1. Part 1 : Understanding Money And Its Management (Ch1,2,3,4) Part 2: Evaluating Business and Engineering Assets (5,6,7)

2. CHAPTER 5 Present Worth Analysis

3. Payback period

4. Chapter Opening Story – Calculating Costs in the Clouds
Nature of Project: Fuel savings through flight route-mapping software
Investment cost: $100 million
Expected savings: $20 million per year
Is it worth investing in the project?
United Airlines

5. Ultimate Questions
Is it worth investing in a new aviation software to save $20 million per year, say over 10 years?
How long does it take to recover the initial investment?
What kind of interest rate should be used in evaluating business investment opportunities?

6. Loan versus Project Cash Flows

7. Describing Project Cash Flows
Year
(n)
Cash Inflows
(Benefits)
(B)
Cash Outflows
(Costs)
(C)
Net
Cash Flows
(B – C)
$1,800,000
-$1,800,000 1
908,000 2 … 7
1,268,000

8. Payback Period Principle:
How fast can I recover my initial investment?
Method:
Based on cumulative cash flow (or accounting profit)
Screening Guideline:
If the payback period is less than or equal to some specified payback period, the project would be considered for further analysis.
Weakness:
Does not consider the time value of money

9. Example 5.1 Payback Period
Current cost
(% saved)
Savings
Setup
$ ,000
70%
$ ,500
Scrap/rework
$ ,530
85%
$ ,751
Operators
$ ,000
100%
Fixturing
$ ,000
$ ,250
Programming time
$ ,000
60%
$ ,000
Floor space
$ ,000
65%
$ ,750
Maintenace
$ ,000
$ ,000
Coolant
$ ,000
50%
$ ,500
Inspection
$ ,000
Documentation
$ ,000
$ ,500
Expediting
$ ,000
75%
$ ,750
Total Annual Saving
$908,000

10. Example 5.1 Payback Period
N Cash Flow Cum. Flow 1 2 3 4 5 6 7
-$1,800,000
$454,000
$681,000
$908,000
$1,268,000
-$1,800,000
-$1,346,000
-$665,000
$243,000
$1,151,000
$2,059,000
$2,967,000
$4,235,000
50%
75%
Payback period should occurs somewhere
between N = 2 and N = 3 or 2.73 years

11. Figure 5-2 Cumulative Project Cash Flows over Time

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

13. Example 5.2 Discounted Payback Period
3.53
years
* Cost of funds = (Unrecovered beginning balance) X (interest rate)

14. Key Points
Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability.

15. Present Worth Analysis

16. Net Present Worth Decision Rule

17. Evaluation of a single project
Determine the interest rate (required interest rate, minimum attractive interest rate MARR)
Estimate the service life of the project
Estimate the cash inflow for each period of the service life
Determine the cash outflow for each period over the service life
Determine the net cash flows for each period
Find the present worth of each net cash flow at MARR
NPW?

18. Example 5.3 EoY Cash flow $-1,800,000 1 $454,000 2 $681,000 3 $908,000
$-1,800,000 1
$454,000 2
$681,000 3
$908,000 4 5 6 7
$1,268,000

19. Excel Solution using NPV FunctionA B C 1
Period
Cash Flow 2
($1,800,000) 3
$454,000 4
$681,000 5
$908,000 6 7 8 9
$1,268,000 10
PW(15%)=
$1,546,571
=NPV(15%,B3:B9)+B2

20. Present Worth Amounts at Varying Interest Rates

21. Figure 5-5 – Net Present Worth Profile

22. An Excel Worksheet to Illustrate the Process of Computing the NPW

23. Figure 5-6 Elements of Required Return (MARR)

24. Can you explain what a typical NPW really means?
Suppose the company has $1,800,000. It has two options.
Option 1:Take the money out and invest it in the project or
Option 2: Leave the money in the firm’s treasury that can be placed in investments that yield a return equal to the MARR. These funds may be viewed as an investment pool.
Let’s see what the consequences are for each option.

25. Figure 5-7 The Concept of Investment Pool

26. Project Balances

27. Net Future Worth and Project Balance Diagram

28. Key Parameters in Project Balance Diagram and Their Economic Interpretations
The area of negative project balance -The exposure to financial risk
The discounted payback period – The time period when the project recovers the initial investment plus the cost of the funds
The area of positive project balance -The profit potential after recovering the initial investment
The net future worth – The net surplus at the time of project completion

29. Capitalized Equivalent Worth

30. Example 5.4 Given: A = $2M, i = 8%, and N = ∞ Find: CE(8%) Solution: ∞ ∞
CE(8%) = $25,000,000

31. Comparing Mutually Exclusive Alternatives

32. Common Terminologies Used
Mutually Exclusive Projects
Alternative vs. Project
Do-Nothing Alternative

33. Revenue versus Service Projects
Revenue Projects
Projects whose revenues depend on the choice of alternatives
Service Projects
Projects whose revenues do not depend on the choice of alternative but must product the same amount of output (revenue)

34. Analysis Period versus Required Service period
The time span over which the economic effects of an investment will be evaluated (study period or planning horizon).
Required Service Period
The time span over which the service of an equipment (or investment) will be needed.

35. Fundamental Principle in Comparing Mutually Exclusive Alternatives
Principle: Projects must be compared over an equal time span (or analysis period).
Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period.

36. Example 5.5: Analysis Period Equals Project Lives
The net cost of not replacing the old system is now $71,384( = $222,937 - $151,533)
Since the new system costs only $28,570, the replacement should be made now.

37. Analysis period differs from Project Lives
Case 1: Project lives longer than the analysis period (Example 5.6)
Case 2: Project lives shorter than the analysis period (Example 5.7)

38. Case 1: Project Lives Longer than Analysis Period
Estimate the salvage value at the end of the required service period.
Compute the PW for each project over
the required service period.
Select the least cost alternative.

39. Example 5.6 Present Worth Comparison Case 1: Project lives longer than the analysis period
Allan company got the permission to harvest southern pines from one of the timberland area. They are considering purchasing a machine, which has the ability to hold, saw, and place trees in bunches to go down to the landing area. The operation on this timberland area must be completed in three years. Allan company could speed up the operation, but doing so is not desirable, as the market demand of the timber does not justify such speed. There are two possible models of machines that Allan could purchase for this job: Model A is a two year old used equipment, where as Model B is a brand new machine.
Model A costs $205,000 and has a life of 10,000 hours before it will require any major overhaul. The operating cost will run $50,000 per year, for 2,000 hours of operation. At this operational rate, the unit will be operable for five years, and at the end of that time it is estimated that the salvage value will be $75,000.
The more efficient Model B costs $275,000, has a life of 14,000 hours before requiring any major overhaul, and cost $32,500 to operate, for 2,000 hours per year in order to complete the job within three years. The estimated salvage value of Model B at the end of seven years is $95,000.

40. Since the life of either model exceeds the required service period of three years, Allan company has to assume some things about the used equipment at the end of that time. Therefore, the engineers at Allan estimate that, after three years, the Model A unit could be sold for $130,000 and Model B unit for $180,000. Allan summarized the resulting cash flows (in thousands of dollars) for the project as follows:
Here, red color figures represent the estimated salvage values at the end of the analysis period (end of year 3). Assume the firm’s MARR is 15%, which option is more acceptable? N
MODEL A
MODEL B
-$205,000
-$275,000 1
-50,000
-32,500 2 3
130, ,000
180, ,500 4 5
75, ,000 6 7
95, ,500

41. Example 5.6: Project’s Life is Longer than Analysis Period
PW(15%) = -$230,852
PW(15%) = -$233,684

42. Case 2: Project’s Life is shorter than Analysis Period
Come up with replacement projects
that match or exceed the required
service period.
Compute the PW for each project over
the required service period.
Select the least cost project.

43. Example 5.7 Present Worth Comparison Case 2: Project lives shorter than the analysis period
Phoenix Manufacturing Company is planning to modernize one of its distributions located outside Denver, Colorado. Two options to move goods in the distribution center have been under consideration: A conveyor system and forklift trucks. The firm expects that the distribution center will be operational for the next 10 years, and then it will be converted into factory outlet. The conveyor system would last eight years, whereas the forklift trucks would last only six years. The two options will be designed differently, but will have identical capacities and will do exactly the same job. The expected cash flows for the two options, including maintenance costs, salvage values are as follows. N
CONVEYOR SYSTEM
LIFT TRUCK
-$68,000
-$40,000 1
-13,000
-15,000 2 3 4 5 6
-15, ,000 7 8
-13, ,000

44. SOLUTION
Since each option has a shorter life than the required service period (10 years), we need to make explicit assumption of how the service requirement is to be met.
If the company goes with conveyor system, it will spend $18,000 to overhaul the system to extend its service life beyond eight years. The expected salvage value of the system at the end of the required service period (10years) will be $6,000. The annual operating and maintenance costs will be $13,000.
If the company goes with the lift truck option, the company will consider leasing a comparable lift truck that has annual lease payment of $8,000, payable at the beginning of each year, with an annual operating cost of $15,000 for the remaining required service period.
The anticipated cash flows for both models under this scenario are shown in the table describes the cash flows associated with each option. Note that both alternatives now have the same required service period of 10 years. N
CONVEYOR SYSTEM
LIFT TRUCK
-$68,000
-$40,000 1
-13,000
-15,000 2 3 4 5 6
-15, , ,000 7
-15, ,000 8
-13, ,000 9 10
-13, ,000

45. N
CONVEYOR SYSTEM
LIFT TRUCK
-$68,000
-$40,000 1
-13,000
-15,000 2 3 4 5 6
-15, ,000 7 8
-13, ,000 N
CONVEYOR SYSTEM
LIFT TRUCK
-$68,000
-$40,000 1
-13,000
-15,000 2 3 4 5 6
-15, , ,000 7
-15, ,000 8
-13, ,000 9 10
-13, ,000
Figure Comparison for mutually exclusive service projects with unequal lives when the required service period is longer than the individual project life

47. PW (12%) Conveyor = - $68,000 - $13,000 (P/A, 12%, 10) - $18,000 (P/F, 12%, 8)
CONVEYOR SYSTEM
LIFT TRUCK
-$68,000
-$40,000 1
-13,000
-15,000 2 3 4 5 6
-15, , ,000 7
-15, ,000 8
-13,000 – 18,000 9 10
-13, ,000
Figure Comparison for mutually exclusive service projects with unequal lives when the required service period is longer than the individual project life

48. PW (12%) Lift Trucks = - $40,000 - $15,000 (P/A, 12%, 10) - $8,000 (P/A, 12%, 4) (P/F, 12%, 5)
Since these projects are service projects, the lift truck option is better choice. N
CONVEYOR SYSTEM
LIFT TRUCK
-$68,000
-$40,000 1
-13,000
-15,000 2 3 4 5 6
-15, , ,000 7
-15, ,000 8
-13,000 – 18,000 9 10
-13, ,000

49. SOLVED PROBLEMS
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55. HOMEWORK PROBLEMS DUE DATE IS Monday, April 25th 2011 class time
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