IRR , computation , Newton's Method

Project valuation may require finding the Internal Rate of Return, or the discount rate at which a project's present value of future cash flows equals to zero. This is the maximum rate of return achievable, and should be greater than the inflation rate. a simple converging guessing method has been illustrated under the Project Valuation chapter. The use of Newton's method was left as an exercise, and a possible solution is shown below. Newton's method finds the x-axis intersection of the gradient line found at an approximate IRR, which will give a better rate to try for IRR, and iterations will find an acceptable IRR where the NPV is small enough to be equal to zero.

background for derivative function of npv:

Axiom: for bx^a ,

d ( bx^a) / dx = ab x^a-1

Hence, if,

npv = CF0 / R⁰ + CF1 / R¹ + .. + CFn / Rⁿ,

or npv = CF0 + CF1 x R^-1 + CF2 x R^-2 + .. + CFn x R^-n,

then ,

d(npv)/dR = 0 x CF0 x R ^-1 + - CF1x R ^-2 + - 2 CF2x R ^-3 ... + -n x CFn x R^{-n - 1}.

#copyright SJT, GNU 2014.

# Newton's method

# graph is y-axis NPV and x-axis is rate, so find where NPV = 0, or the project rate function crosses the x-axis.

# Do this by finding where the slope line of the function curve at any given R crosses the x-axis, which gives a

# better R' which is closer to where the function curve crosses the x-axis, and this is done iteratively until

# a close enough R' is found where the npv is practically 0.

def irr(cfs):

def npv(cfs, r):

R = 1. + r/100.

return reduce( lambda x,(i,y): x + y * R**-i , enumerate(cfs), 0)

def deriv_npv( cfs, r):

R = 1. + r/ 100.

return reduce ( lambda x, (i,y): x - i * y * R **(-i-1) , enumerate( cfs), 0)

r = 10.

lim = 0.1 # ten cents is practically zero

while True:

np_v = npv( cfs, r)

print "np_v", np_v

if abs(np_v ) < lim:

break

dnpv = deriv_npv( cfs, r)

r = r - np_v / dnpv

return r

print irr( [ -30000, 3000, 4000, 16000, 15000, 5000] )