Question

A regular furnace costs $900. A high-efficiency furnace costs $1500. The average monthly bill with the regular furnace is expected to be $80. The average monthly bill with the high-efficiency furnace is expected to be $50. How long until the high-efficiency furnace pays for itself? Assume an interest rate of 1% per month

Answer 1

For deciding the number of months till high efficiency furnace pays for itself.

Cash flow associated with regular furnace = cash flow associated with high efficiency furnace.

Remember that present discounted value (PV) of any future cash flow (FV) at interest rate r received at the end of nth month is given by -

$PV = \frac{FV}{(1+r)^n}$

$\Rightarrow 900 + \frac{80}{1.01}+ \frac{80}{(1.01)^2}+....+ \frac{80}{(1.01)^n} = 1500 + \frac{50}{1.01}+ \frac{50}{(1.01)^2}+....+ \frac{50}{(1.01)^n}$

$\Rightarrow \frac{80-50}{1.01}+ \frac{80-50}{(1.01)^2}+....+ \frac{80-50}{(1.01)^n} = 600$

$\Rightarrow \frac{30}{1.01}+ \frac{30}{(1.01)^2}+....+ \frac{30}{(1.01)^n} = 600$

$\Rightarrow \frac{1}{1.01}+ \frac{1}{(1.01)^2}+....+ \frac{1}{(1.01)^n} = 600/30 = 20$

Sum of a finite G.P. with n terms - first term "a" and common ratio "r" is given by (when r < 1)

$S = \frac{a(1-r^n)}{1-r} = \frac{0.99(1-0.99^n)}{0.01} = 99(1-0.99^n)$ where r = 1/1.01 = 0.99

Hence, equating LHS with RHS, we get:

$99(1-0.99^n) = 20 \Rightarrow 1-0.99^n = 20/99 = 0.202\Rightarrow 0.99^n = 1-0.202$

$\Rightarrow 0.99^n = 0.798\Rightarrow n ln(0.99) = ln(0.798)$

$\Rightarrow n = ln(0.798)/ln(0.99) = 22.45$

Hence by the end of 22 months, high efficiency furnace would have paid it for itself.