i was trying to solve this transportation problem in cplex
enter image description here
here is my OPL code
int p=...;
int q=...;
range i=1..p;
float a[i];
range j=1..q;
float b[j];
float c[i][j];
dvar boolean x[i][j];
minimize
sum(l in i,m in j)x[l][m]*c[l][m];
subject to {
forall (l in i)
sum(m in j) x[l][m] <= a[l];
forall (m in j)
sum(l in i) x[l][m] >= b[m];
}
enter image description here
this is my .dat
I keep getting this error " Data element "a" is already set".
since you declare a in the .dat you should replace
float a[i];
by
float a[i]=...;
Related
I want to add constraints to my Cplex model, that ensures that a bunch of arrays are pairwise different. That is, at least one entry should differ in the two.
(To clarify: The IloNumVarArray h represents an n x m matrix and the constraints should ensure that no two rows are identical)
My code below has two errors (at least) that I can't seem to solve:
- First, there is 'no suitable conversion function from IloNumVar to IloNum',
- Second, it is not allowed to use the != operator to compare IloNumArrays.
IloNumVarArray h(env, n*m);
IloNumArray temp1(env, m);
IloNumArray temp2(env, m);
for (int i = 0; i < n - 1; i++) {
temp1.clear();
temp2.clear();
for (int k = 0; k < n - i; k++)
for (int j = 0; j < m; j++) {
temp1[j] = h[j + i * m];
temp2[j] = h[j + (i + k) * m];
}
model.add(temp1 != temp2);
}
So how can I change temp1 and temp2 such that it is possible to copy from h, and compare the two?
(or do it completely different)
I am quite new to Cplex and I would appreciate any help/suggestions
you could use logical constraints.
Let me give you an example in OPL CPLEX that you could adapt to C++
int n=3;
int m=2;
range N=1..n;
range M=1..m;
float epsilon=0.0001;
dvar float temp1[N][M] in 0..10;
dvar float temp2[N][M] in 0..10;
minimize sum(i in N,j in M) (temp1[i][j]+temp2[i][j]);
subject to
{
// at least for one (i,j) the 2 arrays are different
1<=sum(i in N,j in M) (abs(temp1[i][j]-temp2[i][j])>=epsilon);
}
So essentially, I'm trying to code a small block that should create or generate coefficients for polynomial n-degree that can be represented through vector which is a=[a0, a1..an] given the basic formula is
Well, the issue is when doing a function to get value of polynomial P from point "x" it returns value from entirely using Horner's Rule which it's not the same result as intended although not sure which one I should put on. The math basic explanation tells me at least something out of:
E.g: n=2; (A[i] stack terms by 4, 2, 1) and calculates P(value of x) = 4 * x ^ 0 – 2 * x ^ 1 + 1 * x ^ 2 = 4 – 2x + x2 = x2 – 2x + 4 = 4
With other words , can't find the culprit when for "x" value is meant to go through "i" variable numbered wrong by exponent and the result gets output for P(0)=7 while it shouldn't be and concrete as in P(0) = 0 ^ 2 – 2 * 0 + 4 = 4
Here's a little snippet went through so far, I would appreciate if someone could point me in the right direction.
double horner(const double&x, const int& n, const double& nn) {
if (n < 0)
return nn;
else {
double m; cin>>m;
return horner(x, n-1, nn*x+m);
}
}
int main() {
int n;double x;
cout << "n=";cin >> n;
cout << "x=";cin >> x;
cout << "P(0)=" << horner(x, n, 0);
return 0;
}
Edit: My brain farted for a moment somewhere while was coding and continuously revising the case, I forgot to mention what exactly are the parts of each variables for the program to avoid confusion yes, so:
n, polynomial grade;
p, polynomial coefficient;
x, the point in which evaluates;
And here given the grade equation to polynomial
which any negative and positive input terms adding by exponent to these coefficients are the steps that holds the result exception, hence the reason Horner's rule that it reduces the number of multiplication operations.
Edit: After few hours, managed to fix with polynomial evaluating issue, the only question remains how I'd suppose to generate coefficients using method std::vector ?
float honer(float p[], int n, float x)
{
int i;
float val;
val = p[n];
for (i = n - 1; i >= 0; i--)
val = val * x + p[i];
return val;
}
int main()
{
float p[20]; // Coefficient of the initial polynomial
int n; // Polynomial degree -n
float x; // Value that evaluates P -> X
cout << "(n) = ";
cin >> n;
for (int i = n; i >= 0; i--)
{
cout << "A[" << i << "]=";
cin >> p[i];
}
cout << "x:= ";
cin >> x;
cout << "P(" << x << ")=" << honer(p, n, x);
//Result (input):
//n: 2,
//P[i]: 4, -2, 1 -> x: 0, 1
// Result (output):
//P() = 4
}
Expect some certain output scenarios given below input if assigned:
I want to find the sum up to the 'n'th term for the following series:
(1/2)+((1*3)/(2*4))+((1*3*5)/(2*4*6))....
So, I wrote the following program in c++ :
#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int main()
{
int p=1, k=1, n=0;
float h=0;
cout<<"Enter the term: ";
cin>>n;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=i; j++)
{
p*=((2*j)-1);
k*=(2*j);
}
h+=(p/k);
p=1;
k=1;
}
cout<<"The sum is : "<<h;
return 0;
getch();
}
However, the output of the program always gives me '0'. I can't figure out the problem with the program.
N.B. I'm new to programming.
The problem here is that you haven't declared p and k as float or doubleor explicitly cast them as such before the calculation and assignment to h.
What's happening is for every iteration of the loop p < k (by nature of the problem) since p and k are both declared as int, p / k = 0. So you're just summing 0 for every iteration.
Either declare p and k as float or double or do this:
h += ((float) p) / ((float) k)
Also, for this specific problem I assume you're looking for precision, so be wary and look into that as well Should I use double or float?
implicit conversion and type casting are a trap where all newbies fall.
in the instruction:
h += p/k;
the compiler performs an integer division first, then a promotion of the result to floating point type.
and since:
p < k ; for all i,j < n
then:
res = (p / k) < 1 => truncates to 0; // by integer division
thus:
sum(1->n) of p/k = sum (1->n) 0 = 0;
finally:
h = conversion to float of (0) = 0.0f;
that's why you have the result of 0.0f at the end.
the solution:
1- first of all you need to use the natural type for floating point of c++ which is "double" (under the hood c++ promotes float to double, so use it directly).
2- declare all your variable as double, except the number of terms n:
3- the number of terms is never negative, you need to express that in your code by declaring it as an unsigned int.
4- if you do step 3, make sure to catch overflow errors, that is if the user enters a negative number your risk to have a very big number in "n", expel : n =-1 converts to 0xffffffff positive number.
5- engineer your code sometimes is better.
6- include only the headers that you need, and avoid a importing any namespace in your global namespace.
here is how i think you should write your program.
#include <iostream>
double sum_serie(unsigned int n)
{
double prod = 1.0, sum = 0.0;
for (double c=1; c<=n ; c++)
{
prod *= ( ( 2*c ) - 1 ) / ( 2*c ); // remark the parenthesis
sum += prod;
}
return sum;
}
int main()
{
unsigned int n = 0;
int temp = 0;
std::cout << " enter the number of terms n: ";
std::cin >> temp;
if (temp > 0)
n = temp; // this is how you catch overflow
else
{
std::cout << " n < 0, no result calculated " << std::endl;
return 0;
}
std::cout << " the result is sum = " << sum_serie(n) << std::endl;
return 0;
}
I know that the question was about the implicit conversion and casting in C++, but even the way of writing a code can show you what bugs you have in it, so try to learn a proper way of expressing your ideas into code, debugging comes natural afterward.
Good Luck
So I can calculate a sum of sequence without accuracy E.
int t=1, x, k;
float sum, a, result, factorial=1, E=0.0001;
for(k=0;k<=(n);k++){
while(t<=n){
factorial*=t;
t++;
}
sum=(pow(-x,k))/factorial;
sum+=sum;
//while(fabs(sum-???)<E){
// result=sum;
//}
}
So I know sum of sequence sum(k). But to calculate with accurace E, I must know sum of previous elements sum(k-1). How to get sum(k-1) from for loop?
Sorry for english.
is this a taylor series for e ^ (-x) ? if so you've written it out wrong. i don't think what you've got will converge.
http://www.efunda.com/math/taylor_series/exponential.cfm
e ^ (-x) is 1 + (-x) + (-x)^2/2! + (-x)^3/3! + ...
double calculate_power_of_e(double xx, double accuracy) {
double sum(1.0);
double term(1.0);
for (long kk=1; true; ++kk) {
term *= (-xx) / kk;
sum += term;
if (fabs(term) < accuracy)
break;
}
return sum;
}
printf("e^(-x)" = %.4f\n", calculate_power_of_e(5.0, .0001));
First a remark about the power formula that you apply: according to wikipedia you should add the terms pow(-x,k)/(k!) and not pow(-x,k)/(n!).
This leads to a small optimisation of your code: as k! = k * (k-1)! we can avoid the inner while loop and a lot of useless multiplications.
By the way, there is also an error in the way you build the sum: you always erase the previous result, and then add a second time the current term.
Once this is corrected, you just have to take care of an additional variable:
double myexpo(double x, int n=100) {
int k;
double sum = 1.0, pvsum, factorial = 1.0, E = 0.0001;
for (k = 1; k <= (n); k++){ // start with 1
pvsum = sum;
factorial *= k; // don't calculate factorial for 0.
sum += (pow(-x, k)) / factorial;
if (k > 1 && fabs(sum - pvsum) < E) { // check if diff is small enough
cout << k << " iterations" << endl;
break; // interupt the for loop if it's precise enough
}
}
return sum; // at the end of the loop sum is the best approximation
}
You can test this function with this:
double x;
do {
cout << "Enter number: ";
cin >> x;
cout << myexpo(x) << endl;
cout << exp(-x) << endl;
} while (x > 0);
Remark: I'd suggest to either use double or to use the f suffix for the float litterals (e.g. 0.001f), even if it works as is.
Check when the absolute value of the term becomes smaller than your desired accuracy.
double sum = 0, x = 1, k = 0, E = 0.0001, fact = 1;
while(true){
double term = pow(-x, k) / fact;
if(fabs(term) < E)
break;
sum += term;
fact *= (++k);
}
printf("e^(-x) = %.4f", sum);
When the term is insignificant compare to 1.0, stop looping.
By using recursion, and |x| is not too big, the smallest terms are summed first.
e(x) = 1 + x/1! + x*x/2! + x*x*x/3! + ...
double my_exp_term(double x, double term, unsigned n) {
if (term + 1.0 == 1.0) return term;
n++;
return term + my_exp_term(x, term*x/n, n);
}
double my_exp(double x) {
return 1.0 + my_exp_term(x, x, 1);
}
double y = my_exp(-1);
Exponential function
i want to implement crank-nicolson method in c++ according to that :
−ru(i−1,j+1) + 2(1 + r)u(i,j+1) − ru(i+1,j+1) = 2(1 − r)u(i,j) + r
(u(i−1,j)+ u(i+1,j) )
I use gauss-jordan method to solve the system but i can't figure how to implement the above formula.
const double pi=3.14159265;
double f (double x){
return sin(pi*x);
}
using namespace std;
//gauss-jordan
double* gauss(int n ,double **a){
double factor;
double *b,*x;
x=new double[n];
b=new double[n];
for (int k=1;k<=n;k++) {
for (int i=1;i<=n;i++) {
if (i!=k) {
factor=a[i][k]/a[k][k];
for (int j=1; j<=n; j++ ) {
a[i][j]=a[k][j]*factor-a[i][j];
}
b[i]=b[k]*factor -b[i];
}
}
}
for (int i=1;i<=n;i++) {
x[i]=b[i]/a[i][i];
}
return x;
}
int main()
{
int i,j,n,m,xd,td;
double h,k,r;
double **t,**p;
//----- initialize double pointer------------------
double **u;
u=new double *[m];
for (int o=1;o<=m;o++){
u[o]=new double [n];
}
//----- end of initialization-----------------------
cout <<"\nEnter the value of x dimension : \n";
cin >> xd;
cout <<"\nEnter the step size h for x dimension : \n";
cin >>h;
cout <<"\nEnter the value of time dimension : \n";
cin >>td;
cout<<"\nEnter the step k of time dimension : \n";
cin >>k;
n=xd/h -1.0;
m=td/k -1.0;
cout <<"\n\nThe internal elements of x dimension are :"<<n;
cout <<"\nThe internal elements of t dimension are :"<<m;
r=k/(h*h);
cout <<"\nThe r value is : "<<r;
//initial conditions
for (j=0;j<=m;j++){
u[0][m]=0;
u[10][m]=0;
}
//get the function
for (i=1;i<n;i++){
u[i][0]=f(i*h);
}
//apply crank-nicolson
for (i=1;i<n;i++){
for (j=1;j<n;j++){
-r*u[i-1][j+1] +2.0*(1.0+r)*u[i][j+1] -r*u[i+1][j+1]=2.0*(1.0-r)*u[i][j] +r*(u[i-1][j]+u[i+1][j]);
} // here i can't figure the steps i must follow in order for this to work
//-----delete double pointer-------------
for(int o=1;o<m;o++){
delete [] u[o];
delete [] u;
}
//---------------------------------------
return 0;
}
I am assuming that the variable j represents the time steps. In order to implement Crank-Nicolson, you have to pose the problem as a system of linear equations and solve it. The matrix corresponding to the system will be of tridiagonal form, so it is better to use Thomas' algorithm rather than Gauss-Jordan.
The linear system will be of the form A x = b, with x being the vector (..., u(i-1, j+1), u(i, j+1), u(i+1, j+1), ...) and b being the vector (..., r u(i−1,j), 2(1 − r)u(i,j), r u(i+1,j), ...). The i-th row of the matrix A will be of the form
(0, ..., 0, −r, 2(1 + r), −r, 0, ..., 0). You will have to be careful with the first and last rows, where you'll have to substitute the boundary conditions for your problem.
A good reference for finite difference methods and Crank-Nicolson in particular is the book by John Strikwerda.
Hope this helps.
C++ is not an equation solving system. = does not have the meaning of equals as it does have in math, but mean assignment.
As such having anything complex like you have on the left side makes no sense, what you probably want to do is to solve the equation so there is a single variable that is being assigned to, possibly with an equation solving program.