Numerical Root Finding and Optimization

AUTHOR:

  • William Stein (2007): initial version

  • Nathann Cohen (2008): Bin Packing

Functions and Methods

sage.numerical.optimize.binpacking(items, maximum, k=1, solver=None, verbose=None, integrality_tolerance=0)[source]

Solve the bin packing problem.

The Bin Packing problem is the following :

Given a list of items of weights \(p_i\) and a real value \(k\), what is the least number of bins such that all the items can be packed in the bins, while ensuring that the sum of the weights of the items packed in each bin is at most \(k\) ?

For more informations, see Wikipedia article Bin_packing_problem.

Two versions of this problem are solved by this algorithm :

  • Is it possible to put the given items in \(k\) bins ?

  • What is the assignment of items using the least number of bins with the given list of items ?

INPUT:

  • items – list or dict; either a list of real values (the items’ weight), or a dictionary associating to each item its weight

  • maximum – (default: 1) the maximal size of a bin

  • k – integer (default: None); number of bins:

    • When set to an integer value, the function returns a partition of the items into \(k\) bins if possible, and raises an exception otherwise.

    • When set to None, the function returns a partition of the items using the least possible number of bins.

  • solver – (default: None) specify a Mixed Integer Linear Programming (MILP) solver to be used. If set to None, the default one is used. For more information on MILP solvers and which default solver is used, see the method solve of the class MixedIntegerLinearProgram.

  • verbose – integer (default: 0); sets the level of verbosity. Set to 0 by default, which means quiet.

  • integrality_tolerance – parameter for use with MILP solvers over an inexact base ring; see MixedIntegerLinearProgram.get_values()

OUTPUT:

A list of lists, each member corresponding to a bin and containing either the list of the weights inside it when items is a list of items’ weight, or the list of items inside it when items is a dictionary. If there is no solution, an exception is raised (this can only happen when k is specified or if maximum is less than the weight of one item).

EXAMPLES:

Trying to find the minimum amount of boxes for 5 items of weights \(1/5, 1/4, 2/3, 3/4, 5/7\):

sage: from sage.numerical.optimize import binpacking
sage: values = [1/5, 1/3, 2/3, 3/4, 5/7]
sage: bins = binpacking(values)                                                 # needs sage.numerical.mip
sage: len(bins)                                                                 # needs sage.numerical.mip
3

>>> from sage.all import *
>>> from sage.numerical.optimize import binpacking
>>> values = [Integer(1)/Integer(5), Integer(1)/Integer(3), Integer(2)/Integer(3), Integer(3)/Integer(4), Integer(5)/Integer(7)]
>>> bins = binpacking(values)                                                 # needs sage.numerical.mip
>>> len(bins)                                                                 # needs sage.numerical.mip
3

Checking the bins are of correct size

sage: all(sum(b) <= 1 for b in bins)                                            # needs sage.numerical.mip
True

>>> from sage.all import *
>>> all(sum(b) <= Integer(1) for b in bins)                                            # needs sage.numerical.mip
True

Checking every item is in a bin

sage: b1, b2, b3 = bins                                                         # needs sage.numerical.mip
sage: all((v in b1 or v in b2 or v in b3) for v in values)                      # needs sage.numerical.mip
True

>>> from sage.all import *
>>> b1, b2, b3 = bins                                                         # needs sage.numerical.mip
>>> all((v in b1 or v in b2 or v in b3) for v in values)                      # needs sage.numerical.mip
True

And only in one bin

sage: sum(len(b) for b in bins) == len(values)                                  # needs sage.numerical.mip
True

>>> from sage.all import *
>>> sum(len(b) for b in bins) == len(values)                                  # needs sage.numerical.mip
True

One way to use only three boxes (which is best possible) is to put \(1/5 + 3/4\) together in a box, \(1/3+2/3\) in another, and \(5/7\) by itself in the third one.

Of course, we can also check that there is no solution using only two boxes

sage: from sage.numerical.optimize import binpacking
sage: binpacking([0.2,0.3,0.8,0.9], k=2)                                        # needs sage.numerical.mip
Traceback (most recent call last):
...
ValueError: this problem has no solution

>>> from sage.all import *
>>> from sage.numerical.optimize import binpacking
>>> binpacking([RealNumber('0.2'),RealNumber('0.3'),RealNumber('0.8'),RealNumber('0.9')], k=Integer(2))                                        # needs sage.numerical.mip
Traceback (most recent call last):
...
ValueError: this problem has no solution

We can also provide a dictionary keyed by items and associating to each item its weight. Then, the bins contain the name of the items inside it

sage: values = {'a':1/5, 'b':1/3, 'c':2/3, 'd':3/4, 'e':5/7}
sage: bins = binpacking(values)                                                 # needs sage.numerical.mip
sage: set(flatten(bins)) == set(values.keys())                                  # needs sage.numerical.mip
True

>>> from sage.all import *
>>> values = {'a':Integer(1)/Integer(5), 'b':Integer(1)/Integer(3), 'c':Integer(2)/Integer(3), 'd':Integer(3)/Integer(4), 'e':Integer(5)/Integer(7)}
>>> bins = binpacking(values)                                                 # needs sage.numerical.mip
>>> set(flatten(bins)) == set(values.keys())                                  # needs sage.numerical.mip
True
sage.numerical.optimize.find_fit(data, model, initial_guess=None, parameters=None, variables=None, solution_dict=False)[source]

Finds numerical estimates for the parameters of the function model to give a best fit to data.

INPUT:

  • data – a two dimensional table of floating point numbers of the form \([[x_{1,1}, x_{1,2}, \ldots, x_{1,k}, f_1], [x_{2,1}, x_{2,2}, \ldots, x_{2,k}, f_2], \ldots, [x_{n,1}, x_{n,2}, \ldots, x_{n,k}, f_n]]\) given as either a list of lists, matrix, or numpy array.

  • model – either a symbolic expression, symbolic function, or a Python function. model has to be a function of the variables \((x_1, x_2, \ldots, x_k)\) and free parameters \((a_1, a_2, \ldots, a_l)\).

  • initial_guess – (default: None) initial estimate for the parameters \((a_1, a_2, \ldots, a_l)\), given as either a list, tuple, vector or numpy array. If None, the default estimate for each parameter is \(1\).

  • parameters – (default: None) a list of the parameters \((a_1, a_2, \ldots, a_l)\). If model is a symbolic function it is ignored, and the free parameters of the symbolic function are used.

  • variables – (default: None) a list of the variables \((x_1, x_2, \ldots, x_k)\). If model is a symbolic function it is ignored, and the variables of the symbolic function are used.

  • solution_dict – boolean (default: False); if True, return the solution as a dictionary rather than an equation

EXAMPLES:

First we create some data points of a sine function with some “random” perturbations:

sage: set_random_seed(0)
sage: data = [(i, 1.2 * sin(0.5*i-0.2) + 0.1 * normalvariate(0, 1))             # needs sage.symbolic
....:         for i in xsrange(0, 4*pi, 0.2)]
sage: var('a, b, c, x')                                                         # needs sage.symbolic
(a, b, c, x)

>>> from sage.all import *
>>> set_random_seed(Integer(0))
>>> data = [(i, RealNumber('1.2') * sin(RealNumber('0.5')*i-RealNumber('0.2')) + RealNumber('0.1') * normalvariate(Integer(0), Integer(1)))             # needs sage.symbolic
...         for i in xsrange(Integer(0), Integer(4)*pi, RealNumber('0.2'))]
>>> var('a, b, c, x')                                                         # needs sage.symbolic
(a, b, c, x)

We define a function with free parameters \(a\), \(b\) and \(c\):

sage: model(x) = a * sin(b * x - c)                                             # needs sage.symbolic

>>> from sage.all import *
>>> __tmp__=var("x"); model = symbolic_expression(a * sin(b * x - c)                                             ).function(x)# needs sage.symbolic

We search for the parameters that give the best fit to the data:

sage: find_fit(data, model)                                                     # needs sage.symbolic
[a == 1.21..., b == 0.49..., c == 0.19...]

>>> from sage.all import *
>>> find_fit(data, model)                                                     # needs sage.symbolic
[a == 1.21..., b == 0.49..., c == 0.19...]

We can also use a Python function for the model:

sage: def f(x, a, b, c): return a * sin(b * x - c)
sage: fit = find_fit(data, f, parameters=[a, b, c], variables=[x],              # needs sage.symbolic
....:                solution_dict = True)
sage: fit[a], fit[b], fit[c]                                                    # needs sage.symbolic
(1.21..., 0.49..., 0.19...)

>>> from sage.all import *
>>> def f(x, a, b, c): return a * sin(b * x - c)
>>> fit = find_fit(data, f, parameters=[a, b, c], variables=[x],              # needs sage.symbolic
...                solution_dict = True)
>>> fit[a], fit[b], fit[c]                                                    # needs sage.symbolic
(1.21..., 0.49..., 0.19...)

We search for a formula for the \(n\)-th prime number:

sage: # needs sage.libs.pari
sage: dataprime = [(i, nth_prime(i)) for i in range(1, 5000, 100)]
sage: find_fit(dataprime, a * x * log(b * x),                                   # needs sage.symbolic
....:          parameters=[a, b], variables=[x])
[a == 1.11..., b == 1.24...]

>>> from sage.all import *
>>> # needs sage.libs.pari
>>> dataprime = [(i, nth_prime(i)) for i in range(Integer(1), Integer(5000), Integer(100))]
>>> find_fit(dataprime, a * x * log(b * x),                                   # needs sage.symbolic
...          parameters=[a, b], variables=[x])
[a == 1.11..., b == 1.24...]

ALGORITHM:

Uses scipy.optimize.leastsq() which in turn uses MINPACK’s lmdif and lmder algorithms.

sage.numerical.optimize.find_local_maximum(f, a, b, tol=1.48e-08, maxfun=500)[source]

Numerically find a local maximum of the expression \(f\) on the interval \([a,b]\) (or \([b,a]\)) along with the point at which the maximum is attained.

Note that this function only finds a local maximum, and not the global maximum on that interval – see the examples with find_local_maximum().

See the documentation for find_local_maximum() for more details and possible workarounds for finding the global minimum on an interval.

EXAMPLES:

sage: f = lambda x: x*cos(x)
sage: find_local_maximum(f, 0, 5)
(0.561096338191..., 0.8603335890...)
sage: find_local_maximum(f, 0, 5, tol=0.1, maxfun=10)
(0.561090323458..., 0.857926501456...)
sage: find_local_maximum(8*e^(-x)*sin(x) - 1, 0, 7)                             # needs sage.symbolic
(1.579175535558..., 0.7853981...)

>>> from sage.all import *
>>> f = lambda x: x*cos(x)
>>> find_local_maximum(f, Integer(0), Integer(5))
(0.561096338191..., 0.8603335890...)
>>> find_local_maximum(f, Integer(0), Integer(5), tol=RealNumber('0.1'), maxfun=Integer(10))
(0.561090323458..., 0.857926501456...)
>>> find_local_maximum(Integer(8)*e**(-x)*sin(x) - Integer(1), Integer(0), Integer(7))                             # needs sage.symbolic
(1.579175535558..., 0.7853981...)
sage.numerical.optimize.find_local_minimum(f, a, b, tol=1.48e-08, maxfun=500)[source]

Numerically find a local minimum of the expression f on the interval \([a,b]\) (or \([b,a]\)) and the point at which it attains that minimum. Note that f must be a function of (at most) one variable.

Note that this function only finds a local minimum, and not the global minimum on that interval – see the examples below.

INPUT:

  • f – a function of at most one variable

  • a, b – endpoints of interval on which to minimize \(f\)

  • tol – the convergence tolerance

  • maxfun – maximum function evaluations

OUTPUT:

  • minval – (float) the minimum value that \(f\) takes on in the interval \([a,b]\)

  • x – (float) the point at which \(f\) takes on the minimum value

EXAMPLES:

sage: f = lambda x: x*cos(x)
sage: find_local_minimum(f, 1, 5)
(-3.28837139559..., 3.4256184695...)
sage: find_local_minimum(f, 1, 5, tol=1e-3)
(-3.28837136189098..., 3.42575079030572...)
sage: find_local_minimum(f, 1, 5, tol=1e-2, maxfun=10)
(-3.28837084598..., 3.4250840220...)
sage: show(plot(f, 0, 20))                                                      # needs sage.plot
sage: find_local_minimum(f, 1, 15)
(-9.4772942594..., 9.5293344109...)

>>> from sage.all import *
>>> f = lambda x: x*cos(x)
>>> find_local_minimum(f, Integer(1), Integer(5))
(-3.28837139559..., 3.4256184695...)
>>> find_local_minimum(f, Integer(1), Integer(5), tol=RealNumber('1e-3'))
(-3.28837136189098..., 3.42575079030572...)
>>> find_local_minimum(f, Integer(1), Integer(5), tol=RealNumber('1e-2'), maxfun=Integer(10))
(-3.28837084598..., 3.4250840220...)
>>> show(plot(f, Integer(0), Integer(20)))                                                      # needs sage.plot
>>> find_local_minimum(f, Integer(1), Integer(15))
(-9.4772942594..., 9.5293344109...)

Only local minima are found; if you enlarge the interval, the returned minimum may be larger! See Issue #2607.

sage: # needs sage.symbolic
sage: f(x) = -x*sin(x^2)
sage: find_local_minimum(f, -2.5, -1)
(-2.182769784677722, -2.1945027498534686)

>>> from sage.all import *
>>> # needs sage.symbolic
>>> __tmp__=var("x"); f = symbolic_expression(-x*sin(x**Integer(2))).function(x)
>>> find_local_minimum(f, -RealNumber('2.5'), -Integer(1))
(-2.182769784677722, -2.1945027498534686)

Enlarging the interval returns a larger minimum:

sage: # needs sage.symbolic
sage: find_local_minimum(f, -2.5, 2)
(-1.3076194129914434, 1.3552111405712108)

>>> from sage.all import *
>>> # needs sage.symbolic
>>> find_local_minimum(f, -RealNumber('2.5'), Integer(2))
(-1.3076194129914434, 1.3552111405712108)

One work-around is to plot the function and grab the minimum from that, although the plotting code does not necessarily do careful numerics (observe the small number of decimal places that we actually test):

sage: # needs sage.plot sage.symbolic
sage: plot(f, (x, -2.5, -1)).ymin()
-2.182...
sage: plot(f, (x, -2.5, 2)).ymin()
-2.182...

>>> from sage.all import *
>>> # needs sage.plot sage.symbolic
>>> plot(f, (x, -RealNumber('2.5'), -Integer(1))).ymin()
-2.182...
>>> plot(f, (x, -RealNumber('2.5'), Integer(2))).ymin()
-2.182...

ALGORITHM:

Uses scipy.optimize.fminbound() which uses Brent’s method.

AUTHOR:

  • William Stein (2007-12-07)

sage.numerical.optimize.find_root(f, a, b, xtol=1e-12, rtol=8.881784197001252e-16, maxiter=100, full_output=False)[source]

Numerically find a root of f on the closed interval \([a,b]\) (or \([b,a]\)) if possible, where f is a function in the one variable. Note: this function only works in fixed (machine) precision, it is not possible to get arbitrary precision approximations with it.

INPUT:

  • f – a function of one variable or symbolic equality

  • a, b – endpoints of the interval

  • xtol, rtol – the routine converges when a root is known to lie within xtol of the value return. Should be \(\geq 0\). The routine modifies this to take into account the relative precision of doubles. By default, rtol is 4*numpy.finfo(float).eps, the minimum allowed value for scipy.optimize.brentq(), which is what this method uses underneath. This value is equal to 2.0**-50 for IEEE-754 double precision floats as used by Python.

  • maxiter – integer; if convergence is not achieved in maxiter iterations, an error is raised. Must be \(\geq 0\).

  • full_output – boolean (default: False); if True, also return object that contains information about convergence

EXAMPLES:

An example involving an algebraic polynomial function:

sage: R.<x> = QQ[]
sage: f = (x+17)*(x-3)*(x-1/8)^3
sage: find_root(f, 0,4)
2.999999999999995
sage: find_root(f, 0,1)  # abs tol 1e-6 (note -- precision of answer isn't very good on some machines)
0.124999
sage: find_root(f, -20,-10)
-17.0

>>> from sage.all import *
>>> R = QQ['x']; (x,) = R._first_ngens(1)
>>> f = (x+Integer(17))*(x-Integer(3))*(x-Integer(1)/Integer(8))**Integer(3)
>>> find_root(f, Integer(0),Integer(4))
2.999999999999995
>>> find_root(f, Integer(0),Integer(1))  # abs tol 1e-6 (note -- precision of answer isn't very good on some machines)
0.124999
>>> find_root(f, -Integer(20),-Integer(10))
-17.0

In Pomerance’s book on primes he asserts that the famous Riemann Hypothesis is equivalent to the statement that the function \(f(x)\) defined below is positive for all \(x \geq 2.01\):

sage: def f(x):
....:     return sqrt(x) * log(x) - abs(Li(x) - prime_pi(x))

>>> from sage.all import *
>>> def f(x):
...     return sqrt(x) * log(x) - abs(Li(x) - prime_pi(x))

We find where \(f\) equals, i.e., what value that is slightly smaller than \(2.01\) that could have been used in the formulation of the Riemann Hypothesis:

sage: find_root(f, 2, 4, rtol=0.0001)
2.0082...

>>> from sage.all import *
>>> find_root(f, Integer(2), Integer(4), rtol=RealNumber('0.0001'))
2.0082...

This agrees with the plot:

sage: plot(f,2,2.01)
Graphics object consisting of 1 graphics primitive

>>> from sage.all import *
>>> plot(f,Integer(2),RealNumber('2.01'))
Graphics object consisting of 1 graphics primitive

The following example was added due to Issue #4942 and demonstrates that the function need not be defined at the endpoints:

sage: find_root(x^2*log(x,2)-1,0, 2)  # abs tol 1e-6
1.41421356237

>>> from sage.all import *
>>> find_root(x**Integer(2)*log(x,Integer(2))-Integer(1),Integer(0), Integer(2))  # abs tol 1e-6
1.41421356237

The following is an example, again from Issue #4942 where Brent’s method fails. Currently no other method is implemented, but at least we acknowledge the fact that the algorithm fails:

sage: find_root(1/(x-1)+1,0, 2)
0.0
sage: find_root(1/(x-1)+1,0.00001, 2)
Traceback (most recent call last):
...
NotImplementedError: Brent's method failed to find a zero for f on the interval

>>> from sage.all import *
>>> find_root(Integer(1)/(x-Integer(1))+Integer(1),Integer(0), Integer(2))
0.0
>>> find_root(Integer(1)/(x-Integer(1))+Integer(1),RealNumber('0.00001'), Integer(2))
Traceback (most recent call last):
...
NotImplementedError: Brent's method failed to find a zero for f on the interval

An example of a function which evaluates to NaN on the entire interval:

sage: f(x) = 0.0 / max(0, x)
sage: find_root(f, -1, 0)
Traceback (most recent call last):
...
RuntimeError: f appears to have no zero on the interval

>>> from sage.all import *
>>> __tmp__=var("x"); f = symbolic_expression(RealNumber('0.0') / max(Integer(0), x)).function(x)
>>> find_root(f, -Integer(1), Integer(0))
Traceback (most recent call last):
...
RuntimeError: f appears to have no zero on the interval
sage.numerical.optimize.minimize(func, x0, gradient=None, hessian=None, algorithm='default', verbose=False, **args)[source]

This function is an interface to a variety of algorithms for computing the minimum of a function of several variables.

INPUT:

  • func – either a symbolic function or a Python function whose argument is a tuple with \(n\) components

  • x0 – initial point for finding minimum

  • gradient – (optional) gradient function. This will be computed automatically for symbolic functions. For Python functions, it allows the use of algorithms requiring derivatives. It should accept a tuple of arguments and return a NumPy array containing the partial derivatives at that point.

  • hessian – (optional) hessian function. This will be computed automatically for symbolic functions. For Python functions, it allows the use of algorithms requiring derivatives. It should accept a tuple of arguments and return a NumPy array containing the second partial derivatives of the function.

  • algorithm – string specifying algorithm to use. Options are 'default' (for Python functions, the simplex method is the default) (for symbolic functions bfgs is the default):

    • 'simplex' – using the downhill simplex algorithm

    • 'powell' – use the modified Powell algorithm

    • 'bfgs' – (Broyden-Fletcher-Goldfarb-Shanno) requires gradient

    • 'cg' – (conjugate-gradient) requires gradient

    • 'ncg' – (newton-conjugate gradient) requires gradient and hessian

  • verbose – boolean (default: False); print convergence message

Note

For additional information on the algorithms implemented in this function, consult SciPy’s documentation on optimization and root finding.

EXAMPLES:

Minimize a fourth order polynomial in three variables (see the Wikipedia article Rosenbrock_function):

sage: vars = var('x y z')                                                       # needs sage.symbolic
sage: f = 100*(y-x^2)^2 + (1-x)^2 + 100*(z-y^2)^2 + (1-y)^2                     # needs sage.symbolic
sage: minimize(f, [.1,.3,.4])  # abs tol 1e-6                                   # needs sage.symbolic
(1.0, 1.0, 1.0)

>>> from sage.all import *
>>> vars = var('x y z')                                                       # needs sage.symbolic
>>> f = Integer(100)*(y-x**Integer(2))**Integer(2) + (Integer(1)-x)**Integer(2) + Integer(100)*(z-y**Integer(2))**Integer(2) + (Integer(1)-y)**Integer(2)                     # needs sage.symbolic
>>> minimize(f, [RealNumber('.1'),RealNumber('.3'),RealNumber('.4')])  # abs tol 1e-6                                   # needs sage.symbolic
(1.0, 1.0, 1.0)

Try the newton-conjugate gradient method; the gradient and hessian are computed automatically:

sage: minimize(f, [.1, .3, .4], algorithm='ncg')  # abs tol 1e-6                # needs sage.symbolic
(1.0, 1.0, 1.0)

>>> from sage.all import *
>>> minimize(f, [RealNumber('.1'), RealNumber('.3'), RealNumber('.4')], algorithm='ncg')  # abs tol 1e-6                # needs sage.symbolic
(1.0, 1.0, 1.0)

We get additional convergence information with the \(verbose\) option:

sage: minimize(f, [.1, .3, .4], algorithm='ncg', verbose=True)                  # needs sage.symbolic
Optimization terminated successfully.
...
(0.9999999..., 0.999999..., 0.999999...)

>>> from sage.all import *
>>> minimize(f, [RealNumber('.1'), RealNumber('.3'), RealNumber('.4')], algorithm='ncg', verbose=True)                  # needs sage.symbolic
Optimization terminated successfully.
...
(0.9999999..., 0.999999..., 0.999999...)

Same example with just Python functions:

sage: def rosen(x):  # The Rosenbrock function
....:    return sum(100.0r*(x[1r:]-x[:-1r]**2.0r)**2.0r + (1r-x[:-1r])**2.0r)
sage: minimize(rosen, [.1,.3,.4]) # abs tol 3e-5
(1.0, 1.0, 1.0)

>>> from sage.all import *
>>> def rosen(x):  # The Rosenbrock function
...    return sum(100.0*(x[1:]-x[:-1]**2.0)**2.0 + (1-x[:-1])**2.0)
>>> minimize(rosen, [RealNumber('.1'),RealNumber('.3'),RealNumber('.4')]) # abs tol 3e-5
(1.0, 1.0, 1.0)

Same example with a pure Python function and a Python function to compute the gradient:

sage: # needs numpy
sage: def rosen(x):  # The Rosenbrock function
....:    return sum(100.0r*(x[1r:]-x[:-1r]**2.0r)**2.0r + (1r-x[:-1r])**2.0r)
sage: import numpy
sage: if int(numpy.version.short_version[0]) > 1:
....:     numpy.set_printoptions(legacy="1.25")
sage: from numpy import zeros
sage: def rosen_der(x):
....:    xm = x[1r:-1r]
....:    xm_m1 = x[:-2r]
....:    xm_p1 = x[2r:]
....:    der = zeros(x.shape, dtype=float)
....:    der[1r:-1r] = 200r*(xm-xm_m1**2r) - 400r*(xm_p1 - xm**2r)*xm - 2r*(1r-xm)
....:    der[0] = -400r*x[0r]*(x[1r]-x[0r]**2r) - 2r*(1r-x[0])
....:    der[-1] = 200r*(x[-1r]-x[-2r]**2r)
....:    return der
sage: minimize(rosen, [.1,.3,.4], gradient=rosen_der,  # abs tol 1e-6
....:          algorithm='bfgs')
(1.0, 1.0, 1.0)

>>> from sage.all import *
>>> # needs numpy
>>> def rosen(x):  # The Rosenbrock function
...    return sum(100.0*(x[1:]-x[:-1]**2.0)**2.0 + (1-x[:-1])**2.0)
>>> import numpy
>>> if int(numpy.version.short_version[Integer(0)]) > Integer(1):
...     numpy.set_printoptions(legacy="1.25")
>>> from numpy import zeros
>>> def rosen_der(x):
...    xm = x[1:-1]
...    xm_m1 = x[:-2]
...    xm_p1 = x[2:]
...    der = zeros(x.shape, dtype=float)
...    der[1:-1] = 200*(xm-xm_m1**2) - 400*(xm_p1 - xm**2)*xm - 2*(1-xm)
...    der[Integer(0)] = -400*x[0]*(x[1]-x[0]**2) - 2*(1-x[Integer(0)])
...    der[-Integer(1)] = 200*(x[-1]-x[-2]**2)
...    return der
>>> minimize(rosen, [RealNumber('.1'),RealNumber('.3'),RealNumber('.4')], gradient=rosen_der,  # abs tol 1e-6
...          algorithm='bfgs')
(1.0, 1.0, 1.0)
sage.numerical.optimize.minimize_constrained(func, cons, x0, gradient=None, algorithm='default', **args)[source]

Minimize a function with constraints.

INPUT:

  • func – either a symbolic function, or a Python function whose argument is a tuple with \(n\) components

  • cons – constraints. This should be either a function or list of functions that must be positive. Alternatively, the constraints can be specified as a list of intervals that define the region we are minimizing in. If the constraints are specified as functions, the functions should be functions of a tuple with \(n\) components (assuming \(n\) variables). If the constraints are specified as a list of intervals and there are no constraints for a given variable, that component can be (None, None).

  • x0 – initial point for finding minimum

  • algorithm – (optional) specify the algorithm to use:

    • 'default' – default choices

    • 'l-bfgs-b' – only effective if you specify bound constraints; see [ZBN1997]

  • gradient – (optional) gradient function. This will be computed automatically for symbolic functions. This is only used when the constraints are specified as a list of intervals.

EXAMPLES:

Let us maximize \(x + y - 50\) subject to the following constraints: \(50x + 24y \leq 2400\), \(30x + 33y \leq 2100\), \(x \geq 45\), and \(y \geq 5\):

sage: f = lambda p: -p[0]-p[1]+50
sage: c_1 = lambda p: p[0]-45
sage: c_2 = lambda p: p[1]-5
sage: c_3 = lambda p: -50*p[0]-24*p[1]+2400
sage: c_4 = lambda p: -30*p[0]-33*p[1]+2100
sage: a = minimize_constrained(f,[c_1,c_2,c_3,c_4],[2,3])
sage: a
(45.0, 6.25...)

>>> from sage.all import *
>>> f = lambda p: -p[Integer(0)]-p[Integer(1)]+Integer(50)
>>> c_1 = lambda p: p[Integer(0)]-Integer(45)
>>> c_2 = lambda p: p[Integer(1)]-Integer(5)
>>> c_3 = lambda p: -Integer(50)*p[Integer(0)]-Integer(24)*p[Integer(1)]+Integer(2400)
>>> c_4 = lambda p: -Integer(30)*p[Integer(0)]-Integer(33)*p[Integer(1)]+Integer(2100)
>>> a = minimize_constrained(f,[c_1,c_2,c_3,c_4],[Integer(2),Integer(3)])
>>> a
(45.0, 6.25...)

Let’s find a minimum of \(\sin(xy)\):

sage: x,y = var('x y')                                                          # needs sage.symbolic
sage: f(x,y) = sin(x*y)                                                         # needs sage.symbolic
sage: minimize_constrained(f, [(None,None),(4,10)],[5,5])                       # needs sage.symbolic
(4.8..., 4.8...)

>>> from sage.all import *
>>> x,y = var('x y')                                                          # needs sage.symbolic
>>> __tmp__=var("x,y"); f = symbolic_expression(sin(x*y)                                                         ).function(x,y)# needs sage.symbolic
>>> minimize_constrained(f, [(None,None),(Integer(4),Integer(10))],[Integer(5),Integer(5)])                       # needs sage.symbolic
(4.8..., 4.8...)

Check if L-BFGS-B finds the same minimum:

sage: minimize_constrained(f, [(None,None),(4,10)],[5,5],                       # needs sage.symbolic
....:                      algorithm='l-bfgs-b')
(4.7..., 4.9...)

>>> from sage.all import *
>>> minimize_constrained(f, [(None,None),(Integer(4),Integer(10))],[Integer(5),Integer(5)],                       # needs sage.symbolic
...                      algorithm='l-bfgs-b')
(4.7..., 4.9...)

Rosenbrock function (see the Wikipedia article Rosenbrock_function):

sage: from scipy.optimize import rosen, rosen_der
sage: minimize_constrained(rosen, [(-50,-10),(5,10)],[1,1],
....:                      gradient=rosen_der, algorithm='l-bfgs-b')
(-10.0, 10.0)
sage: minimize_constrained(rosen, [(-50,-10),(5,10)],[1,1],
....:                      algorithm='l-bfgs-b')
(-10.0, 10.0)

>>> from sage.all import *
>>> from scipy.optimize import rosen, rosen_der
>>> minimize_constrained(rosen, [(-Integer(50),-Integer(10)),(Integer(5),Integer(10))],[Integer(1),Integer(1)],
...                      gradient=rosen_der, algorithm='l-bfgs-b')
(-10.0, 10.0)
>>> minimize_constrained(rosen, [(-Integer(50),-Integer(10)),(Integer(5),Integer(10))],[Integer(1),Integer(1)],
...                      algorithm='l-bfgs-b')
(-10.0, 10.0)