1. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Annual Equivalent Examples

2. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 2 ANNUAL EQUIVALENT WORTH ANALYSIS THE ANNUAL EQUIVALENT (AE) IS THE ANNUAL PAYMENT THAT IS EQUIVALENT TO THE PRESENT VALUE: IF THE AE IS POSITIVE, WE ACCEPT THE INVESTMENT DECISIONS WITH AE ARE THE SAME AS WITH NPV

3. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 3 ANNUAL EQUIVALENT WORTH ANALYSIS (CONTINUED) REASONS FOR AE: –IT IS CUSTOMARY TO WORK WITH ANNUAL FIGURES IN OTHER FINANCIAL AREAS –NEED UNIT COSTS AND UNIT PROFITS –MAKE OR BUY DECISIONS –USEFUL TO COMPARE UNEQUAL PROJECT LIVES

4. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 4 DEFINITIONS OPERATING COSTS - RECURRING COSTS THAT DO NOT RESULT IN THE ACQUISITION OF AN ASSET –TELEPHONE BILLS CAPITAL COSTS - ONETIME COSTS USED TO PURCHASE ASSETS –LAPTOP PC CAPITAL RECOVERY COST - THE ANNUAL EQUIVALENT OF A CAPITAL INVESTMENT

5. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 5 UNIT COST OR PROFIT FIND THE NUMBER OF UNITS PRODUCED PER YEAR DETERMINE THE CASH FLOW CALCULATE NPW DETERMINE AE DIVIDE AE BY THE NUMBER OF UNITS PRODUCED

6. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 6 Example

7. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 7

8. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 8 You produce300,000 units of a product in each year and want to know the contribution of the investment in the equipment towards the cost of each unit. The original price of the equipment (in year 0) was $250,000, while the salvage value at the end of the 7 year period is $35,000. The production of the units is continuous through years 1 through 7. Your MARR is 17%.

9. ENGINEERING ECONOMICS ISE460 SESSION 13 Annual Equivalent, June 17, 2015 Geza P. Bottlik Page 9 You have to choose among three alternatives. The three have the following service lives – Two, Four and One. The investments are $150,000, $280,000, and $65,000 respectively. The annual maintenance of the investments are $10,000, $7,000 and $30,000. You expect to have this project for 3 years. The salvage value of each investment is determined linearly to zero at the end of service life. MARR is 20%.