Network flow problem with binary in AMPL? - constraints

this is an AMPL model, I'm pretty new in this, so I'm doing a classical problem of logistics, a network flow problem, where I have to find the least expensive way to transport available blood donations in a net of cities where there are different costs in the edges. So I have to minimize its objective function (maybe is better understandable reading the code).
I solve the problem right now, but now I'm facing the second task, where a fixed cost equal to 10 must be paid for each edge used for transporting blood donations (in addition to the shipping costs). For what I have understood, the question is easy. In practice I just have to add 10*numberOfEdgeUsed to the objective funcion. I want to do it in the correct way trying to add a binary variable for every edge, 1 if an edge is used, 0 if not. I'm pretty new to this kind of programmation, and I don't know how to do it.
Any help is welcome. I put just the .mod code, I don't put the .dat file becouse is not necessary.
This is the code of the first task, I have to modify this:
set Cities;
set Origins within (Cities);
set Destinations within (Cities);
set Link within (Cities cross Cities);
param Costs{Link};
param DemSup{Cities};
param fixedCost{(i, j) in Link} = 10;
var y{Link}, binary;
var Ship{Link} >= 0, <= 1000;
minimize Total_Cost: sum{(i,j) in Link} fixedCost[i,j] * y[i,j] + sum {(i,j) in Link} (Costs[i,j] * Ship[i,j]);
subject to Supply {i in Origins}: - sum {(i,k) in Link} Ship[i,k] >= DemSup[i];
subject to Demand {i in Destinations}: sum {(j,i) in Link} Ship[j,i] - sum {(i,k) in Link} Ship[i,k] == DemSup[i];

You need to add the implication: if y=0 then the corresponding link can not be used. This can be formulated as the constraint:
Ship[i,j] <= 1000 * y[i,j]
(or better Ship[i,j] <= Ship[i,j].ub * y[i,j])

Related

When determining time complexity are variables like n etc always given to an input?

The short version of this long post is when determining time complexity are variables like n etc always given to an input? If not, how else can you define variables?
I'm leaving the long version of my question below in case it helps anyone.
NOTE: I'm aware the question has already been asked here but I'm not satisfied with the answers. The accepted answer ignores the part of the question that the recursion essentially creates a balanced binary tree, while the second answer wrongly presumes that the author used the input as the definition of n rather than the number of levels of calls in the binary tree. (although it may be making the correct point that the difference is the definition of n and its possible the author slipped up or just confused me instead)
I'm comparing these two examples on edition 6 of Cracking the Coding Interview
Pages 44-45 (VI Big O Recursive Runtime section)
int f(int n){
if (n <= 1){
return 1;
}
return f(n-1) + f(n-1);
}
In this case the author defined n as the number of levels created through the recursive calls.
Pages 49-50 (VI Big O Example 9)
Assume the input is a balanced binary search tree
int sum(Node node){
if(node == null){
return 0;
}
return sum(node.left) + node.value + sum(node.right);
}
Here the author defines n as the number of nodes in the tree and states that therefore the depth of the tree is log n. (and since 2^logn equals n its O(n)
So here's the number of calls and depth of the tree based on the input in the first example
Input Calls Depth (author started counting from 0, used the term levels)
1 1 0
2 2 1
3 7 2
4 15 3
etc
I'm actually confused why the author was able to choose the depth of the tree as n because in the past I've always seen an input used as n? (it also seems meaningless b/c the depth is the input minus 1) Was the 2nd answer in the question asked here actually correct instead of the author by using the proper definition of n as the input?
In the 2nd example above it seems sensible that n is the number of nodes in the tree and therefore it has the depth of n?
So I guess I'm asking if an input is always the proper criteria for defining n (or whatever term you want to use as the variable)? If not, how else can you define n? If the input is always used to define n I get why the answers would be different. If not, I'd be confused since the recursion in example 1 essentially does create a balanced binary tree which therefore also has a depth of log n.
Based on googling it does seem that n (etc) is supposed to refer to the input.
Explain Time Complexity?
https://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/
https://www.interviewcake.com/article/java/big-o-notation-time-and-space-complexity

Multi-commodity Flow

I have just been working on chapters of questions from the DPV textbook for my exam preparation. For one of them, I'm having some trouble but I have made some progress:
My solution:
I will be using linear programming to maximize x where
SUM {flow_i(s_i, v) for all i where v are nodes connecting to source s_i} >= x * d_i
subject to the constraints
for each edge e , f1+..fk <= c_e
for each edge e, flow_e >= 0
and some more constraints that I'm unsure of
I think I'm on the right track but I need some help getting a bit further. Should I be considering trying to use a super-node to reduce this to a regular max flow problem?
Any help would be great! Thank you.
Yes, one typical approach to multi-source, multi-sink commodity flow problems is to introduce a super-source and one super-sink. Then connect all the sources s1...sk to the super-source. Connect each sink t1,...tk to the super-sink.
Important: Give a very large capacity to all the edges leaving or entering any of the super-nodes.
Objective: Maximize the total throughoput. (Sum of flows over all edges leaving the sources 1..k)
Constraints:
Edge Capacity Constraints:
You already got this right.
for each edge e , f1+..fk <= c_e
*Flow Conservation (Flow-in == Flow_out):*
for each vertex v, sum of flow into v = sum of flow leaving v
Demand Satisfaction:
for each sink t_i, sum of flow into t_i (all edges ending in t_i) >= demand_i
Non-zero flows, which you already have.
Here's one accessible lecture handout that references your specific problem. Specifically, take a look at Example 2 in the handout.

How to find number of items dropped based on individual probabilities?

My goal is to independently calculate the number of items an enemy would drop after it is killed. For example, say there are 50 potions each with a 50% chance of being dropped, I'd like to randomly return a number from 0 to 50, based on independent trials.
Currently, this is the code I'm using:
int droppedItems(int n, float probability) {
int count = 0;
for (int x = 1; x <= n; ++x) {
if (random() <= probability) {
++count;
}
}
return count;
}
Where probability is a number from 0.0 to 1.0, random() returns 0.0 to 1.0, and n is the maximum number of items to be dropped. This is in C++ code, however, I'm actually using Visual Basic 6 - so there's no libraries to help with this.
This code works flawlessly. However, I'd like to optimize this so that if n happens to be 999999, it doesn't take forever (which it currently does).
Use the binomial distribution. Wiki - Binomial Distribution
Ideally, use the libraries for whatever language this pseudocode will be written in. There's no sense in reinventing the wheel unless of course you are trying to learn how to invent a wheel.
Specifically, you'll want something that will let you generate random values given a binomial distribution with a probability of success in any given trial and a number of trials.
EDIT :
I went ahead and did this (in python, since that's where I live these days). It relies on the very nice numpy library (hooray, abstraction!):
>>>import numpy
>>>numpy.random.binomial(99999,0.5)
49853
>>>numpy.random.binomial(99999,0.5)
50077
And, using timeit.Timer to check execution time:
# timing it across 10,000 iterations for 99,999 items per iteration
>>>timeit.Timer(stmt="numpy.random.binomial(99999,0.5)", setup="import numpy").timeit(10000)
0.00927[... seconds]
EDIT 2 :
As it turns out, there isn't a simple way to implement a random number generator based off of the binomial distribution.
There is an algorithm you can implement without library support which will generate random variables from the binomial distribution. You can view it here as a PDF
My guess is that given what you want to use it for (having monsters drop loot in a game), implementing the algorithm is not worth your time. There's room for fudge factor here!
I would change your code like this (note: this is not a binomial distribution):
Use your current code for small values, say n up to 100.
For n greater than one hundred, calculate the value of count for
100 using your current algorithm and then multiply the result by
n/100.
Again, if you really want to figure out how to implement the BTPE algorithm yourself, you can - I think the method I give above wins in the trade off between effort to write and getting "close enough".
As #IamChuckB pointed out already, the key word is binomial distribution. When the number of Bernoulli trials (number of items in your example) is large enough, a good approximation is the Poisson distribution, which is much simpler to calculate and draw numbers from (the exact algorithm is spelled out in the linked Wikipedia article).

Simple physics-based movement

I'm working on a 2D game where I'm trying to accelerate an object to a top speed using some basic physics code.
Here's the pseudocode for it:
const float acceleration = 0.02f;
const float friction = 0.8f; // value is always 0.0..1.0
float velocity = 0;
float position = 0;
move()
{
velocity += acceleration;
velocity *= friction;
position += velocity;
}
This is a very simplified approach that doesn't rely on mass or actual friction (the in-code friction is just a generic force acting against movement). It works well as the "velocity *= friction;" part keeps the velocity from going past a certain point. However, it's this top speed and its relationship to the acceleration and friction where I'm a bit lost.
What I'd like to do is set a top speed, and the amount of time it takes to reach it, then use them to derive the acceleration and friction values.
i.e.,
const float max_velocity = 2.0;
const int ticks; = 120; // If my game runs at 60 FPS, I'd like a
// moving object to reach max_velocity in
// exactly 2 seconds.
const float acceleration = ?
const float friction = ?
I found this question very interesting since I had recently done some work on modeling projectile motion with drag.
Point 1: You are essentially updating the position and velocity using an explicit/forward Euler iteration where each new value for the states should be a function of the old values. In such a case, you should be updating the position first, then updating the velocity.
Point 2: There are more realistic physics models for the effect of drag friction. One model (suggested by Adam Liss) involves a drag force that is proportional to the velocity (known as Stokes' drag, which generally applies to low velocity situations). The one I previously suggested involves a drag force that is proportional to the square of the velocity (known as quadratic drag, which generally applies to high velocity situations). I'll address each one with regard to how you would deduce formulas for the maximum velocity and the time required to effectively reach the maximum velocity. I'll forego the complete derivations since they are rather involved.
Stokes' drag:
The equation for updating the velocity would be:
velocity += acceleration - friction*velocity
which represents the following differential equation:
dv/dt = a - f*v
Using the first entry in this integral table, we can find the solution (assuming v = 0 at t = 0):
v = (a/f) - (a/f)*exp(-f*t)
The maximum (i.e. terminal) velocity occurs when t >> 0, so that the second term in the equation is very close to zero and:
v_max = a/f
Regarding the time needed to reach the maximum velocity, note that the equation never truly reaches it, but instead asymptotes towards it. However, when the argument of the exponential equals -5, the velocity is around 98% of the maximum velocity, probably close enough to consider it equal. You can then approximate the time to maximum velocity as:
t_max = 5/f
You can then use these two equations to solve for f and a given a desired vmax and tmax.
Quadratic drag:
The equation for updating the velocity would be:
velocity += acceleration - friction*velocity*velocity
which represents the following differential equation:
dv/dt = a - f*v^2
Using the first entry in this integral table, we can find the solution (assuming v = 0 at t = 0):
v = sqrt(a/f)*(exp(2*sqrt(a*f)*t) - 1)/(exp(2*sqrt(a*f)*t) + 1)
The maximum (i.e. terminal) velocity occurs when t >> 0, so that the exponential terms are much greater than 1 and the equation approaches:
v_max = sqrt(a/f)
Regarding the time needed to reach the maximum velocity, note that the equation never truly reaches it, but instead asymptotes towards it. However, when the argument of the exponential equals 5, the velocity is around 99% of the maximum velocity, probably close enough to consider it equal. You can then approximate the time to maximum velocity as:
t_max = 2.5/sqrt(a*f)
which is also equivalent to:
t_max = 2.5/(f*v_max)
For a desired vmax and tmax, the second equation for tmax will tell you what f should be, and then you can plug that in to the equation for vmax to get the value for a.
This seems like a bit of overkill, but these are actually some of the simplest ways to model drag! Anyone who really wants to see the integration steps can shoot me an email and I'll send them to you. They are a bit too involved to type here.
Another Point: I didn't immediately realize this, but the updating of the velocity is not necessary anymore if you instead use the formulas I derived for v(t). If you are simply modeling acceleration from rest, and you are keeping track of the time since the acceleration began, the code would look something like:
position += velocity_function(timeSinceStart)
where "velocity_function" is one of the two formulas for v(t) and you would no longer need a velocity variable. In general, there is a trade-off here: calculating v(t) may be more computationally expensive than simply updating velocity with an iterative scheme (due to the exponential terms), but it is guaranteed to remain stable and bounded. Under certain conditions (like trying to get a very short tmax), the iteration can become unstable and blow-up, a common problem with the forward Euler method. However, maintaining limits on the variables (like 0 < f < 1), should prevent these instabilities.
In addition, if you're feeling somewhat masochistic, you may be able to integrate the formula for v(t) to get a closed form solution for p(t), thus foregoing the need for a Newton iteration altogether. I'll leave this for others to attempt. =)
Warning: Partial Solution
If we follow the physics as stated, there is no maximum velocity. From a purely physical viewpoint, you've fixed the acceleration at a constant value, which means the velocity is always increasing.
As an alternative, consider the two forces acting on your object:
The constant external force, F, that tends to accelerate it, and
The force of drag, d, which is proportional to the velocity and tends to slow it down.
So the velocity at iteration n becomes: vn = v0 + n F - dvn-1
You've asked to choose the maximum velocity, vnmax, that occurs at iteration nmax.
Note that the problem is under-constrained; that is, F and d are related, so you can arbitrarily choose a value for one of them, then calculate the other.
Now that the ball's rolling, is anyone willing to pick up the math?
Warning: it's ugly and involves power series!
Edit: Why doe the sequence n**F** in the first equation appear literally unless there's a space after the n?
velocity *= friction;
This doesn't prevent the velocity from going about a certain point...
Friction increases exponentially (don't quote me on that) as the velocity increases, and will be 0 at rest. Eventually, you will reach a point where friction = acceleration.
So you want something like this:
velocity += (acceleration - friction);
position += velocity;
friction = a*exp(b*velocity);
Where you pick values for a and b. b will control how long it takes to reach top speed, and a will control how abruptly the friction increases. (Again, don't do your own research on this- I'm going from what I remember from grade 12 physics.)
This isn't answering your question, but one thing you shouldn't do in simulations like this is depend on a fixed frame rate. Calculate the time since the last update, and use the delta-T in your equations. Something like:
static double lastUpdate=0;
if (lastUpdate!=0) {
deltaT = time() - lastUpdate;
velocity += acceleration * deltaT;
position += velocity * deltaT;
}
lastUpdate = time();
It's also good to check if you lose focus and stop updating, and when you gain focus set lastUpdate to 0. That way you don't get a huge deltaT to process when you get back.
If you want to see what can be done with very simple physics models using very simple maths, take a look at some of the Scratch projects at http://scratch.mit.edu/ - you may get some useful ideas & you'll certainly have fun.
This is probably not what you are looking for but depending on what engine you are working on, it might be better to use a engine built by some one else, like farseer(for C#).
Note Codeplex is down for maintenance.

How to test randomness (case in point - Shuffling)

First off, this question is ripped out from this question. I did it because I think this part is bigger than a sub-part of a longer question. If it offends, please pardon me.
Assume that you have a algorithm that generates randomness. Now how do you test it?
Or to be more direct - Assume you have an algorithm that shuffles a deck of cards, how do you test that it's a perfectly random algorithm?
To add some theory to the problem -
A deck of cards can be shuffled in 52! (52 factorial) different ways. Take a deck of cards, shuffle it by hand and write down the order of all cards. What is the probability that you would have gotten exactly that shuffle? Answer: 1 / 52!.
What is the chance that you, after shuffling, will get A, K, Q, J ... of each suit in a sequence? Answer 1 / 52!
So, just shuffling once and looking at the result will give you absolutely no information about your shuffling algorithms randomness. Twice and you have more information, Three even more...
How would you black box test a shuffling algorithm for randomness?
Statistics. The de facto standard for testing RNGs is the Diehard suite (originally available at http://stat.fsu.edu/pub/diehard). Alternatively, the Ent program provides tests that are simpler to interpret but less comprehensive.
As for shuffling algorithms, use a well-known algorithm such as Fisher-Yates (a.k.a "Knuth Shuffle"). The shuffle will be uniformly random so long as the underlying RNG is uniformly random. If you are using Java, this algorithm is available in the standard library (see Collections.shuffle).
It probably doesn't matter for most applications, but be aware that most RNGs do not provide sufficient degrees of freedom to produce every possible permutation of a 52-card deck (explained here).
Here's one simple check that you can perform. It uses generated random numbers to estimate Pi. It's not proof of randomness, but poor RNGs typically don't do well on it (they will return something like 2.5 or 3.8 rather ~3.14).
Ideally this would be just one of many tests that you would run to check randomness.
Something else that you can check is the standard deviation of the output. The expected standard deviation for a uniformly distributed population of values in the range 0..n approaches n/sqrt(12).
/**
* This is a rudimentary check to ensure that the output of a given RNG
* is approximately uniformly distributed. If the RNG output is not
* uniformly distributed, this method will return a poor estimate for the
* value of pi.
* #param rng The RNG to test.
* #param iterations The number of random points to generate for use in the
* calculation. This value needs to be sufficiently large in order to
* produce a reasonably accurate result (assuming the RNG is uniform).
* Less than 10,000 is not particularly useful. 100,000 should be sufficient.
* #return An approximation of pi generated using the provided RNG.
*/
public static double calculateMonteCarloValueForPi(Random rng,
int iterations)
{
// Assumes a quadrant of a circle of radius 1, bounded by a box with
// sides of length 1. The area of the square is therefore 1 square unit
// and the area of the quadrant is (pi * r^2) / 4.
int totalInsideQuadrant = 0;
// Generate the specified number of random points and count how many fall
// within the quadrant and how many do not. We expect the number of points
// in the quadrant (expressed as a fraction of the total number of points)
// to be pi/4. Therefore pi = 4 * ratio.
for (int i = 0; i < iterations; i++)
{
double x = rng.nextDouble();
double y = rng.nextDouble();
if (isInQuadrant(x, y))
{
++totalInsideQuadrant;
}
}
// From these figures we can deduce an approximate value for Pi.
return 4 * ((double) totalInsideQuadrant / iterations);
}
/**
* Uses Pythagoras' theorem to determine whether the specified coordinates
* fall within the area of the quadrant of a circle of radius 1 that is
* centered on the origin.
* #param x The x-coordinate of the point (must be between 0 and 1).
* #param y The y-coordinate of the point (must be between 0 and 1).
* #return True if the point is within the quadrant, false otherwise.
*/
private static boolean isInQuadrant(double x, double y)
{
double distance = Math.sqrt((x * x) + (y * y));
return distance <= 1;
}
First, it is impossible to know for sure if a certain finite output is "truly random" since, as you point out, any output is possible.
What can be done, is to take a sequence of outputs and check various measurements of this sequence against what is more likely. You can derive a sort of confidence score that the generating algorithm is doing a good job.
For example, you could check the output of 10 different shuffles. Assign a number 0-51 to each card, and take the average of the card in position 6 across the shuffles. The convergent average is 25.5, so you would be surprised to see a value of 1 here. You could use the central limit theorem to get an estimate of how likely each average is for a given position.
But we shouldn't stop here! Because this algorithm could be fooled by a system that only alternates between two shuffles that are designed to give the exact average of 25.5 at each position. How can we do better?
We expect a uniform distribution (equal likelihood for any given card) at each position, across different shuffles. So among the 10 shuffles, we could try to verify that the choices 'look uniform.' This is basically just a reduced version of the original problem. You could check that the standard deviation looks reasonable, that the min is reasonable, and the max value as well. You could also check that other values, such as the closest two cards (by our assigned numbers), also make sense.
But we also can't just add various measurements like this ad infinitum, since, given enough statistics, any particular shuffle will appear highly unlikely for some reason (e.g. this is one of very few shuffles in which cards X,Y,Z appear in order). So the big question is: which is the right set of measurements to take? Here I have to admit that I don't know the best answer. However, if you have a certain application in mind, you can choose a good set of properties/measurements to test, and work with those -- this seems to be the way cryptographers handle things.
There's a lot of theory on testing randomness. For a very simple test on a card shuffling algorithm you could do a lot of shuffles and then run a chi squared test that the probability of each card turning up in any position was uniform. But that doesn't test that consecutive cards aren't correlated so you would also want to do tests on that.
Volume 2 of Knuth's Art of Computer Programming gives a number of tests that you could use in sections 3.3.2 (Empirical tests) and 3.3.4 (The Spectral Test) and the theory behind them.
The only way to test for randomness is to write a program that attempts to build a predictive model for the data being tested, and then use that model to try to predict future data, and then showing that the uncertainty, or entropy, of its predictions tend towards maximum (i.e. the uniform distribution) over time. Of course, you'll always be uncertain whether or not your model has captured all of the necessary context; given a model, it'll always be possible to build a second model that generates non-random data that looks random to the first. But as long as you accept that the orbit of Pluto has an insignificant influence on the results of the shuffling algorithm, then you should be able to satisfy yourself that its results are acceptably random.
Of course, if you do this, you might as well use your model generatively, to actually create the data you want. And if you do that, then you're back at square one.
Shuffle alot, and then record the outcomes (if im reading this correctly). I remember seeing comparisons of "random number generators". They just test it over and over, then graph the results.
If it is truly random the graph will be mostly even.
I'm not fully following your question. You say
Assume that you have a algorithm that generates randomness. Now how do you test it?
What do you mean? If you're assuming you can generate randomness, there's no need to test it.
Once you have a good random number generator, creating a random permutation is easy (e.g. Call your cards 1-52. Generate 52 random numbers assigning each one to a card in order, and then sort according to your 52 randoms) . You're not going to destroy the randomness of your good RNG by generating your permutation.
The difficult question is whether you can trust your RNG. Here's a sample link to people discussing that issue in a specific context.
Testing 52! possibilities is of course impossible. Instead, try your shuffle on smaller numbers of cards, like 3, 5, and 10. Then you can test billions of shuffles and use a histogram and the chi-square statistical test to prove that each permutation is coming up an "even" number of times.
No code so far, therefore I copy-paste a testing part from my answer to the original question.
// ...
int main() {
typedef std::map<std::pair<size_t, Deck::value_type>, size_t> Map;
Map freqs;
Deck d;
const size_t ntests = 100000;
// compute frequencies of events: card at position
for (size_t i = 0; i < ntests; ++i) {
d.shuffle();
size_t pos = 0;
for(Deck::const_iterator j = d.begin(); j != d.end(); ++j, ++pos)
++freqs[std::make_pair(pos, *j)];
}
// if Deck.shuffle() is correct then all frequencies must be similar
for (Map::const_iterator j = freqs.begin(); j != freqs.end(); ++j)
std::cout << "pos=" << j->first.first << " card=" << j->first.second
<< " freq=" << j->second << std::endl;
}
This code does not test randomness of underlying pseudorandom number generator. Testing PRNG randomness is a whole branch of science.
For a quick test, you can always try compressing it. Once it doesn't compress, then you can move onto other tests.
I've tried dieharder but it refuses to work for a shuffle. All tests fail. It is also really stodgy, it wont let you specify the range of values you want or anything like that.
Pondering it myself, what I would do is something like:
Setup (Pseudo code)
// A card has a Number 0-51 and a position 0-51
int[][] StatMatrix = new int[52][52]; // Assume all are set to 0 as starting values
ShuffleCards();
ForEach (card in Cards) {
StatMatrix[Card.Position][Card.Number]++;
}
This gives us a matrix 52x52 indicating how many times a card has ended up at a certain position. Repeat this a large number of times (I would start with 1000, but people better at statistics than me may give a better number).
Analyze the matrix
If we have perfect randomness and perform the shuffle an infinite number of times then for each card and for each position the number of times the card ended up in that position is the same as for any other card. Saying the same thing in a different way:
statMatrix[position][card] / numberOfShuffle = 1/52.
So I would calculate how far from that number we are.

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