1. Lecture No.15 Chapter 5 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

2. Chapter Opening Story – GE’s Healthymagination Project GE Unveils $6 Billion Health-Unit Plan: Goal: Increase the market share in the healthcare sector. Strategies: Develop products that will lower costs, increase access and improve health-care quality. Investment required: $6 billion over six years Desired project outcome: Would help GE’s health-care unit grow at least twice as fast as the broader economy. Contemporary Engineering Economics, 5 th edition, © 2010

3. Ultimate Questions GE’ s Point of View: Would there be enough demand for their products to justify the investment required in new facilities and marketing? What would be the potential financial risk if the actual demand is far less than its forecast or adoption of technology is too slow? If everything goes as planned, how long does it take to recover the initial investment? Contemporary Engineering Economics, 5 th edition, © 2010

4. Bank Loan vs. Project Cash Flows Contemporary Engineering Economics, 5 th edition, © 2010

5. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project Year (n) Cash Inflows (Benefits) Cash Outflows (Costs) Net Cash Flows 00$650,000-$650,000 1215,50053,000162,500 2215,50053,000162,500 ………… 8215,50053,000162,500

6. Cash Flow Diagram for the Computer Process Control Project Contemporary Engineering Economics, 5 th edition, © 2010

7. Payback Period Contemporary Engineering Economics, 5 th edition, © 2010  Principle: How fast can I recover my initial investment?  Method: Based on the cumulative cash flow (or accounting profit)  Screening Guideline: If the payback period is less than or equal to some specified bench-mark period, the project would be considered for further analysis.  Weakness: Does not consider the time value of money

8. Example 5.3 Payback Period Contemporary Engineering Economics, 5 th edition, © 2010 NCash FlowCum. Flow 01234560123456 -$105,000+$20,000 $15,000 $25,000 $35,000 $45,000 $35,000 -$85,000 -$70,000 -$45,000 -$10,000 $35,000 $80,000 $115,000 Payback period should occurs somewhere between N = 3 and N = 4.

9. Contemporary Engineering Economics, 5 th edition, © 2010 $85,000 $15,000 $25,000 $35,000 $45,000 $35,000 0 123456 Years Annual cash flow -100,000 -50,000 0 50,000 100,000 150,000 0123456 Years (n) 3.2 years Payback period Cumulative cash flow ($)

10. Practice Problem How long does it take to recover the initial investment for the computer process control system project in Example 5.1? Contemporary Engineering Economics, 5 th edition, © 2010

11. Discounted Payback Period  Principle: How fast can I recover my initial investment plus interest?  Method: Based on the cumulative discounted cash flow  Screening Guideline: If the discounted payback period (DPP) is less than or equal to some specified bench-mark period, the project could be considered for further analysis.  Weakness: Cash flows occurring after DPP are ignored

12. Contemporary Engineering Economics, 5 th edition, © 2010 Discounted Payback Period Calculation Period (n) Cash Flow (A n ) Cost of Funds (15%)* Ending Cash Balance 0-$85,0000 115,000-$85,000(0.15) = -$12,750-82,750 225,000-$82,750(0.15) = -12,413-70,163 335,000-$70,163(0.15) = -10,524-45,687 445,000-$45,687(0.15) =-6,853-7,540 545,000-$7,540(0.15) = -1,13136,329 635,000$36,329(0.15) = 5,44976,778 * Cost of funds = (Unrecovered beginning balance) X (interest rate)

13. Illustration of Discounted Payback Period Contemporary Engineering Economics, 5 th edition, © 2010

14. Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability. Contemporary Engineering Economics, 5 th edition, © 2010