Engineering Economy

IEN255 Chapter 5 - Annual Equivalent Worth Analysis Another common method Equal payments on an annual basis Convert NPW to AE

Engineering Economy

Annual Equivalent Worth Criterion AE(I) = PW(I)(A/P,I,N) (5.1)

Engineering Economy

Good or bad? (for revenue)

If AW(i) > 0, accept If AW(i) = 0, indifferent If AW(i) < 0, reject

Engineering Economy

Cyclic flow pattern

figure 5.2

Engineering Economy

Why do AE?

Consistency of report formats

Need for unit costs/profits

Unequal project lives indefinite service period

Engineering Economy

Capital vs Operating Costs

Operating Costs - incurred during operation (labor, matls, etc.) recurring costs

Capital Costs - purchasing assets (non-recurring)

Engineering Economy

A Formula

CR(i) = I(A/P,i,N) - S(A/F,i,N) (5.2) If I = initial cost and S = salvage value we know that,

(A/F,i,N) = (A/P,i,N) - I

so rewriting, CR(i) = I(A/P,i,N) - S[(A/P,i,N) - I]

and rearranging CR(i) = (I-S)(A/P,I,N) +iS

Engineering Economy

Example 5.3

figure 5.4

Engineering Economy

Applications - Unit Profit/Cost Calculation

Determine the number of units to be produced each year over the life of the asset

Identify the cash flow series over life calculate NPW Divide equivalent AE by the number of

units to be produced (if # varies each year, you need to convert them to equivalent annual units).

Engineering Economy

Example 5.4

Engineering Economy

Make or Buy

Determine the time span determine the annual quantity of the part (or

product) Obtain the unit cost of purchasing the part

from outside determine (extra) equipment, labor, etc.

required to make the part. Estimate the cash flows with the make option compute AE of producing the part over the

time span compute the unit cost choose minimum cost

Engineering Economy

Example 5.6

Engineering Economy

Break even point (cost reimbursement)

unit of use calculation depreciation - loss in value

figure 5.6

Engineering Economy

Mutually exclusive projects

Analysis period equals project lives

Example 5.8 operating cost per kWh per unit

determine total input power for both motor types determine total kWh per year for both motor types determine annual energy costs for both motor

types determine capital costs for both motor types determine total AE (= capital cost + annual energy

cost) then calculate unit cost per kWh based on output

power determine savings per operating hour for

switching

Engineering Economy

Break even

$-

$2,000

$4,000

$6,000

$8,000

$10,000

$12,000

$14,000

$16,000

010

0020

0030

0040

0050

0060

0070

0080

0087

50

Conventional Motor

PE MotorBreak-evenoperating hours(6742 hours)

Engineering Economy

Analysis periods differs from lives Service of the selected alternative is

required on a continuous basis each alternative will be replace by an

identical alternative

Iff these two criteria are true we can solve for only initial life span. (saves time from doing LCM)

Engineering Economy

Example 5.9

Quick way for example 4.14

figure 5.8

Engineering Economy

Design economics

AE(i) = a +bx +c/x (5.5)

minimize and

b

cx * (5.6)

Engineering Economy

IEN255 Summer’99 Chapter 3, 4 & 5 Test #2

Test#2 Over Chapters 3 (commercial loans (3.6) and bonds (3.7)), 4, and 5.

Don’t forget chapter 5!!!!

1 hour and 30 minutes for the test

done at 3:30!