Popular computer application such as MS Excel has made analytical solutions easier to obtain. In the past, solving a polynomial of higher degree than 3 was both time consuming and cumbersome. The easiness that MS Excel offers has, however, a drawback in terms of our understanding of methods and procedures. Take for example, the computation of an internal rate of return (IRR). Easy done with a financial calculator or with the Excel function =IRR(c0, c1, ..., cn, guess). The easiness is still involved for more complex problems such as obtaining an IRR from a combination of time-limited cash flows and a perpetuity. Using Goal seek does the work.
This semester I am again teaching Corporate Finance based on Principles of Corporate Finance by R.A. Brealey, S.C. Myers and F. Allen. In chapter 5 on pages 139-141 of the 10th Global edition, they are elaborating on why the IRR rule is unreliable for mutually exclusive projects. They present three projects (F, G and H) with F showing the highest IRR but the lowest NPV. In the building up of the argument, they report the IRR and the NPV for each project. Unfortunately, the majority of my students does not see how they got their IRR for the incremental cash flow between G and F. While the IRR for a perpetuity is easy to obtain, an IRR of a project with cash flows over a limited number of years followed by a perpetuity is not.
The cash flows for F are -$9,000 at t=0, $6,000 at t=1, $5,000 at t=2, $4,000 at t=3, and 0 afterwards. There is a unique IRR for F since there is no a shift in the sign of the cash flows. Using a financial calculator, the IRR is 33%. What about if we are left with no other alternatives than solving it by hand? How we go about obtaining the IRR? I will make the problem more complex by looking at the incremental cash flows on page 141 of the same book and edition. But before, let us get the IRR for G, which cash flow is a perpetuity.
The cash flows for G are -$9,000 at t=0 and $1,800 for ever. The IRR is the rate that makes NPV=0. Therefore, from 0=-$9,000+($1,800/IRR), we get IRR = $1,800/$9,000 = 0.2 or 20%.
Comparing F with G in terms of NPV, G is better than F. However, better than F if the discount rate is less than the rate at which F and G are equal. This is the IRR we need to know how to find. It is actually 15.6%. But how it was obtained? That is the question.
A financial calculator does not completely help in this case. Since you obviously will not be allowed to use Excel in an exam except a financial calculator, a better way to get it manually is to use the Newton Raphson Method, which solves equations of the form f(x)=0.
This is an iterative method to find a solution to an equation with a degree greater than 1.
The incremental cash flow (G-F) are 0 at t=0, -4200 at t=1, -3200 at t=2, -2200 at t=3, 1800 at t=4, and 1800 for ever.
The Newton Raphson method is given by x(n+1) = x(n) - (f(x)/f'(x)), where n=0,1,...,N, N is the number of iterations, f(x) is the equation that is set equal to 0, and f'(x) is the first derivative of f(x).
In our case,
f(x) = -4200(1+r)^(-1)+-3200(1+r)^(-2)+-2200(1+r)^(-3)+1800/r(1+r)^3
f'(x) = -4200(1+r)^(-2)+6400(1+r)^(-3)+6600(1+r)^(-4)-1800[((1+r)^3+3r(1+r)^2))/(r(1+r)^3)^2
where r = IRR and a simple expression for the term in bold is (-1800/r^2)((1+4r)/(1+r)^4).
Now let start by setting x0=10% and getting the sum of f(x0) and f'(x0), respectively.
Iteration 1
x0=0.1
sum of f(x0) = 5,407.964
sum of f'(x0) = -159,332
x1 = 0.1 - (5407.964/-159332) = 0.133941
error bound 1 (e1)= x1 - x0 = 0.133941 - 0.1 = 0.033941 which is greater than 0.000001 (the tolerance level)
Iteration 2
x1=0.133941
sum of f(x1) = 1515.4
sum of f'(x1) = -81549.9
x2 = 0.133941 - (1515.4/-81549.9) = 0.152525
e2 = 0.152525 - 0.133941 = 0.018583 which is greater than the tolerance level
Iteration 3
x2=0.152525
sum of f(x2) = 218.403
sum of f'(x2) = -59523.1
x3 = 0.156194
e3 = 0.00369 which is greater than the tolerance level
Iteration 4
x3=0.156194
sum of f(x3) = 6.344567
sum of f'(x3) = -56107.2
x4 = 0.156307
e4 = 0.000113 which is greater than the tolerance level
Iteration 5
x4=0.156307
sum of f(x4) = 0.00573
sum of f'(x4) = -56005.9
x5 = 0.156307
e5 = 0.0000001 which is less than the tolerance level.
As we not getting further the procedure has converged to an IRR that is 15.631%.
I will not ask such a problem in an exam, but my students' self confidence is boosted by getting a sense of how their brains work through the mirror of the Newton Raphson iterative method.
Be blessed!
This semester I am again teaching Corporate Finance based on Principles of Corporate Finance by R.A. Brealey, S.C. Myers and F. Allen. In chapter 5 on pages 139-141 of the 10th Global edition, they are elaborating on why the IRR rule is unreliable for mutually exclusive projects. They present three projects (F, G and H) with F showing the highest IRR but the lowest NPV. In the building up of the argument, they report the IRR and the NPV for each project. Unfortunately, the majority of my students does not see how they got their IRR for the incremental cash flow between G and F. While the IRR for a perpetuity is easy to obtain, an IRR of a project with cash flows over a limited number of years followed by a perpetuity is not.
The cash flows for F are -$9,000 at t=0, $6,000 at t=1, $5,000 at t=2, $4,000 at t=3, and 0 afterwards. There is a unique IRR for F since there is no a shift in the sign of the cash flows. Using a financial calculator, the IRR is 33%. What about if we are left with no other alternatives than solving it by hand? How we go about obtaining the IRR? I will make the problem more complex by looking at the incremental cash flows on page 141 of the same book and edition. But before, let us get the IRR for G, which cash flow is a perpetuity.
The cash flows for G are -$9,000 at t=0 and $1,800 for ever. The IRR is the rate that makes NPV=0. Therefore, from 0=-$9,000+($1,800/IRR), we get IRR = $1,800/$9,000 = 0.2 or 20%.
Comparing F with G in terms of NPV, G is better than F. However, better than F if the discount rate is less than the rate at which F and G are equal. This is the IRR we need to know how to find. It is actually 15.6%. But how it was obtained? That is the question.
A financial calculator does not completely help in this case. Since you obviously will not be allowed to use Excel in an exam except a financial calculator, a better way to get it manually is to use the Newton Raphson Method, which solves equations of the form f(x)=0.
This is an iterative method to find a solution to an equation with a degree greater than 1.
The incremental cash flow (G-F) are 0 at t=0, -4200 at t=1, -3200 at t=2, -2200 at t=3, 1800 at t=4, and 1800 for ever.
The Newton Raphson method is given by x(n+1) = x(n) - (f(x)/f'(x)), where n=0,1,...,N, N is the number of iterations, f(x) is the equation that is set equal to 0, and f'(x) is the first derivative of f(x).
In our case,
f(x) = -4200(1+r)^(-1)+-3200(1+r)^(-2)+-2200(1+r)^(-3)+1800/r(1+r)^3
f'(x) = -4200(1+r)^(-2)+6400(1+r)^(-3)+6600(1+r)^(-4)-1800[((1+r)^3+3r(1+r)^2))/(r(1+r)^3)^2
where r = IRR and a simple expression for the term in bold is (-1800/r^2)((1+4r)/(1+r)^4).
Now let start by setting x0=10% and getting the sum of f(x0) and f'(x0), respectively.
Iteration 1
x0=0.1
sum of f(x0) = 5,407.964
sum of f'(x0) = -159,332
x1 = 0.1 - (5407.964/-159332) = 0.133941
error bound 1 (e1)= x1 - x0 = 0.133941 - 0.1 = 0.033941 which is greater than 0.000001 (the tolerance level)
Iteration 2
x1=0.133941
sum of f(x1) = 1515.4
sum of f'(x1) = -81549.9
x2 = 0.133941 - (1515.4/-81549.9) = 0.152525
e2 = 0.152525 - 0.133941 = 0.018583 which is greater than the tolerance level
Iteration 3
x2=0.152525
sum of f(x2) = 218.403
sum of f'(x2) = -59523.1
x3 = 0.156194
e3 = 0.00369 which is greater than the tolerance level
Iteration 4
x3=0.156194
sum of f(x3) = 6.344567
sum of f'(x3) = -56107.2
x4 = 0.156307
e4 = 0.000113 which is greater than the tolerance level
Iteration 5
x4=0.156307
sum of f(x4) = 0.00573
sum of f'(x4) = -56005.9
x5 = 0.156307
e5 = 0.0000001 which is less than the tolerance level.
As we not getting further the procedure has converged to an IRR that is 15.631%.
I will not ask such a problem in an exam, but my students' self confidence is boosted by getting a sense of how their brains work through the mirror of the Newton Raphson iterative method.
Be blessed!
