Hey can anyone help with this? This is the classic NPV equation:

$$\texttt{NPV = -CapEx} + \sum_{i=0}^n \frac{\texttt{Revenue − Costs}}{(1+\texttt{Discount})^i}$$

For my purposes all the elements are know except costs.

I need to isolate costs in this equation. Is this possible?



  • $\begingroup$ Yes, because the factor (Revenue - Costs) doesn't depend on $i$ and therefore can be fatored out of the sum. You can then solve for Costs using standard algebra (treating the remaining sum as one big constant). $\endgroup$ Nov 2, 2015 at 20:52

1 Answer 1


Your equation is

$NPV = -CapEx + \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad |+CapEX$

$NPV + CapEx = \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad $

Splitting the fraction

$NPV + CapEx = \sum_{i=0}^n \left( \frac{R }{(1+d)^i} -\frac{C }{(1+d)^i} \right) \quad $

Factoring out $\frac{1}{(1+d)^i}$

$NPV + CapEx = \sum_{i=0}^n \frac{1}{(1+d)^i}\left( R-C \right) \quad $

$NPV + CapEx = \left( R-C \right)\cdot \sum_{i=0}^n \frac{1}{(1+d)^i} \quad | \cdot (-1)$

$-NPV - CapEx = \left( -R+C \right)\cdot \sum_{i=0}^n \frac{1}{(1+d)^i} \quad $

Dividing the equation by $\sum_{i=0}^n \frac{1}{(1+d)^i}$

$\frac{-NPV - CapEx}{\sum_{i=0}^n \frac{1}{(1+d)^i}}=-R+C$

$C=R-\frac{NPV + CapEx}{\sum_{i=0}^n \frac{1}{(1+d)^i}}$

It is $\sum_{i=0}^n \frac{1}{(1+d)^i}=\sum_{i=0}^n \left( \frac{1}{1+d}\right)^ i$

This is a partial sum of a geometric series.

Therefore $\sum_{i=0}^n \left( \frac{1}{1+d}\right)^ i=\Large{\frac{1-\left( \frac{1}{1+d}\right)^{n+1}}{1-\frac{1}{1+d}}}$

In total it is $C=R-(NPV+CapEx)\cdot \Large{\frac{1- \frac{1}{1+d}}{1-\left(\frac{1}{1+d}\right)^{n+1}}}$

Solve for R-C

$R-C=(NVP+CapEx)\cdot \Large{\frac{\left( 1- \frac{1}{1+d}\right) }{\left(1-\left(\frac{1}{1+d}\right)^{n+1}\right)}}$

$R-C=(1909+1315)\cdot \Large{\frac{\left(1- \frac{1}{1+0.08}\right)}{\left(1-\left(\frac{1}{1+0.08}\right)^{24}\right)}}=\normalsize 283.5268$

  • $\begingroup$ I´ve made an edit. The formula itself for $R-C$ is not very difficult. But maybe the input into a hand-held calculator is not very comfortable. I add brackets for the nominator and the denominator. They are often forgotten. I hope you will get the same result. $\endgroup$ Nov 6, 2015 at 12:57
  • $\begingroup$ I overcame my issues with implementing this. I have used this working extensively now and it is holding up perfectly. Its really brilliant. Thank you for your time! $\endgroup$
    – Mike
    Nov 6, 2015 at 19:43

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