Company X is considering $2$ mutually exclusive expansion plans. Plan A calls for expenditure of Rs $40$ Million on a large scale integrated plant which will provide an expected cashflow stream of Rs $6.4$ million per year for 20 years. Plan B calls for an expenditure of Rs $12$ million a somewhat less effective labour intensive plant which has an expected cash flow stream of Rs $2.72$ million per year for $20$ years. Calculate IRR for each plan.
$$IRR_a: -40+ 6.4\left[\frac{1-(1+r)^{-20}}{r}\right]=0 \tag{1}$$
$$IRR_b: -12+ 2.72\left[\frac{1-(1+r)^{-20}}{r}\right]=0 \tag{2}$$
I could use the newton raphson and differentiate $(1)$: $$\frac{r[-20(1+r)^{-19}]-[1-(1+r)^{-20}]}{r^2}$$
I tried proceeding by $x_{n+1}=x_n- \frac{f(x_n)}{f'(x_n)}$, but this is just complicating it more. I need a simpler way.
There is an old but practical approach. To find $IRR_a$ you need to find $r$ such that $$\left[\frac{1-(1+r)^{-20}}{r}\right]=\frac{40}{6.4} = 6.25$$ Get a financial table for the present value of an annuity 20 periods long, and look up what is the rate of interest close of 6.25. I find $6.259$ for $r=0.15$. Take this as an approximate answer or as the initial condition for your favorite algorithm (even the bisection method should give you the answer in very few steps).
The computation for $IRR_b$ has 2 periods in the main text and 20 in the formula. Assuming 20 is right, we have $$\left[\frac{1-(1+r)^{-20}}{r}\right]=\frac{12}{2.72} \approx 4.412$$ which compares nicely to a value of $4.442$ for $r=0.22$.