Integrating the Compound Interest Formula

[IForgotCalc2]

I feel stupid for not being able to solve this. Basically, I have a solar electric system with fixed cost, which has electric output (kWh/day) that is worth a monetary value that increases with time.

Here is the equation for when the system will break even:

Initial Cost = (kWh/day) * t * (average value of kWh)

Units: $ = kWh/d * d * $/kWh
Units: $ = kWh * $/kWh
Units: $ = $

average value of kWh ($/kWh) = average of compound interest formula = base price * integral ((1 + i) ^ t)

Putting it all together:

IC = Output * t * base price * integral ((1 + i) ^ t)

I know the IC, Output, and base price, and an example of those values is:

IC = $9,800
Output = 8.736 kWh/day
base price = 0.13

Substituting these in, you can simplify the formula:

9800/(8.736 * 0.13) = t * integral ((1 + i) ^ t)
8629.19 = t * integral ((1 + i) ^ t)

If this were just slightly easier, I could solve it myself, but I'm afraid it's been too long since I took Calc2. Any help is greatly appreciated.

 

[Deleted member 4993]

I feel stupid for not being able to solve this. Basically, I have a solar electric system with fixed cost, which has electric output (kWh/day) that is worth a monetary value that increases with time.

Here is the equation for when the system will break even:

Initial Cost = (kWh/day) * t * (average value of kWh)

Units: $ = kWh/d * d * $/kWh
Units: $ = kWh * $/kWh
Units: $ = $

average value of kWh ($/kWh) = average of compound interest formula = base price * integral ((1 + i) ^ t)

Putting it all together:

IC = Output * t * base price * integral ((1 + i) ^ t)

I know the IC, Output, and base price, and an example of those values is:

IC = $9,800
Output = 8.736 kWh/day
base price = 0.13

Substituting these in, you can simplify the formula:

9800/(8.736 * 0.13) = t * integral ((1 + i) ^ t)
8629.19 = t * integral ((1 + i) ^ t)

If this were just slightly easier, I could solve it myself, but I'm afraid it's been too long since I took Calc2. Any help is greatly appreciated.

Hint:

\(\displaystyle \int a^t dt \ = \ \frac{a^t}{ln(a)} \ + \ C\)

 

[IForgotCalc2]

Hint:

\(\displaystyle \int a^t dt \ = \ \frac{a^t}{ln(a)} \ + \ C\)

Is this right?


\(\displaystyle t * \int (1 + i) ^ t\)

=

\(\displaystyle t * \ \frac{(1 + i)^t}{ln(1 + i)} \\)

=

\(\displaystyle t * (1 + i)^t * \ \frac{1}{ln(1+i)} \\)

=

\(\displaystyle 20.4959 * t * (1+i)^t\)


Plugging it back into the original equation...

\(\displaystyle \ \frac{IC}{Output * base price * ln(1 + i)}\ = t * (1 + i) ^t\)


Question, though. If this is right, and even if I divide both sides by 20.4959 (i.e. ln(1.05)) and assign a value for i, how do I solve, \(\displaystyle t * (1.05)^t = C\), for t?

My math skills have fallen into disrepair.

 

[Deleted member 4993]

Hint:

\(\displaystyle \int a^t dt \ = \ \frac{a^t}{ln(a)} \ + \ C\)

Is this right?


\(\displaystyle t * \int (1 + i) ^ t\)

=

\(\displaystyle t * \ \frac{(1 + i)^t}{ln(1 + i)} \\)

What are you integrating - function of 't' or function of 'i'? In other words, inside the integration sign - do you have 'dt' or 'di'?
=

\(\displaystyle t * (1 + i)^t * \ \frac{1}{ln(1+i)} \\)

=

\(\displaystyle 20.4959 * t * (1+i)^t\)


Plugging it back into the original equation...

\(\displaystyle \ \frac{IC}{Output * base price * ln(1 + i)}\ = t * (1 + i) ^t\)


Question, though. If this is right, and even if I divide both sides by 20.4959 (i.e. ln(1.05)) and assign a value for i, how do I solve, \(\displaystyle t * (1.05)^t = C\), for t?

My math skills have fallen into disrepair.

 

[IForgotCalc2]

What are you integrating - function of 't' or function of 'i'? In other words, inside the integration sign - do you have 'dt' or 'di'?

Integrating over time, t.

Basically, if I get 10kWh of output per day (we'll make it a nice round number instead of the 8.736 kWh/d mentioned above), that output is worth more and more each year: anywhere from a 5% increase to a 10% increase.

In order to find the value of the output each year, assuming a 5% increase in energy prices annually, I could just do...

Value of Output through Year 1: 3650kWh * .13 ($/kWh)
Value of Output through Year 2: Year 1 + 3650kWh * .13 ($/kWh) * 1.05
Value of Output through Year 3: Year 2 + 3650kWh * .13 ($/kWh) * 1.05^2
Value of Output through Year 4: Year 3 + 3650kWh * .13 ($/kWh) * 1.05^3
Value of Output through Year 5: Year 4 + 3650kWh * .13 ($/kWh) * 1.05^4
Value of Output through Year 6: Year 5 + 3650kWh * .13 ($/kWh) * 1.05^5
Value of Output through Year 7: Year 6 + 3650kWh * .13 ($/kWh) * 1.05^6
Value of Output through Year 8: Year 7 + 3650kWh * .13 ($/kWh) * 1.05^7
Value of Output through Year 9: Year 8 + 3650kWh * .13 ($/kWh) * 1.05^8
Value of Output through Year 10: Year 9 + 3650kWh * .13 ($/kWh) * 1.05^9

And also do these calculations with a 10% increase, but the problem is that these prices are adjusted yearly, whereas I would like them to be integration-based and thus 100% accurate.

 

[IForgotCalc2]

\(\displaystyle 20.4959 * t * (1+i)^t\)

This was a little bit of an oopsie, by the way. I didn't explain that I plugged in i = 0.05 into \(\displaystyle \ \frac{1}{ln(1+i)} \\) in order to get the 20.4959 value until the very end of the post.

I also made a massive error here:

\(\displaystyle LHS = t * (1 + i)^t * \ \frac{1}{ln(1+i)} \\)

Plugging it back into the original equation... correctly this time:

\(\displaystyle \ \frac{IC * ln(1 + i)}{Output * base price}\ = t * (1 + i) ^t\)



My apologies. I warned you that I'm rusty!

 

[Deleted member 4993]

\(\displaystyle \int a^t dt \ = \ \frac{a^t}{ln(a)} \ + \ C\)

\(\displaystyle \int t * a^t dt \ = \ t* \frac{a^t}{ln(a)} \ + \ \frac{1}{ln(a)}*\int a^t dt \ = \ \ t* \frac{a^t}{ln(a)} \ + \ \frac{a^t}{[ln(a)]^2} \ + \ C\)

 

[IForgotCalc2]

\(\displaystyle \int a^t dt \ = \ \frac{a^t}{ln(a)} \ + \ C\)

\(\displaystyle \int t * a^t dt \ = \ t* \frac{a^t}{ln(a)} \ + \ \frac{1}{ln(a)}*\int a^t dt \ = \ \ t* \frac{a^t}{ln(a)} \ + \ \frac{a^t}{[ln(a)]^2} \ + \ C\)

Is that Integration by Parts?

Whatever it is, you're amazing! I'll have to test it out, but I get the feeling that there's a 99.999% chance it's right.

 

[IForgotCalc2]

Glad I checked... lol. I caught another mistake on my part (a^t needs to be a^t/365) and there is a small misunderstanding: I am trying to solve \(\displaystyle \int (a)^\ \frac{t}{365}\ dt \\) , not this: \(\displaystyle \int t* (a)^\ \frac{t}{365}\ dt \\). In other words, there is no t inside the integral. It seems like a really easy fix to your solution would be to multiply by t/t, keeping the t/1 inside the integral, making your solution correct, so long as I multiply the integral result by 1/t. Is that right? or am I once again doing it wrong?



By the way, if you are too frustrated with me and this problem at this point... I'll understand if you don't reply... lol. Pretty much all of this is my fault. I can't apologize enough.



Here it is from scratch. When Initial Cost = Value of Accumulated Output, the system is paid off. The value of accumulated output is \(\displaystyle 8.736 (kWh/day) * t (days) * (average value of 1 kWh)\), where the value of a kWh on any day, which we'll call v(t), is gotten from the compound-interest formula (future value = starting value * (1 + interest rate) ^ number of interest periods), or in this case: \(\displaystyle v(t) = 0.13 * (1+i)^\ \frac{t}{365}\\). In order to get the average of the kWh-value function, v(t), it must be integrated: \(\displaystyle 0.13 * \int (1+i)^\ \frac{t}{365}\ dt \\).

Adding this back into the original equation, using a instead of (1 + i), we get:

\(\displaystyle Initial Cost = 9800 (dollars) = (8.736 kWh/day) * (t days) * 0.13 (dollarsperkWh) * \int a^\ \frac{t}{365}\ dt \\),

or

\(\displaystyle \ \frac{9800}{(8.736 * 0.13)}\ = t * \int a^\ \frac{t}{365}\ dt \\).

It is only that integral that I need help with. So long as you can integrate it, I can figure out the rest.



Both my fingers are crossed that I didn't screw it up this time.

 

[Deleted member 4993]

You should not use integral for this problem - you should use summation (of geometric series - as is done for annuity calculation or mortgage calculation).