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)
...
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_maximum_on_interval(f, a, b, tol=1.48e-08, maxfun=500)¶

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

See the documentation for find_minimum_on_interval() for more details.

EXAMPLES:

sage: f = lambda x: x*cos(x)
sage: find_maximum_on_interval(f, 0,5)
(0.561096338191..., 0.8603335890...)
sage: find_maximum_on_interval(f, 0,5, tol=0.1, maxfun=10)
(0.561090323458..., 0.857926501456...)

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

Numerically find the 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.

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_minimum_on_interval(f, 1, 5)
(-3.28837139559..., 3.4256184695...)
sage: find_minimum_on_interval(f, 1, 5, tol=1e-3)
(-3.28837136189098..., 3.42575079030572...)
sage: find_minimum_on_interval(f, 1, 5, tol=1e-2, maxfun=10)
(-3.28837084598..., 3.4250840220...)
sage: show(plot(f, 0, 20))
sage: find_minimum_on_interval(f, 1, 15)
(-9.4772942594..., 9.5293344109...)

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=9.9999999999999998e-13, rtol=4.5000000000000002e-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.

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)
2.9999999999999951
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 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)
2.0082590205656166

This agrees with the plot:

sage: plot(f,2,2.01)

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

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.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-shannon) 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¶

Functions and Methods¶

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