1. Initial Project Screening Method - Payback Period
Lecture No.15
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th 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, 5th 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, 5th edition, © 2010

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

5. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project
XL Chemicals is thinking of installing a computer process control system in one of its process plants.
The plant is used about 40% of the time, or 3,500 operating hours per year, to produce a proprietary demulsification chemical.
During the remaining 60% of the time, it is used to produce other specialty chemicals.
Annual production of the demulsification chemical amounts to 30,000 kilograms, and it sells for $15 per kilogram.
Contemporary Engineering Economics, 5th edition, © 2010

6. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project
The proposed computer process control system will cost $650,000 and is expected to provide the following specific benefits in the production of the demulsification chemical:
First, the selling price of the product could be increased by $2 per kilogram because the product will be of higher purity,
Second, production volumes will increase by 4,000 kilograms per year as a result of higher reaction yields,
Contemporary Engineering Economics, 5th edition, © 2010

7. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project
Third, the number of process operators can be reduced by one per shift, which represents a savings of $25 per hour,
Finally, the new control system would result in additional maintenance costs of $53,000 per year and has an expected useful life of eight years.
Given: The preceding cost and benefit information.
Find: Net cash flow in each year over the life of the
new system.
Contemporary Engineering Economics, 5th edition, © 2010

8. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project
Revenues from the price increases are ,000 kg/year × $2/kg = $60,000/year.
The added production volume at the new pricing adds revenues
4,000 kg/year × $17/kg = $68,000/year
The elimination of one operator results in an annual savings of
3,500 hrs./year × $25/hr = $87,500/year
Contemporary Engineering Economics, 5th edition, © 2010

9. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project
The net benefits in each of the eight years that make up the useful lifetime of the new system are the gross benefits less the maintenance costs:
$60,000 +$68,000+ $87,500-$53,000 =$162,500/year
Contemporary Engineering Economics, 5th edition, © 2010

10. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project
Year
(n)
Cash Inflows
(Benefits)
Cash Outflows
(Costs)
Net
Cash Flows
$650,000
-$650,000 1
215,500
53,000
162,500 2 … 8
Contemporary Engineering Economics, 5th edition, © 2010

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

12. 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
Contemporary Engineering Economics, 5th edition, © 2010

13. Example 5.3 Payback Period
Autonumerics Company has just bought a new spindle machine at a cost of $105,000 to replace one that had a salvage value of $20,000.
The projected annual after-tax savings via improved efficiency, which will exceed the investment cost, are as follow:
Contemporary Engineering Economics, 5th edition, © 2010

14. Example 5.3 Payback Period
Given: Cash flow series as shown in previous figure
Find: Conventional payback period
Contemporary Engineering Economics, 5th edition, © 2010

15. Example 5.3 Payback Period
N Cash Flow Cum. Flow 1 2 3 4 5 6
-$105,000+$20,000
$15,000
$25,000
$35,000
$45,000
-$85,000
-$70,000
-$45,000
-$10,000
$35,000
$80,000
$115,000
Contemporary Engineering Economics, 5th edition, © 2010

16. Example 5.3 Payback Period
N Cash Flow Cum. Flow 1 2 3 4 5 6
-$105,000+$20,000
$15,000
$25,000
$35,000
$45,000
-$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.
Contemporary Engineering Economics, 5th edition, © 2010

17. Cumulative cash flow ($)
$45,000
$45,000
$35,000
$35,000
$25,000
$15,000
Annual cash flow 1 2 3 4 5 6
Years
$85,000
150,000
3.2 years
Payback period
100,000
50,000
Assumption:
cash flows occur
continuously
throughout the
year
Cumulative cash flow ($)
-50,000
-100,000 1 2 3 4 5 6
Years (n)
Contemporary Engineering Economics, 5th edition, © 2010

18. 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, 5th edition, © 2010

19. 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
Contemporary Engineering Economics, 5th edition, © 2010

20. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750
-82,750 2
25,000
-$82,750(0.15) = -12,413
-70,163 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

21. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750 2
25,000 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

22. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750
-82,750 2
25,000
-$82,750(0.15) = -12,413
-70,163 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

23. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750
-82,750 2
25,000
-$82,750(0.15) = -12,413
-70,163 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

24. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750
-82,750 2
25,000
-$82,750(0.15) = -12,413
-70,163 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

25. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750
-82,750 2
25,000
-$82,750(0.15) = -12,413
-70,163 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

26. Discounted Payback Period Calculation
Period (n)
Cash Flow (An)
Cost of Funds
(15%)*
Ending Cash Balance
-$85,000 1
15,000
-$85,000(0.15) = -$12,750
-82,750 2
25,000
-$82,750(0.15) = -12,413
-70,163 3
35,000
-$70,163(0.15) = -10,524
-45,687 4
45,000
-$45,687(0.15) =-6,853
-7,540 5
-$7,540(0.15) = -1,131
36,329 6
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
Contemporary Engineering Economics, 5th edition, © 2010

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

28. Depending on which cash flow assumption adopted,
the project must remain in use about 4.2 years (continuous cash flows),
or 5 years (year-end cash flows)
Contemporary Engineering Economics, 5th edition, © 2010

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