ECTION 9 ENGINEERING ECONOMICS

SECTION ENGINEERING 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 9.4 9.6 9.6 9.7 9.7 9.7 9.8 9.8 9.8 9.9 9.9 9.10 9.11 9.11 9.12 9.12 9.13 9.13 9.14 9.14 9.14 9.15 9.15 9.16 9.16 9.17 9.18 9.18 9.18 9.19 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 9.20 9.21 9.21 9.22 9.23 9.23 9.24 9.24 9.25 9.26 9.26 9.27 9.28 9.28 9.29 9.31 9.31 9.32 9.33 9.33 9.34 9.35 9.36 9.37 9.38 9.40 9.42 9.43 9.43 9.43 9.44 9.45 9.46 9.46 9.46 9.48 9.48 9.49 9.49 9.51 9.52 9.54 9.55 9.56 9.57 9.57 9.58 9.59 9.60 9.61 9.61 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 9.63 9.63 9.64 9.65 9.67 9.69 9.71 9.71 9.77 9.80 9.80 9.82 9.83 9.84 9.84 9.86 9.86 9.87 9.88 9.89 9.90 9.90 9.92 9.93 9.96 9.97 9.98 9.100 9.101 9.102 9.104 9.106 9.106 9.107 9.107 9.108 9.109 9.110 9.112 9.113 9.115 9.116 9.117 9.120 9.122 9.123 9.125 9.126 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 Leavenworth—Principles of Engineering Economy, Ronald Press; Kasmer—Essentials of Engineering 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); Jelen—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 Operating Costs, Bureau of Mines Information Circular 8426, U.S Department of Interior; Ostwald—Cost Estimation for Engineering and Management, Prentice-Hall; American Association of Cost Engineers—Cost Engineers' Notebook; Gass—Linear Programming: Methods and Applications, McGraw-Hill; Hadley—Linear Programming, Addison-Wesley; 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 Theory and Methodology in Science and Engineering, Wiley; Quinn—Probability and Statistics, 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 Engineering, McGraw-Hill; White et al.—Principles of Engineering Economic Analysis, Wiley; Riggs—Engineering Economics, McGraw-Hill; Guenther—Concepts of Statistical Inference, McGraw-Hill; Lindgren—Statistical Theory, Macmillan; Meyer—introductory 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 Management, 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 = 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 period 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 period; 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 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 compoundamount (SPCA) factor; PIS = single-payment present-worth (SPPW) factor Uniformpayment series—SIR = uniform-series compound-amount (USCA) factor; RfS = sinkingfund-payment (SFP) factor; PIR = uniform-series present-worth (USPW) factor; RIP = capital-recovery (CR) factor Uniform-rate series—S/Ri = uniform-rate-series compoundamount (URSCA) factor; PIR1 = uniform-rate-series present-worth (URSPW) factor Basic Equations Simple interest, single payment S = P(l+ni) (1) Compound interest with discrete compounding SPCA = (1 + O" (2) SPPW = (I+/)"" (3) USCA-P+P-1 I (4) SFP= (TTTFTT "8^-^TTTF CR CR '(1 + O" -(l+0«-i URSCA = r"-(\ + iy _ / URSPW=^f " (6) m (7) (8) (9) A uniform-payment series that continues indefinitely is termed a perpetuity For this case, USPW = - (6a) CR = / (7a) Compound interest with continuous compounding SPCA = eJ" (10) where e = base of natural logarithms = 2.71828 SPPW = IT'11 USCA =7rff USPW= \-e~jn * e7— (11) (12) (13) The compound-interest factors for a single payment and for a uniform-payment series can be found by referring to compound-interest tables or by solving the relevant equations by calculator DETERMINATION OF SIMPLE INTEREST A company borrows $4000 at percent per annum simple interest What payment must be made to retire the debt at the end of years? Calculation Procedure: Apply the equation for simple interest This equation is S = P(I + ni) = $4000(1 + 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 percent per annum compounded quarterly What was the principal in the fund at the end of years? Calculation Procedure: 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 = percent/4 = percent per period With a 3-year deposit period, the number n of interest periods is n = x - 12 Apply the SPCA value given in a compound-interest table Look up the SPCA value for the interest rate, percent, and the number of interest periods, 12 Then substitute in S = P(SPCA) = $2600(1.268) = $3296.80 PRESENT WORTH OF SINGLE PAYMENT On January of a certain year, a deposit was made in a fund that earns interest at percent per annum On December 31,7 years later, the principal resulting from this deposit was $1082 What sum was deposited? Calculation Procedure: Apply the SPPW relation Obtain the SPPW factor for / = percent, n = 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 years in a fund earning interest at percent per annum What was the principal in the fund immediately after the fifth deposit was made? Calculation Procedure: Apply the USCA factor Obtain the USCA factor for / = percent, n = 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 years at percent interest To accumulate this sum, XYZ will make five equal annual deposits in a fund that earns interest at percent, the first deposit being made year after negotiation of the loan What is the amount of the annual deposit required? Calculation Procedure: Compute the sum to be paid at the expiration of the loan Obtain the SPCA factor from the interest table for / = percent, n = Then substitute in the relation S = P(SPCA) = $65,000(1.469) = $95,485 Compute the annual deposit corresponding to this future value Obtain the SFP factor from the interest table for / = percent, n = 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 ABC Corp offers an annuity of 12 annual payments of $15,000 each, the first payment to be made 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 10 percent, which offer should be accepted? Calculation Procedure: Compute the present worth of the annuity, using an interest rate of 10 percent Obtain the USPW factor from an interest table for / = 10 percent, n = 12 and substitute in the relation P = R(USPW) = $15,000(6.814) = $102,210 Since the DEF Corp offered an immediate payment of $120,000, its offer is more attractive than the offer made by ABC Corp CAPITAL-RECOVERY DETERMINATION On January of a certain year a company had a bank balance of $58,000 The company decided to allot this money to an improvement program by making a series of equal payments times a year for years, beginning on April of the same year If the account earned interest at percent compounded quarterly, what was the amount of the periodic payment? Calculation Procedure: Compute the true interest rate and number of interest periods Since the annual rate = percent and there are four interest periods per year, the rate per period is / = percent/4 = percent And with a 5-year pay period, the number of interest periods = years (4 periods per year) = 20 periods 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 / = 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 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 = / = percent/4 =1.5 percent In year there are four interest periods for this account Find the compounded value of $1 at the end of year from S = (1 + i)n = (1 + 0.015)4 = $1.06136 Thus, the interest earned by $1 in 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 continue indefinitely if the endowment fund earns interest of percent compounded semiannually? Calculation Procedure: Compute the effective interest rate Using the same procedure as in the previous calculation procedure for $1, we find the effective interest rate ie = (1.02)2 - = 0.04040, or 4.04 percent 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 of year and $1200 on January of year Jones Corp made a partial payment of $700 on January of year It was agreed that the balance of the loan would be discharged by two payments, one on January of year and the other on January of year 6, with the second payment being 50 percent larger than the first If the interest rate is percent, what is the amount of each payment? Calculation Procedure: Construct a line diagram indicating the loan data Figure shows the line diagram for these loans and is typical of the diagrams that can be prepared for any similar set of loans Select a convenient date for evaluating all the sums For this situation, select January of year Mark the valuation date on Fig 1, as shown Evaluate each sum at the date selected Use the applicable interest rate, percent, and the equivalence equation, value of money borrowed = value of money paid Substituting the applicable SPCA factor from the interest table for each of the interest periods involved, orw = 5,« = 3,w = 2, and n — 1, respectively, gives $900(SPCA) + $1200(SPCA) = $700(SPCA) + jc(SPCA) + 1.5*, where Valuation date Receipts Payments FIGURE Time, receipt, and payment diagram x = payment made on January of year and 1.5* = payment made on January of year 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 of a certain year, ABC Corp borrowed $1,450,000 for 12 years at percent interest The terms of the loan obliged the firm to establish a sinking ftind in which the following 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 the loan at the end of the twelfth year The interest rate earned by the sinking fund was percent 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: Prepare a money-time diagram Figure shows a money-time diagram for this situation, where * = deposit made at end of twelfth year Compute the principal of the loan at the end of the twelfth year Use the relation S = P(SPCA) for i = 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 All sums (except x) in units of $1000 Year FIGURE Money-time diagram Present Worth of Future Costs A cost analysis of alternative schemes may be performed by computing the present worth of all expenses incurred in each scheme during a stipulated period called the analysis period This period should encompass an integral number of lives of each asset required under the alternative schemes PRESENT WORTH OF FUTURE COSTS OF ANINSTALLATION A city contemplates increasing the capacity of existing water-transmission lines Two plans are under consideration: Plan A requires construction of a parallel pipeline, flow being maintained by gravity The initial cost is $800,000, and the life is 60 years with an annual operating cost of $1000 Plan B requires construction of a booster pumping station costing $210,000 with a life of 30 years The pumping equipment costs an additional $50,000; it has a life of 15 years and a salvage value of $10,000 The annual operating cost is $35,000 Which is the more economical plan if the interest rate is percent? Calculation Procedure: Construct a money-time diagram of the situation Figure shows the money-time diagram Note that this diagram uses 60 years as the analysis period Record on the money-time diagram the capital expenditures during this 60-year period Compute the total present worth of the payments For plan A, using the USPW factor for n = 60 years, we get PW = $800,000 + $1000(16.161) = $816,160 For plan B, by using the SPPW factor for the payments shown in Fig 5, and the uniform series present-worth factor for the operating cost, PW = $260,000 + $40,000(SPPW) + $250,000(SPPW) + $40,000(SPPW) + $35,000(USPW) $10,000(SPP W) - $260,000 + $40,000(0.4173) + $250,000(0.1741) + $40,000(0.0727) + $35,000(16.161) - $10,000(0.0303) - $888,460 Since the present worth of plan A is less than that of plan B, the scheme for plan A should be adopted because it is more economical Time in yeors Plan A Plan B *Income from disposal of equipment Al I sums in units of $ 1000 FIGURE Money-time diagram Capitalized Cost In computing the present worth of the costs associated with a proposed scheme, it is often advantageous to select an analysis period of infinite duration The present worth of the future costs is then referred to as the capitalized cost of the scheme Since each expenditure recurs indefinitely during the analysis period, the various costs constitute a group of perpetuities Thus, the capitalized cost C0 is Cc = [(P - L)^1](CR) + L + CIi1, or C0 = [(P - L)U1](SFP) + P + CIi1 If an asset is considered to have an infinite life span, these equations reduce to C0 = P + CIi1 In these equations, i = il9n= N DETERMINATION OF CAPITALIZED COST Two methods of conveying water for an industrial plant are being analyzed Method A uses a tunnel, and method B a ditch and flume The costs are as follows: Method A First cost, $ Salvage value, $ Life, years Operating cost, $/year Method B Tunnel Ditch Flume 180,000 Infinite 2,300 50,000 50 2,000 40,000 5,000 15 3,600 Evaluate these two alternatives on the basis of capitalized cost, using a percent interest rate Calculation Procedure: Compute the capitalized cost of the first alternative Since the tunnel has an infinite life, C0 = P + CIi1 = $180,000 + $2300/0.05 = $226,000 Compute the capitalized cost of the second alternative Using the capital-recovery factor for n = 50 years, i = percent, we find for the ditch C0 = ($50,000/0.05)(0.05478) + $2000/0.05 = $94,780 Using a similar procedure for the flume, which has a 15-year life, gives C0 = ($35,000/0.05)(0.09634) + $5000 + $3600/0.05 = $144,440 The total capitalized cost for method B = the sum of the flume and ditch costs, or $239,220 Since method A costs less, it is more economical CAPITALIZED COST OFASSET WITH UNIFORM INTERMITTENT PAYMENTS What is the capitalized cost of a bridge costing $85,000 and having-a 25-year life, a $10,000 salvage value, $400 annual maintenance cost, and repairs at 5-year intervals of $2000, if the interest rate is percent? Calculation Procedure: Convert the assumed repair costs to an equivalent series of uniform annual payments Assume that the repairs are made at the end of every 5-year interval, including the replacement date Using the SFP factor, we see the equivalent series of uniform annual payments R1 = $2000(SFP) for j = percent, n = years, OrU = $2000(0.18097) = $362 Convert the true repair costs to an equivalent series of uniform annual payments Repairs are omitted when the bridge is scrapped at the end of 25 years, thereby saving $2000 in the final 5-year period Convert this amount to an equivalent series of uniform annual payments (i.e., savings) and subtract from the result in step Or, R2 = $2000(SFP) for j = percent, n = 25 years Or, R2 = $2000(0.02095) = $42 Thus the annual cost of the repairs = $362 - $42 = $320 Compute the capitalized cost Using the capital-recovery factor for / = percent, n = 25 years, we get Cc = ($75,000/0.05)(0.07095) + $10,000 + $400/0.05 + $320/0.05 = $130,830 Related Calculations' An alternative solution could be worked as follows: Since the $2000 saving at the end of every 25-year interval coincides in timing with the income of $10,000 from the sale of the old bridge as scrap, this saving can be combined with the salvage value to obtain an effective value of $12,000 for salvage The annual cost of repairs is therefore taken as $362, the value OfU , step Applying the capital-recovery factor gives C0 = [($85,000 - $12,000)/0.05](0.07095) + $12,000 + $400/0.05 + $362/0.05 = $130,830 This agrees with the previously determined value CAPITALIZED COST OF AN ASSET WITH NONUNIFORM INTERMITTENT PAYMENTS A bridge has the same cost data as in the previous calculation procedure except for the repairs, which are as follows: End of year Repair cost, $ 10 15 20 2000 3500 1500 What is the capitalized cost of the bridge if the interest rate is percent? Calculation Procedure: Compute the present worth of the repairs for one life span Use the single-payment present-worth factor for each of the repair periods Or, PW = $2000(0.6139) + $3500(0.4810) + $1500(0.3769) = $3477 Convert the result of step to an equivalent series of uniform annual payments Using the capital-recovery factor, we find the annual cost of repairs = $3477(CR), where i = percent, n = 25 years Or, ca = $3477(0.07095) = $247 Compute the capitalized cost Using the same method as in step of the previous calculation procedure gives Cc = $106,430 + $10,000 + $8000 + $247/0.05 = $129,370 Related Calculations: An alternative way of solving this problem is to combine the present worth of the payments for repairs ($3477) with the initial cost ($85,000) to obtain an equivalent initial cost P' Then, P' = $88,477, and P' -L = $88,477 - $10,000 = $78,477 By applying the capital-recovery factor, Cc = $129,370 as before STEPPED-PROGRAM CAPITALIZED COST A firm plans to build a new warehouse with provision for anticipated growth Two alternative plans are available Plan A First cost, $ Salvage value, $ Life, years Annual maintenance, $ Cost of enlarging structure 10 years hence, $ 100,000 10,000 25 1,400 Plan B 80,000 15,000 30 1,200 first 10 years, 1,800 thereafter 40,000 If money is worth 10 percent, which is the more economical plan? Calculation Procedure: Compute the total present worth of the second plan costs Let P' represent the total present worth of the costs associated with plan B for one life span Using the SPPW for / = 10 percent, n = 10 years, for the cost of enlarging the structure, and the USPW for the annual maintenance after expansion, and the difference between the annual maintenance costs of this structure and the original structure, we get P' = $80,000 + $40,000(0.3855) + $1800(9.427) - $600(6.144) = $108,700 Compute the capitalized cost of each alternative Using the capital-recovery factor for plan A with i = 10 percent, n = 25 years yields C0 = [($100,000 - $10,000)/0.10](0.11017) + $10,000 + $1400/0.10 = $123,150 For plan B, by using the present worth from step 1, Cc = [($108,700 $15,000)/0.10](0.10608) + $15,000 = $114,400 Note that the capital-recovery factor for plan B is for 30 years Since plan B has the lower capitalized cost, it is more economical Cost Comparisons with Taxation and Technological Advances In the preceding material, the costs of alternative proposals were compared by disregarding taxation and assuming that financial and technological conditions remain static The cost analysis is now made more realistic by including the effects of taxation and technological advances Later the effects of inflation also are included CALCULATION OFANNUAL COST ON AFTER-TAX BASIS An asset has the following cost data: First cost, $80,000; life, 10 years; salvage value, $5000; annual operating cost, $3600 The firm that owns the asset is subject to a tax rate of 47 percent, and its investment rate is percent after payment of taxes Compute the after-tax annual cost of this asset if depreciation is allocated by (a) the straight-line method and (b) the sum-of-digits method Calculation Procedure: Compute the annual depreciation charge under the straight-line method The charge is D = ($80,000 - $5000)710 = $7500 Compute the annual cost under straight-line depreciation Most income earned by a corporation is subject to the payment of corporate income tax The effective (or after-tax) income is the difference between the original income and the tax payment pertaining to that income The before-tax investment rate ib = rate of return on an investment as calculated on the basis of original income; the after-tax investment rate ia = rate of return as calculated on the basis of effective income Every cost incurred in operating an asset serves to reduce taxable income and thus the tax payment The effective cost is the difference between the actual expenditure and the tax savings that results from the expenditure The cost of an asset is said to be computed on an after-tax basis if all calculations are based on effective costs and the after-tax investment rate Let t = tax rate and D = annual depreciation charge Where annual operating costs and depreciation charges are uniform, the annual cost A = (P — Z)(CR, n = N9 i = ia) + Li0 + c ( l - ~ Dt The last term represents the tax savings that accrues from the depreciation charge With n = 10, ia = percent, and t = 47 percent, A = ($80,000 - $5000)(0.14903) + $5000(0.08) + $3600(0.53) - $7500(0.47) = $9960 Compute the annual depreciation charges under the sum-of-digits method As given in an earlier calculation procedure, DU — W(N- U+ 1)/0.5[W(W+ I)], where Dv = depreciation charge for t/th year and W= total depreciation With W= $75,000 and W= 10, Z)1 = $13,636, and every depreciation charge thereafter is $1363.64 less than the preceding charge Convert the depreciation charges under the sum-of-digits method to an equivalent uniform depreciation charge, using an percent interest rate Refer to an earlier calculation procedure for converting a uniform-gradient series to an equivalent uniform series The equivalent uniform depreciation charge D = D1 + (g/0(l - «SFP) With D1 = $13,636, g = -$1363.64, i = percent, and n = 10, D = $13,636 + (-$1363.64/0.08)[1 - 10(0.06903)] = $8357 Compute the annual cost under sum-of-digits depreciation Referring to step and taking the difference between the equivalent depreciation charge in the present case and the depreciation charge under the straight-line method, we determine the annual cost A = $9960 - ($8357 - $7500)(0.47) = $9557 Related Calculations: A comparison of the two values of annual cost—$9960 when straight-line depreciation is used and $9557 when sum-of-digits depreciation is used—confirms the statement made in an earlier calculation procedure Since tax savings accrue more quickly under sum-of-digits depreciation than under straight-line depreciation, the former method is more advantageous to the firm In general, a firm seeks to write off an asset rapidly in order to secure tax savings as quickly as possible, thus allowing it to retain more capital for investment For this reason depreciation accounting is subject to stringent regulation by the IRS COST COMPARISON WITH ANTICIPATED DECREASING COSTS Two alternative machines, A and B, are available for a manufacturing operation The life span is years for machine A and years for machine B The equivalent uniform annual cost is estimated to be $16,000 for machine A and $15,000 for machine B However, as a result of advances in technology, the annual cost is expected to decline at a constant rate from one life to the next, the rate of decline being 10 percent for machine A and percent for machine B Applying an investment rate of 12 percent, determine which machine is preferable Calculation Procedure: Select the analysis period, and compute annual costs for this period The cost comparison will be made by the present-worth method The analysis period is 12 years, since this is the lowest common multiple of and The annual costs are as follows: Machine A: first life, $16,000; second life, $16,000(0.90) = $14,400; third life, $14,400(0.90) = $12,960 Machine B: first life, $15,000; second life, $15,000(0.94) = $14,100 Construct a money-time diagram The equivalent uniform annual payments are shown in Fig Compute the present worth of costs for the first analysis period, and identify the more economical machine For machine A, PW = $16,000(USPW, n = 4) + $14,400(USPW, n = 4)(SPPW, n = 4) + $12,960(USPW, n = 4)(SPPW, n = 8) With / = 12 percent, PW = $16,000(3.037) + $14,400(3.037)(0.6355) + $12,960(3.037)(0.4039) = $92,280 For machine B, PW = $15,000(USPW, /1 = 6) + $14,100(USPW, n = 6)(SPPW, n = 6), or PW = $15,000(4.111) + $14,100(4.111)(0.5066) = $91,030 Machine B should be used for the first 12 years because it costs less Year 1st life 2nd life 3rd life Equivalent annual costs for machine A Year 2nd life 1st life Equivalent annual costs for machine B FIGURE Equivalent payments for 12-year analysis period Compute the present worth of costs for the second analysis period The second 12-year period encompasses the fourth, fifth, and sixth lives of machine A and the third and fourth lives of machine B The "present" is the beginning of the second 12-year period The annual cost of machine A during its fourth life is (0.9O)3 times the annual cost during its first life Therefore, the results of step can be applied For machine A, PW = $92,280(0.90)3 - $67,270 For machine B, PW = $91,030(0.94)2 = $80,430 Thus, machine A should be used after the first 12 years Realistically, since the transfer from machine B to machine A can be made at the end of the first 6-year period, the decision should be reviewed at that time in the light of currently available forecasts ECONOMY OF REPLACING AN ASSET WITH AN IMPROVED MODEL A machine has been in use for years, and its cost data for the next years are shown in Table 2, where the years are counted from the present An improved model of this machine has just appeared on the market, and according to estimates it has an optimal life of years with an equivalent uniform annual cost of $9400 No additional improvements are anticipated in the near future If money is worth 10 percent, when will it be most economical to retire the existing machine? Calculation Procedure: Compute an equivalent single end-of-life payment for each prospective remaining life Let R = remaining life of machine, years The annual cost corresponding to every possible value of R will be found by a method that is a variation of that used in an earlier calculation procedure Let CR = operating cost at end of Rth year and FR = equivalent single pay- TABLE Year Salvage value at end, $ Annual operating cost, $ 12,000 8,000 5,000 4,000 3,500 3,000 2,700 2,600 2,500 4,700 5,200 5,800 6,600 7,500 8,500 9,700 10,900 ment at end ofRth year Then F^ = F^1(I + O + CR By retaining the existing machine, the firm forfeits an income of $12,000, and this is equivalent to making a payment of that amount now Then F0 = $12,000; F1 = $12,000(1.10) + $4700 = $17,900; F2 = $17,900(1.10) + $5200 - $24,890; F3 = $24,890(1.10) + $5800 - $33,179; etc The results are shown in Table Compute the annual cost for every prospective remaining life Let AR = annual cost for a remaining life of R years, and LR — salvage value at end of Rth year Then AR = (FR - LjO(SFP, n = R) Thus, with i = 10 percent, A1 = ($17,900 $8000)1 = $9900; A2 = ($24,890 - $5000)(0.47619) = $9471; A3 = ($33,179 $4000)(0.30211) - $8815; etc The results are shown in Table 3 Determine whether the existing machine should be retired now When an asset is purchased and installed, its resale value drops sharply during the early years of its life, and thus the firm incurs a rapid loss of capital during those years The result is that the annual cost for the remaining life of an existing asset is considerably less than the annual cost when the asset was first purchased Table shows that the optimal remaining life of the existing machine is years and TABLE Calculation of Annual Cost Remaining life, years F, $ Z, $ SFP Annual cost, $ 12,000 17,900 24,890 33,179 43,097 54,907 68,897 85,487 104,936 8,000 5,000 4,000 3,500 3,000 2,700 2,600 2,500 1.00000 0.47619 0.30211 0.21547 0.16380 0.12961 0.10541 0.08744 9,900 9,471 8,815 8,532 8,502 8,580 8,737 8,957 the corresponding annual cost is $8502 Since this is less than the annual cost of the new machine ($9400), the existing machine should be retained for at least years Determine precisely when the existing machine should be replaced Since costs increase beyond the fifth year, the existing machine and the improved model will be compared on a year-by-year basis Let BR = cost of retaining existing machine year beyond the Rth year Then BR = LR(l + i) - LR+l + cR+l Thus, B5 = $3000(1.10) $2700 + $8500 = $9100 Since this is less than the annual cost of the new model, the existing machine should not be retired years hence Continuing, we find B6 = $2700(1.10) - $2600 + $9700 = $10,070, which exceeds $9400 Therefore, the existing machine should be retired years hence ECONOMY OF REPLACEMENT UNDER CONTINUING IMPROVEMENTS A newly acquired machine costs $40,000, and it has the salvage values and annual operating costs shown in Table It is anticipated that a new model will become available at the end of each year All future models will have first costs and salvage values identical with those of the present model, but the annual operating cost for a given model will be $600 lower than the corresponding annual operating cost of the preceding model For example, the model that becomes available year hence will have an operating cost of $11,400 for the first year, $12,400 for the second year, etc Applying an interest rate of 10 percent, determine how long this machine should be held Calculation Procedure: Establish the excess operating costs in relation to a 1-year life When the machine is retired, it will be replaced with the model that becomes available at that date First assume that the machine is retired at the end of each year The operating costs for the next years are shown in Table Now assume that the machine is held for years Subtracting the values just found from the values in Table gives the excess operating costs for an 8-year life; these are shown in Table This table also gives the excess operating costs for every prospective life of the machine For example, if the machine is TABLE Year Salvage value at end, $ Annual operating cost, $ 25,000 20,000 17,000 15,000 13,500 12,000 11,000 10,000 12,000 13,000 14,600 16,500 18,800 21,500 24,500 28,000 TABLE Year Operating cost for -year life, $ Excess operating cost for 8-year life, $ 12,000 11,400 10,800 10,200 9,600 9,000 8,400 7,800 O 1,600 3,800 6,300 9,200 12,500 16,100 20,200 held years, the excess operating cost is $1600 for the second year and $3800 for the third year, and these values apply to each subsequent life Since only differences in cost are significant, the prospective lives of the machine will be compared by applying the excess rather than the actual operating costs Compute an equivalent single end-of-life payment for every prospective life Annual costs will be computed by using the same method as in the previous calculation procedure, but applying excess operating costs Thus, with / = 10 percent, F1 = $40,000(1.10) = $44,000; F2 = $44,000(1.10) + $1600 = $50,000; F3 - $50,000(1.10) + $3800 = $58,800; etc The results are shown in Table Compute the annual cost for every prospective life Proceeding as in the previous calculation procedure gives A1 = ($44,000 - $25,000)1 = $19,000; A2 = ($50,000 - $20,000)(0.47619) = $14,286; A3 = ($58,800 - $17,000) (0.30211) = $12,628; etc The results are shown in Table Identify the most economical life of the machine Table reveals that a 4-year life has the minimum annual cost Related Calculations: Each excess annual operating cost shown in Table consists of two parts: a deterioration cost, which is the increase in operating cost due to aging TABLE Calculation of Annual Cost Life, years F, $ L, $ SFP Annual cost, $ 44,000 50,000 58,800 70,980 87,278 108,506 135,456 169,203 25,000 20,000 17,000 15,000 13,500 12,000 11,000 10,000 1.00,000 0.47619 0.30211 0.21547 0.16380 0.12961 0.10541 0.08744 19,000 14,286 12,628 12,062 12,085 12,508 13,119 13,921 of the machine, and an obsolescence cost, which results from the development of an improved model For example, at the end of the fourth year the deterioration cost is $16,500 - $12,000 = $4500, and the obsolescence cost is $600 x = $1800 If the quality of the product declines as the machine ages, the resulting loss of income can be added to the deterioration cost ECONOMY OF REPLACEMENT ON AFTER-TAX BASIS A machine was purchased years ago at a cost of $45,000 It had a life expectancy of years and anticipated salvage value of $3000 It has been depreciated by the sum-of-digits method The net resale value of the machine is $13,000 at present and is expected to be $9000 a year hence The operating cost during the coming year will be $2600 A newly developed machine can be substituted for the existing one According to estimates, this machine will have an optimal life of years with an annual cost of $4800 on an after-tax basis The tax rate is 45 percent for ordinary income and 30 percent for long-term capital gains The desired investment rate on an after-tax basis is percent Determine whether the existing machine should be replaced at present Calculation Procedure: Compute the depreciation charges for the first years Refer to an earlier calculation procedure for sum-of-digits depreciation The charges are Z)1 = $10,500; D2 = $9000; D3 = $7500; D4 = $6000 Compute the book value at the end of the third and fourth years Let BR = book value at end ofRth year Then B3 = $45,000 - ($10,500 + $9,000 + $7,500) - $18,000 and B4 = $18,000 - $6000 - $12,000 Compute the cost of retaining the machine through the fourth year Income that accrues from normal business operations is called ordinary income; other forms of income are called capital gains The difference between the net income that accrues from selling an asset and its book value at the date of sale is a capital gain (or loss) If the asset was held for a certain minimum amount of time, this capital gain (or loss) is subject to a tax rate different from that for ordinary income Let R = age of asset, years The after-tax cost of retaining the asset through the (R + l)st year, as evaluated at the end of that year, is [LR(\ - tc) + BRtc](\ + ia) + cR+l(\ - t0) LR+l(\ - tc) - BR+ltc - DR+lt09 where ia = after-tax investment rate; t0 and tc = tax rate on ordinary income and long-term capital gains, respectively; L = true salvage value; c = annual operating cost; and the subscript refers to the age of the asset The expression in brackets is the income that would be earned if the asset were sold at the end of the Rth year; if the asset is retained, this income is forfeited and becomes part of the cost of retention With ia9 = percent, t0 = 45 percent, tc = 30 percent, L3 = $13,000, L4 = $9000, and C4 = $2600, the cost of retaining the machine through the fourth year is [$13,000(0.70) + $18,000(0.30)](1.08) + $2600(0.55) - $9000(0.70) - $12,000(0.30) - $6000(0.45) = $4490 Determine whether the existing machine should be retired now Since the cost of retaining the machine for additional year ($4490) is less than the annual cost of the new machine ($4800), the existing machine should not be retired at present Effects of Inflation Notational System Here C0 and C1 are the costs of a commodity now and year hence, respectively, and/= annual rate of inflation during the coming year with respect to this commodity Then/= (C1 - C0, or C1 = C0(I +/) Also, Cn = cost of the commodity n years hence If the annual rate of inflation remains constant at/ then Cn = C0(I +/)" = C0(SPCA, / =/) In the subsequent material, it is understood that the given inflation rate applies to the asset under consideration DETERMINATION OF REPLACEMENT COST WITH CONSTANT INFLATION RATE A machine has just been purchased for $60,000 It is anticipated that the machine will be held years, that it will have a salvage value of $4000 as based on current prices, and that the annual rate of inflation during the next years will be percent The machine will be replaced with a duplicate, and the firm will accumulate the necessary capital by making equal end-of-year deposits in a reserve fund that earns percent per annum Determine the amount of the annual deposit Calculation Procedure: Compute the required replacement capital Both the cost of a new machine and the salvage value of the existing machine increase at the given rate Thus, the amount of money the firm must accumulate to buy a new machine is ($60,000 - $4000)(1.07)5 = $56,000(1.403) = $78,568 Compute the annual deposit Use this relation: Annual deposit R = 5(SFP) With i = percent and n = 5, R = $78,568(0.17740) = $13,938 DETERMINATION OF REPLACEMENT COST WITH VARIABLE INFLATION RATE In the preceding calculation procedure, determine the amount of the annual deposit if the annual rate of inflation is expected to be percent for the next years and percent thereafter Calculation Procedure: Compute the required replacement capital Replacement capital = $56,000(1.07)3(1.09)2 = $56,000(1.225)(1.188) = $81,497 Compute the annual deposit From the preceding calculation procedure, annual deposit = $81,497(0.17740) = $14,458 PRESENT WORTH OF COSTS IN INFLATIONARY PERIOD An asset with a first cost of $70,000 is expected to last years and to have the following additional cost data as based on present costs: salvage value, $5000; annual maintenance, $8400; major repairs at the end of the fourth year, $9000 The asset will be replaced with a duplicate when it is retired Using an interest rate of 12 percent and an inflation rate of percent per year, find the present worth of costs of this asset for the first two lives (i.e., for 12 years) Calculation Procedure: Compute the present worth of the capital expenditures for the first life The "present" refers to the beginning of the first life The payment for repairs will be $9000(1.08)4, and the present worth of this payment is $9000(1.08)4(SPPW, n = 4, i = 12 percent) = $9000(1.08)4/(1.12)4 = $7780 Similarly, the present worth of the salvage value is $5000(1.08)6/(1.12)6 = $4020 Thus, the present worth of capital expenditures for the first life is $70,000 + $7780 - $4020 = $73,760 Compute the present worth of maintenance for the first life The annual payments for maintenance constitute a uniform-rate series in which the first payment R1 = $8400(1.08) = $9072 and the ratio of one payment to the preceding payment is r = 1.08 By Eq 9, the present-worth factor of the series is URSPW = [(1.08/1.12)6 - 1]/(1.08 - 1.12) = 4.901 Then present worth of series = ,K1(URSPW) = $9072(4.901) = $44,460 Compute the present worth of costs for the first life Summing the results, we see that present worth = $73,760 + $44,460 = $118,220 Compute the present worth of costs for the second life Since each payment in the second life is (1.08)6 times the corresponding payment in the first life, the value of all payments in the second life, evaluated at the beginning of that life, is (1.08)6 times that for the first life, or $118,220(1.08)6 The present worth of this amount is $118,220(1.08)6/( 1.12)6 = $95,040 Compute the present worth of costs for the first two lives Summing the results yields PW = $118,220 + $95,040 = $213,260 Related Calculations: Let h = [(I +/)/(! + OF, where N= life of asset, years In the standard case, where all annual payments as based on present costs are equal and no extraordinary intermediate payments occur, the present worth of costs for the first life is P - Lh + c(\ +/)(/* - !)/(/- 0» where P = initial cost; L = salvage value as based on present costs; c - annual payment for operation, maintenance, etc., as based on pres- ent costs Where extraordinary payments occur, simply add the present worth of these payments, as was done in the present case with respect to repairs at the end of the fourth year In the special case where/= f, the present worth of costs for the first life is P - L + Nc COSTCOMPARISON WITH ANTICIPATED INFLATION Two alternative machines have the following cost data as based on present costs: First cost, $ Salvage value, $ Life, years Annual maintenance, $ Machine A Machine B 45,000 3,000 8,000 80,000 2,000 6,000 Determine which machine is more economical, using an interest rate of 10 percent and annual inflation rate of percent Calculation Procedure: Establish the method of cost comparison The present-worth method is suitable here Select an analysis period of 12 years, which encompasses three lives of machine A and two lives of machine B Compute the present worth of costs of machine A for the first life Refer to the equation given at the conclusion of the preceding calculation procedure Set h = (1.07/1.1O)4 = 0.89529 By reversing the sequence in the last two terms of the equation, present worth = $45,000 - $3000(0.89529) + $8000(1.07)0 - 0.89529)7(0.10 - 0.07) = $72,190 Compute the present worth of costs of machine A for the first three lives Refer to step of the preceding calculation procedure Thus, PW = $72,190[1 + (1.07/1.1O)4 + (1.07/1.1O)8] = $72,190(2.69684) = $194,680 Compute the present worth of costs of machine B for the first Ufe Set h = (1.07/1.1O)6 = 0.84712 The present worth of costs for the first life = $80,000 $2000(0.84712) + $60000.07)0 ~ 0.84712)7(0.10 - 0.07) = $111,020 Compute the present worth of costs of machine B for the first two lives PW = $111,020(1 + 0.84712) = $205,070 Determine which machine is preferable Machine A has the lower cost and so is preferable ENDOWMENT WITH ALLOWANCE FOR INFLATION An endowment fund is to provide perpetual annual payments to a research institute The first payment, to be made year hence, will be $10,000 Each subsequent payment will be percent more than the preceding payment, to allow for inflation If the interest rate of the fund is percent per annum, what amount must be deposited in the fund now? Verify the result Calculation Procedure: Compute the amount to be deposited The payments form a uniform-rate series, and the amount to be deposited = P = present worth of series Refer to Eq for the present-worth factor When r < + i and n is infinite, URSPW = 1/(1 + i - r) With i = percent and r = 1.02, URSPW = 1/(1.07 - 1.02) = 20 Then P = ^1URSPW) = $10,000(20) - $200,000 Prove that this deposit will provide an endless stream of payments The proof consists in finding the rate at which the principal in the fund is growing At the end of the first year, principal = $200,000(1.07) - $10,000 •= $204,000 The rate of increase in principal = ($204,000 - $200,000)/$200,000 = percent per year Similarly, at the end of the second year, principal = $204,000(1.07) - $10,000(1.02) = $208,080 The rate of increase in principal = ($208,080 - $204,000)/$204,000 = percent per year Thus, the end-of-year principal expands at the same rate as the payments, and so the payments can continue indefinitely Related Calculations: If the interest period of the fund differs from the payment period, it is necessary to use the interest rate corresponding to the payment period For example, assume that the interest rate is percent per annum compounded quarterly The corresponding annual (or effective) rate is i = (1.0175)4 - = 7.186 percent, and URSPW = 17(1.07186 - 1.02) = 19.283 The amount to be deposited = $192,830 Note that if r > +1, URSPW becomes infinite as n becomes infinite Thus, if the interest rate of the fund is percent per annum, it is impossible to allow the payments to increase by percent or more Evaluation of Investments PREMIUM-WORTH METHOD OF INVESTMENT EVALUATION A firm contemplates investing in a depleting asset and has a choice between two enterprises Project A requires the investment of $57,500; project B requires the investment of $63,000 The forecast end-of-year dividends are as follows: Year Project A, $ Project B, $ 10,000 15,000 25,000 20,000 10,000 15,000 25,000 30,000 20,000