Internal rate of return
1 Answer
Using the formula for geometric series (https://en.wikipedia.org/wiki/Geometric_series) we get
$$NPV= -A+\sum_{t=0}^{T} \frac{EZU}{(1+EZF)^t}=-A+EZU(\frac{1-(1+EZF)^{T+1}}{1-(1+EZF)}) $$
\[=-A+\frac{EZU}{EZF}[(1+EZF)^{T+1}-1].\]
Hence
\[\frac{(1+EZF)^{T+1}-1}{EZF}=\frac{(NPV+A)}{EZU}.\]
This equation can not be explicitly solved in terms of $EZF$ and one needs to solve it numerically.
Erdos
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