Financial Modelling in Python

P1: JYS

c04 JWBK378-Fletcher May 23, 2009 4:21 Printer: Yet to come

Basic Mathematical Tools 43

the expected result is x = [2.97142857 20.2 14].

>>> from numpy import *

>>> from numpy.linalg import svd

>>> A = matrix(array(

... [[1.75, 1.5, -2.5],

... [0.0, -0.5, 0.65],

... [0.0, 0.0, 0.25]], float))

>>> A

matrix([[ 1.75, 1.5 , -2.5 ],

[ 0. , -0.5 , 0.65],

[ 0. , 0. , 0.25]])

>>> b = array([0.5, -1.0, 3.5])

>>> b

array([ 0.5, -1. , 3.5])

>>> u, w, v = svd(A)

>>> x = singular

value decomposition back substitution(u, w, v, b)

>>> x = matrix(x).transpose() # column vector

>>> x

matrix([[ 2.97142857],

[ 20.2 ],

[ 14. ]])

"""

if len(u.shape) <>2:

raise RuntimeError, "Expected ’u’ to be a matrix"

if len(w.shape) <>1:

raise RuntimeError, "Expected ’w’ to be a column vector"

if len(v.shape) <>2:

raise RuntimeError, "Expected ’v’ to be a matrix"

if len(b.shape) <>1:

raise RuntimeError, "Expected ’b’ to be a column vector"

m = u.shape[0]

n = u.shape[1]

if w.shape[0] <>n:

raise RuntimeError, "’w’ column vector has incorrect size"

if b.shape[0] <>m:

raise RuntimeError, "’b’ column vector has incorrect size"

if v.shape[0] <> n or v.shape[1] <>n:

raise RuntimeError, "’v’ matrix has incorrect size"

tmp = numpy.zeros(n)

for j in range(n):

s = 0.0

if w[j] <>0:

for i in range(m):

s += u[i, j]*b[i]

s /= w[j]

tmp[j] = s

x = numpy.zeros(n)