Title: IEN255 Chapter 4 - Present Worth Analysis
1IEN255 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?
2Measures of investment worth
- Payback period
- Cash flow equivalence
- present worth
- future worth
- annual worth (chap 5)
- rate of return (chap 6)
- (tax concerns later)
3Loan vs Project cash flow
4Example 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
5Payback period
- How long does it take to recoup investment?
- Most common measure
- Used for initial screening
6Example 4.2
7Example 4.3
8Payback period - Pros and Cons
- Pro
- simple
- minimize further analysis (screen all projects)
- Cons
- no time value of money
- no consideration of length of investment
9Two competing projects
10Present 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
11Good or bad?
- If PW(i) gt 0, accept
- If PW(i) 0, indifferent
- If PW(i) lt 0, reject
12Example 4.5
13Investment pool (borrowed funds)
- place to get funds for projects within a company
- In pool gt 75000(F/P, 15, 3) 114,066
- Project 119,470 - 114,066 5404
- Bring back to present 3553
- fig 4.5
14Variations (future worth)
- NFW net future worth
- If FW(i) gt 0, accept
- If FW(i) 0, indifferent
- If FW(i) lt 0, reject
15Example 4.6
16Capitalized equivalent method
- Perpetual service life
- capitalized cost
- PW(I) A(P/A,I,N??) A/i (4.3)
- Projects life is extremely long
17Mutually exclusive alternatives
- buying vs leasing
- is a single alternative mutually exclusive? (do
nothing) - revenue vs service projects
- analysis period
- figure 4.11
18Analysis period equals project lives
19Analysis period differs from project lives
- life is longer than analysis period
- figure 4.12
- solution pg 215
20Projects life is shorter than analysis period
- what to do at tend?
- replacement projects
- fig 4.13
21Analysis period coincides with longest project
life
22Lowest common multiple of project lives
23Note table 4.3
24IEN255 Summer99 Chapter 3, 4 5 HW2
- 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 5.15.6 5.11 5.17 5.20 5.28 5.32
5.34 5.38 5.42