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?

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 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

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

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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

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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; 3.78Chapter 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*