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**Engineering Economics Outline Overview**

CASH FLOW, p. 599 TIME VALUE OF MONEY, p. 600 EQUIVALENCE, p. 602 ENGINEERING ECONOMICS, p. 599 Choosing among engineering alternatives often involves analysis of monetary and intangible factors. CASH FLOW, p. 600 Assigning a time schedule to a series of financial transactions clarifies the costs and benefits of the overall situation. TIME VALUE OF MONEY, p. 602 Money received sooner is worth more than money received later; money spent sooner is more expensive than money spent later. EQUIVALENCE, p. 602 Applying an appropriate rate of interest to an amount of money makes it possible to compare the value or cost of money received or spent at different times. Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

COMPOUND INTEREST, p. 603 Symbols and Functional Notation Single Payment Formulas Uniform Payment Series Formulas Uniform Gradient Continuous Compounding COMPOUND INTEREST, p. 603 Symbols and Functional Notation Formulas may be used to find present or future value for a single payment, or present or future value or a periodic payment amount for a series of equal payments, or a series of payments that change each time by a fixed amount (table p. 603). Single Payment Formulas These formulas establish the relationship between a single present and future sum, at a given interest rate and length of time (formulas p. 604). Uniform Payment Series Formulas These formulas establish the relationships between a present or future sum and a series of equal payments to add up to or pay off that sum, at a given interest rate and length of time (figure and formulas pp ). Uniform Gradient These formulas establish the relationships between a present or future sum, or a series of equal payments, and a series of payments which change by a uniform amount each time interval and which add up to or pay off that sum or series of equal payments, at a given interest rate and length of time (figures and formulas pp ). Continuous Compounding Different formulas apply in the above situations if the interest rate involved is compounded continuously (table p. 609). Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

NOMINAL AND EFFECTIVE INTEREST, p. 610 Non-Annual Compounding NOMINAL AND EFFECTIVE INTEREST, p. 610 An interest rate calculated on a yearly basis without compounding is the nominal rate; the rate including the effect of compounding is the effective rate. Non-Annual Compounding Compounding interest more than once a year increases the effective interest rate (formula p. 610). Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

SOLVING ENGINEERING ECONOMICS PROBLEMS, p. 611 Criteria Present Worth Appropriate Problems Infinite Life and Capitalized Cost Future Worth or Value SOLVING ENGINEERING ECONOMICS PROBLEMS, p. 611 Choosing among engineering alternatives often involves analysis of monetary and intangible factors; monetary factors usually include a rate of return on investment. Criteria Engineering problems may involve a fixed budget or other resource inputs, a fixed result or output, or neither may be fixed. Present Worth Alternatives may be compared by optimizing the present worth of the non-fixed components of an engineering problem. Appropriate Problems Present worth analysis usually involves money which will be spent or received in the future; comparison of alternatives should cover equal periods of time. Infinite Life and Capitalized Cost The present worth of a cost which must be paid periodically forever is the periodic payment amount divided by the interest rate (formulas p. 613). Future Worth or Value Alternatives may be compared by optimizing the future worth of the non-fixed components of an engineering problem. Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

SOLVING ENGINEERING ECONOMICS PROBLEMS Annual Cost, p. 615 Criteria Application of Annual Cost Analysis Rate of Return Analysis Two Alternatives Benefit-Cost Analysis Breakeven Analysis Annual Cost This method involves equivalent uniform annual costs or benefits. Criteria Alternatives may be compared by optimizing the annual worth of the non-fixed components of an engineering problem. Application of Annual Cost Analysis Alternatives may be compared on a continuing need basis, or for a term with salvage value factored in. Rate of Return Analysis If an engineering problem can be stated in terms of cash flow where cost and benefits are equal in present, annual, or future worth, the interest rate in the worth calculation defines the rate of return. Two Alternatives A choice between alternatives should be based on whether the difference between the initial costs of the alternatives yields an acceptable rate of return. Benefit-Cost Analysis If the present worth of benefits of a project are greater than the present worth of costs, it is generally considered acceptable; the choice between two projects should be made on the basis of the benefit-cost ratio of the difference in initial cost between the projects (formula p. 619). Breakeven Analysis In engineering economics, the set of conditions under which the costs of two alternatives are equal is the breakeven point. Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

BONDS, p. 620 Bond Value Bond Yield PAYBACK PERIOD, p. 621 BONDS, p. 620 A bond is a document promising repayment of a debt in a specified manner. Bond Value The value of a bond is the present worth of its promised repayment. Bond Yield The yield of a bond is the interest rate at which its value equals its cost. PAYBACK PERIOD, p. 621 The time required for the profit or other monetary benefit of a project to equal its cost is the payback period. Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

VALUATION AND DEPRECIATION, p. 622 Notation Straight Line Depreciation Double Declining-Balance Depreciation Modified Accelerated Cost Recovery System Depreciation Half-Year Convention VALUATION AND DEPRECIATION, p. 622 Accounting for the costs of capital assets has tax implications that affect economic analyses. Notation A depreciation schedule allocates the cost of a capital asset over the term of its useful life. Straight Line Depreciation This method allocates an equal amount of the cost of the depreciable value of an asset each year (formulas p. 622). Double Declining-Balance Depreciation These methods shift more of the depreciation to the earlier years of the schedule (formulas p. 622). Modified Accelerated Cost Recovery System Depreciation This system starts with double or 150% declining balance depreciation, depending on the type of asset, and switches to straight-line when that method gives a larger deduction. Half-Year Convention To simplify record keeping, for accounting purposes all assets are treated as being put into or taken out of service in the middle of the year (table p. 623). Copyright Kaplan AEC Education, 2008

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**Engineering Economics Outline Overview Continued**

INFLATION, p. 624 Effect of Inflation on Rate of Return INFLATION, p. 624 A general rise in the price of goods and services decreases the worth of future payments. Effect of Inflation on Rate of Return Inflation may affect calculation of rate of return, depending on whether future benefits also increase in value, and on tax consequences. Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money An investment is estimated to return $5,000 per year for the next 12 years. If the investor wishes to obtain a 9% return per year, the most he should pay for this investment is closest to (a) $30,000 (b) $40,000 (c) $35,000 (d) $45,000 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution P = A (P/A, 9%, 12) = 5,000 (7.1607) = $35,803Â Answer: (c) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money A college fund is established for a 5 year old boy, with the objective of having $60,000 upon his 18th birthday. If deposits into an account paying 5% per year are made on each birthday starting with his 6th birthday and ending with his 18th, how much must each deposit be? (a) $3400 (b) $3800 (c) $4200 (d) $4600 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution There are 13 equal deposits. A = F (A/F, 5%, 13) = 60,000 ( ) = $3,388 Answer: (a) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money A credit card company charges 12% compounded monthly on the unpaid balance. This is equivalent to an effective annual interest rate of most nearly (a) 12% (b) 12.3% (c) 12.7% (d) 12.9% Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution ie = (1 + r/m)m â€“ 1 = ( /12)12 â€“ 1 = = 12.7% Answer: (c) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money An appliance store advertises that a refrigerator can be purchased for $900 cash, or no money down and 24 equal monthly payments of $40. If you paid for the refrigerator with the 24 monthly payment plan, the effective yearly interest rate on your purchase is closest to Â (a) 5.5% (b) 5.8% (c) 6.2% (d) 6.5% Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution P = A (P/A, i, 24) 900 = 40 (P/A, i, 24) (P/A, i, n) = Looking at the tables this is true for i approximately 0.5% per month iyear = (1 + imonth)12 â€“ 1 = ( )12 â€“ 1 = = 6.17% Answer: (c) Copyright Kaplan AEC Education, 2008

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**Time Value of Money Problems**

The annual maintenance costs for a drilling machine are estimated at $750 the first year, increasing by $50 every year to $800 the second year, $850 the third year, and so on. Assuming a useful life of 8 years and a 6% interest rate, the equivalent uniform annual cost (EUAC) for maintenance costs over the useful life is most nearly (a) $800 (b) $900 (c) $1000 (d) $1100 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution EUAC = (A/G, 6%, 8) = (3.1952) = $910 Answer: (b) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money A store offers an extended warranty for $350 to cover all parts and labor repair costs on a plasma TV over a four year period. The standard warranty only covers repairs during the first year. At a 9% interest rate the minimum equal annual repair costs over years 2 through 4 that makes the extended warranty equally desirable is (a) $150 (b) $175 (c) $200 (d) $225 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution A = 350 (F/P, 9%, 1) (A/P, 9%, 3) = 350 (1.09) ( ) = $151 Answer: (a) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money Two alternatives are being considered: A B Initial cost 55,000 60,000 Operating and Maintenance cost 12,000 9,000 Salvage value 2,000 15,000 Useful life 4 6 At an interest rate of 8% the EUAC of Alternative A is most nearly (a) $28,200 (b) $29,200 (c) $30,200 (d) $31,200 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution EUAC = 55,000(A/P, 8%, 4) + 12, ,000(A/F, 8%, 4) = 55,000(.30192) + 12, ,000( ) = $28,162 Answer: (a) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money A paving contractor can either purchase or lease a road grader. The purchase cost is $89,000. The lease plan is for five equal lease payments payable in advance (i.e., the first lease payment is at the start of the lease). Excluding all operating and maintenance costs and given a MARR of 14%, the maximum lease payment is closest to (a) $22,250 (b) $22,500 (c) $22,750 (d) $23,000 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution A = [89,000 (P/F, 14%, 1)} (A/P, 14%, 5) = {89,000 (0.8772)} ( ) = $22,740 Answer: (c) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Time Value of Money A company purchases a plastic injection system that will save the company $43,000 during the first year of operation, decreasing by $2,000 every year to $41,000 the second year, $39,000 the third year, and so forth. Given a MARR of 10% per year, the present worth of the savings over the 4-year life of the machine is closest to (a) $125,500 (b) $126,500 (c) $127,500 (d) $128,500 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution P = 43,000 (P/A, 10%, 4) â€“ 2,000 (P/G, 10%, 4) = 43,000 (3.1699) â€“ 2,000 (4.3781) = $127,550 Answer: (c) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Interest Rate A credit card company charges 0.05% interest per day on the outstanding balance. The effective annual interest on charges made on this card is nearest to (a) 16% (b) 18% (c) 20% (d) 22% Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution iyear = ( )365 â€“ 1 = = 20.0% Answer: (c) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Interest Rate A company is considering purchasing a new machine for $650,000 that will increase the firmâ€™s net income by $150,000 per year over the next 5 years. If the company wishes to obtain a 15% return on its investment, the minimum salvage value of the machine at the end of the 5-year useful life should be closest to (a) $275,000 (b) $295,000 (c) $315,000 (d) $335,000 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution S = 650,000 (F/P, 15%, 5) â€“ 150,000 (F/A, 10%, 5) = 650,000(2.0114) â€“ 150,000(6.7424) = $296,050 Answer: (b) Copyright Kaplan AEC Education, 2008

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**Comparison of Alternatives**

A company is considering two mutually exclusive alternative projects to enhance its production facility. The respective financial estimates for each project are as follows Project A Project B Initial Cost 75,000 105,000 Annual Savings 16,000 24,000 Salvage Value 9,000 If the useful life of Project A is 4 years, with a MARR of 15%, the useful life in years of Project B that makes both projects equally desirable is most nearly (a) 4 (b) 5 (c) 6 (d) 7 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution EUACA = 75,000(A/P, 15%, 4) - 16, ,000(A/F, 15%, 4) = 75,000(.35027) - 16, ,000( ) = $8,468 EUACB = 8,468 = 105,000(A/P, 15%, n) - 24,000 (A/P, 15%, n) = (8, ,000)/105,000 = From table look-up, the value of n that most nearly makes the above relation true is 5. Answer: (b) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Bonds An investor is considering purchasing a bond with a face value of $20,000 and 10 years left to mature. The bond pays 8% interest payable quarterly. If he wishes to get a 3% per quarter return, the most he should pay for the bond is closest to (a) $15,400 (b) $16,000 (c) $16,400 (d) $16,800 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution Since the bond pays 8% compounded quarterly, its effective interest rate is 2% per 3 months. Interest payment = i(Face value) = 0.02(20,000) = $400/three months P = 400(P/A, 3%, 40) + 20,000(P/F, 3%, 40) = 400( ) + 20,000(0.3066) = $15,377 Answer: (a) Copyright Kaplan AEC Education, 2008

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**Benefit to Cost Analysis**

A county is considering the following project Initial Cost $22,500,000 Maintenance $525,000 per year Savings $5,300,000 per year Given a useful life of 12 years and an interest rate of 8%, the benefit to cost ratio is closest to (a) 0.67 (b) 1.01 (c) 1.51 (d) 1.67 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution PWcost = 22,500, ,000 (P/A, 8%, 12) = 22,500, ,000 (7.5361) = $26,456,453 PWbenefit = 5,300,000 (P/A, 8%, 12) = 5,300,000 (7.5361) = $39,941,330 B/C = PWbenefit/ PWcost = 39,941,330/26,456,453 = 1.51 Answer: (c) Copyright Kaplan AEC Education, 2008

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**Benefit to Cost Analysis**

A proposed change to highway design standards is expected to reduce the number of vehicle crashes by 9,200 per year, but have initial cost of $150,000,000 and annual costs of $25,000,000. Given an interest rate of 10% and a study period of 8 years, the average cost of each vehicle crash in order that the benefit-to-cost ratio be 1.0 is closest to (a) $5700 (b) $6700 (c) $8700 (d) $9700 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution In order that B/C = 1.0 PWcost = PWbenefit PWbenefit = 150,000, ,000,000 (P/A, 10%, 8) = 150,000, ,000,000 (5.3349) = $283,372,500 EUACbenefit = $283,372,500 (A/P, 10%, 8) = $283,372,500 ( ) = $53,115,341 $equivalent/crash = 53,115,341/9,200 = $5,773 Answer: (a) Copyright Kaplan AEC Education, 2008

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**Payback, Breakeven Analysis, and Capitalized Cost**

An engineering department is considering purchase of an advanced computational fluid dynamics software system to enhance productivity. The initial cost of the software is $55,000 but is expected to result in efficiency savings of $25,000 the first year, with this amount decreasing by $5,000 per year thereafter. The payback period for the software is closest to (a) 2 (b) 2.67 (c) 3 (d) 3.67 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution Costs = 55,000 The payback period is the time when total income to date is equal to the total costs. Costs = Income = 25, , ,000 + â€¦ Since the income is stated as $25,000 per year, one can assume that savings occur uniformly throughout the year. Therefore, the payback period is 2.67 years. Answer: (b) Copyright Kaplan AEC Education, 2008

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**Payback, Breakeven Analysis and Capitalized Cost**

A company is considering two alternative forklifts with equal useful lives and the following characteristics A B Initial Cost 25,000 32,000 Total Annual Costs 4350 2500 Given an interest rate of 10%, the service life in years at which both machines have the same equivalent uniform annual cost (EUAC) is most nearly (a) 5 (b) 7 (c) 9 (d) 11 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution EUACA = 25,000 (A/P, 10%, n) + 4,350 EUACB = 32,000 (A/P, 10%, n) + 2,500 EUACA = EUACB 25,000 (A/P, 10%, n) + 4,350 = 32,000 (A/P, 10%, n) + 2,500 (A/P, 10%, n) = (4,350 â€“ 2,500) / 7,000 = From table look-up, the value of n that most nearly makes the above relation true is 15. Answer: (a) Copyright Kaplan AEC Education, 2008

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**Payback, Breakeven Analysis and Capitalized Cost**

An alumnus wishes to endow a scholarship to her alma mater that will provide $12,000 per year in perpetuity. Although the donation will be given today, the first scholarship will be given in 3 years (i.e., at time t = 3 years). At an interest rate of 8%, the amount she will need to donate is closest to (a) $119,600 (b) $122,600 (c) $125,600 (d) $128,600 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution P = (A/i) (P/F, 8%, 2) = (12,000/0.08) (0.8573) = (150,000) (0.8573) = $128,595 Answer: (d) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Depreciation A company purchases a plastic extrusion machine for $95,000. If this machine has an estimated salvage value of $10,000 at the end of its five-year useful life and recovery period, the second year straight line depreciation is closest to $13,000 $15,000 $17,000 $19,000 Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution Dt = D2 = (95,000 â€“ 10,000)/5 = $17,000 Answer: (c) Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Inflation A compact car costs approximately $21,000 today. If a comparable car cost $15,000 ten years ago, the average annual inflation in compact car prices over the past ten years is closest to (a) 2.6% (b) 3.0% (c) 3.4% (d) 3.8% Copyright Kaplan AEC Education, 2008

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**Copyright Kaplan AEC Education, 2008**

Solution 21,000 = 15,000 (1 + f)10 f = (21,000/15,000)0.1 â€“ 1 = = 3.4% Answer: (c) Copyright Kaplan AEC Education, 2008

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