copyright kaplan aec education, 2008 engineering economics outline overview cash flow, p. 599 time...

Copyright Kaplan AEC Education, 2008

Engineering EconomicsOutline Overview

• CASH FLOW, p. 599

• TIME VALUE OF MONEY, p. 600

• EQUIVALENCE, p. 602


• COMPOUND INTEREST, p. 603– Symbols and Functional Notation– Single Payment Formulas– Uniform Payment Series Formulas– Uniform Gradient– Continuous Compounding

Engineering EconomicsOutline Overview Continued


• NOMINAL AND EFFECTIVE INTEREST, p. 610– Non-Annual Compounding



• SOLVING ENGINEERING ECONOMICS PROBLEMS, p. 611– Criteria– Present Worth

• Appropriate Problems

• Infinite Life and Capitalized Cost

– Future Worth or Value



• 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



• BONDS, p. 620– Bond Value– Bond Yield

• PAYBACK PERIOD, p. 621



• VALUATION AND DEPRECIATION, p. 622– Notation– Straight Line Depreciation– Double Declining-Balance Depreciation– Modified Accelerated Cost Recovery System

Depreciation• Half-Year Convention



• INFLATION, p. 624– Effect of Inflation on Rate of Return



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


Solution

P = A (P/A, 9%, 12) = 5,000 (7.1607) = $35,803

Answer: (c)


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

Time Value of Money


Solution

There are 13 equal deposits.

A = F (A/F, 5%, 13) = 60,000 (0.05646) = $3,388

Answer: (a)


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%

Time Value of Money


Solution

ie = (1 + r/m)m – 1 = (1 + 0.12/12)12 – 1 = 0.127 = 12.7%

Answer: (c)


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

(a)5.5%(b) 5.8%(c)6.2%(d) 6.5%

Time Value of Money


Solution

P = A (P/A, i, 24)900 = 40 (P/A, i, 24)(P/A, i, n) = 22.50 Looking at the tables this is true for i approximately 0.5% per month

iyear = (1 + imonth)12 – 1 = (1 + 0.005)12 – 1 = 0.0617 = 6.17%

Answer: (c)


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

Time Value of Money Problems


Solution

EUAC = 750 + 50 (A/G, 6%, 8) = 750 + 50 (3.1952) = $910

Answer: (b)


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

Time Value of Money


Solution

A = 350 (F/P, 9%, 1) (A/P, 9%, 3) = 350 (1.09) (0.39505) = $151

Answer: (a)


Two alternatives are being considered:

Time Value of Money

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


Solution

EUAC = 55,000(A/P, 8%, 4) + 12,000 - 2,000(A/F, 8%, 4) = 55,000(.30192) + 12,000 - 2,000(0.22192) = $28,162

Answer: (a)


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

Time Value of Money


Solution

A = [89,000 (P/F, 14%, 1)} (A/P, 14%, 5) = {89,000 (0.8772)} (0.29128) = $22,740

Answer: (c)


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

Time Value of Money


Solution

P = 43,000 (P/A, 10%, 4) – 2,000 (P/G, 10%, 4) = 43,000 (3.1699) – 2,000 (4.3781) = $127,550

Answer: (c)


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%


Solution

iyear = (1 + 0.0005)365 – 1 = 0.2002 = 20.0%

Answer: (c)


Interest Rate A company is considering purchasing a new machine for $650,000 that will increase the firm’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


Solution

S = 650,000 (F/P, 15%, 5) – 150,000 (F/A, 10%, 5)

= 650,000(2.0114) – 150,000(6.7424) = $296,050

Answer: (b)


Comparison of AlternativesA 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 0

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


SolutionEUACA = 75,000(A/P, 15%, 4) - 16,000 - 9,000(A/F, 15%, 4)

= 75,000(.35027) - 16,000 - 9,000(0.20027)

= $8,468

EUACB = 8,468 = 105,000(A/P, 15%, n) - 24,000

(A/P, 15%, n) = (8,468 + 24,000)/105,000 = 0.3092

From table look-up, the value of n that most nearly makes the above relation true is 5.

Answer: (b)


BondsAn 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


SolutionSince 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(23.1148) + 20,000(0.3066) = $15,377

Answer: (a)


Benefit to Cost AnalysisA county is considering the following project

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

Initial Cost $22,500,000

Maintenance $525,000 per year

Savings $5,300,000 per year


SolutionPWcost = 22,500,000 + 525,000 (P/A, 8%, 12)

= 22,500,000 + 525,000 (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)


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


SolutionIn order that B/C = 1.0PWcost = PWbenefit PWbenefit = 150,000,000 + 25,000,000 (P/A, 10%, 8)

= 150,000,000 + 25,000,000 (5.3349) = $283,372,500

EUACbenefit = $283,372,500 (A/P, 10%, 8) = $283,372,500 (0.18744) = $53,115,341

$equivalent/crash = 53,115,341/9,200 = $5,773

Answer: (a)


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


SolutionCosts = 55,000 The payback period is the time when total income to date is equal to the total costs.

Costs = Income = 25,000 + 20,000 + 15,000 + …

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)


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


SolutionEUACA = 25,000 (A/P, 10%, n) + 4,350EUACB = 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 – 2,500) / 7,000 = 0.2643From table look-up, the value of n that most nearly makes the above relation true is 15.

Answer: (a)


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


SolutionP = (A/i) (P/F, 8%, 2) = (12,000/0.08) (0.8573) = (150,000) (0.8573) = $128,595

Answer: (d)


DepreciationA 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

(a) $13,000

(b) $15,000

(c) $17,000

(d) $19,000


Solution

Dt = D2 = (95,000 – 10,000)/5 = $17,000

Answer: (c)


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%


Solution

21,000 = 15,000 (1 + f)10

f = (21,000/15,000)0.1 – 1 = 0.034 = 3.4%

Answer: (c)

copyright kaplan aec education, 2008 engineering economics outline overview cash flow, p. 599 time...

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