# How to calculate Net Present Value?

So I have this question :

Your boss asked you to evaluate a project with an infinite life.

Sales and costs project to$$\1,000$$ and $$\500$$ per year, respectively. (Assume sales and costs occur at the end of the year, i.e., profit of $$\500$$ at the end of year one.)

There is no depreciation and the tax rate is $$30 \%$$. The real required rate of return is $$10\%$$. The inflation rate is $$4\%$$ and is expected to be $$4 \%$$ forever. Sales and costs will increase at the rate of inflation. If the project costs $$\3,000$$, what is the NPV?

1. $$\500.00$$

2. $$\1629.62$$

3. $$\365.38$$

4. $$\472.22$$

On the answer sheet it states that

$$NPV = -3000+ \cfrac{(1000-50)(1-0.30)}{0.10}$$,

which will give me a result of $$\500$$.

The only thing i do not understand is why is it divided by the real rate of return $$0.10$$, and not $$1+r$$ real?

I have done various exercises where I always divide by $$1+r$$.

The present value of the series of cash flows is as follows;

$$-3000+\frac{(1000-500)\cdot (1-0.3)}{1.1^1}+\frac{(1000-500)\cdot (1-0.3)}{1.1^2}+\frac{(1000-500)\cdot (1-0.3)}{1.1^3}+\frac{(1000-500)\cdot (1-0.3)}{1.1^4}+\ldots$$

$$=-3000+\sum_{k=1}^{\infty}\frac{(1000-500)\cdot (1-0.3)}{1.1^k}$$

For simplicity let $$C=(1000-500)\cdot (1-0.3)$$. The infinite sum is $$\sum\limits_{k=1}^{\infty}\frac{C}{1.1^k}$$

We can look at the partial sum of the geometric series $$\sum\limits_{k=1}^{n}\frac{C}{1.1^k}=C\cdot \frac{1}{1.1}\cdot \frac{1-\left(\frac{1}{1.1} \right)^n}{1-\frac{1}{1.1}}$$

Now we can expand the term by $$1.1$$ (blue terms)

$$=C\cdot \frac{\color{blue}{1.1}}{1.1}\cdot \frac{ 1-\left(\frac{1}{1.1} \right)^n}{\color{blue}{1.1}\cdot \left(1-\frac{1}{1.1}\right)}=C\cdot \frac{ 1-\left(\frac{1}{1.1} \right)^n}{1.1-1}=C\cdot \frac{ 1-\left(\frac{1}{1.1} \right)^n}{0.1}$$

Finally $$n$$ goes to infinty. $$\left(\frac{1}{1.1} \right)$$ is smaller than $$1$$. It decreases when n increases. Therefore

$$\lim_{n \to \infty} C\cdot \frac{ 1-\left(\frac{1}{1.1} \right)^n}{0.1}= C\cdot \frac{ 1-0}{0.1}=\frac{C}{0.1}$$

I think it is clear from where the $$0.1$$ comes.