# Solving a partial sum....

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?

Thanks,

Mike

• 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). Commented Nov 2, 2015 at 20:52

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

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

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

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

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

• 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. Commented Nov 6, 2015 at 12:57
• 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!
– Mike
Commented Nov 6, 2015 at 19:43