text: • Jun 5th 2008, 08:49 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, 09:29 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 $\frac{40204.50-40000}{40000}=.005 = .5\%$ If you figured from just the extra investment made ($40,000-$15,000=$25,000), then the ROI would be
  
  $\frac{40204.50-25000}{40000}=.38=38\%$
  • Jun 5th 2008, 12:00 PM
  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:
  
  $
  FV = PV*(1+i)^n
  $
  
  So here you have
  $
  7500 = 5000*(1+0.05)^n
  $
  
  or
  $
  1.5 = 1.05^n
  $
  
  Now solve for n. The way to do that is using logarithms. Take the log of both sides:
  
  $
  log(1.5) = log(1.05^n)
  $
  
  Recall from highschool that $log(a^b) = b \cdot log(a)$, so you have
  
  $
  log(1.5) = n \cdot log(1.05)$
  , and hence
  $
  n = \frac {log(1.5)} {log(1.05)} = 8.3\ \text{years.}
  $
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