In life cycle cost analysis, all anticipated costs are discounted to their present worth and the life cycle cost is the addition of all present worth values. The cash flow for each year can be calculated, and the life cycle cost can be found by discounting each annual cash flow to its present value and finding the sum of these discounted cash flows. Life cycle costing requires that all costs are projected into the future and the results obtained from such an analysis depend extensively on the predictions of these future costs.
 |
 |
 |
In general, the present worth (or discounted cost) of an investment or cost (C) at the end of year (n) at a discount rate of (d) and interest rate of (i) is obtained by combining Eqs. (12.11) and (12.13):
Equation (12.14) gives the present worth of a future cost or expenditure at the end of (n) years when the cost or expenditure at the end of first year is (C). This equation is useful for estimating the present worth of any one payment of a series of inflating payments. Therefore, in a series of annual payments, if the first payment is €1,000, due to inflation, say, at a rate of 5%, the sixth payment will be €1,276.28, which worth only €761 today at a discount rate of 9%. This is obtained by Eq. (12.14):
1,000(1 + 0.05)5 _
(1 + 0.09)6
 |
 |
![Подпись: C [PWF(n, i, d)]](/img/1192/image1609.png "image1609"){kind=small} |
 |
Equation (12.14) gives the present worth of a single future payment. Summing up all the present worth values of (n) future payments results in the total present worth (TPW), given by
 |
 |
 |
where PWF(n, i, d) = present worth factor, given by
The solution of Eq. (12.16) is as follows.
 |
 |
 |
If i = d,
 |
 |
 |
 |
 |
If i Ф d,
Equation (12.14) can easily be incorporated into a spreadsheet with the parameters (d) and (i) entered into separate cells and referencing them in the formulas as absolute cells. In this way, a change in either (d) or (i) causes automatic recalculation of the spreadsheet. If the PWF(n, i, d) is multiplied by the first of a series of payments made at the end of each year, the result is the sum of (n) payments discounted to the present with a market discount rate (d). The factor PWF(n, i, d) can be obtained with Eqs. (12.17) or (12.18), depending on the values of (i) and (d), or from the tables in Appendix 8, which tabulate PWF(n, i, d) for the most usual range of parameters.
If the first payment is $600, find the present worth of a series of 10 payments, which are expected to inflate at a rate of 6% per year, and the market discount rate is 9%.
Solution
 |
 |
 |
 |
 |
From Eq. (12.18),
Therefore, the present worth is 600 X 8.1176 = $4,870.56.
 |
 |
 |
 |
 |
A mortgage payment is the annual value of money required to cover the funds borrowed at the beginning to install the system. This includes payment of interest and principal. An estimation of the annual mortgage payment can be found by dividing the amount borrowed by the present worth factor (PWF). The PWF is estimated by using the inflation rate equal to 0 (equal payments) and with the market discount rate equal to the mortgage interest rate. The PWF can be obtained from tables (see Appendix 8) or calculated by the following equation, which is obtained from Eq. (12.18):
where
dm = mortgage interest rate (%).
nL = number of years of equal installments for the loan.
 |
 |
 |
Therefore, if the mortgage principal is M, the periodic payment is
The initial cost of a solar energy system is $12,500. If this amount is paid with a 20% down payment and the balance is borrowed at a 9% interest for 10 years, calculate the annual payments and interest charges for a market discount rate of 7%. Also estimate the present worth of the annual payments.
Solution
 |
 |
|
 |
 |
 |
 |
 |
The actual amount borrowed is $10,000, which is the total present worth of all mortgage payments. The annual mortgage payment is estimated with Eq. (12.19):
Therefore, annual mortgage payment = 10,000/6.4177 = $1,558.20.
The annual mortgage payment includes a principal payment and interest charges. Year after year, as the principal remaining on the loan reduces, the interest charge decreases accordingly. The estimation needs to be carried out for every year.
For year 1,
Interest payment = 10,000 X 0.09 = $900
Principal payment = 1,558.20 — 900 = $658.20
Principal remaining at the end of year 1 = 10,000 — 658.20 = $9,341.80
900
(1 + 0.07)1
= $841.12
For year 2,
Interest payment = 9,341.80 X 0.09 = $840.76
Principal payment = 1,558.20 — 840.76 = $717.44
Principal remaining at the end of year 1 = 9,341.80 — 717.44 = $8,624.36
840.76
(1 + 0.07)2 $734.35
These calculations are repeated for all other years. The results are shown in Table 12.3.
|
Table 12.3 Calculations for Remaining Years in Example 12.3
|
 |
 |
 |
 |
As can be understood from Example 12.3, the calculations can be carried out very easily with the help of a spreadsheet program. Alternatively, the total present worth of all payments can be calculated from the following equation:
(12.21)
where nmin = the minimum of nL and period of economic analysis.
It should be noted that the period of economic analysis may not coincide with the term of mortgage; for example, the economic analysis may be performed for 20 years, which is the usual life of solar water heating systems, but the loan is to be paid in the first 10 years.
Calculate the total present worth of interest paid (PW,) in Example 12.3. Solution
The various PWF values may be obtained from the tables of Appendix 8, as follows:
PWF(nmin, 0, d) = PWF(10, 0, 0.07) = 7.0236
PWF(nL, 0, dm) = PWF(10, 0, 0.09) = 6.4177
PWF(nmin, dm, d) = PWF(10, 0.09, 0.07) = 10.172
 |
This is effectively the same answer as the one obtained before. [18] [19]
To understand the method better, various aspects of the economic analysis are examined separately and mainly through examples. In this way, the basic ideas of life cycle analysis are clarified. It should be noted that the costs specified in the various examples that follow are arbitrary and have no significance. Additionally,
these costs vary widely according to the type and size of system, location where the solar energy system is installed, laws and other conditions of the country or region, international fuel prices, and international material prices, such as for copper and steel, which affect the cost of the solar equipment.
