每日归档： 2017年11月5日

the question is posted in follow page as well. https://stackoverflow.com/questions/47119580/how-to-get-the-maximum-result-with-quadratic-programming-on-cvxopt

I want to use CVXOPT to solve a quadratic programming.

'''
                        Value       =2      =3     =5        
                    Value*return    >=9    >=9      no requirement       
                   Value*Duration   <=5    <=9      no requirement
      Value  return  Duration           
        4       5       2           x11      x12   x13    
        6       4       4           x21      x22   x23 

the matrix is like about, I want to get x11, x12,…x23.

the subject is as following:

G*x<=h
-5*x11         -4*x21      <=-9
      -5*x12         -4*x22<=-9
                           <=0
 2*x11         +4*x21      <=5
       2*x12        +4*x22 <=9
                           <=0


A*x=b

x11+x12+x13           =4
           x21+x22+x23=6
x11       +x21        =2
    x12       +x22    =3
        x13       +x23=5

I want get the target Max (x11^2+x12^2+x13^2+x21^2+x22^2+x23^2)

I wrote the following code:

import cvxopt as op
p_tg=op.matrix(0.0,(6,6))
p_tg[::7]=1
P = 2*op.matrix(p_tg)

q = op.matrix( [0.0,0,0,0,0,0])

G=op.matrix([[-5.0,0,0,2,0,0],
             [0,-5,0,0,2,0],
             [0,0,0,0,0,0],
             [-4,0,0,4,0,0],
             [0,-4,0,0,4,0],
             [0,0,0,0,0,0]])

h=op.matrix([-9.0,-9,0,5,9,0])  

A=op.matrix([[1.0,0,1,0,0],
             [1,0,0,1,0],
             [1,0,0,0,1],
             [0,1,1,0,0],
             [0,1,0,1,0],
             [0,1,0,0,1]])              
b=op.matrix([4,6,2.0,3,5])

sol=op.solvers.qp(P, q, G, h, A, b)

print(sol['x'])

of course, the above code get the minimum, not the maximum.

how to get the max result? I try to p_tg[::7]=-1, while it report error.

raise ValueError(“Rank(A) < p or Rank([P; A; G]) < n”)

ValueError: Rank(A) < p or Rank([P; A; G]) < n

# -*- coding: utf-8 -*-
“””
Created on Sun Nov 5 08:03:26 2017
http://cvxopt.org/examples/tutorial/qp.html
“””

#1. Solving a linear program
”’ Linear programs can be specified via the solvers.lp() function.
As an example, we can solve the problem

minimize 2*x1+x2
subject to
-x1+x2<=1
x1+x2>=2
x2>=0
x1-2*x2<=4

the st is equal to
-x1+x2<=1
-x1-x2<=-2
-x2<=0
x1-2*x2<=4

”’

import cvxopt as op
A = op.matrix([ [-1.0, -1.0, 0.0, 1.0],
[1.0, -1.0, -1.0, -2.0] ])
b = op.matrix( [ 1.0, -2.0, 0.0, 4.0 ])
c = op.matrix( [ 2.0, 1.0 ])
sol=op.solvers.lp(c,A,b)
print(sol[‘x’])

#%%2. Solving a quadratic program

”’
minimize 2*x1**2+x2**2+x1*x2+x1+x2
st
x1>=0
x2>=0
x1+x2=1

the object and st are equal to

[x1, x2] *[[2,.5],[.5,1]] *[[x1],[x2]]

st
-x1 <=0
-x2<=0
x1+x2 =1

op.solvers.qp(P, q, G=None, h=None, A=None, b=None, solver=None,
kktsolver=None, initvals=None, **kwargs)

Solves a quadratic program
minimize (1/2)*x’*P*x + q’*x
subject to G*x <= h
A*x = b.
Input arguments.
P is a n x n dense or sparse ‘d’ matrix with the lower triangular part of P stored in the lower triangle. Must be positive semidefinite.
q is an n x 1 dense ‘d’ matrix.
G is an m x n dense or sparse ‘d’ matrix.
h is an m x 1 dense ‘d’ matrix.
A is a p x n dense or sparse ‘d’ matrix.
b is a p x 1 dense ‘d’ matrix or None.
solver is None or ‘mosek’.
The default values for G, h, A and b are empty matrices with zero rows.
”’

P = 2*op.matrix([ [2, .5],
[.5, 1] ])
q = op.matrix( [1.0, 1.0])

G = op.matrix([[-1.0,0.0],
[0.0,-1.0]])
h = op.matrix( [0.0,0.0])

A = op.matrix( [1.0, 1.0], (1,2)) # with 1 row, 2 columns
b = op.matrix(1.0)
sol=op.solvers.qp(P, q, G, h, A, b)

print(sol[‘x’])