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Cyrille gravatar image

how to calculate the internal rate of return

According to Keynes the internal rate of return is the r value which solve 0=I+A1/(1+r)+A2/(1+r)2+...+A1/(1+r)n for I and the list of the Ai given .

It is evident that we gain in in changing variables for α=1/(1+r) to obtain : 0=A1α+A2α2+...+A1αn. And in a last operation after that we have find a solution to solve on r α=1/(1+r) or directly to calculate r=(1/α)1.

My frist attempt to solve this problem whose solution is 7.7138 (after Excel) is the following. First I plot the rhs of the equation to see if there is a solution. It seems to exist in the good interval.

fr =-250+100*x+150*x^2+200*x^3+250*x^4+300*x^5
plot(fr,(x,0,1))

Then I try

sol=fr.find_root(0.001, 1)       
show(sol)
show((1/sol)-1)

which returns sol=0.6380683030615047 and r=1sol1=0.567230334435854. But this solution differs too muxh from Excel's one. I do not understand from where comes the problem.

how to calculate the internal rate of return

According to Keynes the internal rate of return is the r value which solve 0=I+A1/(1+r)+A2/(1+r)2+...+A1/(1+r)n for I and the list of the Ai given .

It is evident that we gain in in changing variables for α=1/(1+r) to obtain : 0=A1α+A2α2+...+A1αn. And in a last operation after that we have find a solution to solve on r α=1/(1+r) or directly to calculate r=(1/α)1.

My frist attempt to solve this problem whose solution is 7.7138 (after Excel) is the following. First I plot the rhs of the equation to see if there is a solution. It seems to exist in the good interval.

fr =-250+100*x+150*x^2+200*x^3+250*x^4+300*x^5
plot(fr,(x,0,1))

Then I try

sol=fr.find_root(0.001, 1)       
show(sol)
show((1/sol)-1)

which returns sol=0.6380683030615047 and r=1sol1=0.567230334435854. But this solution differs too muxh much from Excel's one. I do not understand from where comes the problem.