26
06/23/22 06/23/22 rd rd 1 Engineering Economic Engineering Economic Analysis Analysis Chapter 15 Selection of a MARR

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

Embed Size (px)

Citation preview

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

04/21/2304/21/23 rdrd 11

Engineering Economic AnalysisEngineering Economic Analysis

Chapter 15 Selection of a MARR

Page 2: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 22

Sources of CapitalSources of Capital

Money Generated from the Operations of the Firm

Profit

Depreciation

External Sources of Money

Loans

Mortgage Bonds

Choice of Source of Funds

Page 3: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Preference of CapitalPreference 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.

04/21/2304/21/23 rdrd 33

Page 4: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Equity FinancingEquity FinancingEquity 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.

04/21/2304/21/23 rdrd 44

Page 5: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Debt FinancingDebt FinancingDebt 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.

04/21/2304/21/23 rdrd 55

Page 6: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

MARR FactorsMARR 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

04/21/2304/21/23 rdrd 66

Page 7: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 77

Cost of Borrowed MoneyCost of Borrowed Money

Interest Rate Prime Financial strength of borrower Duration

Cost of Capital

Common stock RoR Mortgage bonds Bank loans

Page 8: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Weighted Average Cost of CapitalWeighted 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.

04/21/2304/21/23 rdrd 88

Page 9: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 99

Example 15-1Example 15-1

ROR Annual

$20 million Bank loan 9% $1.8M 20 Mortgage bonds 7 1.4 60 Common stock 11 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%

Page 10: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 1010

Table 15-1 Budget $1.2MTable 15-1 Budget $1.2M

Project Cost ($K) Estimated RoR %

1 150 30

2 50 45

3 50 38

4 100 40

5 200 35

6 100 28

7 200 18

8 250 25

9 300 20

10 300 10

11 400 15

12 Unlimited 8

RoR's

45

40

38

35

30

28

25

20

18

15

10

8

Opportunity Cost

Page 11: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 1111

Selecting a MARRSelecting 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

Page 12: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Problem 15-2Problem 15-2

A B C

First cost $100 $50 $25UAB 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 < MARR < 4.47 Choose A4.47 < MARR < 9.61 Choose B9.61 < MARR < 20 Choose CMARR > 20% Choose Do Nothing

04/21/2304/21/23 rdrd 1212

Page 13: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 1313

Example 15-3Example 15-3

n 0 1-10 11-20 14.05% 10% 15%A -80 10 20 0 28.83 -5 NPWB -80 13.86 10 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.

Page 14: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 1414

Problem 15-7Problem 15-7

Budget = $70KProject First Cost Benefit IRR (%)

1 $20K $11K (UIRR 20 11 3 0) 29.922 30K 14K (UIRR 30 14 3 0) 18.913 10K 6K (UIRR 10 6 3 0) 36.314 5K 2.4K (UIRR 5 2.4 3 0) 20.715 25K 13K (UIRR 25 13 3 0) 26.016 15K 7K (UIRR 15 7 3 0) 18.917 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%

Page 15: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Problem 15-8 with $500K BudgetProblem 15-8 with $500K Budget

Project First Cost UAB Life IRR

1 2 $200K $50K 15 (UIRR 200 50 15) 24.00%

2 5 300K 70K 10 (UIRR 300 70 10) 19.36

3 1 100K 40K 5 (UIRR 100 40 5) 28.65

4 3 50K 12.5K 10 (UIRR 50 12.5 10) 21.41

5 7 250K 75K 5 (UIRR 250 75 5) 15.24

6 4 150K 32K 20 (UIRR 150 32 20) 20.85

7 6 400K 125K 5 (UIRR 400 125 5) 16.99

04/21/2304/21/23 rdrd 1515

Page 16: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

04/21/2304/21/23 rdrd 1616

Project First Cost UAB Life IRR

E 40,000 11,933 5 15.00% E2 = RATE(life, UAB, -first cost)

H 60,000 12692 8 13.44%

C 30,000 9878 4 12.00%

G 35,000 6794 8 10.97%

I 75,000 14058 8 10.00%

B 20,000 6173 4 9.00%

D 25,000 6261 5 8.00% MARR

A 15,000 4429 4 7.01%

F 50,000 11,550 5 5.00%

Budget = 260,000 for first 6 projects.

Problem 15-9

Page 17: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Problem 16-11Problem 16-11

Parabolic Benefit/Cost equation:

PWB2 – 22PWC + 44 = 0; find PWC for optimal size project.

Let y = PWB and x = PWC.

Then y2 - 22x + 44 = 0; Solve for slope and set slope to 1.

2yy' - 22 = 0;

y' = 11/y = 1 => y = 11 and x = 7.5

112 -22x +44 = 0 => x = 7.5 = PWC

04/21/2304/21/23 rdrd 1717

Page 18: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Problem 16-26Problem 16-26

Conventional B/C and incremental analysis C-A

04/21/2304/21/23 rdrd 1818

MARR = 12% A B C

First Cost $9500 $18,500 $22,000

Annual savings 3200 5000 9800

Annual costs 1000 2750 6400

Salvage value 6000 4200 14000

Life 15 15 15

A *** B C

PW numerator $21,794.77 $34,054.32 $66,746.47

PW denominator 152,14.69 36,462.55 63,031.79

B/C ratio 1.43 0.93 1.06

Page 19: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Determining the MARRDetermining the MARRSuppose 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

04/21/2304/21/23 rdrd 1919

Project First Cost

UAB IRR (%)

1 10K $12K 20

2 10K $11.5K 15

3 10K $11K 10

4 10K $10.8K 8

5 10K $10.7K 7

6 10K $10.4K 4

Page 20: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

ExampleExample

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%

04/21/2304/21/23 rdrd 2020

Page 21: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

ExampleExample

N A B C D

0 -$2000 -$3000 -$1000

1 1000 4000 1400 -$1000

2 1000 -100 1090

3 1000

IRR (%) 23.38 33.33 32.45 9

04/21/2304/21/23 rdrd 2121

With budget at $3500 and lend out remaining funds at 10%, and goal is to maximize future worth at n = 3.FWA(10%) = $648, FWB(10%) = $847; FWC(10%) = $190.08; FWD(10%) = -$11 A & C for $838 + 500(F/P, 10%, 3) = $1503.58

B 847 + 500(F/P, 10%, 3) = $1512.50

Page 22: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

ExampleExample

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.

04/21/2304/21/23 rdrd 2222

Page 23: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Example Debt CapitalExample 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

04/21/2304/21/23 rdrd 2323

Page 24: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Debt CapitalDebt 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%.

04/21/2304/21/23 rdrd 2424

Page 25: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

InflationInflation

1. A machine costing $2550 4 years ago now costs $3930with 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

04/21/2304/21/23 rdrd 2525

Page 26: 12/17/2015rd1 Engineering Economic Analysis Chapter 15  Selection of a MARR

Cost of CapitalCost 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.

04/21/2304/21/23 rdrd 2626

Source Amount Price Total Weight

Common stock

250M 29.30 $7,235E6 83.36%

Preferred stock

20M 40.50 $810M 9.22%

Bond 150M $1.0175 $152,625K 1.74%

Loan $500M $1.00 $500M 5.69%

$8.787.625 100%