ECTION 9 ENGINEERING ECONOMICS

SECTION 9ENGINEERING ECONOMICS MAX KURTZ, P.E.Consulting Engineer CALCULATION OF INTEREST, PMNCIPAL, AND PAYMENTS Determination of Simple Interest Compound Interest; Future Value of Sing

SECTION 9ENGINEERING ECONOMICS MAX KURTZ, P.E.

Consulting Engineer

CALCULATION OF INTEREST, PMNCIPAL, AND PAYMENTS

Determination of Simple Interest

Compound Interest; Future Value of Single Payment

Present Worth of Single Payment

Principal in Sinking Fund

Determination of Sinking-Fund Deposit

Present Worth of a Uniform Series

Capital-Recovery Determination

Effective Interest Rate

Perpetuity Determination

Determination of Equivalent Sums

Analysis of a Nonuniform Series

Uniform Series with Payment Period Different from Interest Period

Uniform-Gradient Series: Conversion to Uniform Series

Present Worth of Uniform-Gradient Series

Future Value of Uniform-Rate Series

Determination of Payments under Uniform-Rate Series

Continuous Compounding

Future Value of Uniform Series with Continuous Compounding

Present Worth of Continuous Cash Flow of Uniform Rate

Future Value of Continuous Cash Flow of Uniform Rate

DEPRECIATION AND DEPLETION

Straight-Line Depreciation

Straight-Line Depreciation with Two Rates

Depreciation by Accelerated Cost Recovery System

Sinking-Fund Method: Asset Book Value

Sinking-Fund Method: Depreciation Charges

Fixed-Percentage (Declining-Balance) Method

Combination of Fixed-Percentage and Straight-Line Methods

Constant-Unit-Use Method of Depreciation

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Declining-Unit-Use Method of Depreciation

Sum-of-the-Digits Method of Depreciation

Combination of Time- and Use-Depreciation Methods

Effects of Depreciation Accounting on Taxes and Earnings

Depletion Accounting by the Sinking-Fund Method

Income from a Depleting Asset

Depletion Accounting by the Unit Method

COST COMPARISONS OF ALTERNATIVE PROPOSALS

Determination of Annual Cost of an Asset

Minimum Asset Life to Justify a Higher Investment

Comparison of Equipment Cost and Income Generated

Selection of Relevant Data in Annual-Cost Studies

Determination of Manufacturing Break-Even Point

Cost Comparison with Nonuniform Operating Costs

Economics of Equipment Replacement

Annual Cost by the Amortization (Sinking-Fund-Depreciation) Method

Annual Cost by the Straight-Line-Depreciation Method

Present Worth of Future Costs of an Installation

Determination of Capitalized Cost

Capitalized Cost of Asset with Uniform Intermittent Payments

Capitalized Cost of an Asset with Nonuniform Intermittent Payments

Stepped-Program Capitalized Cost

Calculation of Annual Cost on After-Tax Basis

Cost Comparison with Anticipated Decreasing Costs

Economy of Replacing an Asset with an Improved Model

Economy of Replacement under Continuing Improvements

Economy of Replacement on After-Tax Basis

EFFECTS OF INFLATION

Determination of Replacement Cost with Constant Inflation Rate

Determination of Replacement Cost with Variable Inflation Rate

Present Worth of Costs in Inflationary Period

Cost Comparison with Anticipated Inflation

Endowment with Allowance for Inflation

EVALUATION OF INVESTMENTS

Premium- Worth Method of Investment Evaluation

Valuation of Corporate Bonds

Rate of Return on Bond Investment

Investment-Rate Calculation as Alternative to Annual-Cost Calculation

Allocation of Investment Capital

Allocation of Capital to Two Investments with Variable Rates of Return

Allocation of Capital to Three Investments by Dynamic Programming

Economic Level of Investment

Relationship between Before-Tax and After-Tax Investment Rates

Apparent Rates of Return on a Continuing Investment

True Rate of Return on a Completed Investment

Average Rate of Return on Composite Investment

Rate of Return on a Speculative Investment

Investment at an Intermediate Date (Ambiguous Case)

Payback Period of an Investment

Payback Period to Yield a Given Investment Rate

Benefit-Cost Analysis

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ANALYSIS OF BUSINESS OPERATIONS

Linear Programming to Maximize Income from Joint Products

Allocation of Production among Multiple Facilities with Nonlinear Costs

Optimal Product Mix with Nonlinear Profits

Dynamic Programming to Minimize Cost of Transportation

Optimal Inventory Level

Effect of Quantity Discount on Optimal Inventory Level

Project Planning by the Critical-Path Method

Project Planning Based on Available Workforce

STATISTICS, PROBABILITY, AND THEIR APPLICATIONS

Determination of Arithmetic Mean, Median, and Standard Deviation

Determination of Arithmetic Mean and Standard Deviation of

Grouped Data

Number of Ways of Assigning Work

Formation of Permutations Subject to a Restriction

Formation of Combinations Subject to a Restriction

Probability of a Sequence of Events

Probability Associated with a Series of Trials

Binomial Probability Distribution

Pascal Probability Distribution

Poisson Probability Distribution

Composite Event with Poisson Distribution

Normal Distribution

Application of Normal Distribution

Negative-Exponential Distribution

Sampling Distribution of the Mean

Estimation of Population Mean on Basis of Sample Mean

Decision Making on Statistical Basis

Probability of Accepting a False Null Hypothesis

Decision Based on Proportion of Sample

Probability of Accepting an Unsatisfactory Shipment

Device with Negative-Exponential Life Span

Correspondence between Poisson Failure and Negative-Exponential

Life Span

Probability of Failure during a Specific Period

System with Components in Series

System with Components in Parallel

System with Identical Components in Parallel

Analysis of Composite System by Conventional Method

Analysis of Composite System by Alternative Method

Analysis of System with Safeguard by Conventional Method

Analysis of System with Safeguard by Alternative Method

Optimal Inventory to Meet Fluctuating Demand

Finding Optimal Inventory by Incremental-Profit Method

Simulation of Commercial Activity by the Monte Carlo Technique

Linear Regression Applied to Sales Forecasting

Standard Deviation from Regression Line

Short-Term Forecasting with a Markov Process

Long-Term Forecasting with a Markov Process

Verification of Steady-State Conditions for a Markov Process

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REFERENCES: Kurtz—Handbook of Engineering Economics, McGraw-Hill; Barish

and Kaplan—Economic Analysis for Engineering and Managerial Decision Making, McGraw-Hill; DeGarmo et al.—Engineering Economy, Macmillan; Grant and Leaven- worth—Principles of Engineering Economy, Ronald Press; Kasmer—Essentials of Engi- neering Economics, McGraw-Hill; Smith—Engineering Economy, Iowa State University Press; Cissell—Mathematics of Finance, Houghton Mifflin; Clifton and Fyffe—Project Feasibilty Analysis, Wiley; Sullivan and Claycombe—Fundamentals of Forecasting, Reston; Weston and Brigham—Essentials of Managerial Finance, Dryden Press; Lock— Engineer's Handbook of Management Techniques, Grove Press (London, England); Je- len—Project and Cost Engineers' Handbook, American Association of Cost Engineers; Kharbanda—Process Plant and Equipment Cost Estimation, Vivek Enterprises (Bombay, India); Johnson and Peters—A Computer Program for Calculating Capital and Operat- ing Costs, Bureau of Mines Information Circular 8426, U.S Department of Interior; Ost- wald—Cost Estimation for Engineering and Management, Prentice-Hall; American As- sociation of Cost Engineers—Cost Engineers' Notebook; Gass—Linear Programming: Methods and Applications, McGraw-Hill; Hadley—Linear Programming, Addison-Wes- ley; Bellman and Dreyfus—Applied Dynamic Programming, Princeton University Press; Hadley—Nonlinear and Dynamic Programming, Addison-Wesley; Allen—Probability and Statistics, and Queuing Theory, Academic Press; Cross and Harris—Fundamentals

of Queuing Theory, Wiley; Beightler—Foundations of Optimization, Prentice-Hall; Blum and Rosenblatt—Probability and Statistics, W B Saunders; Brownlee—Statistical Theo-

ry and Methodology in Science and Engineering, Wiley; Quinn—Probability and tics, Harper & Row; Newnan—Engineering Economic Analysis, Engineering Press; Park—Cost Engineering, Wiley; Taylor—Managerial and Engineering Economy, VNR; Mishan—Cost-Benefit Analysis, Praeger; Jelen and Black—Cost and Optimization Engi- neering, McGraw-Hill; White et al.—Principles of Engineering Economic Analysis, Wi- ley; Riggs—Engineering Economics, McGraw-Hill; Guenther—Concepts of Statistical Inference, McGraw-Hill; Lindgren—Statistical Theory, Macmillan; Meyer—introducto-

Statis-ry Probability and Statistical Applications, Addison-Wesley; Renwick—Introduction to Investments and Finance, Macmillan; O'Brien—CPM in Construction Management, McGraw-Hill; Gupta and Cozzolino—Fundamentals of Operations Research for Man- agement, Holden-Day.

Calculation of Interest, Principal, and Payments

Symbols and Abbreviations

General: With discrete compounding, i = interest rate per period, percent; n = number of

interest periods With continuous compounding, y = nominal annual interest rate, percent;interest period = 1 year

Simple and compound interest—single payment: P = value of payment at beginning of first interest period, also termed present worth of payment; S = value of payment at end of wth interest period, also termed future value of payment.

Compound interest—uniform-payment series: R = sum paid at end of each interest riod for n periods; P = value of payments at beginning of first interest period, also termed present worth of payments; S = value of payments at end of nth interest period, also termed future value of payments.

Compound interest—uniform-gradient series: Rm = payment at end of mth interest riod; g = constant difference between given payment and preceding payment, also termed gradient of series Then Rm = RI + (m— l)g Also, P and S have the same meaning as for

pe-uniform-payment series

Compound interest—uniform-rate series: Rm = payment at end of mth interest period;

r = constant ratio of given payment to preceding payment Then Rm = R^"1'1, and P and S

have the same meaning as for uniform-payment series

Compound-interest factors: Single payment—SIP = single-payment amount (SPCA) factor; PIS = single-payment present-worth (SPPW) factor Uniform- payment series—SIR = uniform-series compound-amount (USCA) factor; RfS = sinking- fund-payment (SFP) factor; PIR = uniform-series present-worth (USPW) factor; RIP = capital-recovery (CR) factor Uniform-rate series—S/Ri = uniform-rate-series compound- amount (URSCA) factor; PIR1 = uniform-rate-series present-worth (URSPW) factor.

compound-Basic Equations

Simple interest, single payment

S = P(l+ni) (1) Compound interest with discrete compounding

SPCA = (1 + O" (2)SPPW = (I+/)"" (3)

ISFP= (TTTFTT <5>

USPW = - (6a)

CR = / (7a)

Compound interest with continuous compounding

SPCA = eJ" (10) where e = base of natural logarithms = 2.71828

by calculator

DETERMINATION OF SIMPLE INTEREST

A company borrows $4000 at 6 percent per annum simple interest What payment must

be made to retire the debt at the end of 5 years?

Calculation Procedure:

Apply the equation for simple interest

This equation is S = P(I + ni) = $4000(1 + 5 x 0.06) = $5200.

Note: See the introduction to this section for the symbols used.

COMPOUND INTEREST; FUTURE VALUE OF

SINGLEPAYMENT

The sum of $2600 was deposited in a fund that earned interest at 8 percent per annumcompounded quarterly What was the principal in the fund at the end of 3 years?

Calculation Procedure:

1 Compute the true interest rate and number of interest periods

Since there are four interest periods per year, the interest rate i per period is i = 8 cent/4 = 2 percent per period With a 3-year deposit period, the number n of interest peri-

per-ods is n = 3 x 4 - 12.

2 Apply the SPCA value given in a compound-interest table

Look up the SPCA value for the interest rate, 2 percent, and the number of interest ods, 12 Then substitute in S = P(SPCA) = $2600(1.268) = $3296.80

peri-PRESENT WORTH OF SINGLE PAYMENT

On January 1 of a certain year, a deposit was made in a fund that earns interest at 6 cent per annum On December 31,7 years later, the principal resulting from this depositwas $1082 What sum was deposited?

per-Calculation Procedure:

Apply the SPPW relation

Obtain the SPPW factor for / = 6 percent, n = 1 years from the interest table Thus P = 5(SPPW) = $1082(0.6651) = $719.64.

PRINCIPAL IN SINKING FUND

To accumulate capital for an expansion program, a corporation made a deposit of

$200,000 at the end of each year for 5 years in a fund earning interest at 4 percent per num What was the principal in the fund immediately after the fifth deposit was made?

an-Calculation Procedure:

Apply the USCA factor

Obtain the USCA factor for / = 4 percent, n = 5 from the interest table Substitute in the relation S = R(USCA) = $200,000(5.416) - $1,083,200.

DETERMINATION OF SINKING-FUND

DEPOSIT

The XYZ Corporation borrows $65,000, which it is required to repay at the end of 5 years

at 8 percent interest To accumulate this sum, XYZ will make five equal annual deposits

in a fund that earns interest at 3 percent, the first deposit being made 1 year after tion of the loan What is the amount of the annual deposit required?

negotia-Calculation Procedure:

1 Compute the sum to be paid at the expiration of the loan

Obtain the SPCA factor from the interest table for / = 8 percent, n = 5 Then substitute in the relation S = P(SPCA) = $65,000(1.469) = $95,485.

2 Compute the annual deposit corresponding to this future value

Obtain the SFP factor from the interest table for / = 3 percent, n = 5 and substitute in the

relation ^ - S(SFP) - $95,485(0.18835) = $17,985

PRESENT WORTH OF A UNIFORM SERIES

An inventor is negotiating with two firms for assignment of rights to a patent The ABCCorp offers an annuity of 12 annual payments of $15,000 each, the first payment to bemade 1 year after sale of the patent The DEF Corp proposes to buy the patent by making

an immediate lump-sum payment of $120,000 If the inventor can invest the capital at 10percent, which offer should be accepted?

Calculation Procedure:

1 Compute the true interest rate and number of interest periods

Since the annual rate = 4 percent and there are four interest periods per year, the rate perperiod is / = 4 percent/4 = 1 percent And with a 5-year pay period, the number of interestperiods = 5 years (4 periods per year) = 20 periods

2 Compute the uniform payment, i.e., capital recovery

The present worth of the sum is $58,000 Obtain the CR factor from an interest table for /

= 1 percent, n = 20 and substitute in the relation R = P(CR) = $58,000(0.05542) =

$3214.36

EFFECTIVE INTEREST RATE

An account earns interest at the rate of 6 percent per annum, compounded quarterly.Compute the effective interest rate to four significant figures

Calculation Procedure:

Compute the interest earned by $1 per year

With four interest periods per year, the interest rate per period = / = 6 percent/4 =1.5 cent In 1 year there are four interest periods for this account

per-Find the compounded value of $1 at the end of 1 year from S = (1 + i)n = (1 + 0.015)4

= $1.06136 Thus, the interest earned by $1 in 1 year = $106136 - 1.00,000 = $0.06136.Hence, the effective interest rate = 6.136 percent

PERPETUITY DETERMINATION

What sum must be deposited to provide annual payments of $10,000 that are to continueindefinitely if the endowment fund earns interest of 4 percent compounded semiannually?

Calculation Procedure:

1 Compute the effective interest rate

Using the same procedure as in the previous calculation procedure for $1, we find the

ef-fective interest rate ie = (1.02)2 - 1 = 0.04040, or 4.04 percent

2 Apply the USPW relation

The endowment or principal required = P = payment//e, or P = $10,000/0.0404 =

$247,525

DETERMINATION OF EQUIVALENT SUMS

Jones Corp borrowed $900 from Brown Corp on January 1 of year 1 and $1200 on ary 1 of year 3 Jones Corp made a partial payment of $700 on January 1 of year 4 It wasagreed that the balance of the loan would be discharged by two payments, one on January

Janu-1 of year 5 and the other on January Janu-1 of year 6, with the second payment being 50 cent larger than the first If the interest rate is 6 percent, what is the amount of each pay-ment?

per-Calculation Procedure:

1 Construct a line diagram indicating the loan data

Figure 1 shows the line diagram for these loans and is typical of the diagrams that can beprepared for any similar set of loans

2 Select a convenient date for evaluating all the sums

For this situation, select January 1 of year 6 Mark the valuation date on Fig 1, as shown

3 Evaluate each sum at the date selected

Use the applicable interest rate, 6 percent, and the equivalence equation, value of moneyborrowed = value of money paid Substituting the applicable SPCA factor from the inter-

est table for each of the interest periods involved, orw = 5,« = 3,w = 2, and n — 1,

respec-tively, gives $900(SPCA) + $1200(SPCA) = $700(SPCA) + jc(SPCA) + 1.5*, where

FIGURE 1 Time, receipt, and payment diagram.

x = payment made on January 1 of year 5 and 1.5* = payment made on January 1 of year

6 Substituting, we get $900(1.338) + $1200(1.191) = $700(1.124) + 1.06* + 1.5*; x =

$721.30 Hence, 1.5* = $1081.95

Related Calculations: Note that this procedure can be used for more than two

loans and for payments of any type that retire a debt

ANALYSIS OFA NONUNIFORM SERIES

On January 1 of a certain year, ABC Corp borrowed $1,450,000 for 12 years at 6 percentinterest The terms of the loan obliged the firm to establish a sinking ftind in which thefollowing deposits were to be made: $200,000 at the end of the second to the sixth years;

$250,000 at the end of the seventh to the eleventh years; and one for the balance of theloan at the end of the twelfth year The interest rate earned by the sinking fund was 3 per-cent Adverse financial conditions prevented the firm from making the deposit of

$200,000 at the end of the fifth year What was the amount of the final deposit?

Calculation Procedure:

1 Prepare a money-time diagram

Figure 2 shows a money-time diagram for this situation, where * = deposit made at end oftwelfth year

2 Compute the principal of the loan at the end of the twelfth year

Use the relation S = P(SPCA) for i = 6 percent, n = 12 Obtain the SPCA value from an interest table, and substitute in the above relation, or S= $1,450,000(2.012) = $2,917,400.

Payments

Receipts

Valuationdate

All sums (except x) in units of $1000

FIGURE 2 Money-time diagram.

Year

3 Set up an expression for the principal In the sinking fund at the

end of the twelfth year

From Fig 2, principal = $200,000(USCA, n = S)(SPCA, n = 8) + $200,000(SPCA, n = 6) + $250,000(USCA, n = S)(SPCA, n = 1) + x With an interest rate of 3 percent, principal

= $200,000(3.091)(1.267) + $200,000(1.194) + $250,000(5.309)(1.030) + x, or principal

= $2,389,100+ jc

4 Compute the final deposit

Equate the principal in the sinking fund to the principal of the loan: $2,389,100 + x =

$2,917,400 Thus, jc = $528,300

UNIFORM SERIES WITH PAYMENT PERIOD

DIFFERENT FROM INTEREST PERIOD

Deposits of $2000 each were made in a fund earning interest at 4 percent per annum pounded quarterly The interval between deposits was 18 months What was the balance

com-in the account immediately after the fifth deposit was made?

Calculation Procedure:

1 Compute the actual interest rate

Replace the interest rate I3 for the quarterly period with an equivalent rate /18 for the month period Or, I18 = (1 +13)" - 1 = (1.01)6 - 1 = 6.15 percent

18-2 Compute the USCA value

Apply the equation USCA = [(I + /)" - 1]/*, or USCA - [(1.0615)5 - 1]/0.0615 = 5.654.

3 Compute the principal in the fund

Use the relation S = /2(USCA) = $2000(5.654) = $11,308.

UNIFORM-GRADIENT SERIES: CONVERSION

TO UNIFORM SERIES

A loan was to be amortized by a group of six end-of-year payments forming an ascendingarithmetic progression The initial payment was to be $5000, and the difference betweensuccessive payments was to be $400, as shown in Fig 3 But the loan was renegotiated to

FIGURE 3 Diagram showing changed payment plan.

Poyments under

original terms

Poyments under

new terms

provide for the payment of equal rather than uniformly varying sums If the interest rate

of the loan was 8 percent, what was the annual payment?

Calculation Procedure:

1 Apply the equivalent-uniform-series equation

Let R1 = initial payment in a uniform-gradient series; g = difference between successive payments; n = number of payments; Re = periodic payment in an equivalent uniform se- ries Then R6 = R1 + (gli)(\ - «SFP) Substituting with ^1 = $5000, g = $400, n = 6, and

i = 8 percent, we find R6 = $5000 + ($400/0.08)(1 - 6 x 0.13632) = $5911.

2 As an alternative, use the uniform-gradient conversion

(UGC) factor

With n = 6 and / = 8 percent, UGC = 2.28 Then, Re = $5000 + $400(2.28) = $5912.

PRESENT WORTH OF UNIFORM-GRADIENT

Calculation Procedure:

1 Apply the relation of step 1 of the previous

calculation procedure

The relation referred to converts a uniform-gradient series to an equivalent uniform

se-ries Thus, with g = -$500, n = 7, and i = 6 percent, we get Re = $8000 + (-$500/0.06) x

(1-7 x 0.11914) = $6617

2 Compute the present worth of the equivalent annuity

Use the relation P = R6(USPW) = $6617(5.582) = $36,936.

FUTURE VALUE OF UNIFORM-RATE SERIES

A deposit was made in a fund at the end of each year for 8 consecutive years The first posit was $1000, and each deposit thereafter was 25 percent more than the preceding de-posit If the interest rate of the fund was 7 percent per annum, what was the principal inthe fund immediately after the eighth deposit was made?

de-Calculation Procedure:

1 Compute the URSCA value

A uniform-rate series is a set of payments made at equal intervals in which the payments

form a geometric progression (i.e., the ratio of a given payment to the preceding payment

is constant) In this instance, the deposits form a uniform-rate series because each deposit

is 1.25 times the preceding deposit Apply Eq 8: URSCA = [r" - (1 + /)"]/(> - / - 1), where r = ratio of given payment to preceding payment, n = number of payments, and i = interest rate for the payment period With r = 1.25, n = 8, and i = 7 percent, URSCA =

[(1.25)8 - (1.07)8]/(1.25 - 0.07 - 1) = 23.568

2 Compute the future value of the set of deposits

Use the relation S = ^1(URSCA), where ^1 = first payment Then S = $1000(23.568) =

of the first and sixth payments

Calculation Procedure:

1 Compute the URSPW value of the uniform-rate series

Apply Eq 9: URSPW = {[r/(l + i)]n - 1 }/(r - i - 1), where the symbols are as defined in the previous calculation procedure With r = 0.95, n = 6, and i = 9 percent, URSPW =

[(0.95/1.09)6 - 1]/(0.95 - 0.09 - 1) - 4.012

2 Find the amount of the first payment

Use the relation P = ,K1(URSPW), where ^1 = first payment Then $30,000 = /^(4.012),

or #! = $7477.60

3 Find the amount of the sixth payment

Use the relation Rm = Rf "^1, where Rm = mth payment Then ^6 = $7477.60(0.95)5 =

Apply the continuous compounding equation

Use the relation SPCA - e>n, where e = base of the natural logarithm system = 2.71828 , j = nominal interest rate, n = number of years Substituting gives SPCA = (2.718)°30

- 1.350 ThCnS = P(SPCA) = $1000(1.350) = $1350

FUTURE VALUE OF UNIFORM SERIES

WITH CONTINUOUS COMPOUNDING

An inventor received a royalty payment of $25,000 at the end of each year for 7 years.The royalties were invested at 12 percent per annum compounded continuously Whatwas the inventor's capital at the expiration of the 7-year period?

Calculation Procedure:

1 Compute the USCA value

Apply Eq 12: USCA - (ejn - V)I(ef- 1), where n - number of annual payments in form-payment series and e and j are as defined in the previous calculation procedure Thus, with /i = 7 andy = 12 percent, USCA = (e0-84 - l)/(e°-12 - 1) = 10.325

uni-2 Compute the future value of the series

Set S = ,R(USCA) = $25,000(10.325) = $258,100.

Related Calculations: Note that if interest were compounded annually at 12

per-cent, the USCA value would be 10.089

PRESENT WORTH OF CONTINUOUS CASH

FLOW OF UNIFORM RATE

An investment syndicate is contemplating purchase of a business that is expected to yield

an income of $200,000 per year continuously and at a constant rate for the next 5 years Ifthe syndicate wishes to earn 18 percent on its investment, what is the maximum price itshould offer for the business?

Calculation Procedure:

1 Compute the present-worth factor

Apply the equation CFPW = (1 - e~Jn)/j, where CFPW = present-worth factor for a uous cash flow of uniform rate and n = number of years of the flow Where the cash flow

is continuous, it is understood that the given interest or investment rate is based on

contin-uous compounding Thus, with n = 5 and 7 = 18 percent, CFPW = (1 - £T°-90)/0.18 =3.297

2 Compute the present worth of the income

Set P = jR(CFPW), where R = annual cash-flow rate Then P = $200,000(3.297) =

Calculation Procedure:

1 Compute the cash-flow rate

Where money is invested daily, the cash flow may be considered to be continuous for allpractical purposes Assume that deposits are made every day of the year The cash-flow

rate R = $30(365) = $10,950 per year.

2 Compute the future-value factor

Apply the equation CFFV - (e jn - I)Ij 9 where CFFV = future-value factor for a

continu-ous cash flow of uniform rate Thus, with n = 1.5 and j = 1 4 percent, CFFV = (e021 1)70.14=1.669

-3 Compute the future value of the money invested

SetS = R(CWV) = $10,950(1.669) = $18,280.

Depreciation and Depletion

Notational System

Here D 17 = depreciation charge for Uth year; D = annual depreciation; ^D 17 = cumulative

depreciation at end of Uth year = D 1 + D 2 + D 3 + • • • + D Uy where the subscript numbers

refer to the year numbers; P 0 = original cost of asset; P 11 = book value of asset at end of Uth year = P 0 - 2,D 17 ; IRS = Internal Revenue Service; L = salvage value; W = wearing

value, or total depreciation = P0 - L\ N = longevity or life of asset, years.

STRAIGHT-LINE DEPRECIATION

The initial cost of a machine, including its installation, is $15,000 The IRS life of thismachine is 10 years The estimated salvage value of the machine is $1000, and the cost ofdismantling the machine is estimated to be $200 Using straight-line depreciation, what isthe annual depreciation charge? What is the book value of the machine at the end of theseventh year?

Calculation Procedure:

1 Compute the annual depreciation charge

When straight-line depreciation is used, the annual depreciation charge is constant,

D = WIN Since P0 = $15,000, L = $1000 - $200 = $800, W= $15,000 - $800 = $14,200, W= 10 ThenD = $14,200/10 = $1420.

2 Compute the book value of the machine at the end

of the seventh year

^D 1 = 1D = 7($1420) = $9940 Then P 1 = $15,000 - $9940 = $5060.

STRAIGHT-LINE DEPRECIATION

WITH TWO RATES

An asset having an initial cost of $30,000 has a life expectancy of 15 years and an mated salvage value of $5000 What are the depreciation charges under a modifiedstraight-line method in which 60 percent of the total depreciation is considered to occurduring the first 5 years of the life of the asset?

esti-Calculation Procedure:

1 Proportion the total wearing value of the asset

Divide the asset's life span into the two specified intervals, and proportion the total

wear-ing value between them Thus, W= $30,000 - $5000 = $25,000; N= 15; W1 = first-period wearing value = 0.60($25,000) = $15,000; W2 = second-period wearing value =

0.40($25,000) = $10,000

2 Compute the annual depreciation charge

For the first 5 years, D = $15,000/5 = $3000 For the next 10 years, D = $10,000/10 =

Re-Calculation Procedure:

1 Compute the depreciation charges

The ACRS for allocating depreciation was adopted by the federal government in 1981,but it subsequently underwent several modifications ACRS was designed to allow a firm

to write off an asset rapidly, the expectation being that industry would thus be encouraged

to modernize its plants and facilities

The salient features of ACRS are as follows: Each asset is assigned a cost-recoveryperiod during which depreciation is to be charged, and this period is independent of its es-timated longevity; the estimated salvage value is ignored; the depreciation for a givenyear is computed by multiplying the first cost of the asset by a specified depreciation fac-tor; the initial depreciation charge occurs in the year the asset is placed in service, and thedepreciation charge for that year is independent of the specific date at which this place-ment occurs Since many assets are placed in service relatively late in the year, the allow-able depreciation charge for the first year is low compared with that for the second year,

and depreciation is charged for one year beyond the cost-recovery period If the salvagevalue that accrues from disposal of the asset exceeds the book value of the asset at thatdate, the excess is subject to taxation.

The depreciation charges are recorded in the accompanying table

2 Compute the end-of-year book values

The results are recorded in the accompanying table Currently, the federal governmentrecognizes only the straight-line method and ACRS for allocating depreciation However,many state governments still recognize other methods Moreover, a firm may wish tocompute depreciation by some other method for its private records as a means of obtain-ing a more accurate appraisal of its annual profit

Depreciation Book value Year charge, $ at year end, $

A factory constructed at a cost of $9,000,000 has an anticipated salvage value of

$400,000 at the end of 30 years What is the book value of this factory at the end of thetenth year if depreciation is charged by the sinking-fund method with an interest rate of 5percent?

Calculation Procedure:

1 Compute the cumulative depreciation

This method of depreciation accounting assumes that when the asset is retired, it is placed by an exact duplicate and that replacement capital is accumulated by making uni-

re-form end-of-year deposits in a reserve fund The cumulative depreciation ^Dv is fore equated to the principal in the fund at the end of the Uth year Or, ^Dv = JF(SFPXUSCA) So W = $9,000,000 - $400,000 = $8,600,000, SFP = 0.01505 for

there-30 years, ( 7 - 1 0 years, and i = 5 percent, SAo = $8,600,000(0.01505)(12.578) =

$1,628,000

2 Compute the book value

At the end of 10 years, the book value P10 = PQ- SZ)10 = $9,000,000 - $1,628,000 =

$7,372,000

SINKING-FUND METHOD:

DEPRECIATION CHARGES

An asset costing $20,000 is expected to remain serviceable for 5 years and to have a vage value of $3000 Compute the depreciation charges, using the sinking-fund methodand an interest rate of 4 percent

sal-Calculation Procedure:

1 Compute the annual sinking-fund payment

Use the relation R = JF(SFP) With W= $20,000 - $3000 - $17,000, N= 5 years, and i =

4 percent, SFP = 0.18463 Then R = $17,000(0.18463) = $3139.

2 Compute the annual depreciation charges

Use the relation D 17 = R(SPCA), or D 1 = $3139(1.000) = $3139; D 2 = $3139(1.040) =

$3265; D 3 = $3139(1.082) = $3396; D 4 = $3139(1.125) = $3531; D 5 = $3139(1.170) =

$3673 Then SD5 = $17,004

FIXED-PERCENTAGE

(DECLINING-BALANCE) METHOD

An asset cost $5000 and has a life expectancy of 6 years and an estimated salvage value

of $800 Construct a depreciation schedule for this asset, using the fixed-percentagemethod

Calculation Procedure:

1 Compute the rate of depreciation

Use the relation h = 1 - (L/P 0 ) l/N , where h = rate of depreciation Substituting gives h = 1

- (800/5000)1/6 - 0.2632, or 26.32 percent

2 Compute the end-of-year book value

Use the relation D 1 = hP 0 = 0.2632($5000) = $1316 Then P 1 =P 0 -D 1 = $5000 - $1316

= $3684 Likewise, D 2 = 0.2632($3684) = $969.63; P 2 = $3684 - $969.63 = $2714.37 In

a similar manner, D 3 = $714.42, P 3 = $1999.95, D 4 = $526.39, P 4 = $1473.56, D 5 =

$387.84, P 5 = $1085.72, D 6 = $285.76, P 6 = $799.96.

COMBINATION OF FIXED-PERCENTAGE

AND STRAIGHT-LINE METHODS

An asset cost $20,000 and has a life of 8 years and a salvage value of $1000 The IRS mits use of the double-declining-balance method to charge depreciation Compute the de-preciation charges

per-Calculation Procedure:

1 Compute the rate of depreciation

Under the double-declining-balance method, depreciation is initially charged on a

fixed-percentage basis, with 2IN as the rate of depreciation Thus, rate of depreciation = 2/8 =

3 Compute the depreciation for the transfer study

Assume that the transfer in depreciation accounting from the fixed-percentage to thestraight-line method is made at the end of a particular year Calculate the annual depreci-

ation charge D' that applies for the remaining life of the asset.

For example, at the end of the third year the book value is $8437, and the depreciation

that remains to be charged during the last 5 years is $7437 Then D' = $7437/5 = $1487.

Record the values found in this manner in Table 1

4 Determine the transfer date

To establish the transfer date, compare each value of D' with the depreciation charge that

will occur in the following year if the fixed-percentage method is used This comparisonshows that the method should be revised at the end of the fifth year because after that timethe fixed-percentage method results in a smaller depreciation charge The depreciation

charges (Table 1) are thus D1 = $5000; D2 = $3750; D3 = $2813; D4 = $2109; D5 = $1582; D6 = D1 = D8 = $1249.

CONSTANT-UNIT-USE METHOD

OF DEPRECIATION

A machine cost $38,000 and has a life of 5 years and a salvage value of $800 The duction output of this machine in units per year is: first year, 2000; second year, 2500;

pro-TABLE 1 Depreciation by the Double-Declining-Balance Method

Depreciation Book value

Year charge, $ at year end, $ D', $

third year, 2250; fourth year, 1750; fifth year, 1500 units If the depreciation is ascribable

to use rather than the effects of time, and the units produced are of uniform quality, whatare the annual depreciation charges?

Calculation Procedure:

1 Determine the depreciation charge per production unit

Proportion the wearing value on the basis of annual production Since W = $38,000

-$800 = $37,200 and 10,000 units are produced in 5 years, the depreciation charge per duction unit = $37,200/10,000 = $3.72

pro-2 Compute the annual depreciation charge

Since the annual depreciation charge is a function of the production rate, take the product

of the depreciation charge per production unit and the annual production Or, D1 =

Calculation Procedure:

1 Compute the number of depreciation units

The depreciation units are related to the annual production by applying the assigned ity rates Thus

qual-Year Depreciation units

2 Proportion the wearing value

Consider the number of depreciation units as the criterion Or, depreciation charge per preciation unit = $37,200/15,500 = $2.40

de-3 Compute the annual depreciation

Take the product of the depreciation charge per depreciation unit and the annual

depreci-ation units Or, D1 = $2.40(4000) = $9600; likewise, D2 = $11,400; D3 = $8100; D4 =

$4500; D5 = $3600 Taking the sum of these charges, we see that the total depreciation =

Calculation Procedure:

1 Compute the machine wearing value

The wearing value W, or total depreciation = $15,000 - $1000 = $14,000.

2 Compute the annual depreciation

Use the relation Dv = W(N- U+ 1)/0.5 [#(#+ I)], where U= year number Thus, for U=

1, D1 - $3500 Likewise, for U= 2, D2 = $3000; for U= 3, D3 = $2500; for U= 4, D4 =

$2000; for U= 5, D5 = $1500; for U= 6, D6 = $1000; for U= 7, D1 = $500.

COMBINATION OF TIME- AND

USE-DEPRECIATION METHODS

A machine cost $38,000 and has a life of 5 years and a salvage value of $800 Studiesshow that one-third of the total depreciation stems from the effects of time and two-thirdsstems from use Compute the annual depreciation charges if time depreciation is based onsum of the digits and use depreciation on a production basis with all units of equal quali-

ty Use the same production as in the third previous procedure

Calculation Procedure:

1 Divide the wearing value into its two elements

Knowing the respective depreciation proportions, let the subscripts t and u refer to time and use, respectively Also, W = $38,000 - $800 = $37,200, and Wt = ^($37,200) =

$12,400; Wu = 2/3($37,200) = $24,800

2 Compute the annual depreciation charge

For the first year, Dn = W1NI[N(N + 1/2] = $12,400(5)/[5(6/2)] - $4133 Also, Dul =($24,800/10,000 units)(2000 units the first year) = $4960 Thus, the total depreciation forthe first year is D1 = $4133 + $4960 = $9093

EFFECTS OF DEPRECIATION ACCOUNTING

ON TAXES AND EARNINGS

The QRS Corp purchased capital equipment for use in a 5-year venture The equipmentcost $240,000 and had zero salvage value If the income tax rate was 52 percent and theannual income from the investment was $83,000 before taxes and depreciation, what wasthe average rate of earnings if the profits after taxes were invested in tax-free bonds yield-ing 3 percent? Compare the results obtained when depreciation is computed by thestraight-line and sum-of-the-digits methods

Calculation Procedure:

1 Compute the taxable income

With straight-line depreciation, the depreciation charge is $240,000/5 = $48,000 per year.Then the taxable income = $83,000 - $48,000 = $35,000, because depreciation is fullydeductible from gross income

2 Compute the annual tax payment

With a tax rate of 52 percent, the annual tax payment, excluding other deductions, is0.52($35,000) = $18,200

3 Compute the net income

The net cash income = gross income - tax payment, if there are no other expenses Or, netincome = $83,000 - $18,200 = $64,800

4 Determine the capital accumulated by investing the

net income in bonds

Use the USCA factor for j = 3 percent, n = 5 years Or, S = tf (USCA) = $64,800(5.309) =

$344,000

5 Compute the average earnings rate on the venture

Use the relation SPCA = (1 + /)", where SPCA = $344,000/$240,000 = (1+ O5 i = 7.47

percent

6 Compute the sum-of-the-digits annual depreciation

Using the previously developed procedure for sum-of-the-digits depreciation chargesgives D1 = $80,000; D 2 = $64,000; Z)3 = $48,000; D 4 = $32,000; D 5 = $16,000.

7 Compute the annual tax and net Income

Using the same method as in steps 2 and 3, we find the annual net income R is RI =

$81,440; ^2 - $73,120; ^3 = $64,800; ^4 = $56,480; R 5 = $48,160

8 Determine the capital accumulated

Use the respective SPCA values for i = 3 percent and years 1 through 5 for the income earned in each year Or, S = $81,440(1.126) + $73,120(1.093) + $64,800(1.061) +

$56,480(1.030) + $48,160 = $346,700

B Compare the average earnings rate on the venture

By the method of step 5, $346,700/$240,000 = (1 + O5; i = 7-63 percent.

The computed interest rates apply to a composite investment-the purchase and tion of the capital equipment and the purchase of bonds The total income accruing fromthe first element is $324,000, regardless of the depreciation method used However, thetiming as well as the amount of this income is important

opera-The straight-line method produces a uniform annual depreciation charge, tax payment,and net income Under the sum-of-digits method, these amounts are nonuniform; the net

income is highest in the first year and then gradually declines Therefore, the interestearned through the purchase of bonds is higher if the firm adopts the sum-of-the-digitsmethod.

If the interest rate associated with the second element of this composite investmenthad been higher, say 4 or 5 percent, the disparity between the two average returns wouldhave been correspondingly higher

DEPLETION ACCOUNTING BY

THE SINKING-FUND METHOD

An oil field is anticipated to yield an annual income, before depletion allowances, of

$120,000 The field will be dry after 5 years, at which time the land will have a residualvalue of $60,000 If a firm desires a return of 10 percent on its investment, what is themaximum amount it should invest in this oil field? Use a 4 percent interest rate for thesinking fund

Calculation Procedure:

1 Determine the replacement cost of the asset

In this method of depletion accounting, it is assumed that the firm deposits a portion ofthe annual income in a reserve fund to accumulate the capital needed to replace the asset

Let C denote the investment required Then the replacement cost r=C- $60,000 for this

venture

2 Compute the annual deposit required

Let d = annual deposit required Then d = r(SFP) for this venture, or any similar situation With i = 4 percent, n = 5, d= (C- $60,000)(0.18463) = 0.18463C- 11,077.80.

3 Compute the investment required

Set the residual income equal to 10 percent of the investment and solve for C Or,

$120,000-(0.18463C- 11,077.80) = 0.1OC; C= $460,520.

Related Calculations: Note that this method can be applied to any situation

where there is a gradual depletion of a valuable, profit-generating asset Further, 'themethod given here is homologous to the sinking-fund method of depreciation accounting

INCOME FROM A DEPLETING ASSET

An oil field purchased for $800,000 is expected to be dry at the end of 4 years If the sale value of the land is $20,000, what annual income is required to yield an investmentrate of 8 percent? Use a sinking-fund rate of 3 percent

re-Calculation Procedure:

1 Compute the annual deposit required to accumulate

the replacement capital

The replacement cost = $800,000 - $20,000 = $780,000 = r Use the relation annual posit d = r(SFP) With i = 3 percent, n = 4,d= $780,000(0.23903) = $186,440.