I have an LP in CPLEX LP format, in file LP.tmp
Maximize
obj: x0 + 2 x1 + 3 x2 + 4 x3 + 5 x4 + 7 x5
Subject To
c0: 1 x0 + 1 x1 + 1 x2 + 1 x3 + 1 x4 + 1 x5 + 1 x6 + 1 x7 + 1 x8 = 0
End
On this I call soplex -X -x -o0 -f0 -v3 LP.tmp
This is obviously unbounded, but calling Soplex gives me the answer (with some other lines).
SoPlex status : problem is solved [optimal]
Solving time (sec) : 0.00
Iterations : 0
Objective value : 0.00000000e+00
Primal solution (name, value):
All other variables are zero (within 1.0e-16). Solution has 0 nonzero entries.
Background: Originally, I had objective 0, but box constraints, and I always got infeasible. So I reduced everything, until I arrived at the above.
What am I doing wrong?
All variables are non-negative by default in the lp file format, see https://www.ibm.com/support/knowledgecenter/SSSA5P_12.5.0/ilog.odms.cplex.help/CPLEX/FileFormats/topics/LP.html
Therefore, your constraint fixes all variables to 0. As soon as you change the coefficient of any of the variables but x5 to -1 or add a bounds section where you define it to be free, e.g., x1 free, SoPlex claims unboundedness and provides a valid primal ray.
This model is not unbounded. There are implicit bounds of 0 on all variables, so the only feasible and hence optimal solution is the one SoPlex returns.
In the .lp data format, all variables are non-negative by default.
Related
I encountered this problem practicing for an upcoming national competition. The problem goes as follows: You need to create a mixture of two ingredients being in relation to 1:1. You are given N different mixtures, each having its own weight Wi, and its relation in the mixture between the ingredients Mi, Ti (Each value, N, Wi, Mi, and Ti, will be less than 100). We need to find the biggest possible weight of the final mixture, keeping the relation to 1:1. We can take from each given mixture how much we want, we don't necessarily need to take the whole mixture, we can take some portion of it.
So with the given relation 1:1 in the final mixture, we know that we need to have an equal amount of weight from both ingredients possible. After that I need to know if I take K grams of some mixture, how much weight that is for ingredients A and B. So I came up with the following formula:
Let W be the weight in grams, and M and T be the relation between the ingredients respectively. If we want to take K (K <= W) grams we have the following:
Weight of ingredient A = M * (K / (M+T))
Weight of ingredient B = T * (K / (M+T))
#include <bits/stdc++.h>
using namespace std;
class state{
public:
int weight;
int A;
int B;
};
int n;
vector<state> arr;
double ans= 0;
void f(double weight_A, double weight_B, int idx){
if(weight_A == weight_B)
ans = max(ans, weight_A + weight_B);
if(idx >= n)
return;
int weight = arr[idx].weight, relA = arr[idx].A, relB = arr[idx].B;
for(int K = 0; K <= weight; K++){
f(weight_A + relA * (K * 1.0/(relA + relB)), weight_B + relB * (K * 1.0/(relA + relB)), idx+1);
}
}
int main(){
cin>>n;
for(int i = 0; i < n; i++){
state in;
cin>>in.weight>>in.A>>in.B;
arr.push_back(in);
}
f(0.0, 0.0, 0);
cout<<fixed<<setprecision(8);
cout<<ans<<endl;
}
The problem I encountered was that we don't necessarily need to take integer weights, some times to achieve the maximum possible weight of the final product we need to take decimal weights. Let's take a look at this example:
5
14 3 2
4 1 3
4 2 2
6 6 1
10 4 3
We have N = 5, and in each row are given 3 integers, Wi, Mi, and Ti. The weight of the ith mixture and its relation. My solution for this example gives 20.0000, and the correct solution for the above example is 20.85714286. Looking back my initial idea won't work because of the decimal numbers. I suppose there is some formula but I can't figure it out, can anyone help?
This is a Linear Programming problem, so you can solve it by constructing the problem in standard form, and then solve it with an optimization algorithm, like the simplex algorithm.
The objective is to maximize the quantity of medicine (from the original problem), that is the sum of quantities taken from each jar (I'll call the quantities x1, x2, ...).
The quantities are bounded to be lower than the weight Wi available in each jar.
The constraint is that the total amount of honey (first ingredient) is equal to the total amount of tahini (second ingredient). This would mean that:
sum(Mi/(Mi+Ti)*xi) = sum(Ti/(Mi+Ti)*xi)
You can take the second summation to the LHS and get:
sum((Mi-Ti)/(Mi+Ti)*xi) = 0
In order to get integer multipliers just multiply everything by the least common multiple of the denominators lcm(Mi+ti) and then divide by the gcd of the coefficients.
Using your example, the constraint would be:
(3-2)/(3+2) x1 + (1-3)/(1+3) x2 + (2-2)/(2+2) x3 + (6-1)/(6+1) x4 + (4-3)/(4+3) x5 = 0
that is
1/5 x1 -2/4 x2 + 0/4 x3 + 5/7 x4 + 1/7 x5 = 0
Multiply by the lcm(5,4,4,7,7)=140:
28 x1 -70 x2 + 0 x3 + 100 x4 + 20 x5 = 0
divide by 2:
14 x1 -35 x2 +0 x3 + 50 x4 + 10 x5 = 0
We are ready to solve the problem. Let's write it in CPLEX format:
maximize
quantity: x1 + x2 + x3 + x4 + x5
subject to
mix: 14 x1 -35 x2 +0 x3 + 50 x4 + 10 x5 = 0
bounds
x1 <= 14
x2 <= 4
x3 <= 4
x4 <= 6
x5 <= 10
end
Feed it to GLPK:
#include <stdio.h>
#include <stdlib.h>
#include <glpk.h>
int main(void)
{
glp_prob *P;
P = glp_create_prob();
glp_read_lp(P, NULL, "problem.cplex");
glp_adv_basis(P, 0);
glp_simplex(P, NULL);
glp_print_sol(P, "output.txt");
glp_delete_prob(P);
return 0;
}
And the output is:
Problem:
Rows: 1
Columns: 5
Non-zeros: 4
Status: OPTIMAL
Objective: quantity = 20.85714286 (MAXimum)
No. Row name St Activity Lower bound Upper bound Marginal
------ ------------ -- ------------- ------------- ------------- -------------
1 mix NS 0 0 = 0.0714286
No. Column name St Activity Lower bound Upper bound Marginal
------ ------------ -- ------------- ------------- ------------- -------------
1 x1 B 2.85714 0 14
2 x2 NU 4 0 4 3.5
3 x3 NU 4 0 4 1
4 x4 NL 0 0 6 -2.57143
5 x5 NU 10 0 10 0.285714
Karush-Kuhn-Tucker optimality conditions:
KKT.PE: max.abs.err = 0.00e+00 on row 0
max.rel.err = 0.00e+00 on row 0
High quality
KKT.PB: max.abs.err = 0.00e+00 on row 0
max.rel.err = 0.00e+00 on row 0
High quality
KKT.DE: max.abs.err = 0.00e+00 on column 0
max.rel.err = 0.00e+00 on column 0
High quality
KKT.DB: max.abs.err = 0.00e+00 on row 0
max.rel.err = 0.00e+00 on row 0
High quality
End of output
Of course given your input you should construct the problem in memory and feed it to the simplex algorithm without going through a file. Additionally, there's no need to get integer coefficients, it was just to allow a nicer problem formulation.
I am new to CPLEx and trying to set up my first problem. What I want to do set up a LP to minimise the sum of absolute deviations. I have set up the below as a start (based on googling possibilities). This is only a single deviation. I thought I would get this to work and then add to ti. It loads ok but won't solve. Can anyone shed some light on where I need to go next?
Minimize
obj: y1pos + y1neg
Subject To
c1: x0 + x1 + x2 + x3 = 1
c2: y1pos - y1neg + 451320 x0 + 500870 x1 + 483425 x2 + 447330 x3 = 58999
Bounds
0 <= x0 <= 1
0 <= x1 <= 1
0 <= x2 <= 1
0 <= x3 <= 1
y1pos >= 0
y1neg <= 0
End
As Erwin Kalvelagen suggested, changing y1neg <= 0 to y1neg >= 0 was the answer since our our error factor in our constraint is y1pos - y1neg which we want to minimise.
Objective function => x1 - 2x2
Subject to =>
x2 <= 5
x1 - x2 >= 2
x1 ,x2, x3 >= 0
Maximize?
convert to standard form :
Maximize -> -x1 + 2x2
Subject to ->
x2 <= 5
-x1 + x2 <= -2
convert to slack form :
Z = -x1 + 2x2
x3 = 5 - x2
x4 = -2 +x1 -x2
Basic solution (0,0,5,-2)
Can I found optimal solution in here? If not why?
I am comparing the values for shadow price (pi) calculated with gurobi and pulp. I get different values for the same input and I am not sure how to do it with pulp. Here is the lp file that I use:
Minimize
x[0] + x[1] + x[2] + x[3]
Subject To
C[0]: 7 x[0] >= 211
C[1]: 3 x[1] >= 395
C[2]: 2 x[2] >= 610
C[3]: 2 x[3] >= 97
Bounds
End
For the above lp file, gurobi gives me shadow prices:
[0.14285714285714285, 0.3333333333333333, 0.5, 0.5]
and with pulp I get:
[0.14285714, 0.33333333, 0.5, 0.5]
But If I execute the following lp model:
Minimize
x[0] + x[1] + x[2] + x[3] + x[4]
Subject To
C[0]: 7 x[0] + 2 x[4] >= 211
C[1]: 3 x[1] >= 395
C[2]: 2 x[2] + 2 x[4] >= 610
C[3]: 2 x[3] >= 97
Bounds
End
With gurobi I get:
[0.0, 0.3333333333333333, 0.5, 0.5]
and with pulp I get:
[0.14285714, 0.33333333, 0.5, 0.5]
The correct value is the one that gurobi returns (I think ?).
Why I get the same shadow prices with pulp for different models ? How I can get the same results as gurobi ?
(I did not supply the source code because the question will be too long, I think the lp models are enough)
In the second example, there are two dual solutions that are optimal: the one PuLP gives you, and the one you get by calling Gurobi directly. The unique optimal primal solution is [0.0, 131.67, 199.5, 48.5, 105.5], which makes the slacks for all the constraints are 0 in the optimal primal solution. For c[0] if you reduce the right hand side, you get no reduction in the objective, but if you increase it, the cheapest way to make the constraint feasible is by increasing x[0]. Gurobi only guarantees that you will produce an optimal primal and dual solution. The specific optimal solution you get is arbitrary.
The first example is just a precision issue.
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I'm doing research in math and I need to perform the following tasks:
Given an unspecified set of vectors, determine which ones can sum to identity vector (1, 1, ..., 1)
For example, consider the vectors
x1 = (1 0 0)
x2 = (0 1 1)
x3 = (1 0 0)
If you add vectors x1 and x2 together, you get (1, 1, 1).
Or for a bigger example
x1 = (1 0 0 0)
x2 = (0 1 1 0)
x3 = (0 0 1 1)
x4 = (0 0 0 1)
If you add vectors 1, 2, and 4 together, you get (1, 1, 1, 1).
I need an algorithm that can do this in general.
The second task... given an unspecified set of numbers, determine which ones sum to 1.
For example:
x1 = 0.2
x2 = 0.5
x3 = 0.6
x4 = 0.4
x5 = 0.3
x6 = 0.1
x1 + x2 + x5 = 0.2 + 0.5 + 0.3 = 1
But also, x1 + x4 + x5 + x6 = 1
I need to be able to program the computer to do one of the above tasks to further my research.
For the first task, it is a variation of the subset sum problem, which can be solved by dynamic programming. The difference between the subset sum problem and your problem is that subset sum problem only have one dimension. In your case, you need to extend its dynamic programming approach to work with multiple dimensions.
For your second task, it is identical to the subset sum problem.
Yeah, research...
How about starting with brute force: create all the 2^N subsets of your set, and for each such candidate make the sum, if it matches the target, store it.