Why doesn't (1+r)^t cancel when on top and bottom of a fraction and discount rate question

Question

A bit of a silly question really but me and my boss can't figure it out even though we've both got maths degrees :p

We're working out the LCOE (levelised cost of electricity) for a project that we're working on (https://en.wikipedia.org/wiki/Cost_of_electricity_by_source#Levelized_cost_of_electricity) and there's 2 problems we're having

1) I don't understand why the discount rate $(1+r)^t$ is on the denominator. Discount rate is all about how the value of money goes down in time right like £100 now might only be worth £75 in 4 years so howcome there is a discount rate for the power generated on the bottom? 1000 MWhs today is surely the same as 1000 MWhs in 4 years time? Is this even a Mathematics question? Sorry if not

2) Since $(1+r)^t$ is on both the top and bottom then howcome both of the $(1+r)^t $ don't cancel out? This is something really obvious isn't it? I'm kicking myself 'cause I can't think of it.

Thanks for your time

Answer 1

They're inside the summation and depend on the summation, take this small example

$$\frac{\sum\limits_{t=1}^2 \frac{A_t}{(1+r)^t}}{\sum\limits_{t=1}^2\frac{B_t}{(1+r)^t}} = \frac{\frac{A_1}{(1+r)}+\frac{A_2}{(1+r)^2}}{\frac{B_1}{(1+r)}+\frac{B_2}{(1+r)^2}} $$ and you can probably see from here that you cannot cancel all powers of $(1+r)$. You could e.g. cancel a single $(1+r)$ and you would get $$\frac{A_1+\frac{A_2}{(1+r)}}{B_1+\frac{B_2}{(1+r)}}$$

Answer 2

Pretend $n = 2$. Then $$ \text{LCOE} = \frac{\frac{I_1+M_1+F_1}{(1+r)^1} + \frac{I_2+M_2+F_2}{(1+r)^2}}{\frac{E_1}{(1+r)^1} + \frac{E_2}{(1+r)^2}} $$ Now can you cancel the $(1+r)^1$ and $(1+r)^2$?