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Calculating the capitalized costs

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Feb 21, 2014 #2

    Ray Vickson

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    When is the first $200,000 payment made? How does that affect the calculation?
     
    Last edited: Feb 21, 2014
  4. Feb 21, 2014 #3
    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: Feb 21, 2014
  5. Feb 21, 2014 #4

    Ray Vickson

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    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}
    + 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.
     
  6. Feb 22, 2014 #5
    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.
     
  7. Feb 22, 2014 #6

    Ray Vickson

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    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 )
    = \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.
     
  8. Feb 23, 2014 #7
    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.
     
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