FEE_ISM_Ch10_3e_OK

Fundamentals of Engineering Economics, 3rd ed. © 2012

Chapter 10 Project Cash Flow Analysis 10.1) a) b) c) d) e) f) g) h) i) j)

Wages paid to temporary workers: Variable cost Property taxes on factory building: Fixed cost Property taxes on administrative building: Fixed cost Sales commission: Variable cost Electricity for machinery and equipment in a plant: Variable cost Heat and air-conditioning for a plant: Fixed cost Salaries paid to design engineers: Fixed cost Regular maintenance on machinery and equipment: Fixed cost Basic raw materials used in production: Variable cost Factory fire insurance: Fixed cost

10.2) a) b) c) d) e) f) g) h) i) j) k)

Raw material costs: Product cost Income taxes paid:Period cost Interest expenses on borrowed funds:Period cost Wages incurred in producing products:Product cost Fire insurance premium paid on factory buildings:Period cost Electricity bill for the warehouse operation:Period cost Salary paid for engineers:Product cost Material handling cost related to production:Product cost Salary paid for plant manager:Product cost Leasing expense for fork-lift trucks in warehouse operation:Period cost Mortgage payments on factory buildings:Period cost

10.3) a) b) c) d) e) f) g) h) i) j) k) l) m) n)

Paint shop superintendent’s salary: Fixed cost Labor costs in assembling a product: Variable cost Rent on a factory building:Fixed cost RFID units embedded in final product during shipping: Variable cost Depreciation on machinery: Variable cost Lubricants used for machines: Variable cost CPU chips used in notebook production: Variable cost Paint used in automobile production: Variable cost Janitorial and custodian salaries:Fixed cost Coffee beans used in packaging roasted coffee: Variable cost Sugar used in ice cream production: Variable cost Electricity for operation of machines: Variable cost Electricity for heating and cooling the factory building:Fixed cost Glue used in electronic board production: Variable cost Page | 1

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.4) a) 6 b) 11 c) 5(Note: It is tempting to select “1”, but the graphs are drawn on cumulative basis) d) 4 e) 2 f) 10 g) 3 h) 7 i) 9 10.5) (a)

(b)

Contribution M arg in per unit = 250 - 80 = $170 170 Contribution M arg in percentage = �100 = 68% Ans. 250 50, 000 B.E.P. = = 294.11 �295units 250 - 80

10.6) (a)

B.E.P. = $450, 000

F 200, 000 = P -V P -V P P

P -V = 0.444 �44.4% Ans. P (b )

Contribution M arg in per unit = 0.444 =

P - V P - 12 = P P

P = 12 / 0.56 = $21.42 per unit Ans. 10.7) (a)

2


(b) (c )

F 50, 000 = = $66666.66 Ans. V 1 - 0.25 1P New Variable cos t and selling price ratio = V / 0.94 P �0.25 / 0.94 = 0.2659 F 50, 000 B.E.P. = = = $68115.94 Ans. V 1 - 0.2659 1P B.E.P. =

10.8) (a)

Line A has the largest contribution margin So, $255, 000 - $3(20, 000) = $195, 000 remains unchanged. For the line B, $195, 000 = 97,500 $2 units should be produced. Therefore, 20,000 units of line A and 97,500 units of line Bneed to be produced to breakeven.

(b)

The total contribution margin:

$3(20,000)  $2(100,000)  $1(80,000) = $340,000. The operating income: $340, 000 - $255, 000 = $85, 000 . (c)

The operating income:

$3(20, 000)  $2(80, 000)  $1(100, 000) - $255, 000 = $65, 000 . The new break-even point:  The fixed cost $255, 000 - 20,000($3) = $195, 000 remains  

unchanged. The demand of Line B is 80,000 units and the fixed cost $195,000 - $2(80,000) = $35,000remains unchanged. The break-even point for line C is $35,000 / $1 = 35,000units. Therefore, the new break-even is 20,000 units of line A, 80,000 units of line B, and 35,000units of line C.

10.9) a) Marginal tax rates: Without project Taxable income $350,000 Income taxes $119,000

With project $530,000 $180,200

marginal tax rate without the project = 34% marginal tax rate with the project = 34% b) Average tax rates: 3


without the project = $119,000/$350,000 = 34% with the project = $180,200/$530,000 = 34%

10.10) a) Marginal tax rates with the project:

b) Average tax rates

10.11) Incremental tax rate calculation: Year 1 Year 2 Revenue $200,000 $200,000 Operating Costs $100,000 $100,000 Depreciation $10,000 $16,000 Taxable income $90,000 $84,000

4

Fundamentals of Engineering Economics, 3rd ed. © 2012 Comments: Note that the marginal tax rates over the project life remain unchanged because the additional income from the new project is not large enough to push the company into a higher tax bracket. 10.12) Taxable income from the project during year 1: D1 = 0.20($100, 000) = $20, 000 Taxableincome = $80, 000 - $20, 000 = $60, 000 a)& b)Increment in income tax due to the project during year 1: Year 1 $195,000 $59,300

Taxable income without project Income taxes Taxable income with project Income taxes

$255,000 $82,700

Incremental taxable income Incremental income taxes Incremental tax rate

$60,000 $23,400 39%

10.13) Incremental tax calculations: a) Additional taxable income due to project: Annual revenue Operating cost Depreciation Taxable income

Year 1 $90,000 $25,000 $16,667 $48,333

Year 2 $90,000 $25,000 $22,222 $42,778

Year 3 $90,000 $25,000 $3,704 $61,296

b) Additional income tax calculation:

5

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.14) (a) Taxable gain:

Ordinary gains = proceeds from old equipment - book value = $23, 000 - $20, 000 = $3, 000 (b) Taxable income:

Note: Income taxes = $113,900 + 0.34($1,125,300- $335,000) = $382,602 Note: Ordinary gains are not included in this calculation, even though these gains will be treated as ordinary income. Of course, these figures can be included to find the total tax liabilities. (c) Marginal and average tax rates: Marginal tax rate = 34% Average (effective) tax rate = $382, 602 / $1,125,300 = 34%

(d) Net cash flow:

6


10.15) (a) Income tax liability:

Note 1: book loss = $60,000 - $75,000 = ($15,000) Note 2: Income taxes = $113,900 + 0.34($575,000- $335,000) = $195,500 (b) Operating income: Taxable operating income Income taxes Net operating income

$590,000 $200,600 $389,400

Note 1: The loss from disposal of the asset is not a part of operating activities, so it is not included in the operating income calculation. Note 2: Income taxes = $113,900 + 0.34($590,000- $335,000) = $200,600 (c) Net cash flow:

7


10.16) (a), (b), and (c)


10.17)

Investment in industrial robot:


10.18)

X = $60, 000( P / A,15%,10) = $301,128

10.21) Investment in energy management system: N = 9 years


10.22) Investment decision based on after-tax IRR:


10.23)

The required lease payment should be $16,651 per year, payable at the end of each year. If the ACLC schedules each lease payment to be made at the beginning of each year, the required lease payment should be much lower, or precisely $15,137 per year.


10.24) Investment decision based on after-tax IRR:


10.25) Investment in a new trench excavator:


10.26) Tucson Solar Company: (a)

(b)

AE(12%) = ( $62, 469 ) �( A / P /12%, 6) = $15,194


10.27) Investment in 3-D computerized car-styling system


10.28) Investment in Mazda automatic screw machine:

Since PW(15%) > 0, accept the investment.

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.29) a) Equal repayment of the principal: n 0 1 2 3 4 5 6

Loan Interest $36,000 $30,000 $24,000 $18,000 $12,000 $6,000

Repayment Loan Principal Balance $300,000 $50,000 $250,000 $50,000 $200,000 $50,000 $150,000 $50,000 $100,000 $50,000 $50,000 $50,000 0

b) Equal repayment of the interest: n 0 1 2 3 4 5 6

Loan Interest $36,000 $36,000 $36,000 $36,000 $36,000 $36,000

Repayment Loan Principal Balance $300,000 $300,000 $300,000 $300,000 $300,000 $300,000 $300,000 0

c) Equal annual installment:

A = $300, 000( A / P,12%, 6) = $72,968 n 0 1 2 3 4 5 6

Loan Interest $36,000 $31,564 $26,595 $26,000 $21,031 $14,798 $7,818

Repayment Loan Principal Balance $300,000 $36,968 $263,032 $41,404 $221,628 $46,373 $175,255 $51,937 $123,318 $58,170 $65,148 $65,148 0

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.30)

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.31) (a), (b), and (c)


10.32) Income statement approach: (a), (b), and (c)


10.33) (a)

(b) This is a service project. The equivalent annual cost is

AEC(18%) = $1,318, 770( A / P,18%,5) = $421, 713.40


10.34)(a) and (b)


10.35) (a) with no borrowed funds:

(b) With borrowed funds:


(c) Which alternative to choose? The debt financing option is more attractive.

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.36) Net cash flow

10.37)

PW(18%) = -$3,500  $6,343( P / F ,18%,1)  L  $9, 454( P / F ,18%,15) = $22,134 > 0 Accept the investment.


Fundamentals of Engineering Economics, 3rd ed. © 2012 10.38) 

Option 1: Lease (a lease paid at the start of each period) PW(12%)lease = -$144, 000(1 - 0.40) ( 1  ( P / A,12%, 29) ) = -$779, 484



Option 2: Purchase -

Note 1: It is assumed that the property is placed in service during January. D1 & D30 = (11.5 /12)(1/ 39)($650, 000) = $15,972 D2 to D29 = (12 /12)(1/ 39)($650, 000) = $16, 667

-

Note 2: Property tax calculation: ($800,000)(0.05) = $40,000

PW(12%)purchase = -$931,551 

Option 3: Remodel


-

Note 1: Depreciation base: Remodeling cost = $300,000 D1 & D30 = (11.5 /12)(1/ 39)($300, 000) = $7,372 D2 to D29 = (12 /12)(1/ 39)($300, 000) = $7, 692

-

Note 2: Cost basis for property tax calculation: Land + building + remodeling cost = $660,000

PW(12%)remodel = -$494, 425 Option 3 is the least costly alternative.

10.39) Comparison by annual equivalent cost (all units in thousand dollars):

Fundamentals of Engineering Economics, 3rd ed. © 2012 Plant A Plant B Book Value (n = 20) Salvage Value Taxable gains Gains tax (39%)

$380.61 $423.80

Plant C $470.56

$853.00 $949.80 $1,054.60 $472.39 $526 $584.04 $184.23 $205.14

$227.78

Net Proceeds from sale $668.77 $744.66

$826.82

Plant A  Capital recovery cost with return:

A1 = ($8,530 - $668.77)( A / P,12%,20)  $668.77(0.12) = $1,132.70 

After-tax O&M cost: A2 = (1 - 0.39)($1,964) = $1,198.04



Depreciation tax shield: A3 = 0.39($8,530) 0.0375( P / F ,12%,1)  L ( A / P,12%, 20) = $172.22



Total equivalent annual cost: A = $1,132.70  $1,198.04 - $172.22 = $2,158.52



Unit cost: $2,158,520 = $0.04317 / kWh 50,000,000kWh

Plant B  Capital recovery cost with return: A1 = ($9, 498 - $744.66)( A / P,12%, 20)  $744.66(0.12) = $1, 261.25 

After-tax O&M cost:

A2 = (1- 0.39)($1,744) = $1,063.84 

Depreciation tax shield:

Fundamentals of Engineering Economics, 3rd ed. © 2012 A3 = 0.39($9, 498) 0.0375( P / F ,12%,1)  L ( A / P,12%, 20) = $191.76 

Total equivalent annual cost:

A = $1, 261.25  $1, 063.84 - $191.76 = $2,133.33 

Unit cost: $2,133,330 = $0.04267 / kWh 50, 000, 000kWh

Plant C 

Capital recovery cost with return: A1 = ($10,546 - $826.82)( A / P,12%, 20)  $826.82(0.12) = $1, 400.41



After-tax O&M cost: A2 = (1 - 0.39)($1, 632) = $995.52



Depreciation tax shield: A3 = 0.39($10,546) 0.0375( P / F ,12%,1)  L ( A / P,12%, 20) = $212.92



Total equivalent annual cost:

A = $1, 400.41  $1,995.52 - $212.92 = $3,183.01 

Unit cost: $3,183, 010 = $0.06366 / kWh 50, 000, 000kWh

Plant B is the most economical.

10.40)

Fundamentals of Engineering Economics, 3rd ed. © 2012 (a) H&H’s cost of leasing in present worth: after-tax lease expense = (1 - 0.40)($11,000) = $6,600 PW(15%) lease = -$6, 600 - $6, 600( P / A,15%,3) = -$21, 670 (b) H&H’s cost of owning in present worth: 

PW of after-tax maintenance expenses: P1 = -$1, 200(1 - 0.40)( P / A,15%, 4) = -$2, 055



PW of after-tax loan repayment

P2 = -$13,169( P / A,15%, 4) = -$37,597 

PW of tax credit (shield) on depreciation and interest:

n Dn In 1 $8, 000 $4,800 2 $12,800 $3, 796 3 $7, 680 $2, 671 4 $2,304 $1, 411

Combined Tax Savings $12,800(0.40) = $5,120 $16,596(0.40) = $6, 638 $10,351(0.40) = $4,140 $3, 715(0.40) = $1, 486

P3 = $5,120( P / F ,15%,1)  $6, 638( P / F ,15%, 2) $4,140( P / F ,15%,3)  $1, 486( P / F ,15%, 4) = $13, 043



PW of net proceeds from sale:

Fundamentals of Engineering Economics, 3rd ed. © 2012 total depreciation amount = $30,784 book value = $9,216 taxable gain = $10,000 - $9,216 = $784 gains tax = (0.40)($784) = $314 net proceeds from sale = $10,000 - $314 = $9,686 P4 = $9,686(P / F ,15%, 4) = $5,538

PW (15%) buy = P1  P2  P3  P4 = -$21, 071 (c) Should the truck be leased or purchased? The borrow–buy option is a better choice.


10.41) Note: Since the operating revenues will be the same for both options, we will only consider the cost of ownership. (a) PW (incremental) cost of owning the equipment: 

PW of after-tax O&M

P1 = -$50, 000(1 - 0.40)( P / A,15%, 4) = -$85, 649 

PW of after-tax loan repayment:

P2 = -$37,857( P / A,15%, 4) = -$108, 080 

PW of tax credit (shield) on depreciation and interest:


n Dn In 1 $24,000 $12,000

Combined Tax Savings $36,000(0.40) = $14,400

2 $38,400 3 $23,040 4 $6,912

$47,817(0.40) = $19,126 $29,610(0.40) = $11,844 $10,353(0.40) = $4,141

$9,414 $6,570 $3,441

P3 = $14, 400( P / F ,15%,1)  $19,126( P / F ,15%, 2) $11,844( P / F ,15%,3)  $4,141( P / F ,15%, 4) = $37,139



PW of net proceeds from sale: total depreciation amount = $92,352 book value = $27,648 taxable gain = $20,000 - $27,648 = ($7,648) loss credit = (0.40)($7,648) = $3,059 net proceeds from sale = $20,000 + $3,059 = $23,059 P4 = $23,059(P / F ,15%, 4) = $13,184

PW (15%)buy = P1  P2  P3  P4 = -$143, 406

Input Tax Rate(%) = MARR(%) = 0

Output PW(15%) = ($143,405)

40 15 1

2

3

4

Income Statement Revenues (savings) Expenses: O&M Depreciation Debt interest (10%)

$0

$0

$0

$0

$50,000 24,000 12,000

$50,000 38,400 9,414

$50,000 23,040 6,570

$50,000 6,912 3,441

Taxable Income Income Taxes (40%)

(86,000) (34,400)

(97,814) (79,610) (39,126) (31,844)

(60,353) (24,141)

Net Income

($51,600) ($58,688) ($47,766) ($36,212)

Fundamentals of Engineering Economics, 3rd ed. © 2012 Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment Net Cash Flow

(51,600) 24,000

(58,688) (47,766) 38,400 23,040

(36,212) 6,912

(120,000) 20,000 3,059 120,000 (25,857)

(28,442) (31,286)

(34,415)

$0 ($53,457) ($48,730) ($56,012) ($40,656)

(b) PW (incremental) cost of leasing the equipment: 

PW of after-tax operating cost:

P1 = $40,000(1 - 0.40)( P / A,15%, 4) = $68,519 

PW of after-tax leasing P2 = $44, 000(1 - 0.40)  $44, 000(1 - 0.40)( P / A,15%,3) = $86, 677 P = P1  P2 = $155,196

(c) Should ICI buy or lease the equipment? The buying option is a better choice. 10.42) (a) OMC’ PW cost of leasing (payments at start of year): PW(15%)leasing = $22, 000(0.60)( P / A,15%,3)*(1.15) =$34,658.98 (b) OMC’ PW cost of owning: 

PW of after-tax maintenance expenses:


P1 = $6, 000(1 - 0.40)( P / A,15%,3) = $8, 219.52 

PW cost of after-tax loan repayment:

P2 = $40,386( P / A,15%,3) = $92, 209.32 

PW of tax credit (shield) on depreciation and interest: n

Dn

In

Combined Tax Savings

1 $13,861 $11,640 $25,501(0.40) = $10,200 2 $23,755 $8,191 $31,946(0.40) = $12,778 3

$8,483

$4,327

$12,810(0.40) = $5,124

P3 = $10,200(P / F,15%,1)  $12,778(P / F,15%,2) $5,124(P / F,15%,3) = $21,901



PW of net proceeds from sale: total depreciation amount = $46,099 book value = $50,901 taxable loss = $45,000 - $50,901 = -$5,901 tax credit = (0.40)($5,901) = $2,360 net proceeds from sale = $45,000 + $2,360 = $47,360 P4 = $47,360(P / F,15%,3) = $31,140

PW(15%) buy = P1  P2 - P3 - P4 = $47,387.84


Fundamentals of Engineering Economics, 3rd ed. © 2012 10.43) (a) and (b)

Sample calculation:  

O & M Expense in year 1: $53,800(1 + 0.05) = $56,490 Salvage value in year 2: $27,000(1 + 0.05)2=$29,768



Note that both depreciation and interest expenses are not responsive to inflation.


10.44) (a) and (b)



(c) Present value gain (or loss) due to inflation: 0.18 - 0.05 = 12.38% 1  0.05 = $92, 781

i� = PW(12.38%) no inflation

PW(18%) with inflation = $98, 771 present value gain = $98, 771 - $92, 781 =$5,990 (d) Present value gain due to borrowing:

n 0 1 2

Net Financing cost Principal Interest(A/T) $130,000 -$7,020 -$130,000 -$7,020

NET Loan flow $130,000 -$7,020 -$137,020

Note: Interest payment (before tax) = $130,000(0.09) = $11,700 Interest payment (after-tax) = $11,700(1 – 0.40) = $7,020

PW(18%) Loan = $130, 000 - $7, 020( P / F ,18%,1) -$137, 020( P / F ,18%, 2) = $25, 642.79

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.45) (a), (b)

(c). The project is acceptable

Fundamentals of Engineering Economics, 3rd ed. © 2012 10.46)


10.47)


10.48) (a) Real after-tax yield on bond investment:




Nontaxable municipal bond:

 imunicipal = 

0.09 - 0.03 = 5.825% 1  0.03

Taxable corporate bond:

 imunicipal =

0.12(1 - 0.3) - 0.03 = 5.245% 1  0.03

(b) Given i = 6%, and f = 3%

 isavings = 2.91% i > 2.91% i > 2.91% Since municipal and corporate , both bond investments are better than the savings account. Now to compare two mutually exclusive bond investment alternatives, we need to perform an incremental analysis. n 0 1 2 3 4 5

After- tax Cash F low Municipal Corporate Incremental -$10,000 -$10,000 $0 $900 $840 -$60 $900 $840 -$60 $900 $840 -$60 $900 $840 -$60 $900 $840 -$60

We cannot find the rate of return on incremental investment, as returns from municipal bond dominate those from corporate bond in every year. Municipal bond is a clear choice for any value of MARR.


10.49) (a), (b), and (c)

Select B.


10.50) (a) & (b) Actual and constant dollar analysis:

(c) Given f = 8%, i = 15%

i�=

0.15 - 0.08 = 6.48% 1  0.08 (Inflation-free MARR)

Since IRR’> 6.48%, the project is a profitable one.


10.51) (a) & (b) Project cash flows in actual and constant dollars:


(c) The effects of project financing under inflation:


(d) The present value loss due to inflation:

Present value loss = $136,553 - $140,656 = ($4,103) (e) Required additional before-tax annual revenue in actual dollars (equal amount) to make-up the inflation loss.

$4,103( A / P,18%, 6) = $1,955 1 - 0.40


10.52)(a), (b), and (c) (Unit:$000)

Note 1: In a strict sense, capital gains are only realized for the sale of land. Note 2: It is assumed that the building will be disposed of at the end of December of the 12th year.


10.53) (a) The net after-tax cash flows for each financing option: 

Option 1: Equity Financing (Retained earnings)




Option 2: Debt Financing at 12%




Option 3: Lease Financing

Fundamentals of Engineering Economics, 3rd ed. © 2012 (b) Vermont’s PW cost of owning the equipment by borrowing: 

PW of total after-tax revenue:

P1 = $174,000(1 - 0.39)( P / A,18%, 6) = $371, 236 

PW cost of working capital drain:

P2 = $25,000 - $25, 000( P / F ,18%, 6) 

= $15,739 PW cost of operating expense: P3 = $22, 000(1 - 0.39)( P / A,18%,6) = $46,938



PW cost of owning by borrowing:

Net cost = -$214, 469  P1 - P2 - P3 = $94, 090 (c) Vermont’s PW cost of leasing the equipment: 

PW cost of after-tax leasing

P = $55, 000(1 - 0.39)  $55, 000(1 - 0.39)( P / A,18%,5) = $138, 467

(d) Buy the tipping machine.


10.54) (a),(b),(c) & (d): Assumption: The building will be placed in service in January.

Note: If the firm decides not to invest in the project, the firm could write off the R&D expenditure. The amount of write-off will be (0.38)($8,000,000) = $3,040,000. If the firm decides to undertake this project, then an opportunity cost of $3,040,000 will be incurred.


(e)

(f)


10.55) (a) The net cash flow from the cogeneration project with bond financing:



(b). The maximum annual lease amount that ACC is willing to pay is $907,673.

10.56)


(a) & (b) The project cash flows and IRR with no inflation:

(c) & (d)


(e). The economic gain in present worth due to inflation = $108,411- $90,992 = $17,419.

FEE_ISM_Ch10_3e_OK

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