Nonlinear Convex Optimization¶
In this chapter we consider nonlinear convex optimization problems of the form
The functions are convex and twice differentiable and the linear inequalities are generalized inequalities with respect to a proper convex cone, defined as a product of a nonnegative orthant, secondorder cones, and positive semidefinite cones.
The basic functions are cp
and
cpl
, described in the sections
Problems with Nonlinear Objectives and Problems with Linear Objectives. A simpler interface for geometric
programming problems is discussed in the section Geometric Programming.
In the section Exploiting Structure we explain how custom solvers can be
implemented that exploit structure in specific classes of problems.
The last section
describes the algorithm parameters that control the solvers.
Problems with Nonlinear Objectives¶

cvxopt.solvers.
cp
(F[, G, h[, dims[, A, b[, kktsolver]]]])¶ Solves a convex optimization problem
(1)¶
The argument
F
is a function that evaluates the objective and nonlinear constraint functions. It must handle the following calling sequences.F()
returns a tuple (m
,x0
), where is the number of nonlinear constraints and is a point in the domain of .x0
is a dense real matrix of size (, 1).F(x)
, withx
a dense real matrix of size (, 1), returns a tuple (f
,Df
).f
is a dense real matrix of size (, 1), withf[k]
equal to . (If is zero,f
can also be returned as a number.)Df
is a dense or sparse real matrix of size ( + 1, ) withDf[k,:]
equal to the transpose of the gradient . If is not in the domain of ,F(x)
returnsNone
or a tuple (None
,None
).F(x,z)
, withx
a dense real matrix of size (, 1) andz
a positive dense real matrix of size ( + 1, 1) returns a tuple (f
,Df
,H
).f
andDf
are defined as above.H
is a square dense or sparse real matrix of size (, ), whose lower triangular part contains the lower triangular part ofIf
F
is called with two arguments, it can be assumed that is in the domain of .
The linear inequalities are with respect to a cone defined as a Cartesian product of a nonnegative orthant, a number of secondorder cones, and a number of positive semidefinite cones:
with
Here denotes a symmetric matrix stored as a vector in column major order.
The arguments
h
andb
are real singlecolumn dense matrices.G
andA
are real dense or sparse matrices. The default values forA
andb
are sparse matrices with zero rows, meaning that there are no equality constraints. The number of rows ofG
andh
is equal toThe columns of
G
andh
are vectors inwhere the last components represent symmetric matrices stored in column major order. The strictly upper triangular entries of these matrices are not accessed (i.e., the symmetric matrices are stored in the
'L'
type column major order used in theblas
andlapack
modules).The argument
dims
is a dictionary with the dimensions of the cones. It has three fields.dims['l']
:, the dimension of the nonnegative orthant (a nonnegative integer).
dims['q']
:, a list with the dimensions of the secondorder cones (positive integers).
dims['s']
:, a list with the dimensions of the positive semidefinite cones (nonnegative integers).
The default value of
dims
is{'l': h.size[0], 'q': [], 's': []}
, i.e., the default assumption is that the linear inequalities are componentwise inequalities.The role of the optional argument
kktsolver
is explained in the section Exploiting Structure.cp
returns a dictionary that contains the result and information about the accuracy of the solution. The most important fields have keys'status'
,'x'
,'snl'
,'sl'
,'y'
,'znl'
,'zl'
. The possible values of the'status'
key are:'optimal'
In this case the
'x'
entry of the dictionary is the primal optimal solution, the'snl'
and'sl'
entries are the corresponding slacks in the nonlinear and linear inequality constraints, and the'znl'
,'zl'
and'y'
entries are the optimal values of the dual variables associated with the nonlinear inequalities, the linear inequalities, and the linear equality constraints. These vectors approximately satisfy the KarushKuhnTucker (KKT) conditionswhere .
'unknown'
This indicates that the algorithm terminated before a solution was found, due to numerical difficulties or because the maximum number of iterations was reached. The
'x'
,'snl'
,'sl'
,'y'
,'znl'
, and'zl'
entries contain the iterates when the algorithm terminated.
cp
solves the problem by applyingcpl
to the epigraph form problemThe other entries in the output dictionary of
cp
describe the accuracy of the solution and are copied from the output ofcpl
applied to this epigraph form problem.cp
requires that the problem is strictly primal and dual feasible and thatfor all and all positive .
 Example: equality constrained analytic centering
The equality constrained analytic centering problem is defined as
The function
acent
defined below solves the problem, assuming it is solvable.from cvxopt import solvers, matrix, spdiag, log def acent(A, b): m, n = A.size def F(x=None, z=None): if x is None: return 0, matrix(1.0, (n,1)) if min(x) <= 0.0: return None f = sum(log(x)) Df = (x**1).T if z is None: return f, Df H = spdiag(z[0] * x**2) return f, Df, H return solvers.cp(F, A=A, b=b)['x']
 Example: robust leastsquares
The function
robls
defined below solves the unconstrained problemwhere .
from cvxopt import solvers, matrix, spdiag, sqrt, div def robls(A, b, rho): m, n = A.size def F(x=None, z=None): if x is None: return 0, matrix(0.0, (n,1)) y = A*xb w = sqrt(rho + y**2) f = sum(w) Df = div(y, w).T * A if z is None: return f, Df H = A.T * spdiag(z[0]*rho*(w**3)) * A return f, Df, H return solvers.cp(F)['x']
Example: analytic centering with cone constraints
from cvxopt import matrix, log, div, spdiag, solvers def F(x = None, z = None): if x is None: return 0, matrix(0.0, (3,1)) if max(abs(x)) >= 1.0: return None u = 1  x**2 val = sum(log(u)) Df = div(2*x, u).T if z is None: return val, Df H = spdiag(2 * z[0] * div(1 + x**2, u**2)) return val, Df, H G = matrix([ [0., 1., 0., 0., 21., 11., 0., 11., 10., 8., 0., 8., 5.], [0., 0., 1., 0., 0., 10., 16., 10., 10., 10., 16., 10., 3.], [0., 0., 0., 1., 5., 2., 17., 2., 6., 8., 17., 7., 6.] ]) h = matrix([1.0, 0.0, 0.0, 0.0, 20., 10., 40., 10., 80., 10., 40., 10., 15.]) dims = {'l': 0, 'q': [4], 's': [3]} sol = solvers.cp(F, G, h, dims) print(sol['x']) [ 4.11e01] [ 5.59e01] [7.20e01]
Problems with Linear Objectives¶

cvxopt.solvers.
cpl
(c, F[, G, h[, dims[, A, b[, kktsolver]]]])¶ Solves a convex optimization problem with a linear objective
c
is a real singlecolumn dense matrix.F
is a function that evaluates the nonlinear constraint functions. It must handle the following calling sequences.F()
returns a tuple (m
,x0
), wherem
is the number of nonlinear constraints andx0
is a point in the domain of .x0
is a dense real matrix of size (, 1).F(x)
, withx
a dense real matrix of size (, 1), returns a tuple (f
,Df
).f
is a dense real matrix of size (, 1), withf[k]
equal to .Df
is a dense or sparse real matrix of size (, ) withDf[k,:]
equal to the transpose of the gradient . If is not in the domain of ,F(x)
returnsNone
or a tuple (None
,None
).F(x,z)
, withx
a dense real matrix of size (, 1) andz
a positive dense real matrix of size (, 1) returns a tuple (f
,Df
,H
).f
andDf
are defined as above.H
is a square dense or sparse real matrix of size (, ), whose lower triangular part contains the lower triangular part ofIf
F
is called with two arguments, it can be assumed that is in the domain of .
The linear inequalities are with respect to a cone defined as a Cartesian product of a nonnegative orthant, a number of secondorder cones, and a number of positive semidefinite cones:
with
Here denotes a symmetric matrix stored as a vector in column major order.
The arguments
h
andb
are real singlecolumn dense matrices.G
andA
are real dense or sparse matrices. The default values forA
andb
are sparse matrices with zero rows, meaning that there are no equality constraints. The number of rows ofG
andh
is equal toThe columns of
G
andh
are vectors inwhere the last components represent symmetric matrices stored in column major order. The strictly upper triangular entries of these matrices are not accessed (i.e., the symmetric matrices are stored in the
'L'
type column major order used in theblas
andlapack
modules.The argument
dims
is a dictionary with the dimensions of the cones. It has three fields.dims['l']
:, the dimension of the nonnegative orthant (a nonnegative integer).
dims['q']
:, a list with the dimensions of the secondorder cones (positive integers).
dims['s']
:, a list with the dimensions of the positive semidefinite cones (nonnegative integers).
The default value of
dims
is{'l': h.size[0], 'q': [], 's': []}
, i.e., the default assumption is that the linear inequalities are componentwise inequalities.The role of the optional argument
kktsolver
is explained in the section Exploiting Structure.cpl
returns a dictionary that contains the result and information about the accuracy of the solution. The most important fields have keys'status'
,'x'
,'snl'
,'sl'
,'y'
,'znl'
,'zl'
. The possible values of the'status'
key are:'optimal'
In this case the
'x'
entry of the dictionary is the primal optimal solution, the'snl'
and'sl'
entries are the corresponding slacks in the nonlinear and linear inequality constraints, and the'znl'
,'zl'
, and'y'
entries are the optimal values of the dual variables associated with the nonlinear inequalities, the linear inequalities, and the linear equality constraints. These vectors approximately satisfy the KarushKuhnTucker (KKT) conditions'unknown'
This indicates that the algorithm terminated before a solution was found, due to numerical difficulties or because the maximum number of iterations was reached. The
'x'
,'snl'
,'sl'
,'y'
,'znl'
, and'zl'
entries contain the iterates when the algorithm terminated.
The other entries in the output dictionary describe the accuracy of the solution. The entries
'primal objective'
,'dual objective'
,'gap'
, and'relative gap'
give the primal objective , the dual objective, calculated asthe duality gap
and the relative gap. The relative gap is defined as
and
None
otherwise. The entry with key'primal infeasibility'
gives the residual in the primal constraints,where is the point returned by
F()
. The entry with key'dual infeasibility'
gives the residualcpl
requires that the problem is strictly primal and dual feasible and thatfor all and all positive .
 Example: floor planning
This example is the floor planning problem of section 8.8.2 in the book Convex Optimization:
This problem has 22 variables
5 nonlinear inequality constraints, and 26 linear inequality constraints. The code belows defines a function
floorplan
that solves the problem by callingcp
, then applies it to 4 instances, and creates a figure.import pylab from cvxopt import solvers, matrix, spmatrix, mul, div def floorplan(Amin): # minimize W+H # subject to Amink / hk <= wk, k = 1,..., 5 # x1 >= 0, x2 >= 0, x4 >= 0 # x1 + w1 + rho <= x3 # x2 + w2 + rho <= x3 # x3 + w3 + rho <= x5 # x4 + w4 + rho <= x5 # x5 + w5 <= W # y2 >= 0, y3 >= 0, y5 >= 0 # y2 + h2 + rho <= y1 # y1 + h1 + rho <= y4 # y3 + h3 + rho <= y4 # y4 + h4 <= H # y5 + h5 <= H # hk/gamma <= wk <= gamma*hk, k = 1, ..., 5 # # 22 Variables W, H, x (5), y (5), w (5), h (5). # # W, H: scalars; bounding box width and height # x, y: 5vectors; coordinates of bottom left corners of blocks # w, h: 5vectors; widths and heigths of the 5 blocks rho, gamma = 1.0, 5.0 # min spacing, min aspect ratio # The objective is to minimize W + H. There are five nonlinear # constraints # # wk + Amink / hk <= 0, k = 1, ..., 5 c = matrix(2*[1.0] + 20*[0.0]) def F(x=None, z=None): if x is None: return 5, matrix(17*[0.0] + 5*[1.0]) if min(x[17:]) <= 0.0: return None f = x[12:17] + div(Amin, x[17:]) Df = matrix(0.0, (5,22)) Df[:,12:17] = spmatrix(1.0, range(5), range(5)) Df[:,17:] = spmatrix(div(Amin, x[17:]**2), range(5), range(5)) if z is None: return f, Df H = spmatrix( 2.0* mul(z, div(Amin, x[17::]**3)), range(17,22), range(17,22) ) return f, Df, H G = matrix(0.0, (26,22)) h = matrix(0.0, (26,1)) G[0,2] = 1.0 # x1 <= 0 G[1,3] = 1.0 # x2 <= 0 G[2,5] = 1.0 # x4 <= 0 G[3, [2, 4, 12]], h[3] = [1.0, 1.0, 1.0], rho # x1  x3 + w1 <= rho G[4, [3, 4, 13]], h[4] = [1.0, 1.0, 1.0], rho # x2  x3 + w2 <= rho G[5, [4, 6, 14]], h[5] = [1.0, 1.0, 1.0], rho # x3  x5 + w3 <= rho G[6, [5, 6, 15]], h[6] = [1.0, 1.0, 1.0], rho # x4  x5 + w4 <= rho G[7, [0, 6, 16]] = 1.0, 1.0, 1.0 # W + x5 + w5 <= 0 G[8,8] = 1.0 # y2 <= 0 G[9,9] = 1.0 # y3 <= 0 G[10,11] = 1.0 # y5 <= 0 G[11, [7, 8, 18]], h[11] = [1.0, 1.0, 1.0], rho # y1 + y2 + h2 <= rho G[12, [7, 10, 17]], h[12] = [1.0, 1.0, 1.0], rho # y1  y4 + h1 <= rho G[13, [9, 10, 19]], h[13] = [1.0, 1.0, 1.0], rho # y3  y4 + h3 <= rho G[14, [1, 10, 20]] = 1.0, 1.0, 1.0 # H + y4 + h4 <= 0 G[15, [1, 11, 21]] = 1.0, 1.0, 1.0 # H + y5 + h5 <= 0 G[16, [12, 17]] = 1.0, 1.0/gamma # w1 + h1/gamma <= 0 G[17, [12, 17]] = 1.0, gamma # w1  gamma * h1 <= 0 G[18, [13, 18]] = 1.0, 1.0/gamma # w2 + h2/gamma <= 0 G[19, [13, 18]] = 1.0, gamma # w2  gamma * h2 <= 0 G[20, [14, 19]] = 1.0, 1.0/gamma # w3 + h3/gamma <= 0 G[21, [14, 19]] = 1.0, gamma # w3  gamma * h3 <= 0 G[22, [15, 20]] = 1.0, 1.0/gamma # w4 + h4/gamma <= 0 G[23, [15, 20]] = 1.0, gamma # w4  gamma * h4 <= 0 G[24, [16, 21]] = 1.0, 1.0/gamma # w5 + h5/gamma <= 0 G[25, [16, 21]] = 1.0, gamma # w5  gamma * h5 <= 0.0 # solve and return W, H, x, y, w, h sol = solvers.cpl(c, F, G, h) return sol['x'][0], sol['x'][1], sol['x'][2:7], sol['x'][7:12], sol['x'][12:17], sol['x'][17:] pylab.figure(facecolor='w') pylab.subplot(221) Amin = matrix([100., 100., 100., 100., 100.]) W, H, x, y, w, h = floorplan(Amin) for k in range(5): pylab.fill([x[k], x[k], x[k]+w[k], x[k]+w[k]], [y[k], y[k]+h[k], y[k]+h[k], y[k]], facecolor = '#D0D0D0') pylab.text(x[k]+.5*w[k], y[k]+.5*h[k], "%d" %(k+1)) pylab.axis([1.0, 26, 1.0, 26]) pylab.xticks([]) pylab.yticks([]) pylab.subplot(222) Amin = matrix([20., 50., 80., 150., 200.]) W, H, x, y, w, h = floorplan(Amin) for k in range(5): pylab.fill([x[k], x[k], x[k]+w[k], x[k]+w[k]], [y[k], y[k]+h[k], y[k]+h[k], y[k]], 'facecolor = #D0D0D0') pylab.text(x[k]+.5*w[k], y[k]+.5*h[k], "%d" %(k+1)) pylab.axis([1.0, 26, 1.0, 26]) pylab.xticks([]) pylab.yticks([]) pylab.subplot(223) Amin = matrix([180., 80., 80., 80., 80.]) W, H, x, y, w, h = floorplan(Amin) for k in range(5): pylab.fill([x[k], x[k], x[k]+w[k], x[k]+w[k]], [y[k], y[k]+h[k], y[k]+h[k], y[k]], 'facecolor = #D0D0D0') pylab.text(x[k]+.5*w[k], y[k]+.5*h[k], "%d" %(k+1)) pylab.axis([1.0, 26, 1.0, 26]) pylab.xticks([]) pylab.yticks([]) pylab.subplot(224) Amin = matrix([20., 150., 20., 200., 110.]) W, H, x, y, w, h = floorplan(Amin) for k in range(5): pylab.fill([x[k], x[k], x[k]+w[k], x[k]+w[k]], [y[k], y[k]+h[k], y[k]+h[k], y[k]], 'facecolor = #D0D0D0') pylab.text(x[k]+.5*w[k], y[k]+.5*h[k], "%d" %(k+1)) pylab.axis([1.0, 26, 1.0, 26]) pylab.xticks([]) pylab.yticks([]) pylab.show()
Geometric Programming¶

cvxopt.solvers.
gp
(K, F, g[, G, h[, A, b]])¶ Solves a geometric program in convex form
where
and the vector inequality denotes componentwise inequality.
K
is a list of + 1 positive integers withK[i]
equal to the number of rows in .F
is a dense or sparse real matrix of size(sum(K), n)
.g
is a dense real matrix with one column and the same number of rows asF
.G
andA
are dense or sparse real matrices. Their default values are sparse matrices with zero rows.h
andb
are dense real matrices with one column. Their default values are matrices of size (0, 1).gp
returns a dictionary with keys'status'
,'x'
,'snl'
,'sl'
,'y'
,'znl'
, and'zl'
. The possible values of the'status'
key are:'optimal'
In this case the
'x'
entry is the primal optimal solution, the'snl'
and'sl'
entries are the corresponding slacks in the nonlinear and linear inequality constraints. The'znl'
,'zl'
, and'y'
entries are the optimal values of the dual variables associated with the nonlinear and linear inequality constraints and the linear equality constraints. These values approximately satisfy'unknown'
This indicates that the algorithm terminated before a solution was found, due to numerical difficulties or because the maximum number of iterations was reached. The
'x'
,'snl'
,'sl'
,'y'
,'znl'
, and'zl'
contain the iterates when the algorithm terminated.
The other entries in the output dictionary describe the accuracy of the solution, and are taken from the output of
cp
.gp
requires that the problem is strictly primal and dual feasible and thatfor all and all positive .
As an example, we solve the small GP of section 2.4 of the paper A Tutorial on Geometric Programming. The posynomial form of the problem is
with variables , , .
from cvxopt import matrix, log, exp, solvers
Aflr = 1000.0
Awall = 100.0
alpha = 0.5
beta = 2.0
gamma = 0.5
delta = 2.0
F = matrix( [[1., 1., 1., 0., 1., 1., 0., 0.],
[1., 1., 0., 1., 1., 1., 1., 1.],
[1., 0., 1., 1., 0., 0., 1., 1.]])
g = log( matrix( [1.0, 2/Awall, 2/Awall, 1/Aflr, alpha, 1/beta, gamma, 1/delta]) )
K = [1, 2, 1, 1, 1, 1, 1]
h, w, d = exp( solvers.gp(K, F, g)['x'] )
Exploiting Structure¶
By default, the functions cp
and
cpl
do not exploit problem
structure. Two mechanisms are provided for implementing customized solvers
that take advantage of problem structure.
 Providing a function for solving KKT equations
The most expensive step of each iteration of
cp
is the solution of a set of linear equations (KKT equations) of the form(2)¶
where
The matrix depends on the current iterates and is defined as follows. Suppose
where
Then is a blockdiagonal matrix,
with the following diagonal blocks.
The first block is a positive diagonal scaling with a vector :
This transformation is symmetric:
The second block is a positive diagonal scaling with a vector :
This transformation is symmetric:
The next blocks are positive multiples of hyperbolic Householder transformations:
where
These transformations are also symmetric:
The last blocks are congruence transformations with nonsingular matrices:
In general, this operation is not symmetric, and
It is often possible to exploit problem structure to solve (2) faster than by standard methods. The last argument
kktsolver
ofcp
allows the user to supply a Python function for solving the KKT equations. This function will be called asf = kktsolver(x, z, W)
. The argumentx
is the point at which the derivatives in the KKT matrix are evaluated.z
is a positive vector of length it + 1, containing the coefficients in the 1,1 block .W
is a dictionary that contains the parameters of the scaling:W['dnl']
is the positive vector that defines the diagonal scaling for the nonlinear inequalities.W['dnli']
is its componentwise inverse.W['d']
is the positive vector that defines the diagonal scaling for the componentwise linear inequalities.W['di']
is its componentwise inverse.W['beta']
andW['v']
are lists of length with the coefficients and vectors that define the hyperbolic Householder transformations.W['r']
is a list of length with the matrices that define the the congruence transformations.W['rti']
is a list of length with the transposes of the inverses of the matrices inW['r']
.
The function call
f = kktsolver(x, z, W)
should return a routine for solving the KKT system (2) defined byx
,z
,W
. It will be called asf(bx, by, bz)
. On entry,bx
,by
,bz
contain the righthand side. On exit, they should contain the solution of the KKT system, with the last component scaled, i.e., on exit,The role of the argument
kktsolver
in the functioncpl
is similar, except that in (2), Specifying constraints via Python functions
In the default use of
cp
, the argumentsG
andA
are the coefficient matrices in the constraints of (2). It is also possible to specify these matrices by providing Python functions that evaluate the corresponding matrixvector products and their adjoints.If the argument
G
ofcp
is a Python function, thenG(u, v[, alpha = 1.0, beta = 0.0, trans = 'N'])
should evaluates the matrixvector productsSimilarly, if the argument
A
is a Python function, thenA(u, v[, alpha = 1.0, beta = 0.0, trans = 'N'])
should evaluate the matrixvector productsIn a similar way, when the first argument
F
ofcp
returns matrices of first derivatives or second derivativesDf
,H
, these matrices can be specified as Python functions. IfDf
is a Python function, thenDf(u, v[, alpha = 1.0, beta = 0.0, trans = 'N'])
should evaluate the matrixvector productsIf
H
is a Python function, thenH(u, v[, alpha, beta])
should evaluate the matrixvector product
If
G
,A
,Df
, orH
are Python functions, then the argumentkktsolver
must also be provided.
As an example, we consider the unconstrained problem
where is an by matrix with less than . The Hessian of the objective is diagonal plus a lowrank term:
We can exploit this property when solving (2) by applying the matrix inversion lemma. We first solve
and then obtain
The following code follows this method. It also uses BLAS functions for matrixmatrix and matrixvector products.
from cvxopt import matrix, spdiag, mul, div, log, blas, lapack, solvers, base
def l2ac(A, b):
"""
Solves
minimize (1/2) * A*xb_2^2  sum log (1xi^2)
assuming A is m x n with m << n.
"""
m, n = A.size
def F(x = None, z = None):
if x is None:
return 0, matrix(0.0, (n,1))
if max(abs(x)) >= 1.0:
return None
# r = A*x  b
r = b
blas.gemv(A, x, r, beta = 1.0)
w = x**2
f = 0.5 * blas.nrm2(r)**2  sum(log(1w))
# gradf = A'*r + 2.0 * x ./ (1w)
gradf = div(x, 1.0  w)
blas.gemv(A, r, gradf, trans = 'T', beta = 2.0)
if z is None:
return f, gradf.T
else:
def Hf(u, v, alpha = 1.0, beta = 0.0):
# v := alpha * (A'*A*u + 2*((1+w)./(1w)).*u + beta *v
v *= beta
v += 2.0 * alpha * mul(div(1.0+w, (1.0w)**2), u)
blas.gemv(A, u, r)
blas.gemv(A, r, v, alpha = alpha, beta = 1.0, trans = 'T')
return f, gradf.T, Hf
# Custom solver for the Newton system
#
# z[0]*(A'*A + D)*x = bx
#
# where D = 2 * (1+x.^2) ./ (1x.^2).^2. We apply the matrix inversion
# lemma and solve this as
#
# (A * D^1 *A' + I) * v = A * D^1 * bx / z[0]
# D * x = bx / z[0]  A'*v.
S = matrix(0.0, (m,m))
v = matrix(0.0, (m,1))
def Fkkt(x, z, W):
ds = (2.0 * div(1 + x**2, (1  x**2)**2))**0.5
Asc = A * spdiag(ds)
blas.syrk(Asc, S)
S[::m+1] += 1.0
lapack.potrf(S)
a = z[0]
def g(x, y, z):
x[:] = mul(x, ds) / a
blas.gemv(Asc, x, v)
lapack.potrs(S, v)
blas.gemv(Asc, v, x, alpha = 1.0, beta = 1.0, trans = 'T')
x[:] = mul(x, ds)
return g
return solvers.cp(F, kktsolver = Fkkt)['x']
Algorithm Parameters¶
The following algorithm control parameters are accessible via the
dictionary solvers.options
. By default the dictionary
is empty and the default values of the parameters are used.
One can change the parameters in the default solvers by adding entries with the following key values.
'show_progress'
True
orFalse
; turns the output to the screen on or off (default:True
).'maxiters'
maximum number of iterations (default:
100
).'abstol'
absolute accuracy (default:
1e7
).'reltol'
relative accuracy (default:
1e6
).'feastol'
tolerance for feasibility conditions (default:
1e7
).'refinement'
number of iterative refinement steps when solving KKT equations (default:
1
).
For example the command
>>> from cvxopt import solvers
>>> solvers.options['show_progress'] = False
turns off the screen output during calls to the solvers. The tolerances
abstol
, reltol
and feastol
have the
following meaning in cpl
.
cpl
returns with status 'optimal'
if
where is the point returned by F()
, and
where
The functions cp
and
gp
call cpl
and hence use the
same stopping criteria (with for gp
).