1. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Project Screening Method – Payback Period Lecture No.15 Chapter 5 Contemporary Engineering Economics Copyright © 2016

2. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Chapter Opening Story: Building a New Football Stadium for Colorado State University CSU unveils a new $246 million stadium expansion plan. Goal Attract better football players to win more games. Attract more out-of-state students Desired project outcome Raise the university profile and increase tuition revenue to make up the shortfall from state funding.

3. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Ultimate Questions CSU ’ s Point of View – Would there be enough demand for their football tickets 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 tuition revenue from out-of-state students is too low? – If everything goes as planned, how long does it take to recover the initial investment?

4. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Bank Loan vs. Project Cash Flows

5. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved 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. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Cash Flow Diagram for the Computer Process Control Project

7. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Payback Period  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. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 5.3: Payback Period 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, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved $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. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Practice Problem How long does it take to recover the initial investment for the computer process control system project in Example 5.1?

11. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved 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, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved 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. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Illustration of Discounted Payback Period

14. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Summary Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability. To consider the project profitability, we need to consider the time value of money of project cash flows over the entire life of the project.