ien255 chapter 4 - present worth analysis
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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?. Measures of investment worth. Payback period - PowerPoint PPT PresentationTRANSCRIPT
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Engineering Economy
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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Engineering Economy
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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Engineering Economy
Loan vs Project cash flow
Figure 4.1
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Engineering Economy
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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Engineering Economy
Payback period
How long does it take to recoup investment?
Most common measure
Used for initial screening
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Engineering Economy
Example 4.2
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Engineering Economy
Example 4.3
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Engineering Economy
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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Engineering Economy
Two competing projects
Table 4.1
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Engineering Economy
Present worth analysis
MARR = minimum acceptable rate of return
MARR is a management decision estimate
service lifecash 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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Engineering Economy
Good or bad?
If PW(i) > 0, accept If PW(i) = 0, indifferent If PW(i) < 0, reject
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Engineering Economy
Example 4.5
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Engineering Economy
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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Engineering Economy
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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Engineering Economy
Example 4.6
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Engineering Economy
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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Engineering Economy
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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Engineering Economy
Analysis period equals project lives
table solution on pg 212
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Engineering Economy
Analysis period differs from project lives life is longer than analysis period
figure 4.12
solution pg 215
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Engineering Economy
Project’s life is shorter than analysis period what to do at tend? replacement projects
fig 4.13
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Engineering Economy
Analysis period coincides with longest project life
fig 4.14
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Engineering Economy
Lowest common multiple of project lives
figure 4.15
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Engineering Economy
Note table 4.3
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Engineering Economy
IEN255 Summer’99 Chapter 3, 4 & 5 HW#2
Homework Assignment:Chapter 3#’s 3.66; 3.73; 3.78Chapter 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*