## Example

Clean Car Corporation (CCC), a car wash business in Vienna, Virginia, bought an automated car washing system for \$500,000. The cost of operation of this system (labor, material, overhead, administration, etc.) is \$100,000 per year increasing by \$5,000 each year. The income from the operation of this system is \$250,000 per year decreasing due to loss of productivity by 10% per year. The resale value of this system at the end of years 1 to 10 is \$350,000, \$250,000, \$200,000, \$170,000, and remains at \$170,000 for the rest of the 10

years. If CCC has an 8% MARR, what is the optimum economical life of this system? Do not consider depreciation and tax.

Solution:

Step 1:

We start by calculating the EUAW if CCC keeps the system for only one year. Net income = 250,000-100,000 = 150,000. The cash flow for this situation is

Step 2:

EUAC of initial cost = - SOOOOO (A/P, S.I) = - S40000 EUAB of Operation = 150000 EUAB of reiale = 350000

EUAW for a Qjtt-vgAr life â€” 40000

Step 1 repeat

Step 2 repeat

EUAC Of initial cost = - 500000 (A/P, %,!) = - 2&M00

EUAB of operation - [15Q0Q0 (P/F, 8,1) + (250000 * 0.9 - 105000){P/F, \$. 2)] (A/P, 8, 2) -

We can of course continue doing this for 3, 4, 5 year life as we did for Example 14.1, and obtain the economic life. An easier way is to use the already available spreadsheet program EL. The solution is shown below.