The `scipy.optimize` package provides several commonly used
optimization algorithms. A detailed listing is available:
`scipy.optimize` (can also be found by `help(scipy.optimize)`).

The module contains:

- Unconstrained and constrained minimization of multivariate scalar
functions (
`minimize`) using a variety of algorithms (e.g. BFGS, Nelder-Mead simplex, Newton Conjugate Gradient, COBYLA or SLSQP) - Global (brute-force) optimization routines (e.g.,
`anneal`,`basinhopping`) - Least-squares minimization (
`leastsq`) and curve fitting (`curve_fit`) algorithms - Scalar univariate functions minimizers (
`minimize_scalar`) and root finders (`newton`) - Multivariate equation system solvers (
`root`) using a variety of algorithms (e.g. hybrid Powell, Levenberg-Marquardt or large-scale methods such as Newton-Krylov).

Below, several examples demonstrate their basic usage.

The `minimize` function provides a common interface to unconstrained
and constrained minimization algorithms for multivariate scalar functions
in `scipy.optimize`. To demonstrate the minimization function consider the
problem of minimizing the Rosenbrock function of variables:

The minimum value of this function is 0 which is achieved when

Note that the Rosenbrock function and its derivatives are included in
`scipy.optimize`. The implementations shown in the following sections
provide examples of how to define an objective function as well as its
jacobian and hessian functions.

In the example below, the `minimize` routine is used
with the *Nelder-Mead* simplex algorithm (selected through the `method`
parameter):

```
>>> import numpy as np
>>> from scipy.optimize import minimize
```

```
>>> def rosen(x):
... """The Rosenbrock function"""
... return sum(100.0*(x[1:]-x[:-1]**2.0)**2.0 + (1-x[:-1])**2.0)
```

```
>>> x0 = np.array([1.3, 0.7, 0.8, 1.9, 1.2])
>>> res = minimize(rosen, x0, method='nelder-mead',
... options={'xtol': 1e-8, 'disp': True})
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 339
Function evaluations: 571
```

```
>>> print res.x
[ 1. 1. 1. 1. 1.]
```

The simplex algorithm is probably the simplest way to minimize a fairly well-behaved function. It requires only function evaluations and is a good choice for simple minimization problems. However, because it does not use any gradient evaluations, it may take longer to find the minimum.

Another optimization algorithm that needs only function calls to find
the minimum is *Powell*‘s method available by setting `method='powell'` in
`minimize`.

In order to converge more quickly to the solution, this routine uses the gradient of the objective function. If the gradient is not given by the user, then it is estimated using first-differences. The Broyden-Fletcher-Goldfarb-Shanno (BFGS) method typically requires fewer function calls than the simplex algorithm even when the gradient must be estimated.

To demonstrate this algorithm, the Rosenbrock function is again used. The gradient of the Rosenbrock function is the vector:

This expression is valid for the interior derivatives. Special cases are

A Python function which computes this gradient is constructed by the code-segment:

```
>>> def rosen_der(x):
... xm = x[1:-1]
... xm_m1 = x[:-2]
... xm_p1 = x[2:]
... der = np.zeros_like(x)
... der[1:-1] = 200*(xm-xm_m1**2) - 400*(xm_p1 - xm**2)*xm - 2*(1-xm)
... der[0] = -400*x[0]*(x[1]-x[0]**2) - 2*(1-x[0])
... der[-1] = 200*(x[-1]-x[-2]**2)
... return der
```

This gradient information is specified in the `minimize` function
through the `jac` parameter as illustrated below.

```
>>> res = minimize(rosen, x0, method='BFGS', jac=rosen_der,
... options={'disp': True})
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 51
Function evaluations: 63
Gradient evaluations: 63
>>> print res.x
[ 1. 1. 1. 1. 1.]
```

The method which requires the fewest function calls and is therefore often the fastest method to minimize functions of many variables uses the Newton-Conjugate Gradient algorithm. This method is a modified Newton’s method and uses a conjugate gradient algorithm to (approximately) invert the local Hessian. Newton’s method is based on fitting the function locally to a quadratic form:

where is a matrix of second-derivatives (the Hessian). If the Hessian is positive definite then the local minimum of this function can be found by setting the gradient of the quadratic form to zero, resulting in

The inverse of the Hessian is evaluated using the conjugate-gradient method. An example of employing this method to minimizing the Rosenbrock function is given below. To take full advantage of the Newton-CG method, a function which computes the Hessian must be provided. The Hessian matrix itself does not need to be constructed, only a vector which is the product of the Hessian with an arbitrary vector needs to be available to the minimization routine. As a result, the user can provide either a function to compute the Hessian matrix, or a function to compute the product of the Hessian with an arbitrary vector.

The Hessian of the Rosenbrock function is

if with defining the matrix. Other non-zero entries of the matrix are

For example, the Hessian when is

The code which computes this Hessian along with the code to minimize the function using Newton-CG method is shown in the following example:

```
>>> def rosen_hess(x):
... x = np.asarray(x)
... H = np.diag(-400*x[:-1],1) - np.diag(400*x[:-1],-1)
... diagonal = np.zeros_like(x)
... diagonal[0] = 1200*x[0]**2-400*x[1]+2
... diagonal[-1] = 200
... diagonal[1:-1] = 202 + 1200*x[1:-1]**2 - 400*x[2:]
... H = H + np.diag(diagonal)
... return H
```

```
>>> res = minimize(rosen, x0, method='Newton-CG',
... jac=rosen_der, hess=rosen_hess,
... options={'avextol': 1e-8, 'disp': True})
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 19
Function evaluations: 22
Gradient evaluations: 19
Hessian evaluations: 19
>>> print res.x
[ 1. 1. 1. 1. 1.]
```

For larger minimization problems, storing the entire Hessian matrix can
consume considerable time and memory. The Newton-CG algorithm only needs
the product of the Hessian times an arbitrary vector. As a result, the user
can supply code to compute this product rather than the full Hessian by
giving a `hess` function which take the minimization vector as the first
argument and the arbitrary vector as the second argument (along with extra
arguments passed to the function to be minimized). If possible, using
Newton-CG with the Hessian product option is probably the fastest way to
minimize the function.

In this case, the product of the Rosenbrock Hessian with an arbitrary vector is not difficult to compute. If is the arbitrary vector, then has elements:

Code which makes use of this Hessian product to minimize the
Rosenbrock function using `minimize` follows:

```
>>> def rosen_hess_p(x,p):
... x = np.asarray(x)
... Hp = np.zeros_like(x)
... Hp[0] = (1200*x[0]**2 - 400*x[1] + 2)*p[0] - 400*x[0]*p[1]
... Hp[1:-1] = -400*x[:-2]*p[:-2]+(202+1200*x[1:-1]**2-400*x[2:])*p[1:-1] \
... -400*x[1:-1]*p[2:]
... Hp[-1] = -400*x[-2]*p[-2] + 200*p[-1]
... return Hp
```

```
>>> res = minimize(rosen, x0, method='Newton-CG',
... jac=rosen_der, hess=rosen_hess_p,
... options={'avextol': 1e-8, 'disp': True})
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 20
Function evaluations: 23
Gradient evaluations: 20
Hessian evaluations: 44
>>> print res.x
[ 1. 1. 1. 1. 1.]
```

The `minimize` function also provides an interface to several
constrained minimization algorithm. As an example, the Sequential Least
SQuares Programming optimization algorithm (SLSQP) will be considered here.
This algorithm allows to deal with constrained minimization problems of the
form:

As an example, let us consider the problem of maximizing the function:

subject to an equality and an inequality constraints defined as:

The objective function and its derivative are defined as follows.

```
>>> def func(x, sign=1.0):
... """ Objective function """
... return sign*(2*x[0]*x[1] + 2*x[0] - x[0]**2 - 2*x[1]**2)
```

```
>>> def func_deriv(x, sign=1.0):
... """ Derivative of objective function """
... dfdx0 = sign*(-2*x[0] + 2*x[1] + 2)
... dfdx1 = sign*(2*x[0] - 4*x[1])
... return np.array([ dfdx0, dfdx1 ])
```

Note that since `minimize` only minimizes functions, the `sign`
parameter is introduced to multiply the objective function (and its
derivative by -1) in order to perform a maximization.

Then constraints are defined as a sequence of dictionaries, with keys
`type`, `fun` and `jac`.

```
>>> cons = ({'type': 'eq',
... 'fun' : lambda x: np.array([x[0]**3 - x[1]]),
... 'jac' : lambda x: np.array([3.0*(x[0]**2.0), -1.0])},
... {'type': 'ineq',
... 'fun' : lambda x: np.array([x[1] - 1]),
... 'jac' : lambda x: np.array([0.0, 1.0])})
```

Now an unconstrained optimization can be performed as:

```
>>> res = minimize(func, [-1.0,1.0], args=(-1.0,), jac=func_deriv,
... method='SLSQP', options={'disp': True})
Optimization terminated successfully. (Exit mode 0)
Current function value: -2.0
Iterations: 4
Function evaluations: 5
Gradient evaluations: 4
>>> print res.x
[ 2. 1.]
```

and a constrained optimization as:

```
>>> res = minimize(func, [-1.0,1.0], args=(-1.0,), jac=func_deriv,
... constraints=cons, method='SLSQP', options={'disp': True})
Optimization terminated successfully. (Exit mode 0)
Current function value: -1.00000018311
Iterations: 9
Function evaluations: 14
Gradient evaluations: 9
>>> print res.x
[ 1.00000009 1. ]
```

All of the previously-explained minimization procedures can be used to solve a least-squares problem provided the appropriate objective function is constructed. For example, suppose it is desired to fit a set of data to a known model, where is a vector of parameters for the model that need to be found. A common method for determining which parameter vector gives the best fit to the data is to minimize the sum of squares of the residuals. The residual is usually defined for each observed data-point as

An objective function to pass to any of the previous minization algorithms to obtain a least-squares fit is.

The `leastsq` algorithm performs this squaring and summing of the
residuals automatically. It takes as an input argument the vector
function and returns the
value of which minimizes
directly. The user is also encouraged to provide the Jacobian matrix
of the function (with derivatives down the columns or across the
rows). If the Jacobian is not provided, it is estimated.

An example should clarify the usage. Suppose it is believed some measured data follow a sinusoidal pattern

where the parameters , and are unknown. The residual vector is

By defining a function to compute the residuals and (selecting an appropriate starting position), the least-squares fit routine can be used to find the best-fit parameters . This is shown in the following example:

```
>>> from numpy import *
>>> x = arange(0,6e-2,6e-2/30)
>>> A,k,theta = 10, 1.0/3e-2, pi/6
>>> y_true = A*sin(2*pi*k*x+theta)
>>> y_meas = y_true + 2*random.randn(len(x))
```

```
>>> def residuals(p, y, x):
... A,k,theta = p
... err = y-A*sin(2*pi*k*x+theta)
... return err
```

```
>>> def peval(x, p):
... return p[0]*sin(2*pi*p[1]*x+p[2])
```

```
>>> p0 = [8, 1/2.3e-2, pi/3]
>>> print array(p0)
[ 8. 43.4783 1.0472]
```

```
>>> from scipy.optimize import leastsq
>>> plsq = leastsq(residuals, p0, args=(y_meas, x))
>>> print plsq[0]
[ 10.9437 33.3605 0.5834]
```

```
>>> print array([A, k, theta])
[ 10. 33.3333 0.5236]
```

```
>>> import matplotlib.pyplot as plt
>>> plt.plot(x,peval(x,plsq[0]),x,y_meas,'o',x,y_true)
>>> plt.title('Least-squares fit to noisy data')
>>> plt.legend(['Fit', 'Noisy', 'True'])
>>> plt.show()
```

Often only the minimum of an univariate function (i.e. a function that
takes a scalar as input) is needed. In these circumstances, other
optimization techniques have been developed that can work faster. These are
accessible from the `minimize_scalar` function which proposes several
algorithms.

There are actually two methods that can be used to minimize an univariate
function: `brent` and `golden`, but `golden` is included only for academic
purposes and should rarely be used. These can be respectively selected
through the *method* parameter in `minimize_scalar`. The `brent`
method uses Brent’s algorithm for locating a minimum. Optimally a bracket
(the *bs* parameter) should be given which contains the minimum desired. A
bracket is a triple such that and . If this is not given, then alternatively two starting points can
be chosen and a bracket will be found from these points using a simple
marching algorithm. If these two starting points are not provided *0* and
*1* will be used (this may not be the right choice for your function and
result in an unexpected minimum being returned).

Here is an example:

```
>>> from scipy.optimize import minimize_scalar
>>> f = lambda x: (x - 2) * (x + 1)**2
>>> res = minimize_scalar(f, method='brent')
>>> print res.x
1.0
```

Very often, there are constraints that can be placed on the solution space
before minimization occurs. The *bounded* method in `minimize_scalar`
is an example of a constrained minimization procedure that provides a
rudimentary interval constraint for scalar functions. The interval
constraint allows the minimization to occur only between two fixed
endpoints, specified using the mandatory *bs* parameter.

For example, to find the minimum of near
, `minimize_scalar` can be called using the interval
as a constraint. The result is
:

```
>>> from scipy.special import j1
>>> res = minimize_scalar(j1, bs=(4, 7), method='bounded')
>>> print res.x
5.33144184241
```

If one has a single-variable equation, there are four different root
finding algorithms that can be tried. Each of these algorithms requires the
endpoints of an interval in which a root is expected (because the function
changes signs). In general `brentq` is the best choice, but the other
methods may be useful in certain circumstances or for academic purposes.

A problem closely related to finding the zeros of a function is the
problem of finding a fixed-point of a function. A fixed point of a
function is the point at which evaluation of the function returns the
point: Clearly the fixed point of
is the root of
Equivalently, the root of is the fixed_point of
The routine
`fixed_point` provides a simple iterative method using Aitkens
sequence acceleration to estimate the fixed point of given a
starting point.

Finding a root of a set of non-linear equations can be achieve using the
`root` function. Several methods are available, amongst which `hybr`
(the default) and `lm` which respectively use the hybrid method of Powell
and the Levenberg-Marquardt method from MINPACK.

The following example considers the single-variable transcendental equation

a root of which can be found as follows:

```
>>> import numpy as np
>>> from scipy.optimize import root
>>> def func(x):
... return x + 2 * np.cos(x)
>>> sol = root(func, 0.3)
>>> sol.x
array([-1.02986653])
>>> sol.fun
array([ -6.66133815e-16])
```

Consider now a set of non-linear equations

We define the objective function so that it also returns the Jacobian and
indicate this by setting the `jac` parameter to `True`. Also, the
Levenberg-Marquardt solver is used here.

```
>>> def func2(x):
... f = [x[0] * np.cos(x[1]) - 4,
... x[1]*x[0] - x[1] - 5]
... df = np.array([[np.cos(x[1]), -x[0] * np.sin(x[1])],
... [x[1], x[0] - 1]])
... return f, df
>>> sol = root(func2, [1, 1], jac=True, method='lm')
>>> sol.x
array([ 6.50409711, 0.90841421])
```

Methods `hybr` and `lm` in `root` cannot deal with a very large
number of variables (*N*), as they need to calculate and invert a dense *N
x N* Jacobian matrix on every Newton step. This becomes rather inefficient
when *N* grows.

Consider for instance the following problem: we need to solve the following integrodifferential equation on the square :

with the boundary condition on the upper edge and
elsewhere on the boundary of the square. This can be done
by approximating the continuous function *P* by its values on a grid,
, with a small grid spacing
*h*. The derivatives and integrals can then be approximated; for
instance . The problem is then equivalent to finding the root of
some function `residual(P)`, where `P` is a vector of length
.

Now, because can be large, methods `hybr` or `lm` in
`root` will take a long time to solve this problem. The solution can
however be found using one of the large-scale solvers, for example
`krylov`, `broyden2`, or `anderson`. These use what is known as the
inexact Newton method, which instead of computing the Jacobian matrix
exactly, forms an approximation for it.

The problem we have can now be solved as follows:

```
import numpy as np
from scipy.optimize import root
from numpy import cosh, zeros_like, mgrid, zeros
# parameters
nx, ny = 75, 75
hx, hy = 1./(nx-1), 1./(ny-1)
P_left, P_right = 0, 0
P_top, P_bottom = 1, 0
def residual(P):
d2x = zeros_like(P)
d2y = zeros_like(P)
d2x[1:-1] = (P[2:] - 2*P[1:-1] + P[:-2]) / hx/hx
d2x[0] = (P[1] - 2*P[0] + P_left)/hx/hx
d2x[-1] = (P_right - 2*P[-1] + P[-2])/hx/hx
d2y[:,1:-1] = (P[:,2:] - 2*P[:,1:-1] + P[:,:-2])/hy/hy
d2y[:,0] = (P[:,1] - 2*P[:,0] + P_bottom)/hy/hy
d2y[:,-1] = (P_top - 2*P[:,-1] + P[:,-2])/hy/hy
return d2x + d2y + 5*cosh(P).mean()**2
# solve
guess = zeros((nx, ny), float)
sol = root(residual, guess, method='krylov', options={'disp': True})
#sol = root(residual, guess, method='broyden2', options={'disp': True, 'max_rank': 50})
#sol = root(residual, guess, method='anderson', options={'disp': True, 'M': 10})
print 'Residual', abs(residual(sol.x)).max()
# visualize
import matplotlib.pyplot as plt
x, y = mgrid[0:1:(nx*1j), 0:1:(ny*1j)]
plt.pcolor(x, y, sol.x)
plt.colorbar()
plt.show()
```

When looking for the zero of the functions ,
*i = 1, 2, ..., N*, the `krylov` solver spends most of its
time inverting the Jacobian matrix,

If you have an approximation for the inverse matrix
, you can use it for *preconditioning* the
linear inversion problem. The idea is that instead of solving
one solves : since
matrix is “closer” to the identity matrix than
is, the equation should be easier for the Krylov method to deal with.

The matrix *M* can be passed to `root` with method `krylov` as an
option `options['jac_options']['inner_M']`. It can be a (sparse) matrix
or a `scipy.sparse.linalg.LinearOperator` instance.

For the problem in the previous section, we note that the function to solve consists of two parts: the first one is application of the Laplace operator, , and the second is the integral. We can actually easily compute the Jacobian corresponding to the Laplace operator part: we know that in one dimension

so that the whole 2-D operator is represented by

The matrix of the Jacobian corresponding to the integral
is more difficult to calculate, and since *all* of it entries are
nonzero, it will be difficult to invert. on the other hand
is a relatively simple matrix, and can be inverted by
`scipy.sparse.linalg.splu` (or the inverse can be approximated by
`scipy.sparse.linalg.spilu`). So we are content to take
and hope for the best.

In the example below, we use the preconditioner .

```
import numpy as np
from scipy.optimize import root
from scipy.sparse import spdiags, kron
from scipy.sparse.linalg import spilu, LinearOperator
from numpy import cosh, zeros_like, mgrid, zeros, eye
# parameters
nx, ny = 75, 75
hx, hy = 1./(nx-1), 1./(ny-1)
P_left, P_right = 0, 0
P_top, P_bottom = 1, 0
def get_preconditioner():
"""Compute the preconditioner M"""
diags_x = zeros((3, nx))
diags_x[0,:] = 1/hx/hx
diags_x[1,:] = -2/hx/hx
diags_x[2,:] = 1/hx/hx
Lx = spdiags(diags_x, [-1,0,1], nx, nx)
diags_y = zeros((3, ny))
diags_y[0,:] = 1/hy/hy
diags_y[1,:] = -2/hy/hy
diags_y[2,:] = 1/hy/hy
Ly = spdiags(diags_y, [-1,0,1], ny, ny)
J1 = kron(Lx, eye(ny)) + kron(eye(nx), Ly)
# Now we have the matrix `J_1`. We need to find its inverse `M` --
# however, since an approximate inverse is enough, we can use
# the *incomplete LU* decomposition
J1_ilu = spilu(J1)
# This returns an object with a method .solve() that evaluates
# the corresponding matrix-vector product. We need to wrap it into
# a LinearOperator before it can be passed to the Krylov methods:
M = LinearOperator(shape=(nx*ny, nx*ny), matvec=J1_ilu.solve)
return M
def solve(preconditioning=True):
"""Compute the solution"""
count = [0]
def residual(P):
count[0] += 1
d2x = zeros_like(P)
d2y = zeros_like(P)
d2x[1:-1] = (P[2:] - 2*P[1:-1] + P[:-2])/hx/hx
d2x[0] = (P[1] - 2*P[0] + P_left)/hx/hx
d2x[-1] = (P_right - 2*P[-1] + P[-2])/hx/hx
d2y[:,1:-1] = (P[:,2:] - 2*P[:,1:-1] + P[:,:-2])/hy/hy
d2y[:,0] = (P[:,1] - 2*P[:,0] + P_bottom)/hy/hy
d2y[:,-1] = (P_top - 2*P[:,-1] + P[:,-2])/hy/hy
return d2x + d2y + 5*cosh(P).mean()**2
# preconditioner
if preconditioning:
M = get_preconditioner()
else:
M = None
# solve
guess = zeros((nx, ny), float)
sol = root(residual, guess, method='krylov',
options={'disp': True,
'jac_options': {'inner_M': M}})
print 'Residual', abs(residual(sol.x)).max()
print 'Evaluations', count[0]
return sol.x
def main():
sol = solve(preconditioning=True)
# visualize
import matplotlib.pyplot as plt
x, y = mgrid[0:1:(nx*1j), 0:1:(ny*1j)]
plt.clf()
plt.pcolor(x, y, sol)
plt.clim(0, 1)
plt.colorbar()
plt.show()
if __name__ == "__main__":
main()
```

Resulting run, first without preconditioning:

```
0: |F(x)| = 803.614; step 1; tol 0.000257947
1: |F(x)| = 345.912; step 1; tol 0.166755
2: |F(x)| = 139.159; step 1; tol 0.145657
3: |F(x)| = 27.3682; step 1; tol 0.0348109
4: |F(x)| = 1.03303; step 1; tol 0.00128227
5: |F(x)| = 0.0406634; step 1; tol 0.00139451
6: |F(x)| = 0.00344341; step 1; tol 0.00645373
7: |F(x)| = 0.000153671; step 1; tol 0.00179246
8: |F(x)| = 6.7424e-06; step 1; tol 0.00173256
Residual 3.57078908664e-07
Evaluations 317
```

and then with preconditioning:

```
0: |F(x)| = 136.993; step 1; tol 7.49599e-06
1: |F(x)| = 4.80983; step 1; tol 0.00110945
2: |F(x)| = 0.195942; step 1; tol 0.00149362
3: |F(x)| = 0.000563597; step 1; tol 7.44604e-06
4: |F(x)| = 1.00698e-09; step 1; tol 2.87308e-12
Residual 9.29603061195e-11
Evaluations 77
```

Using a preconditioner reduced the number of evaluations of the
`residual` function by a factor of *4*. For problems where the
residual is expensive to compute, good preconditioning can be crucial
— it can even decide whether the problem is solvable in practice or
not.

Preconditioning is an art, science, and industry. Here, we were lucky in making a simple choice that worked reasonably well, but there is a lot more depth to this topic than is shown here.

References

Some further reading and related software:

[KK] | D.A. Knoll and D.E. Keyes, “Jacobian-free Newton-Krylov methods”, J. Comp. Phys. 193, 357 (2003). |

[PP] | PETSc http://www.mcs.anl.gov/petsc/ and its Python bindings http://code.google.com/p/petsc4py/ |

[AMG] | PyAMG (algebraic multigrid preconditioners/solvers) http://code.google.com/p/pyamg/ |