• Jun 5th 2008, 07:49 AM
JAMESANANE
Machine A can be bought for $15,000, but machine B, the automated model, will cost$40,000 to buy and install. The automated machine will reduce annual cost by $5743.50 over machine A. Either machine will have a service life of seven years. The rate of return on the extra investment required by buy machine B is • Jun 5th 2008, 08:29 AM masters Quote: Originally Posted by JAMESANANE Machine A can be bought for$15,000, but machine B, the automated model, will cost $40,000 to buy and install. The automated machine will reduce annual cost by$5743.50 over machine A. Either machine will have a service life of seven years. The rate of return on the extra investment required by buy machine B is

Someone with more business saavy might approach this differently, but here goes:

The $40,000 purchase will more than pay for itself in 7 years considering the annual reduction in operating costs (7 X$5,743.50 = $40,204.50). The return on investment (ROI) given just that information would be$\displaystyle \frac{40204.50-40000}{40000}=.005 = .5\%$If you figured from just the extra investment made ($40,000-$15,000=$25,000), then the ROI would be

$\displaystyle \frac{40204.50-25000}{40000}=.38=38\%$
• Jun 5th 2008, 11:00 AM
ebaines
1. The rate of return on the extra investment is merely the extra profit you get from it each year divided by the additional initial investment. So that's 5743/25,000 = 23%

2. The formula for compounding interest is:

$\displaystyle FV = PV*(1+i)^n$

So here you have
$\displaystyle 7500 = 5000*(1+0.05)^n$
or
$\displaystyle 1.5 = 1.05^n$

Now solve for n. The way to do that is using logarithms. Take the log of both sides:

$\displaystyle log(1.5) = log(1.05^n)$

Recall from highschool that $\displaystyle log(a^b) = b \cdot log(a)$, so you have

$\displaystyle log(1.5) = n \cdot log(1.05)$, and hence
$\displaystyle n = \frac {log(1.5)} {log(1.05)} = 8.3\ \text{years.}$