1. Engineering Economy: Eide (chapter 13) & Chase (pages 703-719)
Definition of terms
Concept of Equivalence in Engineering Economy
Simple and Compound Interest Calculation
Engineering Economy Symbols (P, F. A, n, and i)
Cash-flow and End-of-period conventions
Constructing cash-flow diagrams

2. Definitions from Chase, pp 703 - 719
Assets
Fixed and Variable Costs:
Sunk Costs:
Economic Life & Obsolescence:
Book Value: Investment cost - Depreciation
Depreciation & Amortization:
depreciation is allocation of costs to tangible assets (land,equipment, buildings)
amortization is allocation of intangible assets (patents, licenses, franchises, goodwill)

3. Definitions
Fixed Costs - expense that remain constant regardless of varying levels of production output e.g. rent, property tax, depreciation
no costs are truly fixed, WHY ?
Variable Costs - expense that vary (fluctuate) with changes in output levels e.g. materials, labor, energy usage

4. Definitions continued.
Sunk Costs - past expenses that have no salvage value. They are therefore not considered in evaluating investment alternatives. They could be current but fixed costs e.g. rent on building.
Opportunity Costs - benefit or advantage forgone that results from choosing one alternative over another
IBM not getting into pc market sooner
Investment A has $25,000 profit and Investment B has $20,000 profit, opportunity cost for choosing B is $5000.

5. Definitions continued.
Economic Life and Obsolescence
Economic Life - for a piece of equipment, is the period over which it provides the best method for performing its task. When a superior method is developed, the machine becomes obsolete - thus book value becomes a meaningless figure in the accounting records.
Book Value - used in accounting to reflect the value at a period after deducting depreciation from prior periods.

6. Engineering Economy
is a collection of mathematical techniques that simplify economic comparisons.
it provides a rational and systematic approach for evaluating different economic decision
E.G.
purchase of a new manufacturing equipment
evaluating different manufacturing methods in terms of economic value to the company
replacing existing manufacturing equipment or method

7. Alternatives Alternatives are always present for any economic decision
Identifying appropriate alternatives is as important as -if not more important than - evaluating the alternatives
Alternatives evaluation variables:
initial cost; Interest rate (rate of return)
anticipated life of equipment (useful/economic life)
annual maintenance or operating cost
resale or salvage value

8. Basis of Alternative Evaluation
$$$$ is generally the basis for comparison
method with lowest overall cost is usually selected
intangible factors (e.g. effect of process changes, human factors, social and others) must be considered
Example: impact on employee morale; relevance to strategic direction, etc.

9. Time Value of Money
change in the amount of money over time is called time value of money
Impact of Interest:
Interest = total $ accumulated - original investment
OR =amount owed - original loan
% Int. Rate = [$ accrued/original amt] X 100%

10. Example - Interest
You borrowed $10,000 for 1 yr. at 15% interest. Calculate (a) the interest and (b) total amount due after one year
a) Interest = $10,000(0.15) = $1,500
b) Total Due = $10,000 + $1,500 = $11,500
Total Amount Due = P(1 + interest rate)
What is :
Principal amount (P); Period (n)
Multiple periods - simple or compound interest considerations

11. Equivalence basis for engineering economy analysis
combination of time value of money and interest rate generate the concept of equivalence
it means that different sums of money at different times can be equal in economic value

12. Example - Equivalence
You will be indifferent (i.e.. in economic terms) to a gift of $100 today or a gift of $112 a year from now if interest rate is 12% per annum
Because:
Amount after 1 yr.. = (100x0.12) = $112
If interest rate were different, your decision will be different

13. Equivalence continued
Example: $5,000 loan that has to be paid back in a five-year period at 15% interest per year
Plan 1: no int or principle recovered until year 5. Int however accumulates each year and is added to principal
Plan 2: pay annual interest each year and principal is recovered at end of year 5
Plan 3: accrued int. and 20% of principal is paid each yr
Plan 4: Equal payments are made annually with some portion going towards principal recovery
Which plan is better?

14. Example:
Relationship between time-value-of-money and equivalence

15. Simple and Compound Interest
simple interest is calculated using the principal only-i.e. ignoring any int. that was accrued in preceding interest periods
Simple int. = P x n x i
E.G. $1000 borrowed for 3 yrs. at 14% per year simple int.
Int. per yr. = 1000(.14) = $140
Total Int. for 3 yrs. = (1000)(3)(.14) = $420
Amt due in 3 yrs. = = $1420

16. Compound Interest
in this case, the int. for a period is calculated on the principal plus the total int accumulated in previous periods
14% compound int. for 3 yrs.

17. Symbols of Engineering Economy
P = sum/value at time denoted as the present
F = sum at some future time
A = equal end-of-yr amounts of money
n = # of int. periods - months, yrs., weeks, etc.
i = int. rate per interest period - % per yr., % per month, etc.

18. Cash- Flow Diagrams
receipts and disbursements over a span of time is called cash flow - companies and individuals have these.
positive cash flows usually represent receipts
negative cash flow represent disbursements
it is assumed in engineering economy that cash flow occurs at the end of the interest period. This is called the “end-of-period” convention

19. Cash flows continued
simplifying assumption for multiple receipts within a period ?
present is zero time

20. Example
You make 5 deposits of $1000 per yr in a 17% per annum account, how much money will be accumulated immediately after you made the last deposit

21. In-Class Exercise
A company invested $2500 in a new compressor 7 yrs ago, Annual income from the compressor was $750. During the first yr, $100 was spent on maintenance, this cost increased each yr. by $25. The company wants to sell the compressor for salvage at the end of next yr. for $150. Construct the cash flow diagram for the piece of equipment.

22. Example: (1.20)
If you buy a new television set in 1996 for $900, maintain it for three yrs at a cost of $50 per yr, and then sell it for $200 at the end of the 3rd year, show a diagram of your cash flows and label each arrow as P, F, or A with its respective dollar value so that you can find the single amount in 1995 that would be equivalent to all of the cash flows shown. Assume an interest rate of 12% per yr.

23. Derivation of Single-Payment Formulas:
Given:
P = an amt invested at t = 0
F1 = amt accumulated at period = 1
i = interest rate per period
then, at end of period 1
F1 = P + Pi
F1 = P(1+ i)
at the end of period 2,
F2 = F1 + F1x i = P(1+ i) + P(1 + i) = P(1 + 2i + i2)
F2 = P(1 + i)

24. Derivation contd. at end of period 3. In general, F3 = F2 + F2 x i =
= P(1+i)(1+i)2 = P (1 + i)3
In general,
F = P (1 + i)n
P = F [ 1/(1+i)n ]
(1 + i)n = single payment compound amount factor(SPCAF)
[ 1/(1+i)n ] = single payment present-worth factor (SPPWF)

25. Single Payment formulas continued.
Note that single payment formulas are used to find the present or future amount when payment or receipt is involved.

26. Derivation of Uniform-series present-worth series
P = A
uniform-series present-worth factor (USPWF)
A = P
capital-recovery factor (CRF)
A = uniform annual worth over n periods (e.g. years) of a given investment P when interest rate is i

27. Derivation contd. (pages 27 and 28)

28. Derivation of uniform-series compound-amount factor and the Sinking-Fund factor

29. Standard Notation and Use of Int Table
( X/Y, i%, n )
X - represents what you want
Y - represents what is given
i - is interest rate in percent
n - periods

30. Using the Interest Tables
If a woman deposits $600 now, $300 2 yrs. from now and $400 5 yrs from now, How much will she have in her account in 10 yrs. if the interest rate is 5% per annum.

31. Using the tables - interpolating
determine the value of the A/P factor for an interest rate of 7.3% and n of 10 yrs. - (A/P, 7.3%, 10)

32. Calculating unknown periods (yrs)
important in breakeven or pay-back period economic analysis
Example: How long will it take for $1000 invested today to double in value if interest is 5% per annum