## Economic Life and Continuous Replacement

In Part 2 we discussed different definitions of life, among them the economic life. This is the length of time for which the profitability of the system is optimized if it is continuously replaced with a system of exactly the same cost and benefit parameters. If the system is kept in service one year longer or one year shorter, the profitability of the system is not maximized. This is based on the assumption that the net system profitability reduces with time, e.g., the cost of operation increases and the productivity falls. The resale value also reduces as the system ages. We will show that with these assumptions, if we calculate the EUAW and plot it against the years of keeping the system in operation, there is a specific lifetime that the EUAB is a maximum or the EUAC is a minimum. A typical cash flow diagram representing the aging system is shown in Fig. 14.1.

C Oils

[ji itiai Cost

Fig. 14.1 can be broken into three parts as shown in Fig. 14.2

To find the economic life, we have to follow the following steps: Step 1:

Draw the cash flow diagram assuming a life of one year.

Fig. 14.1 can be broken into three parts as shown in Fig. 14.2

Step 2:

Calculate the EUAW for the three parts of a one-year life cash flow

EUAW (C or B> [ (B-C) (P/A, i, ])] (A/Pf i, L) Step 3:

Repeat steps 1 and 2 for every life year up to year n. EUAW (P) = P (A/P, i, n) EUAW (R) = R (A/Fi it n)

The plot of the above three expressions is as presented in Fig. 14.3a. It is assumed, as usually is the case, that as years go by the benefits are reduced and the costs are increased.

Fig. 14.3a

Yea 15

Fig. 14.3a

Step 4:

Plot the total EUAW against a horizontal axis representing the life years 1 to n. The year representing the maximum or minimum is the economic life EL as shown in Fig. 14.3b.

Fig. 14.3b

You should remember from Part 1 that an EUAW of \$An at any year n is the representation of a \$An cash flow for all the years one to n and not the cash flow at year n. Therefore, the optimum situation, i.e., maximum benefit or minimum cost, is obtained if the system is replaced every EL years, EL being the economic life.

N Ijfiir.njfii EL

Example 14.1

Air handling equipment is bought to work with the air conditioning system of a building. The initial cost is \$18,000. The cost of maintenance and operation of this equipment is \$5,100 for the first year increasing by \$1,800 every year. Resale value at the end of the first year is \$12,000 decreasing by \$3,000 each year. The MARR for the owner of this air conditioning system is 20%. The pump output remains constant for 10 years. What is the economic life of this equipment? Do not consider tax and depreciation.

Solution:

Since the output of the pump is constant, the annual benefits from the operation remain constant and will not affect the economic life calculation. The economic life, therefore, is the year of life where the EUAC is a minimum.

One-Year Life

Step 1:

Step 2:

EUAC for Midyear jjfc JsM FIIAB-Tfrtal EUAC = - 14.700

Two-Year Life Step 1 repeat

Step 2 repeat EUAC

FIJAC ofoperati EUA&of resate

Three-Year Life Step 1 repeat Step 2 repeat

EUAC of operation = - [ 5,100 (P/F,20, 1) + 6,900 (P/F,20, 2) + 8,700 (P/F,20, 3)] (A/P, 20T 3) - -

5,700

5,700