $sell)
$buy and $sell represent the number of shares bought and sold of a company respectively.
The problem with this formula is that, even if the intensity is negative (even -5) the ratio term is always added to the price which makes the overall term increase. The prices are refreshed every 10 seconds and with the above formula they keep on increasing and never come down. So, can anyone help me out with this formula so that it varies more closely to the actual stock market?
If I understand correctly, you're trying to define an algorithm to determine a logical next price based on the current price, some market activity, and a random input. This is called a Random Walk, and the linked page is quite informative.
In economics, the "random walk hypothesis" is used to model shares prices and other factors. Empirical studies found some deviations from this theoretical model, especially in short term and long term correlations.
It's difficult for us to provide an exact function for you, since the exact behavior you expect of a function like this is inherently application specific. However it is possible to test the behavior and improve on it, by pulling it out into its own method and tweaking it until you see the behavior you want.
I would suggest pulling this behavior you've defined into an SSCCE (or a unit test, but assuming you don't already have a PHP unit test framework set up, an example will do fine) and creating some test cases, then you can tweak your algorithm in a vacuum and find behavior you like.
Here's some boilerplate to get started:
<?php
function nextPrice($price, $intensity, $buy, $sell, $rand) {
// TODO
}
// Can tweak these values between runs, or put them in a loop if you want
$testPrice = 10.0;
$testBuy = 10000;
$testSell = 10000;
for ($i = -5; $i <= 5; $i++) {
// random float, from http://stackoverflow.com/a/14155720/113632
// set to a constant if you want to isolate the randomness and test other variables
$testRand = mt_rand(0, mt_getrandmax() - 1) / mt_getrandmax();
echo "<p>Intensity: $i - Rand: $testRand = ".
nextPrice($testPrice, $i, $testBuy, $testSell, $testRand)."</p>";
}
?>
Some additional thoughts:
Your $ran definition is definitely flawed, if $intensity is -5 you're executing $ran = rand(5, -10); which generates a warning and doesn't return the value you want. This is likely the root of your issue, as any negative $intensity will essentially set $ran to zero.
Furthermore your $ran definition is biased towards positive numbers, meaning the price is - rather quickly - going to rise even if there's bad news. I'd suggest ensuring your random value is equally likely to lower the stock as raise it, and if you intend for the stock to rise in value over time regardless (which seems like a bad idea to me) set a separate $longTermGrowthFactor that always increases the stock by that factor, separately from the randomness.
Turn on warning reporting in PHP - since you presumably hadn't seen the warnings related to your rand() call, you likely have warnings and other error types turned off, which quite likely means there are other errors hidden in your code you aren't aware of, and without the reporting they're going to be hard to spot.
Use mt_rand() instead of rand(), the latter is deprecated, and mt_rand() is a drop-in replacement providing better randomness.
Related
In the course of testing an algorithm I computed option prices for random input values using the standard pricing function blsprice implemented in MATLAB's Financial Toolbox.
Surprisingly ( at least for me ) ,
the function seems to return negative option prices for certain combinations of input values.
As an example take the following:
> [Call,Put]=blsprice(67.6201,170.3190,0.0129,0.80,0.1277)
Call =-7.2942e-15
Put = 100.9502
If I change time to expiration to 0.79 or 0.81, the value becomes non-negative as I would expect.
Did anyone of you ever experience something similar and can come up with a short explanation why that happens?
I don't know which version of the Financial Toolbox you are using but for me (TB 2007b) it works fine.
When running:
[Call,Put]=blsprice(67.6201,170.3190,0.0129,0.80,0.1277)
I get the following:
Call = 9.3930e-016
Put = 100.9502
Which is indeed positive
Bit late but I have come across things like this before. The small negative value can be attributed to numerical rounding error and / or truncation error within the routine used to compute the cumulative normal distribution.
As you know computers are not perfect and small numerical error always persists in all calculations, in my view therefore the question one should must ask instead is - what is the accuracy of the input parameters being used and therefore what is the error tolerance for outputs.
The way I thought about it when I encountered it before was that, in finance, typical annual stock price return variance are of the order of 30% which means the mean returns are typically sampled with standard error of roughly 30% / sqrt(N) which is roughly of the order of +/- 1% assuming 2 years worth of data (so N = 260 x 2 = 520, any more data you have the other problem of stationarity assumption). Therefore on that basis the answer you got above could have been interpreted as zero given the error tolerance.
Also we typically work to penny / cent accuracy and again on that basis the answer you had could be interpreted as zero.
Just thought I'd give my 2c hope this is helpful in some ways if you are still checking for answers!
I'm trying to write a basic digit counter (an integer is inputted and the number of digits of that integer is outputted) for positive integers. This is my general formula:
dig(x) := Math.floor(Math.log(x,10))
I tried implementing the equivalent of dig(x) in Ruby, and found that when I was computing dig(1000) I was getting 2 instead of 3 because Math.log was returning 2.9999999999999996 which would then be truncated down to 2. What is the proper way to handle this problem? (I'm assuming this problem can occur regardless of the language used to implement this approach, but if that's not the case then please explain that in your answer).
To get an exact count of the number of digits in an integer, you can do the usual thing: (in C/C++, assuming n is non-negative)
int digits = 0;
while (n > 0) {
n = n / 10; // integer division, just drops the ones digit and shifts right
digits = digits + 1;
}
I'm not certain but I suspect running a built-in logarithm function won't be faster than this, and this will give you an exact answer.
I thought about it for a minute and couldn't come up with a way to make the logarithm-based approach work with any guarantees, and almost convinced myself that it is probably a doomed pursuit in the first place because of floating point rounding errors, etc.
From The Art of Computer Programming volume 2, we will eliminate one bit of error before the floor function is applied by adding that one bit back in.
Let x be the result of log and then do x += x / 0x10000000 for a single precision floating point number (C's float). Then pass the value into floor.
This is guaranteed to be the fastest (assuming you have the answer in numerical form) because it uses only a few floating point instructions.
Floating point is always subject to roundoff error; that's one of the hazards you need to be aware of, and actively manage, when working with it. The proper way to handle it, if you must use floats is to figure out what the expected amount of accumulated error is and allow for that in comparisons and printouts -- round off appropriately, compare for whether the difference is within that range rather than comparing for equality, etcetera.
There is no exact binary-floating-point representation of simple things like 1/10th, for example.
(As others have noted, you could rewrite the problem to avoid using the floating-point-based solution entirely, but since you asked specifically about working log() I wanted to address that question; apologies if I'm off target. Some of the other answers provide specific suggestions for how you might round off the result. That would "solve" this particular case, but as your floating operations get more complicated you'll have to continue to allow for roundoff accumulating at each step and either deal with the error at each step or deal with the cumulative error -- the latter being the more complicated but more accurate solution.)
If this is a serious problem for an application, folks sometimes use scaled fixed point instead (running financial computations in terms of pennies rather than dollars, for example). Or they use one of the "big number" packages which computes in decimal rather than in binary; those have their own round-off problems, but they round off more the way humans expect them to.
Why is Shannon's Entropy measure used in Decision Tree branching?
Entropy(S) = - p(+)log( p(+) ) - p(-)log( p(-) )
I know it is a measure of the no. of bits needed to encode information; the more uniform the distribution, the more the entropy. But I don't see why it is so frequently applied in creating decision trees (choosing a branch point).
Because you want to ask the question that will give you the most information. The goal is to minimize the number of decisions/questions/branches in the tree, so you start with the question that will give you the most information and then use the following questions to fill in the details.
For the sake of decision trees, forget about the number of bits and just focus on the formula itself. Consider a binary (+/-) classification task where you have an equal number of + and - examples in your training data. Initially, the entropy will be 1 since p(+) = p(-) = 0.5. You want to split the data on an attribute that most decreases the entropy (i.e., makes the distribution of classes least random). If you choose an attribute, A1, that is completely unrelated to the classes, then the entropy will still be 1 after splitting the data by the values of A1, so there is no reduction in entropy. Now suppose another attribute, A2, perfectly separates the classes (e.g., the class is always + for A2="yes" and always - for A2="no". In this case, the entropy is zero, which is the ideal case.
In practical cases, attributes don't typically perfectly categorize the data (the entropy is greater than zero). So you choose the attribute that "best" categorizes the data (provides the greatest reduction in entropy). Once the data are separated in this manner, another attribute is selected for each of the branches from the first split in a similar manner to further reduce the entropy along that branch. This process is continued to construct the tree.
You seem to have an understanding of the math behind the method, but here is a simple example that might give you some intuition behind why this method is used: Imagine you are in a classroom that is occupied by 100 students. Each student is sitting at a desk, and the desks are organized such there are 10 rows and 10 columns. 1 out of the 100 students has a prize that you can have, but you must guess which student it is to get the prize. The catch is that everytime you guess, the prize is decremented in value. You could start by asking each student individually whether or not they have the prize. However, initially, you only have a 1/100 chance of guessing correctly, and it is likely that by the time you find the prize it will be worthless (think of every guess as a branch in your decision tree). Instead, you could ask broad questions that dramatically reduce the search space with each question. For example "Is the student somewhere in rows 1 though 5?" Whether the answer is "Yes" or "No" you have reduced the number of potential branches in your tree by half.
I have a project where I am asked to develop an application to simulate how different page replacement algorithms perform (with varying working set size and stability period). My results:
Vertical axis: page faults
Horizontal axis: working set size
Depth axis: stable period
Are my results reasonable? I expected LRU to have better results than FIFO. Here, they are approximately the same.
For random, stability period and working set size doesnt seem to affect the performance at all? I expected similar graphs as FIFO & LRU just worst performance? If the reference string is highly stable (little branches) and have a small working set size, it should still have less page faults that an application with many branches and big working set size?
More Info
My Python Code | The Project Question
Length of reference string (RS): 200,000
Size of virtual memory (P): 1000
Size of main memory (F): 100
number of time page referenced (m): 100
Size of working set (e): 2 - 100
Stability (t): 0 - 1
Working set size (e) & stable period (t) affects how reference string are generated.
|-----------|--------|------------------------------------|
0 p p+e P-1
So assume the above the the virtual memory of size P. To generate reference strings, the following algorithm is used:
Repeat until reference string generated
pick m numbers in [p, p+e]. m simulates or refers to number of times page is referenced
pick random number, 0 <= r < 1
if r < t
generate new p
else (++p)%P
UPDATE (In response to #MrGomez's answer)
However, recall how you seeded your input data: using random.random,
thus giving you a uniform distribution of data with your controllable
level of entropy. Because of this, all values are equally likely to
occur, and because you've constructed this in floating point space,
recurrences are highly improbable.
I am using random, but it is not totally random either, references are generated with some locality though the use of working set size and number page referenced parameters?
I tried increasing the numPageReferenced relative with numFrames in hope that it will reference a page currently in memory more, thus showing the performance benefit of LRU over FIFO, but that didn't give me a clear result tho. Just FYI, I tried the same app with the following parameters (Pages/Frames ratio is still kept the same, I reduced the size of data to make things faster).
--numReferences 1000 --numPages 100 --numFrames 10 --numPageReferenced 20
The result is
Still not such a big difference. Am I right to say if I increase numPageReferenced relative to numFrames, LRU should have a better performance as it is referencing pages in memory more? Or perhaps I am mis-understanding something?
For random, I am thinking along the lines of:
Suppose theres high stability and small working set. It means that the pages referenced are very likely to be in memory. So the need for the page replacement algorithm to run is lower?
Hmm maybe I got to think about this more :)
UPDATE: Trashing less obvious on lower stablity
Here, I am trying to show the trashing as working set size exceeds the number of frames (100) in memory. However, notice thrashing appears less obvious with lower stability (high t), why might that be? Is the explanation that as stability becomes low, page faults approaches maximum thus it does not matter as much what the working set size is?
These results are reasonable given your current implementation. The rationale behind that, however, bears some discussion.
When considering algorithms in general, it's most important to consider the properties of the algorithms currently under inspection. Specifically, note their corner cases and best and worst case conditions. You're probably already familiar with this terse method of evaluation, so this is mostly for the benefit of those reading here whom may not have an algorithmic background.
Let's break your question down by algorithm and explore their component properties in context:
FIFO shows an increase in page faults as the size of your working set (length axis) increases.
This is correct behavior, consistent with Bélády's anomaly for FIFO replacement. As the size of your working page set increases, the number of page faults should also increase.
FIFO shows an increase in page faults as system stability (1 - depth axis) decreases.
Noting your algorithm for seeding stability (if random.random() < stability), your results become less stable as stability (S) approaches 1. As you sharply increase the entropy in your data, the number of page faults, too, sharply increases and propagates the Bélády's anomaly.
So far, so good.
LRU shows consistency with FIFO. Why?
Note your seeding algorithm. Standard LRU is most optimal when you have paging requests that are structured to smaller operational frames. For ordered, predictable lookups, it improves upon FIFO by aging off results that no longer exist in the current execution frame, which is a very useful property for staged execution and encapsulated, modal operation. Again, so far, so good.
However, recall how you seeded your input data: using random.random, thus giving you a uniform distribution of data with your controllable level of entropy. Because of this, all values are equally likely to occur, and because you've constructed this in floating point space, recurrences are highly improbable.
As a result, your LRU is perceiving each element to occur a small number of times, then to be completely discarded when the next value was calculated. It thus correctly pages each value as it falls out of the window, giving you performance exactly comparable to FIFO. If your system properly accounted for recurrence or a compressed character space, you would see markedly different results.
For random, stability period and working set size doesn't seem to affect the performance at all. Why are we seeing this scribble all over the graph instead of giving us a relatively smooth manifold?
In the case of a random paging scheme, you age off each entry stochastically. Purportedly, this should give us some form of a manifold bound to the entropy and size of our working set... right?
Or should it? For each set of entries, you randomly assign a subset to page out as a function of time. This should give relatively even paging performance, regardless of stability and regardless of your working set, as long as your access profile is again uniformly random.
So, based on the conditions you are checking, this is entirely correct behavior consistent with what we'd expect. You get an even paging performance that doesn't degrade with other factors (but, conversely, isn't improved by them) that's suitable for high load, efficient operation. Not bad, just not what you might intuitively expect.
So, in a nutshell, that's the breakdown as your project is currently implemented.
As an exercise in further exploring the properties of these algorithms in the context of different dispositions and distributions of input data, I highly recommend digging into scipy.stats to see what, for example, a Gaussian or logistic distribution might do to each graph. Then, I would come back to the documented expectations of each algorithm and draft cases where each is uniquely most and least appropriate.
All in all, I think your teacher will be proud. :)
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