Financial Modelling in Python

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

34 Financial Modelling in Python

widths h

i

and each b is a linear combination of the y

i

. More specifically

b

0

= d

lef t

b

i

=

6

h

i

+h

i+1



y

i+1

− y

i

h

i+1

−

y

i

− y

i−1

h

i



b

n

= d

right

(4.8)

where d

left

and d

right

are constants dependent only on the choice of the value of the derivatives

at the endpoints of the curve.

3

Implementation of this scheme in Python requires a little more

effort than the earlier cases:

class cubic spline(interpolation base):

def

init (self, abscissae, ordinates, a 0 = 0.5, d 0=0, b n=0.5,

d

n=0):

interpolation

base. init (self, abscissae, ordinates)

xs, ys, N = self.abscissae, self.ordinates, self.N

b=[d

0]+(N - 1)*[0]

A

sub, A dia, A sup = N*[0], [2.0] + (N - 1)*[0], [a 0] +

(N - 1)*[0]

for i in range(1, N - 1):

H, h = xs[i + 1]- xs[i], xs[i] - xs[i - 1]

b[i] = (6./(h + H))*(((ys[i + 1] - ys[i])/H) -

((ys[i] - ys[i - 1])/h))

a

i = H/(h + H)

b

i=1.0-ai

A

dia[i], A sup[i], A sub[i] = 2., a i, b i

A

sub[N - 1], A dia[N - 1], b[N - 1] = b n, 2.0, d n

self.C = linear

algebra.solve tridiagonal system(N, A sub, A dia,

A

sup, b)

def

call (self, x):

xs, ys, C = self.abscissae, self.ordinates, self.C

i, j,

, , , = interpolation base.locate(self, x)

h

i = xs[j] - xs[i]

x

low = xs[j] - x

x

low3 = math.pow(x low, 3)

x

high = x - xs[i]

x

high3 = math.pow(x high, 3)

hi

sqrd 6=hi*h i/6.0

return C[i]*x

low3/(6.0*h i)+C[j]*x high3/(6.0*h i)+\

(ys[i]-C[i]*hi

sqrd 6)*x low/h i+(ys[j]- ...