IEN255 - Wichita State University

Download Report

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?
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)
Engineering Economy
Loan vs Project cash flow
 Figure 4.1
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
Engineering Economy
Payback period
 How long does it take to recoup investment?
 Most common measure
 Used for initial screening
Engineering Economy
Example 4.2
Engineering Economy
Example 4.3
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
Engineering Economy
Two competing projects
 Table 4.1
Engineering Economy
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
Engineering Economy
Good or bad?
 If PW(i) > 0, accept
 If PW(i) = 0, indifferent
 If PW(i) < 0, reject
Engineering Economy
Example 4.5
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
Engineering Economy
Variations (future worth)




NFW = net future worth
If FW(i) > 0, accept
If FW(i) = 0, indifferent
If FW(i) < 0, reject
Engineering Economy
Example 4.6
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
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
Engineering Economy
Analysis period equals project lives
 table solution on pg 212
Engineering Economy
Analysis period differs from project
lives
 life is longer than analysis period
 figure 4.12
 solution pg 215
Engineering Economy
Project’s life is shorter than analysis
period
 what to do at tend?
 replacement projects
 fig 4.13
Engineering Economy
Analysis period coincides with
longest project life
 fig 4.14
Engineering Economy
Lowest common multiple of project
lives
 figure 4.15
Engineering Economy
Note table 4.3
Engineering Economy
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*
Engineering Economy