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=1, k=None)¶

Solves 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 put in the bins, while keeping sure that each bin contains a weight of at most $K$ ?

For more informations : http://en.wikipedia.org/wiki/Bin_packing_problem

Two version of this problem are solved by this algorithm :

Is it possible to put the given items in $L$ bins ?
What is the assignment of items using the least number of bins with the given list of items ?

INPUT:

items – A list of real values (the items’ weight)
maximum – The maximal size of a bin
k – 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 number possible of bins.

OUTPUT:

A list of lists, each member corresponding to a box and containing the list of the weights inside it. If there is no solution, an exception is raised (this can only happen when k is specified or if maximum is less that the size 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)
sage: len(bins)
3

Checking the bins are of correct size

sage: all([ sum(b)<= 1 for b in bins ])
True

Checking every item is in a bin

sage: b1, b2, b3 = bins
sage: all([ (v in b1 or v in b2 or v in b3) for v in values ])
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)
Traceback (most recent call last):
...
ValueError: This problem has no solution !

sage.numerical.optimize.find_fit(data, model, initial_guess=None, parameters=None, variables=None, solution_dict=False)¶

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

We define a function with free parameters $a$ , $b$ and $c$ :

sage: model(x) = a * sin(b * x - c)

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

sage: find_fit(data, model)
[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], solution_dict = True)
sage: fit[a], fit[b], fit[c]
(1.21..., 0.49..., 0.19...)

We search for a formula for the $n$ -th prime number:

sage: dataprime = [(i, nth_prime(i)) for i in xrange(1, 5000, 100)]
sage: find_fit(dataprime, a * x * log(b * x), 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)¶

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)
(1.579175535558..., 0.7853981...)

sage.numerical.optimize.find_local_minimum(f, a, b, tol=1.48e-08, maxfun=500)¶

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 self.
tol – the convergence tolerance
maxfun – maximum function evaluations

OUTPUT:

minval – (float) the minimum value that self takes on in the interval $[a,b]$
x – (float) the point at which self 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))
sage: find_local_minimum(f, 1, 15)
(-9.4772942594..., 9.5293344109...)

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

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

Enlarging the interval returns a larger minimum:

sage: find_local_minimum(f, -2.5, 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: plot(f, (x,-2.5, -1)).ymin()
-2.1827...
sage: plot(f, (x,-2.5, 2)).ymin()
-2.1827...

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=4.5e-16, maxiter=100, full_output=False)¶

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.
maxiter – integer; if convergence is not achieved in maxiter iterations, an error is raised. Must be $\geq 0$ .
full_output – bool (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)  # note -- precision of answer isn't very good on some machines.
0.124999...
sage: find_root(f, -20,-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))

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...

This agrees with the plot:

sage: plot(f,2,2.01)

sage.numerical.optimize.linear_program(c, G, h, A=None, b=None, solver=None)¶

Solves the dual linear programs:

Minimize $c'x$ subject to $Gx + s = h$ , $Ax = b$ , and $s \geq 0$ where $'$ denotes transpose.
Maximize $-h'z - b'y$ subject to $G'z + A'y + c = 0$ and $z \geq 0$ .

INPUT:

c – a vector
G – a matrix
h – a vector
A – a matrix
b — a vector
solver (optional) — solver to use. If None, the cvxopt’s lp-solver

is used. If it is ‘glpk’, then glpk’s solver is used.

These can be over any field that can be turned into a floating point number.

OUTPUT:

A dictionary sol with keys x, s, y, z corresponding to the variables above:

sol['x'] – the solution to the linear program
sol['s'] – the slack variables for the solution
sol['z'], sol['y'] – solutions to the dual program

EXAMPLES:

First, we minimize $-4x_1 - 5x_2$ subject to $2x_1 + x_2 \leq 3$ , $x_1 + 2x_2 \leq 3$ , $x_1 \geq 0$ , and $x_2 \geq 0$ :

sage: c=vector(RDF,[-4,-5])
sage: G=matrix(RDF,[[2,1],[1,2],[-1,0],[0,-1]])
sage: h=vector(RDF,[3,3,0,0])
sage: sol=linear_program(c,G,h)
sage: sol['x']
(0.999..., 1.000...)

Next, we maximize $x+y-50$ subject to $50x + 24y \leq 2400$ , $30x + 33y \leq 2100$ , $x \geq 45$ , and $y \geq 5$ :

sage: v=vector([-1.0,-1.0,-1.0])
sage: m=matrix([[50.0,24.0,0.0],[30.0,33.0,0.0],[-1.0,0.0,0.0],[0.0,-1.0,0.0],[0.0,0.0,1.0],[0.0,0.0,-1.0]])
sage: h=vector([2400.0,2100.0,-45.0,-5.0,1.0,-1.0])
sage: sol=linear_program(v,m,h)
sage: sol['x']
(45.000000..., 6.2499999...3, 1.00000000...)
sage: sol=linear_program(v,m,h,solver='glpk')
GLPK Simplex Optimizer...
OPTIMAL SOLUTION FOUND
sage: sol['x']
(45.0..., 6.25, 1.0...)

sage.numerical.optimize.minimize(func, x0, gradient=None, hessian=None, algorithm='default', **args)¶

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'
- 'powell'
- 'bfgs' – (Broyden-Fletcher-Goldfarb-Shanno) requires gradient
- 'cg' – (conjugate-gradient) requires gradient
- 'ncg' – (newton-conjugate gradient) requires gradient and hessian

EXAMPLES:

sage: vars=var('x y z')
sage: f=100*(y-x^2)^2+(1-x)^2+100*(z-y^2)^2+(1-y)^2
sage: minimize(f,[.1,.3,.4],disp=0)
(1.00..., 1.00..., 1.00...)

sage: minimize(f,[.1,.3,.4],algorithm="ncg",disp=0)
(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],disp=0)
(1.00..., 1.00..., 1.00...)

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

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: 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,algorithm="bfgs",disp=0)
(1.00...,  1.00..., 1.00...)

sage.numerical.optimize.minimize_constrained(func, cons, x0, gradient=None, algorithm='default', **args)¶

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 [ZBN97].
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: y = var('y')
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)

Let’s find a minimum of $\sin(xy)$ :

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

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

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

Rosenbrock function, [http://en.wikipedia.org/wiki/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)

REFERENCES:

[ZBN97]

C. Zhu, R. H. Byrd and J. Nocedal. L-BFGS-B: Algorithm 778: L-BFGS-B, FORTRAN routines for large scale bound constrained optimization. ACM Transactions on Mathematical Software, Vol 23, Num. 4, pp.550–560, 1997.

Numerical Root Finding and Optimization¶

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