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12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR.

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Presentation on theme: "12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR."— Presentation transcript:

1 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

2 12/17/2015rd2 Sources of Capital Money Generated from the Operations of the Firm Profit Depreciation External Sources of Money Loans Mortgage Bonds Choice of Source of Funds

3 Preference of Capital Companies prefer Internal to External financing and debt to equity when external financing. You need to raise $1M. Debt or Equity or does it matter? Doesn't matter for fair valued assets. 12/17/2015rd3

4 Equity Financing Equity financing uses retained earnings raised from issuance of stock to finance capital investments. A company needs $10M and decides to sell its common stock. Current price is $30/share, but investment bankers feel the price of $28/share is better because of decreasing demand. Flotation costs (banker's fee, etching fee, lawyers’ fee) is 6% of selling price; thus $26.32 How many shares to sell to raise $10M? Let X be the number of shares sold. Flotation cost is 0.06 * 28 * X = 1.68X Sales proceeds – flotation cost = Net proceeds 28X – 1.68X = $10M => X = 379,940 shares Flotation cost for issuing common stock is 1.68 * 379,940 = $638,300. 12/17/2015rd4

5 Debt Financing Debt financing uses money raised through loans by an issuance of bonds to finance a capital investment. Task is to raise $10M by debt financing. Bond financing ~ floatation cost is 1.8% of the $10M issue with face value of $1000 sold at discount $985. Bond pays annual interest of 12%. Term loan ~ $10M bank loan secured at 11%/year for 5 years. How many $1000 par value bonds to raise $10M? 10,338.38 What is annual payments on bond and what is annual payment on loan? $1,240,605.60 To net $10M; $10M/(1 – 0.018) = $10,183,300 paying $183,300 in floatation costs. But bond sold at 1.5% discount, for bond financing Total number of bonds sold $10,183,300/$985 = 10,338.38 bonds Annual interest is $10,338,380 * 0.12 = $1,240,605.60 paid each year. Term loan ~ $10M(A/P, 11%, 5) = $2,705,703.10 annual payment. 12/17/2015rd5

6 MARR Factors Project risk ~ higher perceived risk, higher MARR Investment opportunity ~ MARR may be lowered to encourage investment. Flexibility is important. Tax structure ~ higher corporate taxes => higher MARR Limited capital ~ tends to increase MARR. Opportunity cost Market rates at other corporations ~ Keep in step. Before-tax MARR = (1 – tax rate) * after-tax MARR 12/17/2015rd6

7 12/17/2015rd7 Cost of Borrowed Money Interest Rate Prime Financial strength of borrower Duration Cost of Capital Common stock RoR Mortgage bonds Bank loans

8 Weighted Average Cost of Capital WACC= (equity fraction)(cost of equity capital) + (debt fraction)(cost of debt capital) where equity capital can be preferred stock, common stock, or retained earnings. 12/17/2015rd8

9 12/17/2015rd9 Example 15-1 RORAnnual $20 millionBank loan9% $1.8M 20Mortgage bonds7 1.4 60Common stock11 6.6 $100 million raised $9.8 M After Tax analysis: Assume tax rate at 40% Bank loan  0.09(1 – 0.4) = 5.4% Mortgage bonds  0.07(1 – 0.4) = 4.2% Dividends and retained earnings are not tax deductible. $20M(5.4%) + 20M(4.2%) + 60M(11%) = $8.52M Cost of capital = 8.52M/100M = 8.52%

10 12/17/2015rd10 Table 15-1 Budget $1.2M ProjectCost ($K)Estimated RoR % 115030 25045 35038 410040 520035 610028 720018 825025 930020 1030010 1140015 12Unlimited8 RoR's 45 40 38 35 30 28 25 20 18 15 10 8 Opportunity Cost

11 12/17/2015rd11 Selecting a MARR Cost of Borrowed Money Cost of Capital Opportunity Cost MARR should be the largest rate of the above costs. Now we need to hedge for uncertainty in the estimates. Probability => Risk

12 Problem 15-2 ABC First cost$100$50$25 UAB 16.27 9.96 5.96 IRR 10% 15% 20% Express the three mutually exclusive alternatives with10-year lives over an unspecified MARR. B-C  (UIRR 25 4 10)  9.61% A-B  (UIRR 50 6.31 10)  4.47% A-C  (UIRR 75 10.31 10)  6.25% 0 20%Choose Do Nothing 12/17/2015rd12

13 12/17/2015rd13 Example 15-3 n01-1011-20 14.05% 10% 15% A-801020 0 28.83 -5 NPW B-8013.8610 0 28.83 1.97 NPW 15.48% B – A $6.97 At a MARR of 10% both are 28.83, both equally desirable, but B is believed to have greater risk. Thus choose A. At MARR 15%, A has negative return, but B is positive; thus choose B.

14 12/17/2015rd14 Problem 15-7 Budget = $70K Project First CostBenefitIRR (%) 1$20K$11K (UIRR 20 11 3 0)  29.92 2 30K 14K(UIRR 30 14 3 0)  18.91 3 10K 6K(UIRR 10 6 3 0)  36.31 4 5K 2.4K(UIRR 5 2.4 3 0)  20.71 5 25K 13K(UIRR 25 13 3 0)  26.01 6 15K 7K(UIRR 15 7 3 0)  18.91 7 40K 21K(UIRR 40 21 3 0)  26.67 3 1 7 5 4 2 6 10 20 40 25 5 30 15 The opportunity cost of capital is (first reject project 5) at 26.1%

15 Problem 15-8 with $500K Budget ProjectFirst CostUABLifeIRR 1 2$200K$50K15(UIRR 200 50 15)  24.00% 2 5 300K 70K10(UIRR 300 70 10)  19.36 3 1 100K 40K 5(UIRR 100 40 5)  28.65 4 3 50K 12.5K10(UIRR 50 12.5 10)  21.41 5 7 250K 75K 5(UIRR 250 75 5)  15.24 6 4 150K 32K20(UIRR 150 32 20)  20.85 7 6 400K125K 5(UIRR 400 125 5)  16.99 12/17/2015rd15

16 12/17/2015rd16 ProjectFirst CostUABLifeIRR E40,00011,933515.00% E2 = RATE(life, UAB, -first cost) H60,00012692813.44% C30,0009878412.00% G35,0006794810.97% I75,00014058810.00% B20,000617349.00% D25,000626158.00% MARR A15,000442947.01% F50,00011,55055.00% Budget =260,000 for first 6 projects. Problem 15-9

17 Problem 16-11 Parabolic Benefit/Cost equation: PWB 2 – 22PWC + 44 = 0; find PWC for optimal size project. Let y = PWB and x = PWC. Then y 2 - 22x + 44 = 0; Solve for slope and set slope to 1. 2yy' - 22 = 0; y' = 11/y = 1 => y = 11 and x = 7.5 11 2 -22x +44 = 0 => x = 7.5 = PWC 12/17/2015rd17

18 Problem 16-26 Conventional B/C and incremental analysis C-A 12/17/2015rd18 MARR = 12%ABC First Cost$9500$18,500$22,000 Annual savings 320050009800 Annual costs 100027506400 Salvage value 6000420014000 Life 15 A ***BC PW numerator$21,794.77$34,054.32$66,746.47 PW denominator 152,14.69 36,462.5563,031.79 B/C ratio1.430.931.06

19 Determining the MARR Suppose cost of capital is 10% (borrowing rate) and lending rate is 6% (opportunity cost) with budget at) $40K b) $60K and c) $0K, determine MARR using ranked projects by their IRR. a) With $40K budget, invest in 1,2,3,4. MARR = 8%. b) With $60K, invest in projects 1,2,3,4,5. Lend $10K at 6%. MARR = 6%. c) Borrow for projects 1 & 2. MARR = 10% lending < MARR < borrowing 12/17/2015rd19 ProjectFirst Cost UABIRR (%) 110K$12K20 210K$11.5K15 310K$11K10 410K$10.8K8 510K$10.7K7 610K$10.4K4

20 Example Calculate the after-tax cost of debt for the following: a)Interest rate is 12%; tax rate is 25% ¾ * 12 = 9% b)Interest rate is 14%; tax rate is 34% 0.66 *14 = 9.24% c) Interest rate is 15%; tax rate is 40% 0.6 * 15 = 9% 12/17/2015rd20

21 Example NABCD 0-$2000-$3000-$1000 1100040001400-$1000 21000-1001090 31000 IRR (%)23.3833.3332.459 12/17/2015rd21 With budget at $3500 and lend out remaining funds at 10%, and goal is to maximize future worth at n = 3. FW A (10%) = $648, FW B (10%) = $847; FW C (10%) = $190.08; FW D (10%) = -$11 A & C for $838 + 500(F/P, 10%, 3) = $1503.58 B  847 + 500(F/P, 10%, 3) = $1512.50

22 Example You need $10M in capital. Capital stock sales  $5M at 13.7% Use of retained earnings  $2M at 8.9% Debt financing with bonds  $3M at 7.5%. Historical D-E mix of 40% from debt costing 7.5% and 60% from equity costing 10%. Calculate historical WACC and current WACC. Historical: 0.6(10) + 0.4(7.5) = 9% Current: (5/10)(13.7) + (2/10)(8.9) + (3/10)(7.5) = 10.88%. After-tax analysis is appropriate. 12/17/2015rd22

23 Example Debt Capital You need $10M in debt capital by issuing 5,000 $1K 10-year bonds paying 8%/year with 50% tax rate. Bonds are discounted 2% for quick sale. Ignore flotation costs. Compute cost of debt capital before and after taxes. 980 = 80(P/A, i%, 10) + 1000(P/F, i%, 10) (UIRR 980 80 10 1000)  8.3% Before tax (UIRR 980 40 10 1000)  4.24% After-tax 12/17/2015rd23

24 Debt Capital You buy a $20K 10-year asset by putting $10K down and borrowing $10K at 6%/year by paying the interest each year and the $10K in year 10. Tax rate is 42%. Compute after tax cost of debt capital. Deduction for interest is $600 at tax rate 42% leaving ATCF 600(1 – 0.42) = $348 (UIRR 10000 348 10 10000)  3.48%. 12/17/2015rd24

25 Inflation 1.A machine costing $2550 4 years ago now costs $3930 with general inflation at 7% per year. Calculate the true percentage increase in the cost of the machine. a) 14.95% b) 54.12% c) 35.11% d) 7% e) 17.58% 2.If you want a 7% inflation-free return on your investment with f = 9% per year, your actual interest rate must be a)16% b) 20% c) 12% d) 15% e) 14% 3.I want $25K per year forever in R$ when I die for my family. Insurer offers 7% per year while inflation is expected to be 4% per year. First payment is 1 year after my death. How much insurance do I need? a) $866K b) $357K c) $625K d) $841K 12/17/2015rd25

26 Cost of Capital 250M shares priced at $29.30 20M preferred stock priced at $40.50 150M selling at $101.75 per 100 500M loan at 4.5% interest. 12/17/2015rd26 SourceAmountPriceTotal Weight Common stock 250M29.30$7,235E683.36% Preferred stock 20M40.50$810M9.22% Bond150M$1.0175$152,625K1.74% Loan$500M $1.00$500M5.69% $8.787.625100%


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