ENMG 400 Problems, Spring 2011

Chapter 1 1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 1 Foundations of Engineering Economy Solutions to Problems 1.1 Time value of money means that there is a certain worth in having money and the worth changes as a function of time. 1.2 Morale, goodwill, friendship, convenience, aesthetics, etc. 1.3 (a) Evaluation criterion is the measure of value that is used to identify “best”. (b) The primary evaluation criterion used in economic analysis is cost. 1.4 Nearest, tastiest, quickest, classiest, most scenic, etc 1.5 If the alternative that is actually the best one is not even recognized as an alternative, it obviously will not be able to be selected using any economic analysis tools. 1.6 In simple interest, the interest rate applies only to the principal, while compound interest generates interest on the principal and all accumulated interest. 1.7 Minimum attractive rate of return is the lowest rate of return (interest rate) that a company or individual considers to be high enough to induce them to invest their money. 1.8 Equity financing involves the use of the corporation’s or individual’s own funds for making investments, while debt financing involves the use of borrowed funds. An example of equity financing is the use of a corporation’s cash or an individual’s savings for making an investment. An example of debt financing is a loan (secured or unsecured) or a mortgage. 1.9 Rate of return = (45/966)(100) = 4.65% 1.10 Rate of increase = [(29 – 22)/22](100) = 31.8% 1.11 Interest rate = (275,000/2,000,000)(100) = 13.75% 1.12 Rate of return = (2.3/6)(100) = 38.3% Chapter 1 2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1.13 Profit = 8(0.28) = $2,240,000 1.14 P + P(0.10) = 1,600,000 1.1P = 1,600,000 P = $1,454,545 1.15 Earnings = 50,000,000(0.35) = $17,500,000 1.16 (a) Equivalent future amount = 10,000 + 10,000(0.08) = 10,000(1 + 0.08) = $10,800 (b) Equivalent past amount: P + 0.08P = 10,000 1.08P = 10,000 P = $9259.26 1.17 Equivalent cost now: P + 0.1P = 16,000 1.1P = 16,000 P = $14,545.45 1.18 40,000 + 40,000(i) = 50,000 i = 25% 1.19 80,000 + 80,000(i) = 100,000 i = 25% 1.20 F = 240,000 + 240,000(0.10)(3) = $312,000 1.21 Compound amount in 5 years = 1,000,000(1 + 0.07)5 = $1,402,552 Simple amount in 5 years = 1,000,000 + 1,000,000(0.075)(5) = $1,375,000 Compound interest is better by $27,552 1.22 Simple: 1,000,000 = 500,000 + 500,000(i)(5) i = 20% per year simple Compound: 1,000,000 = 500,000(1 + i)5 (1 + i)5 = 2.0000 (1 + i) = (2.0000)0.2 i = 14.87% Chapter 1 3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1.23 Simple: 2P = P + P(0.05)(n) P = P(0.05)(n) n = 20 years Compound: 2P = P(1 + 0.05)n (1 + 0.05)n = 2.0000 n = 14.2 years 1.24 (a) Simple: 1,300,000 = P + P(0.15)(10) 2.5P = 1,300,000 P = $520,000 (b) Compound: 1,300,000 = P(1 + 0.15)10 4.0456P = 1,300,000 P = $321,340 1.25 Plan 1: Interest paid each year = 400,000(0.10) = $40,000 Total paid = 40,000(3) + 400,000 = $520,000 Plan 2: Total due after 3 years = 400,000(1 + 0.10)3 = $532,400 Difference paid = 532,400 – 520,000 = $12,400 1.26 (a) Simple interest total amount = 1,750,000(0.075)(5) = $656,250 Compound interest total = total amount due after 4 years – amount borrowed = 1,750,000(1 + 0.08)4 – 1,750,000 = 2,380856 – 1,750,000 = $630,856 (b) The company should borrow 1 year from now for a savings of $656,250 – $630,856 = $25,394 1.27 The symbols are F = ?; P = $50,000; i = 15%; n = 3 1.28 (a) FV(i%,n,A,P) finds the future value, F (b) IRR(first_cell:last_cell) finds the compound interest rate, i (c) PMT(i%,n,P,F) finds the equal periodic payment, A (d) PV(i%,n,A,F) finds the present value, P Chapter 1 4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1.29 (a) F = ?; i = 7%; n = 10; A = $2000; P = $9000 (b) A = ?; i = 11%; n = 20; P = $14,000; F = 0 (c) P = ?; i = 8%; n = 15; A = $1000; F = $800 1.30 (a) PV = P (b) PMT = A (c) NPER = n (d) IRR = i (e) FV = F 1.31 For built-in Excel functions, a parameter that does not apply can be left blank when it is not an interior one. For example, if there is no F involved when using the PMT function to solve a particular problem, it can be left blank because it is an end function. When the function involved is an interior one (like P in the PMT function), a comma must be put in its position. 1.32 (a) Risky (b) Safe (c) Safe (d) Safe (e) Risky 1.33 (a) Equity (b) Equity (c) Equity (d) Debt (e) Debt 1.34 Highest to lowest rate of return is as follows: Credit card, bank loan to new business, corporate bond, government bond, interest on checking account 1.35 Highest to lowest interest rate is as follows: rate of return on risky investment, minimum attractive rate of return, cost of capital, rate of return on safe investment, interest on savings account, interest on checking account. 1.36 WACC = (0.25)(0.18) + (0.75)(0.10) = 12% Therefore, MARR = 12% Select the last three projects: 12.4%, 14%, and 19% 1.37 End of period convention means that the cash flows are assumed to have occurred at the end of the period in which they took place. 1.38 The following items are inflows: salvage value, sales revenues, cost reductions The following items are outflows: income taxes, loan interest, rebates to dealers, accounting services Chapter 1 5 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1.39 The cash flow diagram is: 1.40 The cash flow diagram is: 1.41 Time to double = 72/8 = 9 years 1.42 Time to double = 72/9 = 8 years Time to quadruple = (8)(2) = 16 years 1.43 4 = 72/i i = 18% per year 1.44 Account must double in value five times to go from $62,500 to $2,000,000 in 20 years. Therefore, account must double every 20/5 = 4 years. Required rate of return = 72/4 = 18% per year FE Review Solutions 1.45 Answer is (c) 1.46 2P = P + P(0.05)(n) n = 20 Answer is (d) 0 1 2 3 4 5 6 7 8 F= ? $9000 $3000 $10,000 i = 10% 0 1 2 3 4 5 P = ? $40,000 i = 15% Chapter 1 6 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1.47 Amount now = 10,000 + 10,000(0.10) = $11,000 Answer is (c) 1.48 i = 72/9 = 8 % Answer is (b) 1.49 Answer is (c) 1.50 Let i = compound rate of increase: 235 = 160(1 + i)5 (1 + i)5 = 235/160 (1 + i) = (1.469)0.2 (1 + i) = 1.07995 i = 7.995% = 8.0% Answer is (c) Chapter 1 7 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Extended Exercise Solution F = ? 1. $2000 0 1 2 3 4 $500 $500 $500 $9000 F = [{[–9000(1.08) – 500] (1.08)} – 500] (1.08) + (2000–500) = $–10,960.60 or F = –9000(F/P,8%,3) – 500(F/A,8%,3) + 2000 2. A spreadsheet uses the FV function as shown in the formula bar. F = $–10,960.61. Chapter 1 8 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3. F = [{[–9000(1.08) – 300] (1.08)} – 500] (1.08) + (2000 –1000) = $–11,227.33 Change is 2.02%. Largest maintenance charge is in the last year and, therefore, no compound interest is accumulated by it. 4. The fastest method is to use the spreadsheet function: FV(12.32%,3,500,9000) + 2000 It displays the answer: F = $–12,445.43 Case Study Solution There is no definitive answer to the case study exercises. The following are examples only. 1. The first four steps are: Define objective, information collection, alternative definition and estimates, and criteria for decision-making. Objective: Select the most economic alternative that also meets requirements such as production rate, quality specifications, manufacturability for design specifications, etc. Information: Each alternative must have estimates for life (likely 10 years), AOC and other costs (e.g., training), first cost, any salvage value, and the MARR. The debt versus equity capital question must be addressed, especially if more than $5 million is needed. Alternatives: For both A and B, some of the required data to perform an analysis are: P and S must be estimated. AOC equal to about 8% of P must be verified. Training and other cost estimates (annual, periodic, one-time) must be finalized. Confirm n = 10 years for life of A and B. MARR will probably be in the 15% to 18% per year range. Criteria: Can use either present worth or annual worth to select between A and B. Chapter 1 9 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2. Consider these and others like them: Debt capital availability and cost Competition and size of market share required Employee safety of plastics used in processing 3. With the addition of C, this is now a make/buy decision. Economic estimates needed are:  Cost of lease arrangement or unit cost, whatever is quoted.  Amount and length of time the arrangement is available. Some non-economic factors may be:  Guarantee of available time as needed.  Compatibility with current equipment and designs.  Readiness of the company to enter the market now versus later. Chapter 10 1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 10 Making Choices: the Method, MARR, and Multiple Attributes Solutions to Problems 10.1 The circumstances are when the lives for all alternatives are: (1) finite and equal, or (2) considered infinite. It is also correct when (3) the evaluation will take place over a specified study period. 10.2 Incremental cash flow analysis is mandatory for the ROR method and B/C method. (It is noteworthy that if unequal-life cash flows are evaluated by ROR using an AW-based relation that reflects the differences in cash flows between two alternatives, the breakeven i* will be the same as the incremental i*. (See Table 10.2 and Section 10.1 for comments.) 10.3 Numerically largest means the alternative with the largest PW, AW or FW identifies the selected alternative. For both revenue and service alternatives, the largest number is chosen. For example, $-5000 is selected over $-10,000, and $+100 is selected over $-50. 10.4 (a) Hand solution: After consulting Table 10.1, choose the AW or PW method at 8% for equal lives of 8 years. Computer solution: either the PMT function or the PV function can give single-cell solutions for each alternative. In either case, select the alternative with the numerically largest value of AW or PW. (b) (1) Hand solution: Find the PW for each cash flow series. PW8 = -10,000 + 2000(P/F,18%,8) + (6500 – 4000) (P/A,18%,8) = -10,000 + 2000(0.2660) + 2500(4.0776) = $726 PW10 = -14,000 + 2500(P/F,18%,8) + (10,000 – 5500) (P/A,18%,8) = $5014 PW15 = -18,000 + 3000(P/F,18%,8) + (14,000 – 7000) (P/A,18%,8) = $11,341 Chapter 10 2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. PW20 = -24,000 + 3500(P/F,18%,8) + (20,500 – 11,000) (P/A,18%,8) = $15,668 PW25 = -33,000 + 6000(P/F,18%,8) + (26,500 – 16,000) (P/A,18%,8) = $11,411 Select the 20 cubic meter size. Computer solution: Use the PV function to find the PW in a separate spreadsheet cell for each alternative. Select the 20 cubic meter alternative. (b) (2) Buy another 20 cubic meter truck, not a smaller size, because it is always correct to spend the largest amount that is economically justified. 10.5 (a) Hand solution: Choose the AW or PW method at 0.5% for equal lives over 60 months. Computer solution: Either the PMT function or the PV function can give single-cell solutions for each alternative. (b) The B/C method was the evaluation method in chapter 9, so rework it using AW. Hand solution: Find the AW for each cash flow series on a per household per month basis. Chapter 10 3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. AW1 = 1.25 – 60(A/P,0.5%,60) = 1.25 – 60(0.01933) = 1.25 - 1.16 = $0.09 AW2 = 8.00 - 500(A/P,0.5%,60) = 8.00 – 9.67 = $-1.67 Select program 1. Computer solution: Develop the AW value using the PMT function in a separate cell for each program. Select program 1. 10.6 Long to infinite life alternatives. Examples are usually public sector projects such as dams, highways, buildings, railroads, etc. 10.7 (a) The expected return is 12 - 8 = 4% per year. (b) Retain MARR = 12% and then estimate the project i*. Take the risk-related return expectation into account before deciding on the project. If 12% < i* < 17%, John must decide if the risk is worth less than 5% over MARR = 12%. 10.8 (a) Bonds are debt financing (b) Stocks are always equity (c) Equity (d) Equity loans are debt financing, like house mortgage loan Chapter 10 4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.9 The project that is rejected, say B, and has the next highest ROR measure, i*B, in effect sets the MARR, because it‟s rate of return is a lost opportunity rate of return. Were any second alternative selected, project B would be it and the effective MARR would be i*B. 10.10 Before-tax opportunity cost is the 16.6% forgone rate. Determine the after-tax percentage after the effective tax rate (Te ) is calculated. Te = 0.06 + (0.94)(0.20) = 0.248 After-tax MARR = Before-tax MARR (1- Te ) = 16.6 (1 – 0.248) = 12.48% 10.11 (a) Select 2. It is the alternative investing the maximum available with incremental i* > 9%. (b) Select 3. (c) Select 3. (d) MARR = 10% for alternative 4 is opportunity cost at $400,000 level, since 4 is the first unfunded project due to unavailability of funds. 10.12 Set the MARR at the cost of capital. Determine the rate of return for the cash flow estimates and select the best alternative. Examine the difference between the return and MARR to separately determine if it is large enough to cover the other factors for this selected alternative. (This is different than increasing the MARR before the evaluation to accommodate the factors.) 10.13 (a) MARR may tend to be set lower, based on the success of the last purchase. (b) Set the MARR and then treat the risk associated with the purchase separately from the MARR. 10.14 (a) Calculate the two WACC values. WACC1 = 0.6(12%) + 0.4 (9%) = 10.8% WACC2 = 0.2(12%) + 0.8(12.5%) = 12.4% Use approach 1, with a D-E mix of 40%-60% Chapter 10 5 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.14 (cont) (b) Let x1 and x2 be the maximum costs of debt capital. Alternative 1: 10% = WACC1 = 0.6(12%) + 0.4(x1) x1 = [10% - 0.6(12%)]/0.4 = 7% Debt capital cost would have to decrease from 9% to 7%. Alternative 2: 10% = WACC2 = 0.2(12%) + 0.8(x2) x2 = [10% - 0.2(12%)]/0.8 = 9.5% Debt capital cost would, again, have to decrease; now from 12.5% to 9.5% Chapter 10 6 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.15 The lowest WACC value of 7.9% occurs at the D-E mixes of $10,000 and $25,000 loan. This translates into funding between $75,000 and $90,000 from their own funds. 10.16 WACC = cost of debt capital + cost of equity capital = (0.4)[0.667(8%) + 0.333(10%)] + (0.6)[(0.4)(5%) + (0.6)(9%)] = 0.4[8.667%] + 0.6 [7.4%] = 7.907% 10.17 (a) Compute and plot WACC for each D-E mix. D-E Mix WACC 100-0 14.50% 70-30 11.44 65-35 10.53 50-50 9.70 35-65 9.84 20-80 12.48 0-100 12.50 (b) D-E mix of 50%-50% has the lowest WACC value. Chapter 10 7 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.18 (a) The spreadsheet shows a 50% - 50% mix to have the lowest WACC at 9.70%. (b) Change the debt rate column (C to D) to add the 10% and observe the new plot. Now debt of 35% (D-E of 35-65) has the lowest WACC = 10.18%. Chapter 10 8 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.19 Solve for the cost of debt capital, x. WACC = 10.7% = 0.8(6%) + (1-0.8) (x) x = (10.7 – 4.8)/0.2 = 29.5% The rate of 29.5% for debt capital (loans, bonds, etc.) seems very high. 10.20 Before-taxes: WACC = 0.4(9%) + 0.6(12%) = 10.8% per year After-tax: Insert Equation [10.3] into the before-tax WACC relation. After-tax WACC = (equity)(equity rate) + (debt)(before-tax debt rate)(1–Te ) = 0.4(9%) + 0.6(12%)(1-0.35) = 8.28% per year The tax advantage reduces the WACC from 10.8% to 8.28% per year, or 2.52% per year. 10.21 (a) Face value = $2,500,000 = $2,577,320 0.97 (b) Bond interest = 0.042(2,577,320) = $27,062 every 3 months 4 Dividend quarterly net cash flow = $27,062(1 - 0.35) = $17,590 The rate of return equation per 3-months over 20(4) quarters is: 0 = 2,500,000 – 17,590(P/A,i*,80) – 2,577,320(P/F,i*,80) i* = 0.732% per 3 months (RATE function) Nominal i* = 2.928% per year Effective i* = (1.00732)4 – 1 = 2.96% per year 10.22 (a) Annual loan payment is the cost of the $160,000 debt capital. First, determine the after-tax cost of debt capital. Chapter 10 9 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.22 (cont) Debt cost of capital: before-tax (1-Te) = 9%(1-0.22) = 7.02% Annual interest 160,000(0.0702) = $11,232 Annual principal re-payment = 160,000/15 = $10,667 Total annual payment = $21,899 (b) Equity cost of capital: 6.5% per year on $40,000 is $2600 annually. Set up the spreadsheet with the three series. Equity rate is 6.5%, loan interest rate is 7.02%, and principle re-payment rate is 6.5% since the annual amount will not earn interest at the equity rate of 6.5%. The difference in PW values is: Difference = 200,000 - PW equity lost – PW of loan interest paid – PW of loan principle re-payment not saved as equity = $-26, 916 This means the PW of the selling price in the future must be at least $26,916 more than the current purchase price to make a positive return on the investment, assuming all the current numbers remain stable. Chapter 10 10 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. (c) After-tax WACC = 0.2(6.5%) + 0.8(9%(1-0.22) = 6.916% 10.23 Equity cost of capital is stated as 6%. Debt cost of capital benefits from tax savings. Before-tax bond annual interest = 4 million (0.08) = $320,000 Annual bond interest NCF = 320,000(1 – 0.4) = $192,000 Effective quarterly dividend = 192,000/4 = $48,000 Find quarterly i* using a PW relation. 0 = 4,000,000 - 48,000(P/A,i*,40) - 4,000,000(P/F,i*,40) i* = 1.2% per quarter = 4.8% per year (nominal) Debt financing at 4.8% per year is cheaper than equity funds at 6% per year. (Note: The correct answer is also obtained if the before-tax debt cost of 8% is used to estimate the after-tax debt cost of 8%(1 - 0.4) = 4.8% from Equation [10.3].) 10.24 (a) Bank loan: Annual loan payment = 800,000(A/P,8%,8) = 800,000(0.17401) = $139,208 Principal payment = 800,000/8 = $100,000 Annual interest = 139,208 – 100,000 = $39,208 Tax saving = 39,208(0.40) = $15,683 Effective interest payment = 39,208 – 15,683 = $23,525 Effective annual payment = 23,525 + 100,000 = $123,525 The AW-based i* relation is: 0 = 800,000(A/P,i*,8) –123,525 (A/P,i*,8) = 123,525 = 0.15441 800,000 i* = 4.95% Chapter 10 11 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Bond issue: Annual bond interest = 800,000(0.06) = $48,000 Tax saving = 48,000(0.40) = $19,200 Effective bond interest = 48,000 – 19,200 = $28,800 The AW-based i* relation is: 0 = 800,000(A/P,i*,10) - 28,800 - 800,000(A/F,i*,10) i* = 3.6% (RATE or IRR function) Bond financing is cheaper. (b) Bonds cost 6% per year, which is less than the 8% loan. The answer is the same before-taxes. 10.25 Face value of bond issue = (10,000,000)/ 0.975 = $10,256,410 Annual bond interest = 0.0975(10,256,410) = $1,000,000 Interest net cash flow = $1,000,000(1 - 0.32) = $680,000 The PW-based rate of return equation is: 0 = 10,000,000 – 680,000(P/A,i*,30) –10,256,410(P/F,i*,30) i* = 6.83% per year (Excel RATE function) Bonds are cheaper than the bank loan at 7.5% with no tax advantage. 10.26 Dividend method: Re = DV1/P + g = 0.93/18.80 + 0.015 = 6.44% Chapter 10 12 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CAPM: (The return values are in percents.) Re = Rf + β(Rm - Rf) = 4.5 + 1.19(4.95 – 4.5) = 5.04% CAPM estimate of cost of equity capital is 1.4% lower. 10.27 Debt capital cost: 9.5% for $6 million (60% of total capital) Equity -- common stock: 100,000(32) = $3.2 million or 32% of total capital Re = 1.10/ 32 + 0.02 = 5.44% Equity -- retained earnings: cost is 5.44% for this 8% of total capital. WACC = 0.6(9.5%) + 0.32(5.44%) + 0.08(5.44%) = 7.88% 10.28 Last year CAPM computation: Re = 4.0 + 1.10(5.1 – 4.0) = 4.0 + 1.21 = 5.21% This year CAPM computation: Re = 3.9 + 1.18(5.1 – 3.9) = 3.9 + 1.42 = 5.32% Equity costs slightly more in part because the company‟s stock became more volatile based on an increase in beta. The safe return rate stayed about the same in the switch from US to Euro bonds. 10.29 Determine the effective annual interest rate ia for each plan using the effective interest rate equation in chapter 4. All the dollar values can be neglected. Plan 1: ia for debt = (1 + 0.00583)12 -1 = 7.225% ia for equity = (1 + 0.03)2 - 1 = 6.09% Chapter 10 13 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. WACCA = 0.5(7.225%) + 0.5(6.09%) = 6.66% Plan 2: ia for 100% equity = WACCB = (1 + 0.03)2 - 1 = 6.09% Plan 3: ia for 100% debt = WACCC = (1 + 0.00583)12 -1 = 7.225% Plan 2: 100% equity has the lowest before-tax WACC. 10.30 (a) Equity capital: 50% of capital at 15% per year. Debt capital: 15% in bonds and 35% in loans. Cost of loans: 10.5% per year Cost of bonds: 6% from the problem statement, or determine i*. Bond annual interest per bond = $10,000(0.06) = $600 0 = 10,000 - 600(P/A,i*,15) - 10,000(P/F,i*,15) i* = 6.0% (RATE function) Before-tax WACC = 0.5(15%) + 0.15(6%) + 0.35(10.5%) = 12.075% (b) Use Te = 35% to calculate after-tax WACC with Equation [10.3] inserted into Equation [10.1], as mentioned at the end of Section 10.3 in the text. After-tax WACC = (equity)(equity rate) + (debt)(before-tax debt rate)(1–Te ) = 0.5(15%) + [0.15(6%) + 0.35(