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978-0073523439 Chapter 1
1 Solutions to end–of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 1 Foundations of Engineering Economy Basic Concepts 1.1 Financial units for economically best. 1.2 Morale, goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc. 1.3 Measure of […]
978-0073523439 Chapter 10 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 10 Project Financing and Noneconomic Attributes Working with MARR 10.1 The two primary sources of capital are debt and equity. Debt capital refers to capital obtained by borrowing from outside […]
978-0073523439 Chapter 10 Part 2
After–tax cost of debt = 10.4%(1 – 0.32) = 7.072% After–tax WACC = equity cost + debt cost = (4/15)(7.4%) + (6/15)(4.8%) + 5/15(7.072%) = 1.973 + 1.920 + 2.357 = 6.25% After-tax MARR = 6.25 + 12.5 = 18.75% […]
978-0073523439 Chapter 11 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 11 Replacement and Retention Decisions Foundations of Replacement 11.1 The defender refers to the currently–owned, in-place asset while the challenger refers to the equipment/process that is under consideration as its […]
978-0073523439 Chapter 11 Part 2
By spreadsheet: Select option A (Buy robot Y now) 11.36 (a) Two options; (1) upgrade and retain defender for 3 years, or (2) buy challenger now AW1 = – (40,000 + 79,000)(A/P,15%,3) – 85,000 + 30,000(A/F,15%,3) = -119,000(0.43798) – 85,000 […]
978-0073523439 Chapter 12 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 12 Independent Projects with Budget Limitation Capital Rationing and Independent Projects 12.1 1. Several independent projects are identified with cash flow estimates for each 2. Each project is selected entirely […]
978-0073523439 Chapter 12 Part 2
Copyright © 2018 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. 12 Linear Programming and Capital Budgeting 12.22 To develop the 0–1 ILP formulation, first calculate PWE, since it was not […]
978-0073523439 Chapter 13
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 13 Breakeven and Payback Analysis Breakeven Analysis for a Project 13.1 (a) 89x = 5,000,000 + 45x 44x = 5,000,000 x = 113,636 (b) At 100,000 units: Profit = revenue […]
978-0073523439 Chapter 14
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 14 Effects of Inflation Adjusting for Inflation 14.1 Inflated dollars are converted into constant-value dollars by dividing by one plus the inflation rate per period for however many periods are […]
978-0073523439 Chapter 15
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 15 Cost Estimation and Indirect Cost Allocation Understanding Cost Estimation 15.1 (1) scoping/feasibility, (2) order of magnitude, (3) partially designed, (4) design 60-100% complete, (5) detailed estimate. 15.2 Rent: AOC […]
978-0073523439 Chapter 16 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 16 Depreciation Methods Fundamentals of Depreciation 16.1 Depreciation is a tax-allowed deduction; it reduces the tax burden. 16.2 Unadjusted basis is the installed cost of an asset and includes all […]
978-0073523439 Chapter 16 Part 2
Additional Problems and FE Exam Review Questions 16.43 Answer is (b) 16.44 Depreciation is same for all years in straight line method. D3 = [100,000 – 0.15(100,000)]/5 = $17,000 Answer is (a) 16.45 D = 50,000 – 10,000 = $8000 […]
978-0073523439 Chapter 17 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 17 After-Tax Economic Analysis Terminology and Basic Tax Computations 17.1 NOI = Net Operating Income; GI = Gross Income; Te = Effective Tax Rate: NOPAT = Net Operating Profit After […]
978-0073523439 Chapter 17 Part 2
Alternative Y PWY = -13,000 + 4733(P/F,8%,1) + 5311(P/F,8%,2) + 3770(P/F,8%,3) + 385(P/F,8%,4) + 1200(P/F,8%,4) = $93 0 -13,000 – – – – -13,000 1 5000 4333 667 267 4733 2 5000 5779 -779 -311 5311 3 5000 1925 3075 […]
978-0073523439 Chapter 18 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 18 Sensitivity Analysis and Staged Decisions Sensitivity to Parameter Variation 18.1 $290,000: AW = -850,000(A/P,20%,5) + 290,000 = -850,000(0.33438) + 290,000 = $5777 (ROR > 20%) $325,000: AW = -850,000(A/P,20%,5) […]
978-0073523439 Chapter 18 Part 2
18.31 (a) The subscripts identify the series by probability. PW0.5 = –5000 + 1000(P/A,20%,3) = –5000 + 1000(2.1065) = $–2894 PW0.2 = –6000 + 500(P/F,20%,1) + 1500(P/F,20%,2) + 2000(P/F,20%,3) = –6000 + 500(0.8333) + 1500(0.6944) + 2000(0.5787) = $–3384 PW0.3 […]
978-0073523439 Chapter 19
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 19 More on Variation and Decision Making under Risk Certainty, Risk, and Uncertainty 19.1 (a) Discrete (b) Discrete (c) Continuous (d) Continuous (e) Discrete 19.2 (a) Continuous (assumed) and uncertain […]
978-0073523439 Chapter 2
Solutions to end–of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 2 Factors: How Time and Interest Affect Money Determination of F, P and A 2.1 (1) (F/P, 10%, 7) = 1.9487 (2) (A/P, 12%,10) = 0.17698 […]
978-0073523439 Chapter 3 Part 1
1 Solutions to end-of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 3 Combining Factors and Spreadsheet Functions Present Worth Calculations 3.1 P = 400,000(P/A,10%,15)(P/F,10%,1) = 400,000(7.6061)(0.9091) = $2,765,882 3.2 P = 40,000(P/F,12%,4) = 40,000(0.6355) = $25,420 […]
978-0073523439 Chapter 3 Part 2
3.47 P = 29,000 + 13,000(P/A,10%,3) + 13,000[7/(1 + 0.10)](P/F,10%,3) = 29,000 + 13,000(2.4869) + 82,727(0.7513) = $123,483 3.48 (a) Find P in year –1 and then move to year 5 P-1 = 210,000[6/(1 + 0.08)] = 210,000(0.92593) = $1,166,667 […]
978-0073523439 Chapter 4
Solutions to end–of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 4 Nominal and Effective Interest Rates Nominal and Effective Rates 4.1 (a) week (b) quarter (c) six months 4.2 APR is a nominal interest rate while […]
978-0073523439 Chapter 5
Solutions to end–of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 5 Present Worth Analysis Types of Projects 5.1 (a) The DN alternative (b) Each other 5.2 (a) Service alternatives all have the same revenues; revenue alternatives […]
978-0073523439 Chapter 6
Solutions to end–of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 6 Annual Worth Analysis Annual Worth and Capital Recovery Calculations 6.1 Multiply the FW values by (A/F,i%,n), where n is equal to the LCM or stated […]
978-0073523439 Chapter 7 Part 1
Solutions to end-of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 7 Rate of Return Analysis: One Project Understanding ROR 7.1 (a) Highest possible is infinity (b) Lowest possible is -100% 7.2 Total amount owed = principal […]
978-0073523439 Chapter 7 Part 2
Copyright © 2018 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. 12 FW3 = 3000(F/P,15%,2) + 15,000(F/P,15%,1) = 3000(1.3225) + 15,000(1.1500) = $21,218 Find i′ at which PW0 is equivalent to […]
978-0073523439 Chapter 8
Solutions to end–of–chapter problems Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 8 Rate of Return Analysis: Multiple Alternatives Understanding Incremental ROR 8.1 Alternative B is preferred if the rate of return on the increment of investment between […]
978-0073523439 Chapter 9 Part 1
Engineering Economy, 8th edition Leland Blank and Anthony Tarquin Chapter 9 Benefit/Cost Analysis and Public Sector Economics Understanding B/C Concepts 9.1 The primary purpose of public sector projects is to provide services for the public good at no profit. 9.2 […]
978-0073523439 Chapter 9 Part 2
BS = $144,000,000 Q vs. P: ∆B/C = (96 – 11)/(40 – 10) = 2.83 R vs. P: ∆B/C = (70 – 11)/(50 – 10) = 1.48 S vs. P: ∆B/C = (144 – 11)/(80 – 10) = 1.90 R […]