1 1995 Times Mirror Higher Education Group, Inc. IRWIN Engineering Economy: Eide (chapter 13) & Chase (pages 703-719) u Definition of terms u Concept

1 1995 Times Mirror Higher Education Group, Inc. IRWIN

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


Definitions from Chase, pp 750 - 765 Depreciation & Amortization:

– depreciation is allocation of costs to tangible assets e.g. equipment, buildings

– amortization is allocation of intangible assets (patents, licenses, franchises, goodwill)

– Different depreciation methods {Chase’ page 751-753: e.g. straight line, sum-of-the-years digit (SYD), Declining balance, double declining balance, and depreciation by use methods}

Book Value: Investment cost – Depreciation allocations


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


Definitions continued. Sunk Costs - past expenses that have no salvage

value. They are therefore not considered in evaluating investment alternatives. They could be current fixed costs e.g. rent on building, previous periods’ investment, etc.

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.


Definitions contd. 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.

– Economic life therefore, may be different from useful life of an asset (equipment).

– Book Value - used in accounting to reflect the value at a period after deducting depreciation from prior periods.


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


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 (economic life)– annual maintenance/operating cost or benefit– resale or salvage value


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.


Time Value of Money Change in the amount of money (due to the impact

of interest or cost 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%


Interest: Simple and Compound 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 = Pni– b) Total Due = $10,000 + $1,500 = $11,500

» Total Amount Due (F) = P(1 + interest rate x n) Where Principal amount = P; Period = n

Multiple periods – – Simple interest: F = P (1 + ni)– Compound interest: F = P(1+i)n


Simple and Compound Interest simple interest is calculated using the principal

only-i.e. ignoring any interest that was accrued in preceding interest periods

Simple interest amount = P x n x I = Pni E.G. $1000 borrowed for 3 yrs. at 14% per year

simple interest; find amount due in 3 years.– Int. per yr. = 1000(.14) = $140– Total Int. for 3 yrs. = (1000)(3)(.14) = $420– Amt due at the end of 3 yrs. =1000 + 420 = $1420


Compound Interest in this case, the interest for a period is calculated on

the principal plus the total interest accumulated in previous periods

$1000 @ 14% compound interest for 3 yrs.– Amount due at the end of 3 years = F3 = P(1+i)3

» F3 = 1000(1 + .14)3 = $1481.54

» Engineering/Economic analyses are based on compound interest.


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


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.. = 100 + (100x0.12) = $112

Implies that $100 now, is equivalent to $112 a year from now if the applicable interest rate is 12% per year.


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 interest or principle recovered until year 5. Interest

however accumulates each year and is added to the principal

– Plan 2: pay annual interest each year and principal is recovered at end of year 5

– Plan 3: accrued interest 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? - show transparency


Symbols of variables used in Engineering Economy

P = sum/value at time denoted as the present F = sum at some future time A = equal end-of-year amounts of money n = # of interest periods - months, years, weeks, etc. i = interest. rate per interest period - % per year., %

per month, etc.


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 .....


Cash flows continued simplifying assumption for multiple receipts within

a period ? present is zero time NOTE important relationship between nominal

annual rate and compounding period. – Effective annual rate the “actual” annual percentage rate

(APR) » 12% nominal rate compounded semi-annually yields APR of

{1 + (0.12/2)}2 – 1 = 0.1236 = 12.26% APR» Calculate the APR with a 12% nominal rate and compounded

quarterly.


Cash flow Example You make 5 deposits of $1000 per year in a 17% per

annum account, how much money will be accumulated immediately after you made the last deposit

0 1 2 3 4

i= 17%

A = $1,000

F = ?


In-Class Exercise A company invested $2500 in a new compressor 7

years ago, Annual income from the compressor was $750. During the first year, $100 was spent on maintenance, this cost increased each year. by $25. The company wants to sell the compressor for salvage at the end of next year for $150. Construct the cash flow diagram for the piece of equipment.


Example: (1.20) If you bought a new television set in 1996 for $900,

maintain it for three years at a cost of $50 per year, and then sold 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 1999 that would be equivalent to all of the cash flows shown. Assume an interest rate of 12% per year.


Derivation of Single-Payment Formulas: Given:

– P = an amount invested at t = 0– F1 = amount 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)2 ......


Derivation contd. at end of period 3.

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)


Derivation of Uniform-series present-worth series

– 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

n

n

ii

iAP

)1(

1)1(


Standard Notation and Use of Interest Table ( X/Y, i%, n )

– X - represents what you want– Y - represents what is given– i - is interest rate in percent– n - periods


Class Practice If a woman deposits $600 now, $300 two years

from now and $400 five years from now; how much will she have in her account in 10 years. if the interest rate is 5% per annum.


Class PracticeDetermine the value of the A/P factor for an interest

rate of 7.3% and n of 10 years. - (A/P, 7.3%, 10)


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


Class Practice You have made an investment that will yield $8,000 exactly 6 years from today.

If the current interest rate is 8.3 percent-compounded quarterly, what is your investment worth today?

[Hint: Draw the cash flow diagram and solve the problem]

Two machines are being considered for the same task. Initial cost of Machine A is $18,000 new and is estimated to last 6 years. The cost to replace Machine A will be 4% more each year than it was the year before. It will cost $1,200 per year to operate and maintain Machine A and it will have a trade-in (salvage) of $1,500. Machine B costs $38,000 to buy, will last 12 years, and will have a trade-in value of $2,000. The cost of operating and maintaining Machine B is $700 per year.Compare the two machines using the net present worth (value) method and recommend one of the two machines for the process. Include a cash-flow diagram for each alternative. Assume all interest rate is at 8% per year; also assume machine A will be replaced at the end of the first 6th year.

Documents

1 1995 Times Mirror Higher Education Group, Inc. IRWIN Engineering Economy: Eide (chapter 13) & Chase (pages 703-719) u Definition of terms u Concept