Calculating the capitalized costs

  • Thread starter Thread starter ainster31
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves calculating the capitalized costs associated with constructing an electric power plant, covering annual operating costs, and performing periodic renovations. The contractor must determine the total amount needed in a trust fund that earns interest to cover these expenses indefinitely.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the initial construction cost, annual operating costs, and renovation costs, exploring how these factors contribute to the total capitalized cost.
  • Some participants question the calculations presented, particularly regarding the formulas used for the annual operating costs and renovation costs.
  • There is a consideration of when the renovation costs are incurred and how that impacts the overall calculation.
  • One participant shares a personal test with different initial capital amounts to see how they sustain the expenses over time.
  • Another participant suggests using Excel for calculations, discussing the setup of formulas to track balances over time.

Discussion Status

The discussion is ongoing, with various interpretations and calculations being explored. Some participants have provided insights into their thought processes and assumptions, while others are questioning the validity of certain approaches. There is no explicit consensus on the correct method or final outcome yet.

Contextual Notes

Participants are operating under the assumption that the operating costs are deducted before interest is applied, and they are considering the timing of the renovation costs in relation to the overall financial model. There are also concerns about the accuracy of financial formulas and the implications of different assumptions on the calculations.

ainster31
Messages
158
Reaction score
1

Homework Statement



A municipal contractor has agreed to construct an electric power plant and to deposit sufficient money in a perpetual trust fund to pay a $10,000/year operating cost and to perform a major renovation to the plant every 15 years at a cost of $200,000. The plant itself will initially cost $500,000 to construct. If the trust fund earns 10% interest per year (compounded annually), what is his capitalized cost to construct the plant, to make the future periodic renovations, and to pay the annual operating costs forever?

Homework Equations


The Attempt at a Solution



A = $10,000 /year.
Renovation cost every 15 years = $200,000
Initial cost = $500,000. Interest rate = 10%
Capitalized cost, P0 = 500,000 + (100,000/0.10) + {200,000 (A/P, 10%, 15)/0.10} = 500,000 + 1,000,000 + {200,000(0.1315)/0.10} = $1,763,000

I think I see two mistakes here: (100,000/0.10) should be (10,000/0.10), right?

And 200,000 (A/P, 10%, 15) should be 200,000 (A/F, 10%, 15) too?
 
Physics news on Phys.org
ainster31 said:

Homework Statement



A municipal contractor has agreed to construct an electric power plant and to deposit sufficient money in a perpetual trust fund to pay a $10,000/year operating cost and to perform a major renovation to the plant every 15 years at a cost of $200,000. The plant itself will initially cost $500,000 to construct. If the trust fund earns 10% interest per year (compounded annually), what is his capitalized cost to construct the plant, to make the future periodic renovations, and to pay the annual operating costs forever?

Homework Equations


The Attempt at a Solution



A = $10,000 /year.
Renovation cost every 15 years = $200,000
Initial cost = $500,000. Interest rate = 10%
Capitalized cost, P0 = 500,000 + (100,000/0.10) + {200,000 (A/P, 10%, 15)/0.10} = 500,000 + 1,000,000 + {200,000(0.1315)/0.10} = $1,763,000

I think I see two mistakes here: (100,000/0.10) should be (10,000/0.10), right?

And 200,000 (A/P, 10%, 15) should be 200,000 (A/F, 10%, 15) too?

When is the first $200,000 payment made? How does that affect the calculation?
 
Last edited:
As I understand we need some amount of dollars to be deposited into the trustfund which will, due to interest, keep the initial and periodic expenses covered forever (yikes).

I assume it is built overnight so the 1st annual operation fee applies to the same year and is taken before the annual interest gain and that the 1st major renovation cost applies to 15 years from now , applying to the end of every 15th year taken, again, before the annual interest thing.

Did I understand correctly?

Actually with that understanding, I tested the scheme with $800 000 starting capital and it worked well, 700 000, however, not enough money. It's some kind of game of proportions, there is a sweet spot where the interest indefinitely feeds the expenses.

800 000 grows into nearly $90M in a century's time.
 
Last edited:
lendav_rott said:
As I understand we need some amount of dollars to be deposited into the trustfund which will, due to interest, keep the initial and periodic expenses covered forever (yikes).

I assume it is built overnight so the 1st annual operation fee applies to the same year and is taken before the annual interest gain and that the 1st major renovation cost applies to 15 years from now , applying to the end of every 15th year taken, again, before the annual interest thing.

Did I understand correctly?

Actually with that understanding, I tested the scheme with $800 000 starting capital and it worked well, 700 000, however, not enough money. It's some kind of game of proportions, there is a sweet spot where the interest indefinitely feeds the expenses.

800 000 grows into nearly $90M in a century's time.

Well, I get that the NPV of all payouts is $662947.55, so $700,000 is more than enough. Of course, I made certain assumptions: (1) the $10,000 annual operating cost is paid at the year's end; (2) the first maintenance payment of $200,000 starts in 15 year's time; and (3) the construction cost is paid up front, at time 0. So, the NPV (for r = 10% = 1/10) is
[tex]\text{NPV} = 500\,000 + 10\,000 \sum_{n=1}^{\infty} \frac{1}{(1+r)^n}<br /> + 200\,000 \sum_{n=1}^{\infty} \frac{1}{(1+r)^{15n}} \doteq 662\,497.55[/tex]
Basically, if you deposit $662,947.55 into a bank account paying 10% per annum compound interest, you will be able to pay out the costs of construction (immediately), operating costs (annually at year's end) and maintenance (once every 15 years). Note that if the operating costs are paid at the start of the year (so we need $10,000 for operating right away at time 0) then the PV becomes $10,000 more than the above, and would equal $672,947.55, which is still less than your $700,000.

I like to avoid like the plague the financial formulas containing things like (A/P, 10%, 15), etc. I think it is much better to actually write things out explicitly in standard mathematical notation.
 
I made an excel table quick and dirty for the calculation. Here
The assumptions are that expenses are subtracted before interest gain and the 15 year renovation fee applies to the 16th year, meaning it can run for 15 years and then there is renovation. Once it reaches -xx xxx dollars it returns nothing but 0s for the visual of it.
The only difference is I didn't pay up front for the construction, which is why we have differences.
 
lendav_rott said:
I made an excel table quick and dirty for the calculation. Here
The assumptions are that expenses are subtracted before interest gain and the 15 year renovation fee applies to the 16th year, meaning it can run for 15 years and then there is renovation. Once it reaches -xx xxx dollars it returns nothing but 0s for the visual of it.
The only difference is I didn't pay up front for the construction, which is why we have differences.

I won't download your file because my security system advises against it; I think you would be better to print it out as a pdf file instead of an executable file.

So, I cannot tell exactly what you have done, but I can say how I would do it in EXCEL. First, I would choose units of $10,000 instead of dollars, so the payments would be 50 for construction, ##a = 1## annually for operating and ##m = 20## at the end of each 15-year stretch, all in an environment with an interest rate of ##r = 1/10##. I would leave open some cell, say A1, to hold the starting bank balance. Like you, I would leave out the construction cost of 50, because we would just put it in the bank and draw it out again instantly. So, A1 holds the starting balance that is to cover future costs of operating and maintenance. In the following, I will leave a, m and r as symbolic constants, so their role can be more easily separated and examined; however, we of course use a = 1, m = 20 and r = 0.1 in the spreadsheet.

In another cell, say B1, I would put (1+r)*A1 -a, which will be the bank balance just after the end of year 1 = starting balance for years 2,3,4,... . Then, in cell B2 I would put (1+r)*B1-a (balance at start of year 3), and would copy that formula all the way down to B15. The cell B15 will contain the bank balance at the end of year 15, after paying for operating but before maintenance. I would put in cell B16 the quantity B15-m, which is the starting balance for years 16, 17,... after paying for maintenance. This quantity should be the same as the starting balance in A1, so now I would use the EXCEL Solver tool to solve the equation A1 = B16.

If you take the time to work out algebraically what is happening, you will find that
[tex]B16 = A1 \cdot (1+r)^{15} - m - a [1 + (1+r) + (1+r)^2 + \cdots + (1+r)^{14}].[/tex]
This involves a finite geometric sum, which is doable using
[tex]1 + u + u^2 + \cdots + u^N = \frac{u^{N+1} - 1}{u-1}[/tex]
with ##u = 1+r## and ##N = 14##. We can solve the equation to get
[tex]A1 = \frac{a}{r} + \frac{m}{(1+r)^{15}-1}[/tex]
This ought to be the solution arrived at by the EXCEL Solver tool.

Here comes the magic: the NPV of all payments is simply
[tex]\text{NPV} = a (R + R^2 + R^3 + \cdots) + m (R^{15} + R^{30} + \cdots ) <br /> = \frac{a R}{R-1} + \frac{m R^{15}}{R^{15}-1},[/tex]
where ##R = 1/(1+r)##. If you grind it through you will see that NPV and A1 are equal; that is, the simple NPV calculation has hidden in it all the details of earned interest as income and various payments over time as outgo.
 
Ray Vickson said:
Well, I get that the NPV of all payouts is $662947.55, so $700,000 is more than enough. Of course, I made certain assumptions: (1) the $10,000 annual operating cost is paid at the year's end; (2) the first maintenance payment of $200,000 starts in 15 year's time; and (3) the construction cost is paid up front, at time 0. So, the NPV (for r = 10% = 1/10) is
[tex]\text{NPV} = 500\,000 + 10\,000 \sum_{n=1}^{\infty} \frac{1}{(1+r)^n}<br /> + 200\,000 \sum_{n=1}^{\infty} \frac{1}{(1+r)^{15n}} \doteq 662\,497.55[/tex]
Basically, if you deposit $662,947.55 into a bank account paying 10% per annum compound interest, you will be able to pay out the costs of construction (immediately), operating costs (annually at year's end) and maintenance (once every 15 years). Note that if the operating costs are paid at the start of the year (so we need $10,000 for operating right away at time 0) then the PV becomes $10,000 more than the above, and would equal $672,947.55, which is still less than your $700,000.

I like to avoid like the plague the financial formulas containing things like (A/P, 10%, 15), etc. I think it is much better to actually write things out explicitly in standard mathematical notation.

Sweet. I got the same answer.

I would love to avoid functional notation but my professor and textbook uses that notation so I am forced to use it because it is too much mental work to switch back and forth.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
5K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
10
Views
4K
Replies
13
Views
12K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
7K