Jim Thomlinson said:
Are you sure that is correct.
Yes; and you could confirm that yourself with less than 60 seconds of
effort.
But first, you need to understand my interpretation of the problem for which
I provided a solution. I believe that Kyle is asking: what is the sum of
the cost (or cost saving?) over 30 years if the cost (or cost saving?) the
first year is $0.17 times 21,128 kWh, and the cost (or cost saving?)
increases 5% each year.
If you disagree that that is what Kyle asked for, fine. We simply have a
difference of opinion of the definition of the problem. If you disagree
with Kyle that that's the problem to solve, that's another matter
altogether. I am simply saying that FV solves my interpretation of Kyle's
problem.
To demonstrate....
Let A1 be the cost the first year: =21128*0.17. Let A2 be the cost the
second year: =21128*0.17*(1+5%). But that's just: =A1*(1+5%). Let A3 be
the cost the third year: =21128*0.17*(1+5%)*(1+5%). But that's just:
=A2*(1+5%). So drag A2 down through A30. Then =SUM(A1:A30) is the total
cost (or cost saving?) over 30 years.
Now compare that SUM with FV(5%,30,-0.17*21128,0). QED.
FV is normally used to calculate the future value of an investment.
Fred is better at explaining the concepts behind this. I can only explain
the algebra.
Consider an investment of P dollars at the end of each year, with a growth
rate of 5%. At the end of the second year, the investment value is
P*(1+5%)+P. At the end of the third year, the value is (P*(1+5%)+P)*(1+5%),
which is P*(1+5%)^2 + P*(1+5%) + P. At the end of thirty years, the value
is P*(1+5%)^29 +...+ P*(1+5%) + P. FV(5%,30,-P,0) is the result of the sum.
Now consider my interpretation of Kyle's problem. The cost (or cost
saving?) in the first year is P, where P=kWh*$0.17. The cost in the second
year is P*(1+5%), as demonstrated by the A1:A30 model above. The cost in
the second year is P*(1+5%)^2. The cost in year 30 is P*(1+5%)^29. The
total cost (or cost saving?) over 30 years is P + P*(1+5%) +...+
P*(1+5%)^29.
Look familiar? QED.
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