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CHAPTER 5
Present Worth Analysis
Part 1 : Understanding Money And
Its Management (Ch1,2,3,)
Part 2: Evaluating Business and
Engineering Assets (5,6,7)
Present -Worth Analysis
 Initial Project Screening Methods: Payback period
method
 Present-Worth Analysis
 Methods to Compare Mutually Exclusive Alternatives
Evaluating Business and Engineering Assets
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?
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?
Loan versus Project Cash Flows
Year
(n)
Cash Inflows
(Benefits)
(B)
Cash Outflows
(Costs)
(C)
Net
Cash Flows
(B – C)
0 0 $1,800,000 -$1,800,000
1 908,000 908,000
2 908,000 908,000
… … … …
7 1,268,000 1,268,000
Describing Project Cash Flows
One of the primary concerns of the most business people is
whether, and when, the money invested in a project can be
recovered.
The payback method screens projects on the basis of
how long it takes for net receipts to equal investment
spending?
INITIAL PROJECT SCREENING
METHOD
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
Cumulative Cash Flow
Cumulative Cash Flow –
The cumulative cash flow is the sum of algebraic difference between the
investment and the revenue.
Through the cumulative cash flow the payback period can easily be
determined.
The payback period is the period of time, required for the profit or other
benefits from an investment to equal the cost of investment.
11
Net Cash Flow Diagram Example
 An investment of $10,000 can be made that will produce uniform
annual revenue of $6,000 for 5 years and then have a salvage value
of $2,000 at the end of 5 years. Annual expenses will be $3,000 at the
end of each year for operating and maintaining the project.
S
NR NR NR NR NR
I =10,000
0 1 2 3 4 5 NR = R-E = 5,000 - 3,000 = 2,000
S= 2,000
YEAR
I
NET CASH FLOW DIAGRAM
7K
4K
1K
2 K
S
R R R R R
I =10,000
0 1 2 3 4 5 R = 5,000
E = 3,000
E E E E E S = 2,000
I YEAR
CASH FLOW DIAGRAM
S
6000
6000-3000=3000
5 K
I =10,000
0 1 2 3 4 5 NR = R-E = 3,000
S= 2,000
YEAR
10K
CUMULATIVE CASH FLOW DIAGRAM
Example 5.1 Payback Period
Example 5.1 Payback Period
Current cost (% saved) Savings
Setup $ 335,000 70% $ 234,500
Scrap/rework $ 58,530 85% $ 49,751
Operators $ 220,000 100% $ 220,000
Fixturing $ 185,000 85% $ 157,250
Programming
time $ 80,000 60% $ 48,000
Floor space $ 35,000 65% $ 22,750
Maintenance $ 45,000 60% $ 27,000
Coolant $ 15,000 50% $ 7,500
Inspection $ 120,000 100% $ 120,000
Documentation
$ 5,000 50% $ 2,500
Expediting $ 25,000 75% $ 18,750
Total Annual Saving $908,000
Example 5.1 Payback Period
N Cash Flow Cum. Flow
-$1,800,000
$454,000
$681,000
$908,000
$908,000
$908,000
$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
Payback period should occurs somewhere
between N = 2 and N = 3 or 2.73 years
50% of annual benefit
75% of annual benefit
+20% salvage value
0
1
2
3
4
5
6
7
Figure 5-2 Cumulative Project
Cash Flows over Time
Benefits and Flaws of Payback Screening
 The simplicity is the only appealing quality
 The payback period can produce misleading results.
 Indicates liquidity (riskiness) rather than profitability
 It assumes that no profit is made during the payback period.
 It just measures how long it will take to recover your initial investment.
(suppose the money has been borrowed on certain interest, the method
will not tell how much invested money is contributing towards your
interest expenses)
 It fails to recognize the difference between present and future value of
money…
n Project 1 Project 2
0 -$10000 -$10000
1 $1000 $9000
2 $9000 $1000
3 $1000 $2000
DISCOUNTED PAYBACK PERIOD
 We consider the time value of money – that is,
the cost of funds (interest) used to support
project…
 In other words, we define the discounted-
payback period as the number of years
required to recover the investment from
discounted cash flows.
 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
Discounted Payback Period
* Cost of funds = (Unrecoveredbeginning balance) X (interest rate)
Example 5.2 Discounted Payback Period
3.53
years
Key Points
 Payback periods can be used as a screening tool for
liquidity, but we need a measure of investment worth
for profitability.
Present Worth Analysis
PRESENT WORTH ANALYSIS
 The payback method was widely used for making
investment decisions until 1950.
 When mistakes in this methods were recognized,
business people began to search for methods to improve
project evaluations…
 Discounted cash flow techniques (DCFs) was
developed, which take into account the time value of
money.
PRESENT WORTH ANALYSIS
 One of the DCF’s is the net-present-worth (or
net-present-value) (NPW, or NPV) method.
 A capital investment problem is essentially a
matter of determining whether the anticipated
cash inflows from a proposed project are
sufficiently attractive to invest funds in the
project…
 In developing the NPW criterion, we will use the
concept of cash flow equivalence.
PRESENT WORTH ANALYSIS
 In chapter 2, we observed the most convenient point at which
to calculate equivalent values is often at time zero.
 Under PW criterion, the present worth of all cash inflows
associated with an investment project is compared with the
present worth all cash outflows associated with that project.
 The difference between the present worth of these cash flows,
referred to as the net present worth (NPW), determines
whether the project is or is not an acceptable investment…
 When two or more projects are under consideration, NPW
analysis further allows us to select the best project by
comparing their NPW figures directly.
Evaluation of a single project
Step 1: Determine the interest rate that the firm wishes to earn on its
investments.
Firm always invest the money in its investment pool. This interest rate
often
referred to required rate of return or a minimum attractive rate of return
(MARR). Usually this selection is a policy decision made by the top
management.
Step 2: Estimate the service life of a project.
Step 3: Estimate the cash inflow for each period over the service life.
Step 4: Estimate the cash outflow for each period over the service life…
Step 5: Determine the net cash flow for each period
net cash flow = cash inflow – cash outflow
Net – Present – Worth Criterion
Step 6: Find the present worth of each net cash flow at the MARR. Add up
these present worth figures; their sum is identified as the project’s NPW.
Step 7: In this context, a positive NPW means that the equivalent worth of
the inflows is greater than the equivalent worth of outflows so that the
project makes a profit.
 If NPW( i ) > 0, accept the investment.
 If NPW( i ) = 0, remain indifferent to the investment
 If NPW( i ) < 0, reject the investment
Net Present Worth (NPW) Measure
 Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of
i.
 Decision Rule: Accept the project if the net surplus is positive.
2 3 4 5
0 1
Inflow
Outflow
0
PW(i )inflow
PW(i )outflow
Net surplus
PW (i ) > 0
Example 5.3
Net Present Worth – Uneven Flows
Given: MARR = 15% Find: NPW
End of year Cash
Flow
0
1
2
3
4
5
6
7
-$1,800,000
$454,000
$681,000
$908,000
$908,000
$908,000
$908,000
$1,268,000
Figure 5-4 Calculating the net present value of the Multi-Tasking machine
EoY Cash flow
0 $-1,800,000
1 $454,000
2 $681,000
3 $908,000
4 $908,000
5 $908,000
6 $908,000
7 $1,268,000
Present Worth Amounts at Varying Interest Rates
i (%) NPW( i ) i (%) NPW( i )
0 $4,235,000 22 $866,128
2 $3,726,686 24 $710,125
4 $3,277,023 26 $567,751
6 $2,877,892 28 $437,522
8 $2,522,454 30 $318,140
10 $2,204,931 32 $208,470
12 $1,920,417 34 $107,518
14 $1,664,735 36 $14,407
16 $1,434,319 38 -$71,636
18 $1,226,106 40 -$151,293
20 $1,037,466 42 -$225,170
*Break even interest rate = 36.32%
Figure 5-5 Net present-worth profile described in example 5.3
Figure 5-6 Elements of Required
Return (MARR)
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.
Figure 5-7
The concept of the investment pool,
with the company as a lender and
the project as a borrower
Project Balances The Borrowed fund concept
Net Future Worth and Project
Balance Diagram
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
Comparing Mutually
Exclusive Alternatives
Common Terminologies Used
 Mutually Exclusive Projects
 Alternative vs. Project
 Do-Nothing Alternative
Comparing Mutually Exclusive Projects
Mutually Exclusive Projects
When alternatives are mutually exclusive, any one of them will fulfill the
same
need, and the selection of one alternative implies that the others will be
excluded.
Example of buying versus leasing car.
Alternative vs. Project
When we use terms alternative and project interchangeably to mean
decision option.
41
Revenue Projects generate revenues that depend on
the choice of alternative that we
want to select the alternative with
the largest net gains
Service Projects generate revenues that do not depend
on the choice of project, but must
produce the same amount of output
(revenue) with lower production cost.
Analysis 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.
Analysis Period versus Required
Service period
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.
Example 5.5
Compressor is expected to run 70% of the time that the plant will be in
operation during the upcoming years. Compressor require 260 kW of
electricity at a rate of $0.05 / kWH. Plant runs 250 days a year and 24
hours a day
Option 1:
Continue current operation
Compressor run time will increase by 7% per year for the next 5 years
because of ever-worsening leaks
Option 2:
Replace old piping system, New pipes will cost $28,570
Compressor still run same number of days
Compressor will run 23% less (70%(1-0.23)=53.9%
260X.05 = $13 /H
13 X 24 = $169 /DAY
169 X 250 = $ 78000 / YEAR
78000 X O.70 = $54600
$54600 * .77
COST = $ 42042
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.
For revenue projects, we select the one with largest NPW, so producing product B is more
economical
215000-126000= 89000
289000-168000= 121000
48
Analysis period differs from Project Lives
Case 1: Project lives longer than the
analysis period
Case 2: Project lives shorter than the
analysis period
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.
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
At this operational rate, the unit will be operable for 7 years. The estimated salvage
value of Model B at the end of seven years is $95,000.
 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
0 -$205,000 -$275,000
1 -50,000 -32,500
2 -50,000 -32,500
3 130,000 -50,000 180,000 -32,500
4 -50,000 -32,500
5 75,000 -50,000 -32,500
6 -32,500
7 95,000 - 32,500
Example 5.6: Project’s Life is
Longer than Analysis Period
PW(15%) = -$233,684 PW(15%) = -$230,852
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.
Example 5.8 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
0 -$68,000 -$40,000
1 -13,000 -15,000
2 -13,000 -15,000
3 -13,000 -15,000
4 -13,000 -15,000
5 -13,000 -15,000
6 -13,000 -15,000 + 4,000
7 -13,000
8 -13,000 + 5,000
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
0 -$68,000 -$40,000
1 -13,000 -15,000
2 -13,000 -15,000
3 -13,000 -15,000
4 -13,000 -15,000
5 -13,000 -15,000
6 -13,000 -15,000 + 4,000 - 8,000
7 -13,000 -15,000 - 8,000
8 -13,000 - 18,000 -15,000 - 8,000
9 -13,000 -15,000 - 8,000
10 -13,000 + 6,000 -15,000
Figure 5-12 Comparison for mutually exclusive service projects with unequal lives when the
required service period is longer than the individual project life
N CONVEYOR SYSTEM LIFT TRUCK
0 -$68,000 -$40,000
1 -13,000 -15,000
2 -13,000 -15,000
3 -13,000 -15,000
4 -13,000 -15,000
5 -13,000 -15,000
6 -13,000 -15,000 + 4,000
7 -13,000
8 -13,000 + 5,000
N CONVEYOR SYSTEM LIFT TRUCK
0 -$68,000 -$40,000
1 -13,000 -15,000
2 -13,000 -15,000
3 -13,000 -15,000
4 -13,000 -15,000
5 -13,000 -15,000
6 -13,000 -15,000 + 4,000 - 8,000
7 -13,000 -15,000 - 8,000
8 -13,000 - 18,000 -15,000 - 8,000
9 -13,000 -15,000 - 8,000
10 -13,000 + 6,000 -15,000
Figure 5-12 Comparison for mutually exclusive service projects with unequal lives when the
required service period is longer than the individual project life
N CONVEYOR SYSTEM LIFT TRUCK
0 -$68,000 -$40,000
1 -13,000 -15,000
2 -13,000 -15,000
3 -13,000 -15,000
4 -13,000 -15,000
5 -13,000 -15,000
6 -13,000 -15,000 + 4,000 - 8,000
7 -13,000 -15,000 - 8,000
8 -13,000 – 18,000 -15,000 - 8,000
9 -13,000 -15,000 - 8,000
10 -13,000 + 6,000 -15,000
PW (12%) Conveyor = - $68,000 - $13,000 (P/A, 12%, 10) - $18,000 (P/F, 12%, 8)
+ $6,000 (P/F, 12%,10)
PW (12%) Conveyor = -$146,791
N CONVEYOR SYSTEM LIFT TRUCK
0 -$68,000 -$40,000
1 -13,000 -15,000
2 -13,000 -15,000
3 -13,000 -15,000
4 -13,000 -15,000
5 -13,000 -15,000
6 -13,000 -15,000 + 4,000 - 8,000
7 -13,000 -15,000 - 8,000
8 -13,000 – 18,000 -15,000 - 8,000
9 -13,000 -15,000 - 8,000
10 -13,000 + 6,000 -15,000
PW (12%) Lift Trucks = - $40,000 - $15,000 (P/A, 12%, 10) - $8,000 (P/A, 12%, 4) (P/F, 12%, 5)
+ $4,000 (P/F, 12%,6)
PW (12%) Lift Trucks = -$136,514
Since these projects are service projects, the lift
truck option is better choice.
SOLVED PROBLEMS
7; 13; 14; 32; 42
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present worth analysis.ppt

  • 2. Part 1 : Understanding Money And Its Management (Ch1,2,3,) Part 2: Evaluating Business and Engineering Assets (5,6,7)
  • 3. Present -Worth Analysis  Initial Project Screening Methods: Payback period method  Present-Worth Analysis  Methods to Compare Mutually Exclusive Alternatives Evaluating Business and Engineering Assets
  • 4. 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?
  • 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. Year (n) Cash Inflows (Benefits) (B) Cash Outflows (Costs) (C) Net Cash Flows (B – C) 0 0 $1,800,000 -$1,800,000 1 908,000 908,000 2 908,000 908,000 … … … … 7 1,268,000 1,268,000 Describing Project Cash Flows
  • 8. One of the primary concerns of the most business people is whether, and when, the money invested in a project can be recovered. The payback method screens projects on the basis of how long it takes for net receipts to equal investment spending? INITIAL PROJECT SCREENING METHOD
  • 9. 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
  • 10. Cumulative Cash Flow Cumulative Cash Flow – The cumulative cash flow is the sum of algebraic difference between the investment and the revenue. Through the cumulative cash flow the payback period can easily be determined. The payback period is the period of time, required for the profit or other benefits from an investment to equal the cost of investment.
  • 11. 11 Net Cash Flow Diagram Example  An investment of $10,000 can be made that will produce uniform annual revenue of $6,000 for 5 years and then have a salvage value of $2,000 at the end of 5 years. Annual expenses will be $3,000 at the end of each year for operating and maintaining the project. S NR NR NR NR NR I =10,000 0 1 2 3 4 5 NR = R-E = 5,000 - 3,000 = 2,000 S= 2,000 YEAR I NET CASH FLOW DIAGRAM 7K 4K 1K 2 K S R R R R R I =10,000 0 1 2 3 4 5 R = 5,000 E = 3,000 E E E E E S = 2,000 I YEAR CASH FLOW DIAGRAM S 6000 6000-3000=3000 5 K I =10,000 0 1 2 3 4 5 NR = R-E = 3,000 S= 2,000 YEAR 10K CUMULATIVE CASH FLOW DIAGRAM
  • 13. Example 5.1 Payback Period Current cost (% saved) Savings Setup $ 335,000 70% $ 234,500 Scrap/rework $ 58,530 85% $ 49,751 Operators $ 220,000 100% $ 220,000 Fixturing $ 185,000 85% $ 157,250 Programming time $ 80,000 60% $ 48,000 Floor space $ 35,000 65% $ 22,750 Maintenance $ 45,000 60% $ 27,000 Coolant $ 15,000 50% $ 7,500 Inspection $ 120,000 100% $ 120,000 Documentation $ 5,000 50% $ 2,500 Expediting $ 25,000 75% $ 18,750 Total Annual Saving $908,000
  • 14. Example 5.1 Payback Period N Cash Flow Cum. Flow -$1,800,000 $454,000 $681,000 $908,000 $908,000 $908,000 $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 Payback period should occurs somewhere between N = 2 and N = 3 or 2.73 years 50% of annual benefit 75% of annual benefit +20% salvage value 0 1 2 3 4 5 6 7
  • 15. Figure 5-2 Cumulative Project Cash Flows over Time
  • 16. Benefits and Flaws of Payback Screening  The simplicity is the only appealing quality  The payback period can produce misleading results.  Indicates liquidity (riskiness) rather than profitability  It assumes that no profit is made during the payback period.  It just measures how long it will take to recover your initial investment. (suppose the money has been borrowed on certain interest, the method will not tell how much invested money is contributing towards your interest expenses)  It fails to recognize the difference between present and future value of money… n Project 1 Project 2 0 -$10000 -$10000 1 $1000 $9000 2 $9000 $1000 3 $1000 $2000
  • 17. DISCOUNTED PAYBACK PERIOD  We consider the time value of money – that is, the cost of funds (interest) used to support project…  In other words, we define the discounted- payback period as the number of years required to recover the investment from discounted cash flows.
  • 18.  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 Discounted Payback Period
  • 19. * Cost of funds = (Unrecoveredbeginning balance) X (interest rate) Example 5.2 Discounted Payback Period 3.53 years
  • 20. Key Points  Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability.
  • 22. PRESENT WORTH ANALYSIS  The payback method was widely used for making investment decisions until 1950.  When mistakes in this methods were recognized, business people began to search for methods to improve project evaluations…  Discounted cash flow techniques (DCFs) was developed, which take into account the time value of money.
  • 23. PRESENT WORTH ANALYSIS  One of the DCF’s is the net-present-worth (or net-present-value) (NPW, or NPV) method.  A capital investment problem is essentially a matter of determining whether the anticipated cash inflows from a proposed project are sufficiently attractive to invest funds in the project…  In developing the NPW criterion, we will use the concept of cash flow equivalence.
  • 24. PRESENT WORTH ANALYSIS  In chapter 2, we observed the most convenient point at which to calculate equivalent values is often at time zero.  Under PW criterion, the present worth of all cash inflows associated with an investment project is compared with the present worth all cash outflows associated with that project.  The difference between the present worth of these cash flows, referred to as the net present worth (NPW), determines whether the project is or is not an acceptable investment…  When two or more projects are under consideration, NPW analysis further allows us to select the best project by comparing their NPW figures directly.
  • 25. Evaluation of a single project Step 1: Determine the interest rate that the firm wishes to earn on its investments. Firm always invest the money in its investment pool. This interest rate often referred to required rate of return or a minimum attractive rate of return (MARR). Usually this selection is a policy decision made by the top management. Step 2: Estimate the service life of a project. Step 3: Estimate the cash inflow for each period over the service life. Step 4: Estimate the cash outflow for each period over the service life… Step 5: Determine the net cash flow for each period net cash flow = cash inflow – cash outflow Net – Present – Worth Criterion
  • 26. Step 6: Find the present worth of each net cash flow at the MARR. Add up these present worth figures; their sum is identified as the project’s NPW. Step 7: In this context, a positive NPW means that the equivalent worth of the inflows is greater than the equivalent worth of outflows so that the project makes a profit.  If NPW( i ) > 0, accept the investment.  If NPW( i ) = 0, remain indifferent to the investment  If NPW( i ) < 0, reject the investment
  • 27. Net Present Worth (NPW) Measure  Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i.  Decision Rule: Accept the project if the net surplus is positive. 2 3 4 5 0 1 Inflow Outflow 0 PW(i )inflow PW(i )outflow Net surplus PW (i ) > 0
  • 28. Example 5.3 Net Present Worth – Uneven Flows Given: MARR = 15% Find: NPW End of year Cash Flow 0 1 2 3 4 5 6 7 -$1,800,000 $454,000 $681,000 $908,000 $908,000 $908,000 $908,000 $1,268,000
  • 29. Figure 5-4 Calculating the net present value of the Multi-Tasking machine EoY Cash flow 0 $-1,800,000 1 $454,000 2 $681,000 3 $908,000 4 $908,000 5 $908,000 6 $908,000 7 $1,268,000
  • 30. Present Worth Amounts at Varying Interest Rates i (%) NPW( i ) i (%) NPW( i ) 0 $4,235,000 22 $866,128 2 $3,726,686 24 $710,125 4 $3,277,023 26 $567,751 6 $2,877,892 28 $437,522 8 $2,522,454 30 $318,140 10 $2,204,931 32 $208,470 12 $1,920,417 34 $107,518 14 $1,664,735 36 $14,407 16 $1,434,319 38 -$71,636 18 $1,226,106 40 -$151,293 20 $1,037,466 42 -$225,170 *Break even interest rate = 36.32%
  • 31. Figure 5-5 Net present-worth profile described in example 5.3
  • 32. Figure 5-6 Elements of Required Return (MARR)
  • 33. 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.
  • 34. Figure 5-7 The concept of the investment pool, with the company as a lender and the project as a borrower
  • 35. Project Balances The Borrowed fund concept
  • 36. Net Future Worth and Project Balance Diagram
  • 37. 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
  • 39. Common Terminologies Used  Mutually Exclusive Projects  Alternative vs. Project  Do-Nothing Alternative
  • 40. Comparing Mutually Exclusive Projects Mutually Exclusive Projects When alternatives are mutually exclusive, any one of them will fulfill the same need, and the selection of one alternative implies that the others will be excluded. Example of buying versus leasing car. Alternative vs. Project When we use terms alternative and project interchangeably to mean decision option.
  • 41. 41 Revenue Projects generate revenues that depend on the choice of alternative that we want to select the alternative with the largest net gains Service Projects generate revenues that do not depend on the choice of project, but must produce the same amount of output (revenue) with lower production cost.
  • 42. Analysis 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. Analysis Period versus Required Service period
  • 43. 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.
  • 44. Example 5.5 Compressor is expected to run 70% of the time that the plant will be in operation during the upcoming years. Compressor require 260 kW of electricity at a rate of $0.05 / kWH. Plant runs 250 days a year and 24 hours a day Option 1: Continue current operation Compressor run time will increase by 7% per year for the next 5 years because of ever-worsening leaks Option 2: Replace old piping system, New pipes will cost $28,570 Compressor still run same number of days Compressor will run 23% less (70%(1-0.23)=53.9% 260X.05 = $13 /H 13 X 24 = $169 /DAY 169 X 250 = $ 78000 / YEAR 78000 X O.70 = $54600 $54600 * .77 COST = $ 42042
  • 45. 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.
  • 46.
  • 47. For revenue projects, we select the one with largest NPW, so producing product B is more economical 215000-126000= 89000 289000-168000= 121000
  • 48. 48 Analysis period differs from Project Lives Case 1: Project lives longer than the analysis period Case 2: Project lives shorter than the analysis period
  • 49. 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.
  • 50. 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 At this operational rate, the unit will be operable for 7 years. The estimated salvage value of Model B at the end of seven years is $95,000.
  • 51.  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 0 -$205,000 -$275,000 1 -50,000 -32,500 2 -50,000 -32,500 3 130,000 -50,000 180,000 -32,500 4 -50,000 -32,500 5 75,000 -50,000 -32,500 6 -32,500 7 95,000 - 32,500
  • 52. Example 5.6: Project’s Life is Longer than Analysis Period PW(15%) = -$233,684 PW(15%) = -$230,852
  • 53. 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.
  • 54. Example 5.8 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 0 -$68,000 -$40,000 1 -13,000 -15,000 2 -13,000 -15,000 3 -13,000 -15,000 4 -13,000 -15,000 5 -13,000 -15,000 6 -13,000 -15,000 + 4,000 7 -13,000 8 -13,000 + 5,000
  • 55. 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 0 -$68,000 -$40,000 1 -13,000 -15,000 2 -13,000 -15,000 3 -13,000 -15,000 4 -13,000 -15,000 5 -13,000 -15,000 6 -13,000 -15,000 + 4,000 - 8,000 7 -13,000 -15,000 - 8,000 8 -13,000 - 18,000 -15,000 - 8,000 9 -13,000 -15,000 - 8,000 10 -13,000 + 6,000 -15,000
  • 56. Figure 5-12 Comparison for mutually exclusive service projects with unequal lives when the required service period is longer than the individual project life N CONVEYOR SYSTEM LIFT TRUCK 0 -$68,000 -$40,000 1 -13,000 -15,000 2 -13,000 -15,000 3 -13,000 -15,000 4 -13,000 -15,000 5 -13,000 -15,000 6 -13,000 -15,000 + 4,000 7 -13,000 8 -13,000 + 5,000 N CONVEYOR SYSTEM LIFT TRUCK 0 -$68,000 -$40,000 1 -13,000 -15,000 2 -13,000 -15,000 3 -13,000 -15,000 4 -13,000 -15,000 5 -13,000 -15,000 6 -13,000 -15,000 + 4,000 - 8,000 7 -13,000 -15,000 - 8,000 8 -13,000 - 18,000 -15,000 - 8,000 9 -13,000 -15,000 - 8,000 10 -13,000 + 6,000 -15,000
  • 57.
  • 58. Figure 5-12 Comparison for mutually exclusive service projects with unequal lives when the required service period is longer than the individual project life N CONVEYOR SYSTEM LIFT TRUCK 0 -$68,000 -$40,000 1 -13,000 -15,000 2 -13,000 -15,000 3 -13,000 -15,000 4 -13,000 -15,000 5 -13,000 -15,000 6 -13,000 -15,000 + 4,000 - 8,000 7 -13,000 -15,000 - 8,000 8 -13,000 – 18,000 -15,000 - 8,000 9 -13,000 -15,000 - 8,000 10 -13,000 + 6,000 -15,000 PW (12%) Conveyor = - $68,000 - $13,000 (P/A, 12%, 10) - $18,000 (P/F, 12%, 8) + $6,000 (P/F, 12%,10) PW (12%) Conveyor = -$146,791
  • 59. N CONVEYOR SYSTEM LIFT TRUCK 0 -$68,000 -$40,000 1 -13,000 -15,000 2 -13,000 -15,000 3 -13,000 -15,000 4 -13,000 -15,000 5 -13,000 -15,000 6 -13,000 -15,000 + 4,000 - 8,000 7 -13,000 -15,000 - 8,000 8 -13,000 – 18,000 -15,000 - 8,000 9 -13,000 -15,000 - 8,000 10 -13,000 + 6,000 -15,000 PW (12%) Lift Trucks = - $40,000 - $15,000 (P/A, 12%, 10) - $8,000 (P/A, 12%, 4) (P/F, 12%, 5) + $4,000 (P/F, 12%,6) PW (12%) Lift Trucks = -$136,514 Since these projects are service projects, the lift truck option is better choice.