IEN255 - Wichita State University
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Transcript IEN255 - Wichita State University
IEN255 Chapter 4 - Present Worth
Analysis
Do the product or not?
3 main issues
How much additional investment in plant &
equipment to mfg the product?
How long to recover initial investment
Can we make a profit a $X price?
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Measures of investment worth
Payback period
Cash flow equivalence
present worth
future worth
annual worth (chap 5)
rate of return (chap 6)
(tax concerns later)
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Loan vs Project cash flow
Figure 4.1
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Example 4.1
Purchase cost = $300,000
5000 x 40% x 3 = 6000 productive hours
6,000/60% = 10,000 hours of paid time per year
Avoided cost = 10,000 hours x $25 /hour =
$250,000/year
So, net benefits = ($250000 - $175000) = $75000 per
year
Fig 4.2
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Payback period
How long does it take to recoup investment?
Most common measure
Used for initial screening
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Example 4.2
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Example 4.3
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Payback period - Pros and Cons
Pro
simple
minimize further analysis (screen all projects)
Cons
no time value of money
no consideration of length of investment
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Two competing projects
Table 4.1
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Present worth analysis
MARR = minimum acceptable rate of return
MARR is a management decision
estimate
service life
cash flows (in and out) (if An positive net cash inflow and An
is negative if net cash outflow)
determine net cash flows
find present worth of each net cash flow
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Good or bad?
If PW(i) > 0, accept
If PW(i) = 0, indifferent
If PW(i) < 0, reject
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Example 4.5
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Investment pool (borrowed funds)
place to get funds for projects within a company
In pool => $75000(F/P, 15%, 3) = $114,066
Project = $119,470 - $114,066 = $5404
Bring back to present = $3553
fig 4.5
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Variations (future worth)
NFW = net future worth
If FW(i) > 0, accept
If FW(i) = 0, indifferent
If FW(i) < 0, reject
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Example 4.6
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Capitalized equivalent method
Perpetual service life
capitalized cost
PW(I) = A(P/A,I,N)= A/i (4.3)
Project’s life is extremely long
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Mutually exclusive alternatives
buying vs leasing
is a single alternative mutually exclusive? (do
nothing)
revenue vs service projects
analysis period
figure 4.11
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Analysis period equals project lives
table solution on pg 212
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Analysis period differs from project
lives
life is longer than analysis period
figure 4.12
solution pg 215
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Project’s life is shorter than analysis
period
what to do at tend?
replacement projects
fig 4.13
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Analysis period coincides with
longest project life
fig 4.14
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Lowest common multiple of project
lives
figure 4.15
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Note table 4.3
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IEN255 Summer’99 Chapter 3, 4 & 5
HW#2
Homework Assignment:
Chapter 3
#’s 3.66; 3.73; 3.78
Chapter 4
#’s 4.1; 4.3; 4.7; 4.22; 4.26; 4.34; 4.39; 4.48
Due together (Tues June 29)
Chapter 5 - will not be collected * problems will be done in class,
others will be posted.
#’s 5.1;5.6*; 5.11*; 5.17; 5.20; 5.28*; 5.32; 5.34*; 5.38*; 5.42*
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