SOLUTIONS MANUAL for Fundamentals of Engineering Economics, 4e Chan Park by StudyGuide

Chapter 1 Engineering Economic Decisions 1.1)

Not provided For The Wall Street Journal, go to the Front page to find the section on “What’s News.” This is a section on a brief summary on major headlines of the day’s news. Quickly browse through the news summary to see if there is any news related to business investment. The best places to find the major business news on investment are sections on “BUSINESS”, “MARKETS” or “TECH.”

1.2) Not provided Some of the well-known business publications are: •

Daily Newspapers: o The Wall Street Journal o The New York Times (Business Section) o Financial Times

•

Weekly or Monthly Magazines: o o o o o

BusinessWeek Forbes Money Smart Money Fortune

Chapter 2: Time Value of Money 2.1)= I (iP = ) N (0.06)($2, 000)(5) = $600 2.2)

•

Simple interest: $20,000 = $10,000(1 + 0.075N ) (1 + 0.075N ) = 2 N=

•

1 = 13.33≈ 14 years 0.075

Compound interest: $20,000 = $10,000(1 + 0.07) N (1 + 0.07) N = 2 N = 10.24 ≈ 11years

2.3) •

Simple interest: = I iPN = (0.10)($10, 000)(5) = $5, 000

•

Compound interest: I= P[(1 + i ) N − 1]= $10, 000(1.6105 − 1)= $6,105

2.4) •

Option 1: Compound interest with 8.5%: F = $4,500(1 + 0.085)5 = $4,500(1.5037) = $6, 766.65

•

Option 2: Simple interest with 9%: $4,500(1 += 0.09 × 5) $5, 000(1.45) = $6,525

∴ Option 1 is still better.

2.5)

•

Compound interest:

= F $1, 000(1 + 0.065)5 = $1,370.09 •

Simple interest:

= F $1, 000(1 + 0.068(5)) = $1,340 The compound interest option is better.

2.6) End of Year 0 1 2 3

Principal Repayment

Interest payment

$4,620.50 $4,990.14 $5,389.35

$1,200.00 $830.36 $431.15

2.7) = P $22, 000( = P / F ,5%,5) $22, = 000(0.7835) $17, 237.58

2.8) = F $30, 000( = F / P,9%,3) $30, = 000(1.295) $38,850.87 2.9)

F = $100( F / P,10%,10) + $200( F / P,10%,8) = $688

2.10) F $250, = = 000( F / P, 6%,10) $447, 712 2.11) P = $300, 000( P / F ,8%,10) = $138,958 2.12)

i = 10.5% , two-year discount rate is (1 + 0.105) 2 = 1.221 (or 22.1%)

2.13) (a) F $5, = = 000( F / P, 7%,5) $7, 013

Remaining Balance $15,000.00 $10,379.50 $5,389.36 $0

(b) F $7, = = 250( F / P,9%,15) $26, 408 (c) F $9, = = 000( F / P, 6%,33) $61,565 (d) F $12, = = 000( F / P,5.5%,8) $18, 416

2.14) (a) P $25,500( = = P / F ,12%,8) $10, 299 (b) P $58, = = 000( P / F , 4%,12) $36, 227 (c) P $25, = = 000( P / F , 6%,9) $14, 797 (d) P $35, = = 000( P / F ,9%, 4) $24, 795 2.15) (a)

P = $12, 000( P / F ,13%, 4) = $7,360

(b)

F = $30, 000( F / P,13%,5) = $55, 273

2.16)

= F 3= P P(1 + 0.08) N log 3 = N log(1.08) = N 14.27 → 15 years 2.17)

= F 2= P P(1 + 0.06) N log 2 = N log 1.06 N = 11.896 years (or 12 years) 2.18) •

= F 2= P P(1 + 0.06) N log 2 = N log(1.06) N = 11.90 years  12 years

•

Rule of 72: 72 / 6 = 12 years

394 2.19) = F $1(1.08) = $14, 755, 694, 730, 611

2.20) = P $35, 000( P / F ,9%, 4) + $10, 000( P / F ,9%, 2) = $35, 000(0.7084) + $10, 000(0.8417) = $33, 211 2.21) = P $450, 000( = P / F ,5%,5) 450, = 000(0.7835) $352,575 2.22)

•

Simple interest (John): = I iPN = (0.1)($1, 000)(5) = $500

•

Compound interest (Susan):

I= P (1 + i ) N − 1= $1, 000 (1 + .095)5 − 1 = $574.24 •

2.23) P=

Susan’s balance will be greater by $74 (or $74.24 to be exact)

$2, 000 $800 $1, 000 + + = $3, 230.65 1.11 1.12 1.13

2.24)

$15, 000 $23, 000 $36, 000 $48, 000 + + + = $93,564 1.07 2 1.073 1.07 4 1.075

2.25) F = $2, 000( F / P, 6%,10) + $2,500( F / P, 6%,8) + $3, 000( F / P,6%,6) = $11,822

2.26) P = $3, 000, 000 + $2, 400, 000( P / F ,8%,1) +  +$3, 000, 000( P / F ,8%,10) = $20, 734, 618

Or,

P = $3, 000, 000 + $2, 400, 000( P / A,8%,5) +$3, 000, 000( P / A,8%,5)( P / F ,8%,5) = $20, 734, 618

2.27) P = $9, 000( P / F ,8%, 2) + $6, 000( P / F ,8%,5) + $3, 000( P / F ,8%,7) = $13,550

2.28) $1,000 +

2.29)

2.30)

$1,000 $1,500 $1,210 X + = + 4 3 2 1.1 1.1 1.1 1.1 X = $2,981

$180, 000 = $20, 000( P / A,9%,5) − $10, 000( P / F ,9%,3) + X ( P / F ,9%, 6)180, 000 − 20, 000(3.8897) + 10, 000(0.7722) = X (0.5963) X = $184,350.16

$60, 000 $77, 000 $65, 000 $57, 000 45, 000 + + + + = $212,873.89 1.14 1.142 1.143 1.144 1.145

2.31) (a)

With deposits made at the end of each year

= F $3, = 000( F / A,9%,15) $88, 083

(b) With deposits made at the beginning of each year = F $3, = 000( F / A,9%,15)(1.09) $96, 010

2.32) (a) F $8, = = 000( F / A,11.75%,5) $50,571 (b) F $2, = = 000( F / A, 4.25%,12) $30, 486 (c) F $7, = = 000( F / A, 6.45%, 20) $270,309

= = 000( F / A, 7.75%,12) $74, 793 (d) F $4,

2.33) (a) A $45, = = 000( A / F ,8%,11) $2, 703 (b) A $35, = = 000( A / F , 6%,18) $1,132 (c) A $25, = = 000( A / F , 7%, 6) $3, 495 = = 000( A / F ,11%,13) $458 (d) A $12,

2.34) A $250,= = 000( A / F ,5%,5) $250, = 000(0.1810) $45, 250 (a) = F $5, 000( = F / A,5%, 7) $5, = 000(8.1420) $40, 710 F = $5, 000( F / A,5%, 7)(1.05) (b) = $5, = 000(8.1420)(1.05) $42, 745.50

2.35)

= A $250, = 000( A / F ,5%,5) $45, 244

2.36) $3, 000( F / A, 6%, N ) = $55, 000 N = 13 years

2.37) = A $15, = 000( A / F ,11%,5) $2, 408.55

2.38)

F =$500(1.04)10 + $1, 000(1.04)8 + $1, 000(1.04)6 +$1, 000(1.04) 4 + $1, 000(1.04) 2 + $1, 000 = $6, 625.47

2.39) (a) A $15, = = 000( A / P,3.5%, 6) $2,815.02 (b) A $7,500( = = A / P, 7.5%, 7) $1, 416

2.40)

•

Equal annual payment amount:

= A $20, 000( = A / P,10%,3) $20, = 000(0.4021) $8, 042

•

Loan balance calculation:

End of period 0 1 2 3

Principal Payment $0.00 $6,042.00 $6,646.20 $7,310.82

Interest Payment $0.00 $2,000.00 $1,395.80 $731.18

Remaining Balance $20,000.00 $13,958.00 $7,311.80 $0

Interest payment for the second year = $1,395.80

2.41) (a) = P $1, = 000( P / A, 7.2%,8) $5,925.29 (b) P $4,500( = = P / A,9.5%,12) $31, 427.28 (c) P $1,900( = = P / A,8.25%,13) $14,812.86 = = P / A, 7.75%,8) $111,970.11 (d) P $19,300(

2.42) (a) The capital recovery factor ( A / P, i, N ) for N 35 40

6% 0.0690 0.0665

7% 0.0772 0.0750

To find ( A / P, 6.25%, 38) , first, interpolate for N = 38 :

N 38

6% 0.0675

7% 0.0759

Then, interpolate for i = 6.25% ; (A / P, 6.25%, 38) = 0.0696 :

As compared to the value from the interest formula: (A / P, 6.25%, 38) = 0.0694

(b) The equal payment series present-worth factor ( P / A, i, 85) for i

9% 11.1038

10% 9.9970

Then, interpolate for i = 9.25% : ( P / A, 9.25%, 85) = 10.8271

As compared to the value from the interest formula: ( P / A, 9.25%, 85) = 10.8049

2.43) = F $500(= F / A, 7%,15)(1.07) $500(25.1290)(1.07) = $13, 444.02

2.44) •

Equal annual payment: A = $50, 000( A / P,12%,3) = $20,817.45

•

Interest payment for the second year: End of Year 0 1 2 3

Principal Repayment

Interest payment

$14,817.45 $16,595.54 $18,587.01

$6,000 $4,221.91 $2,230.44

Remaining Balance $50,000 $35,182.55 $18,587.01 0

2.45) P = $35,000(P / A,12%,10) = $197,758 Since $200,000 > $197,758, You should not purchase the equipment.

2.46)

P= −$3, 460 +

250 = 0  i = 7.225% i

2.47)

= P

1, 000 = $10, 000 0.1

2.48) F= F1 + F2 = $20, 000( F / A, 6%,5) + $5, 000( F / G, 6%,5) = $20, 000( F / A, 6%,5) + $5, 000( A / G, 6%,5)( F / A, 6%,5) = $20, 000(5.6371) + $5, 000(1.8836)(5.6371) = $165,833

2.49) = F $10, 000( F / A,8%,5) − $2, 000( F / G,8%,5) = $10, 000( F / A,8%,5) − $2, 000( P / G,8%,5)( F / P, 8%, 5) = $37, 001

2.50)

P= $100 + [$100( F / A,9%, 7) + $50( F / A,9%, 6) +$50( F / A,9%,4) + $50( F / A, 9%, 2)]( P / F , 9%, 7) = $991.32

2.51) = A $15, 000 − $1, 000( A / G, 8%, 12) = $10, 404.25 2.52)

= P $1, 000( P / A, 6%,5) + $250( P / G, 6%,5) = $6,196

Geometric Gradient Series 2.53)

F = $8, 000( P / A1 ,5%, 7%,30)( F / P, 7%,30) = $172,895.56(7.61226) = $1,316,126 2.54) (a) P = $6,000,000( P / A1 ,−10%,12%,7) = $21,372,076 (b) Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate of 10%. Therefore, the annual revenue can be expressed as follows: An = $60(1 + 0.05) n −1100, 000(1 − 0.1) n −1 = $6, 000, 000(0.945) n −1 = $6, 000, 000(1 − 0.055) n −1

This revenue series is equivalent to a decreasing geometric gradient series with g = -5.5%. So, = P $6, 000, 000( P / A1 , −5.5%,12%,7) = $23,847,897

(c) Computing the present worth of the remaining series ( A4 , A5 , A6 , A7 ) at the end of period 3 gives 3 = A4 6, 000, 000(1 − 0.055) = 5, 063, 451.75

= P $5, 063, 451.75( P / A4 , −5.5%,12%, = 4) $14, 269, 627.82

2.55)

P = ∑ An (1 + i ) − n n =1 20

= ∑ (2, 000, 000)n(1.06) n −1 (1.06) − n n =1

20 1.06 n = (2, 000, 000 /1.06)∑ n( ) 1.06 n =1 = $396, 226, 415

2.56)

(a) The withdrawal series would be Period 11 12 13 14 15

Withdrawal $15,000 $15,000(1.08) $15,000(1.08)(1.08) $15,000(1.08)(1.08)(1.08) $15,000(1.08)(1.08)(1.08)(1.08)

Amount $15,000 $16,200 $17,496 $18,896 $20,407

= P10 $15, = 000( P / A1 ,8%,9%,5) $67,556

Assuming that each deposit is made at the end of each year, then: $67,556 = A( F / A,9%,10) A = $4, 446.54

(b) P10 $15, = = 000( P / A1 ,8%, 6%,5) $73, 476 $73, 476 = A( F / A, 6%,10) A = $5,574.47

2.57) = $1, 000, 000 A= ( F / A, 6%,30) A(79.0582)  A = $12,649 should be set aside on the account a) = $1, 000, 000 A= ( P / A, 6%, 20) A(11.4699)  A = $87,185 / year b)

$1, 000, 000 = A1 ( P / A1, 3%, 6%, 20) 1 − (1.03) (1.06 ) = A1 0.06 − 0.03 = $68, 674 / year 20

−20

Equivalence Calculations 2.58) P= [$100( F / A,12%,9) + $50( F / A,12%, 7) + $50( F / A,12%,5)]( P / F ,12%,10) = $740.49

2.59) P (1.08) = + $200 $200( P / F ,8%,1) + $120( P / F ,8%, 2) + $120( P / F ,8%,3) + $300( P / F ,8%, 4) P = $373.92

2.60)

= A( P / A,15%,5) $100( P / A,15%,5) + $20( P / A,15%,3)( P / F ,15%, 2) 3.35216 A = $369.74 A = $110.30

2.61) P1 = $200 + $100( P / A,8%,5) + $50( P / F ,8%,1) +$50( P / F ,8%, 4) + $100( P / F ,8%,5) = $750.37 = P2 X= ( P / A,8%,5) $750.37 X = $187.93

2.62) Establish economic equivalent at N = 8 :

C ( F / A,8%,8) − C ( F / A,8%,2)( F / P,8%,3) = $6,000( P / A,8%,2) 10.6366C − (2.08)(1.2597)C = $6,000(1.7833) 8.0164C = $10,699.80 C = $1,334.73

2.63) The original cash flow series is

0 1 2 3 4

0 6 $900 $800 7 $920 $820 8 $300 $840 9 $300 $860 10 $300 − $500

$880

2.64) $300( F / A,10%,8) + $200( F / A,10%,3) = 2C ( F / P,10%,8) + C ( F / A,10%,7) $4,092.77 = 2C (2.1436) + C (9.4872) C = $297.13

2.65) $25, 000 + $30, 000( P / F ,10%, 6) = C ( P / A,10%,12) + $1, 000( P / A,10%, 6)( P / F ,10%, 6) $41,935 = 6.8137C + $2, 458.43 C = $5, 794

2.66) (b), (d), and (e)

2.67)

$200( F / A,8%,5) − $50( F / P,8%,1) = X ( F / A,8%,5) − ($200 + X )[( F / P,8%, 2) + ( F / P,8%,1)] $1,119.32 = X (5.8666) − ($200 + X )(2.2464) X = $433.29

2.68)

(b)

2.69)

(a)

2.70) P1 = 30, 723( P / F , i %,5) P2 = A( P / A, i %,10)  (1 + i )10 − 1  $50, 000(1 + i ) −5 = $5, 000  10   i (1 + i )  ∴ i =13.06%

2.71)

2= P P(1 + i )5 21/5 = 1 + i i = 14.87% 2.72)

Establishing equivalence at n = 0 $2,000( P / A, i,6) = $2,500( P / A1 ,−25%, i,6)

By Excel software, i = 92.36%

2.73) = $40, 000 $15, 000( F= / P, i,5) $15, 000(1 + i )5 i = 21.67%

2.74)

Option 1: $100, 000( F / A, 7%, 7)( F / P, 7%,13) = 2, 085, 484.95 Option 2: $100, 000( F / A, 7%,13) = 2, 014, 064.29

$100, 000( F / A, i, 7)( F / P, i,13) = $100, 000( F / A, i,13) i = 6.6% Short Case Studies with Excel 2.75)

The equivalent future worth of the prize payment series at the end of Year 20 (or beginning of Year 21) is

F1 = $1,952,381( F / A, 6%, 20) = $1,952,381(36.7856) = $71,819,506.51 The equivalent future worth of the lottery receipts is

= F2 ($36,100, 000 − $1,952,381)( F / P, 6%, 20) = ($36,100, 000 − $1,952,381)(3.2071) = $109,514,828.9 The resulting surplus at the end of Year 20 is = F2 − F1 $109,514,828.9 − $71,819,506.51 = $37, 695,322.4

2.76)

$1, 000( F / P,9.4%,5) + $500( F / A,9.4%,5) (1 + 0.094)5 − 1 ) 0.094 = $1, 000(1.5671) + $500(6.0326) = $4,583.4 = $1, 000((1 + 0.094)5 ) + $500(

$4,583.4( F / P,9.4%, 60) = $4,583.4((1 + 0.094)60 ) = $4,583.4(219.3) = $1, 005,141.21

The main question is whether or not the U.S. government will be able to invest the social security deposits at 9.4% interest over 60 years.

2.77) (a) It is worth $218.20M.

Season

Salary

Present Worth (6%)

2015 $ 6.50 2016 $ 9.00 2017 $ 14.50 2018 $ 25.00 2019 $ 26.00 2020 $ 26.00 2021 $ 29.00 2022 $ 29.00 2023 $ 32.00 2024 $ 32.00 2025 $ 32.00 2026 $ 29.00 2027 $ 25.00 (buy out) $ 10.00

$6.50 $8.49 $12.90 $20.99 $20.59 $19.43 $20.44 $19.29 $20.08 $18.94 $17.87 $15.28 $12.42 $4.97

Sum

$218.20

$ 325.00

(b) With the buyout option, it is better staying on the contract. Without the buyout contract, it is better being a free agent after 2020.

Season

Salary

2015 2016 2017 2018 2019 2020 2021 $ 29.00 2022 $ 29.00 2023 $ 32.00 2024 $ 32.00 2025 $ 32.00 2026 $ 29.00 2027 $ 25.00 (buy out) $ 10.00

Salary for being Free Agent

$30.00 $30.00 $30.00 $30.00 $30.00 $30.00 $30.00

Sum $ 218.00

$210.00

$ 183.00

$177.52

Chapter 3 Understanding Money Management 3.1)

•

Nominal interest rate:

r= 1.3% ×12= 15.6% •

Effective annual interest rate: ia = (1 + 0.013)12 − 1 = 16.77%

3.2)

= i 17.85% ÷= 12 1.4875% (a) Monthly interest rate: Annual effective rate: ia = (1 + 0.014875)12 − 1 = 19.385% (b) $2,500(1 + 0.014875) 2 = $2,574.93

= 3.3) r

10, 000 − 8,800 = 13.64% 8,800

3.4) Assuming weekly compounding: r = 6.89% 0.0689 52 ) − 1 = 0.07128 ia = (1 + 52

3.5) The effective annual interest rate : 1 9.09% = ia e0.087 −=

3.6) Interest rate per week:

= $450 $400(1 + i ) i = 12.5% per week

(a) Nominal interest rate:

r = 12.5% × 52 = 650% (b) Effective annual interest rate

ia = (1+ 0.125)52 − 1 = 45,602% 3.7)

$20, 000 = $520( P / A, i, 48) ( P / A, i, 48) = 38.4615 Use Excel to calculate i : i = 0.9431% per month = r 0.9431*12 = 11.32%

3.8)

$16,000 = $517.78(P / A,i,36) (P / A,i,36) = 30.901155 i = 0.85% per month r = 0.85 × 12 = 10.2%

3.9) The three options : a) ia= r= 6.12% 4

 0.06  b) ia = 1 +  − 1 = 6.136% 4   ia e0.059 −= 1 6.078% c) =

Bank B is the best option.

3.10)

0.06 1 ) − 1 = 0.5% 12 0.06 3 b) i = (1 + ) − 1 = 1.508% 12 0.06 6 c) i = (1 + ) − 1 = 3.038% 12 0.06 12 d) i = (1 + ) − 1 = 6.168% 12

a) i = (1 +

3.11)

iquarter = e0.09/4 − 1 = 0.022755 (or 2.28%)

3.12)

0.06 1 ) − 1 = 0.5% 12 0.06 3 b) i = (1 + ) − 1 = 1.508% 12 0.06 6 c) i = (1 + ) − 1 = 3.038% 12 0.06 12 d) i = (1 + ) − 1 = 6.168% 12

a) i = (1 +

3.13)

$25, 000 = $563.44( P / A, i, 48) ( P / A, i, 48) = 44.3703 i = 0.3256% per month

3.14)

0.11 1 ) − 1 = 11% 1 0.08 2 b) i = (1 + ) − 1 = 8.16% 2 0.095 4 c) i = (1 + ) − 1 = 9.844% 4 0.075 365 d) i = (1 + ) − 1 = 7.788% 365

a) i = (1 +

3.15) (a) 0.082 24 ) = $9,545( F / P, 4.1%, 24) 2 = $25, 037.64

F = $9,545(1 +

(b) 0.06 40 ) = $6,500( F / P,1.5%, 40) 4 = $11, 791.12

F = $6,500(1 +

0.09 96 ) = $42, 000( F / P, 0.75%,96) 12

= $87, 693.83 3.16) (a) = F $10, = 000( F / A, 4%, 20) $297, 781

(b) = F $9, = 000( F / A, 2%, 24) $273, 796.76 (c) = F $5, = 000( F / A, 0.75%,168) $1, 672,590.40 3.17) (a) = A $11, = 000( A / F , 4%, 20) $369.60 (b) = A $3, = 000( A / F ,1.5%, 60) $31.18 (c) = A $48, = 000( A / F , 0.6125%, 60) $484.46

3.18)

(a) Quarterly effective interest rate = 2.25%

= F $10, = 000( F / A, 2.25%, 60) $1, 244,504

(b) Quarterly effective interest rate = 2.267% = F $10, = 000( F / A, 2.267%, 60) $1, 251,976

3.19) •

Equivalent future worth of the receipts:

F1 = $1, 500(F / P, 2%, 4) + $2, 500 = $4,123.65 •

Equivalent future worth of deposits:

= F2 A( F / A, 2%,8) + A( F / P, 2%,8) = 9.7546 A ∴ Letting F1 = F2 and solving for A yields A = $422.74

3.20) (d)

Effective interest rate per payment period i = (1 + 0.01)3 – 1 = 3.03%

$1,000

3.21) (b)

3.22)

A = $70,000( A / F,0.5%,36) = $1,779.54

3.23) •

The balance just before the transfer:

F9 = $22,000(F / P,0.5%,108) + $16,000(F / P,0.5%,72) +$13,500(F / P,0.5%,48) = $77,765.70 Therefore, the remaining balance after the transfer will be $38,882.85. This remaining balance will continue to grow at 6% interest compounded monthly. Then, the balance 6 years after the transfer will be: = F15 $38,882.85( = F / P, 0.5%, 72) $55, 681.96

•

The funds transferred to another account will earn 8% interest compounded quarterly. The resulting balance six years after the transfer will be:

= F15 $38,882.85( = F / P, 2%, 24) $62,540.63

3.24) Establish the cash flow equivalence at the end of 25 years. Let’s define A as the required quarterly deposit amount. Then we obtain the following: A( F / A,1.5%,100) = $80, 000( P / A, 6.136%,15) 228.8030 A = $770,104 A = $3,365.79

3.25)

C ( F / A, 6.168%, 7) + C ( F / P, 0.5%,84) = $1, 600 + $1, 400( F / P, 0.5%,12) + $1, 200( F / P, 0.5%, 24) +$1, 000( F / P, 0.5%,36) =8.437C + 1.52C =1, 600 + 1, 486.35 + 1,352.59 + 1,196.68 9.957C = 5, 635.62 ∴C = $566

3.26)

$200, 000 = $2, 000( P / A,9% /12, N ) N = 186 months N = 15.5 years

3.27) To find the amount of quarterly deposit (A), we establish the following equivalence relationship: 4

 0.06  ia = 1 +  − 1 = 0.06136 4   A( F / A,1.5%, = 60) $60, 000 + $60, 000( P / A, 6.136%,3) A = $219,978 / 96.2147 A = $2, 286.32

3.28) Setting the equivalence relationship at the end of 20 years gives 2

 0.06  isemiannual = 1 +  − 1 = 3.0225% 4   6% A( F / A, ,80) = $40, 000( P / A,3.0225%, 20) 4 152.71A = $593,862.93 A = $3,888.81

3.29) •

Monthly installment amount:

= A $22, = 000( A / P, 0.75%, 60) $456.68

•

The lump-sum amount for the remaining balance:

= P24 $456.68( = P / A, 0.75%,36) $14,361.13 3.30) Given= i

5% = 0.417% per month 12

A = $500,000( A / P,0.417%,120) = $5,303.26

3.31) First compute the equivalent present worth of the energy cost savings during the first operating cycle:

$70

June July Aug.

$80

Dec. Jan. Feb. .

P $70( P / A, 0.5%,3)( P / F , 0.5%,1) + $80( P / A, 0.5%,3)( P / F , 0.5%, 7) = $436.35 Then, compute the total present worth of the energy cost savings over 5 years. P= $436.35 + $436.35( P / F , 0.5%,12) + $436.35( P / F , 0.5%, 24) +$436.35( P / F , 0.5%,36) + $436.35( P / F , 0.5%, 48) = $1,942.55

3.32) •

Option 1 .06 1 ) − 1 = 1.5% 4 $1, = 000( F / A,1.5%, 40)( F / P,1.5%, 60) $132,587

i = (1 + F •

Option 2

.06 4 ) − 1 = 6.136% 4 = F $6, = 000( F / A, 6.136%,15) $141,111 i = (1 +

•

Option 2 – Option 1 = $141,110 – 132,587 = $8,523

•

Select (b)

3.33) Given: r = 7% compounded daily, N = 25 years •

Since deposits are made at year end, find the effective annual interest rate:

ia = (1 + 0.07 / 365)365 − 1 = 7.25% •

Then, find the total amount accumulated at the end of 25 years: F = $3,250( F / A,7.25%,25) + $150(F / G,7.25%,25) = $3,250( F / A,7.25%,25) + $150(P / G,7.25%,25)(F / P,7.25%,25) = $297,016.95

3.34) (a) Quarterly interest rate = 2.25% 3= P P(1 + 0.0225) N log 3 = N log1.0225 N = 49.37quarters = 12.34 years (b) Monthly interest rate = 0.75% 3= P P(1 + 0.0075) N log 3 = N log1.0075 N = 147.03 months = 12.25 years (c) 3 = e0.09 N ln(3) = 0.09 N N = 12.20 years

3.35) (a) Quarterly effective interest rate = 1.5% = F $10, = 000( F / A,1.5%, 60) $962,147

(b) Quarterly effective interest rate = 1.508% = F $10, = 000( F / A,1.508%, 60) $964, 722

3.36) rN = F Pe = $5, 000e(0.09×5) = $7,841.56

3.37) (a) Quarterly effective interest rate = 2.25%

F = $4,000( F / A,2.25%,40) = $255,145

(b) Quarterly effective interest rate = 2.2669% F = $4,000( F / A,2.2669%,40) = $256,093

3.38) = ia e0.086/4= − 1 2.1733% A = $10, 000( A / P, 2.1733%, 20) = $ 621.84

3.39) (a) Monthly effective interest rate = 0.74444% = F $1,500( = F / A, 0.74444%,96) $209,170

(b) Monthly effective interest rate = 0.75% = F $1,500( = F / A, 0.75%,96) $209, 784

3.40) Effective interest rate per month = e0.0975/12 − 1 =0.8158% = A $48, = 000( A / P, 0.8158%, 60) $1, 014.90

3.41) Effective interest rate per quarter = e0.0688/ 4 − 1 = 1.7349% = P $2,500( = P / A,1.7349%, 20) $41,944

3.42) = i e0.0225 = − 1 2.2755%

= F $5, = 000( F / A, 2.2755%, 40) $320, 721 = F $320, = 721( F / P, 2.2755%, 20) $502,990

3.43) 0.12 1 ) − 1 = 1% per month 12 ⋅1 = P $2, = 000( F / P,1%, 2) $2, 040.20 i = (1 +

3.44) • Effective interest rate for Bank A 0.18 4 i = (1 + ) − 1 = 19.252% 4 • Effective interest rate for Bank B 0.175 365 i = (1 + ) − 1 = 19.119% 365 • Select (c)

3.45)

= im 2.9% = = /12 0.2417%,17% /12 1.417% $3, 000( F / P, 0.2417%, 6)( F / P,1.417%, 6) −$100[( F / A, 0.2417%, 6)( F / P, 0.2417%, 6) + ( F / A,1.417%, 6)] = $2, 077.79 3.46) (a) • Bank A: ia = (1 + 0.0155)12 − 1 = 20.27% per year • Bank B: ia = (1 + 0.195 / 12)12 − 1 = 21.34% per year (b) Given = i 6% = / 365 0.01644% per day , find the total cost of credit card usage for each bank over 24 months. We first need to find the effective interest rate per payment period (month—30 days per month):

i = (1 + 0.0001644)30 − 1 = 0.494%

•

Monthly interest payment: Bank A: $300(0.0155) = $4.65/month Bank B: $300(

0.195 ) = $4.875/month 12

We also assume that the $300 remaining balance will be paid off at the end of 24 months. • Bank A: P = $20 + $4.65( P / A,0.494%,24) + $20( P / F,0.494%,12) = $143.85 • Bank B: = P $4.875( = P / A, 0.494%, 24) $93.25 Select Bank B

3.47) (a)

im = (1 +

0.12 1 ) − 1 = 1% 12 ⋅1

(b) $10, 000( = A / P,1%, 48) 10, = 000(0.0263) $263 / month (c) Remaining balance at the beginning of 20th month is $263( = P / A,1%, 29) $263(25.0658) = $6,592.3 1 So, the interest payment for the 20th payment is $6,592.31*1% = $65.92. (d) $263 x 48months = $12,624 So, the total interest paid over the life of the loan is $2,624. 3.48) Loan = repayment schedule: A $20, = 000( A / P, 0.5%,36) $608.44 End of month 0 1 2 13 24 36

3.49)

Interest Payment $0.00 $100.00 $97.46 $68.64 $38.20 $3.03

Principal Payment $0.00 $508.44 $510.98 $539.80 $570.24 $605.41

Remaining Balance $20,000.00 $19,491.56 $18,980.58 $13,188.31 $7,069.38 $0

Given: P = $120,000 , N = 360 months, i = 9% / 12 = 0.75% per month (a)

A = $120,000( A / P,0.75%,360) = $965.55

(b) If r = 9.75% APR after 5 years, then i = 9.75% / 12 = 0.8125% per month. •

The remaining balance after the 60th payment: B60 = $965.55( P / A,0.75%,300) = $115,056.50

•

Then, we determine the new monthly payments as A = $115, 056.50(A / P, 0.8125%, 300) = $1, 025.31

3.50) (a) (i) $10,000( A / P,0.75%,24) (b) (iii) B12 = A( P / A,0.75%,12)

3.51) Given information: = i 9.45% = / 365 0.0259% per day , N = 36 months.

•

Effective monthly interest rate, i = (1 + 0.000259)30 − 1 = 0.78% per month

•

Monthly payment, A = $13,000( A / P,0.78%,36) = $415.58 per month

•

Total interest payment, I = $415.58 × 36 − $13, 000 = $1,960.88

3.52) Given Data: P = $25,000, r = 9% compounded monthly, N = 36 month, and i = 0.75% per month. •

Required monthly payment:

= A $25, = 000( A / P, 0.75%,36) $795

•

The remaining balance immediately after the 20th payment:

= B20 $795( = P / A, 0.75%,16) $11,944.33

3.53) Given Data: P = $250,000 - $50,000 = $200,000. •

Option 1: N = 15 years × 12 = 180 months APR = 4.25%

∴ A $200, = = 000( A / P, 4.25% /12,180) $1,504.56 •

Option 2: N = 30 years × 12 = 360 months APR = 5% ∴ A = $200,000( A / P,5% / 12,360) = $1,073.64 ∴ Difference = $1,504.56 - $1,073.64 = $430.92

3.54) 9% = A $400, = 000( A / P, ,180) $4, 057.07 12 • Total payments over the first 5 years (60 months)

$4,057.07 × 60 = $243,424.20 •

Remaining balance at the end of 5 years: B60 = $4,057.07( P / A,0.75%,120) = $320,271.97

• •

Reduction in principal = $400,000 - $320,271.97 = $79,728.03 Total interest payments = $243, 424.20 − $79,728.03=$163,696.17

3.55) The amount to finance = $300,000 - $45,000 = $255,000 = A $255, = 000( A / P, 0.5%,360) $1,528.85

Then, the minimum acceptable monthly salary (S) should be

= S

A $1,528.85 = = $6,115.42 0.25 0.25

3.56) Given Data: purchase price = $150,000, down payment (sunk equity) = $30,000, interest rate = 0.75% per month, N = 360 months, •

Monthly payment: A = $120,000( A / P,0.75%,360) = $965.55

•

Balance at the end of 5 years ( 60 months):

= B60 $965.55( = P / A, 0.75%,300) $115, 056.50 •

Realized equity = sales price – balance remaining – sunk equity: $185,000 - $115,056.60 - $30,000 = $39,943.50 Note: For tax purpose, we do not consider the time value of money on $30,000 down payment made five years ago.

3.57) Given Data: interest rate = 0.75% per month, each individual has the identical remaining balance prior to their 20th payment, that is,$80,000. With equal remaining balances, all will pay the same interest for the 20th mortgage payment. $80,000(0.0075) = $600 3.58) Given Data: loan amount = $130,000, point charged = 3%, N = 360 months, interest rate = 0.75% per month, actual amount loaned = $126,100: •

Monthly repayment: A = $130,000( A / P,0.75%,360) = $1,046

•

Effective interest rate on this loan $126,100 = $1, 046( P / A, i,360) i = 0.7787% per month

∴ ia = (1 + 0.007787)12 − 1 = 9.755% per year 3.59) (a)

$50,000 = $7,500(P / A,i,5) + $2,500( P / G,i,5) i = 6.914%

(b) P = $50,000 Total payments = $7,500 + $10,000 + … + $17,500 = $62,500 Interest payments = $3,456.87 + … + $1,131.66 = $12,500

End of month 0 1 2 3 4 5

Interest Payment $0.00 $3,456.87 $3,177.34 $2,705.64 $2,028.48 $1,131.66

Principal Payment $0.00 $4,043.13 $6,822.66 $9,794.36 $12,971.52 $16,368.34

Remaining Balance $50,000.00 $45,956.87 $39,134.21 $29,339.85 $16,368.34 $0

3.60)

(a) Amount of dealer financing = $15,458(0.90) = $13,912 A = $13, 912(A / P,11.5% / 12, 60) = $305.96

(b) Assuming that the remaining balance will be financed over 56 months, B4 = $305.96(P / A,11.5% / 12, 56) = $13, 211.54 A = $13, 211.54(A / P,10.5% / 12, 56) = $299.43

I dealer = $305.96 × 4 + $13,211.54 − $13,912 = $523.38 Interest payments to the credit union: I union = $299.43 × 56 − $13, 211.54 = $3,556.54

3.61)

•

The monthly payment to the bank: Deferring the loan payment for 6 months is equivalent to borrowing $16, 000( F / P, 0.75%, 6) = $16, 733.64

To pay off the bank loan over 36 months, the required monthly payment is = A $16, = 733.64( A / P, 0.75%,36) $532.13 per month

•

The remaining balance after making the 16th payment: $532.13( P / A, 0.75%, 20) = $9,848.67

•

The loan company will pay off this remaining balance and will charge $308.29 per month for 36 months. The effective interest rate for this new arrangement is: $9,848.67 = $308.29( P / A, i,36) ( P / A, i,36) = 31.95 i = 0.66% per month

∴= r 0.66% ×12 = 7.92% per year

3.62) = $18, 000 A( P / A, 0.667%,12) + A( P / A, 0.75%,12)( P / F , 0.667%,12) = A(11.4958) + A(11.4349)(0.9234) = 22.05479 A A = $816.15 3.63)

Given: i = 9% / 12 = 0.75% per month, deferred period = 6 months, N = 36 monthly payments, first payment due at the end of 7th month, the amount of initial loan = $15,000 (a) First, find the loan adjustment required for the 6-month grace period. $15,000( F / P,0.75%,6) = $15,687.78

Then, the new monthly payments should be

A = $15,687.78( A / P,0.75%,36) = $498.87

(b) Since there are 10 payments outstanding, the loan balance after the 26th payment is B26 = $498.87( P / A,0.75%,10) = $4,788.95 (c) The effective interest rate on this new financing is $4,788.95 = $186(P / A,i,30) i = 1.0161% per month r = 1.0161% × 12 = 12.1932% ia = (1 + 0.010161)12 − 1 = 12.90%

3.64) (a) Using the bank loan at 9.2% compound monthly Purchase price = $22,000, Down payment = $1,800 A = $20,200( A / P,(9.2 / 12)%,48) = $504.59

(b) Using the dealer’s financing, Purchase price = $22,000, Down payment = $2,000, Monthly payment = $505.33, N = 48 end of month payments. Find the effective interest rate: $505.33 = $20, 000( A / P, i, 48) i = 0.8166% per month APR( = r ) 0.8166%= ×12 9.80%

3.65) • 24-month lease plan: P = ($2,500 + $520) + $500 + $520(P / A,0.5%,23) −$500(P / F,0.5%,24) = $13,884.13 • Up-front lease plan:

P = $12,780 + $500 − $500(P / F,0.5%,24) = $12,836.4 Select the single up-front lease plan. 3.66) Given: purchase price = $155,000, down payment = $25,000 •

Option 1: i 7.5% = = /12 0.625% per month , N = 360 months

•

Option 2: For the assumed mortgage, P1 = $97, 218 , i1 = 5.5% / 12 = 0.458% per month , N1 = 300 months , A1 = $597 per month ; For

the 2nd mortgage P2 = $32,782 , i2 9% = = /12 0.75% per month , N 2 = 120 months (a) For the second mortgage, the monthly payment will be A2 = P2 (A / P,i2 , N 2 ) = $32, 782(A / P, 0.75%,120) = $415.27

$130, 000 $597( P / A, i,300) + $415.27( P / A, i,120) = i = 0.5005% per month = r 0.5005%= ×12 6.006% per year ia = 6.1741%

(b) Monthly payment •

Option 1: A = $130,000( A / P,0.625%,360) = $908.97

•

Option 2: $1,012.27 (= $597 + $415.27) for 120 months, then $597 for remaining 180 months.

Option 1: I = $908.97 × 360 − $130, 000 = $197, 229.20 Option 2: I = $228,932.4 − $130,000 = $98,932.4

(d) Equivalent interest rate:

$908.97( P / A, i,360) = $597( P / A, i,300) + $415.27( P / A, i,120) i = 1.2016% per month r = 1.2016% × 12 = 14.419% per year ia = 15.4114%

3.67)

No answers given

3.68) = P $50( P / A,3%,14) + $1, 000( P / F ,3%,14) = $1, 225.92 3.69) If you left the $15,000 in your savings account, the total balance at the end of 48 months at 8% interest compounded monthly would be = FI $15, = 000( F / P,8%/12, 48) $20, 635 The earned interest during this period is then I = $20, 635 − $15, 000 = $5, 635

Now if you borrowed $15,000 from the dealer at interest 11% compounded monthly over 48 months, the monthly payment would be = A $15, = 000( A / P,11%/12, 48) $388

You can easily find the total interest payment over 48 months under this financing by I= ($388 × 48) − $15, 000= $3, 624

It appears that you save about $2,011 in interest ($5,635 - $3,624). However, reasoning this line neglects the time value of money for the portion of principal payments. Since your money is worth 8%/12 interest per month, you may calculate the total equivalent loan payment over the 48-month period. This is done by calculating the equivalent future worth of the loan payment series. = FII $388( = F / A,8%/12, 48) $21,863.77 Now compare FI with FII . The dealer financing would cost $1,229 more in future dollars at the end of the loan period.

3.70) (a) = A $60, = 000( A / P,13% /12,360) $664 (b)

$60,000 = $522.95(P / A,i,12) +$548.21(P / A,i,12)(P / F,i,12) +$574.62(P / A,i,12)(P / F,i,24) +$602.23(P / A,i,12)(P / F,i,36) +$631.09(P / A,i,12)(P / F,i,48) +$661.24(P / A,i,300)(P / F,i,60) Solving for i by trial and error yields i = 1.0028% ia = (1 + 0.010028)12 − 1 = 12.72%

Comments: With Excel, you may enter the loan payment series and use the IRR(range, guess) function to find the effective interest rate. Assuming that the loan amount ($60,000) is entered in cell A1 and the following loan repayment series in cells A2 through A361, the effective interest rate is found with a guessed value of 11.5/12%: = IRR( = A1: A361, 0.95833%) 0.010028

(c) Compute the mortgage balance at the end of 5 years: • Conventional mortgage: = B60 $664( = P / A,13% /12,300) $58,873.84 • FHA mortgage (not including the mortgage insurance): = B60 $635.28( = P / A,11.5% /12,300) $62, 498.71 (d) Compute the total interest payment for each option: • Conventional mortgage(using either Excel or Loan Analysis Program at the book’s website—http://www.prenhall.com/park):

I = $178,937.97 • FHA mortgage:

I = $163,583.28

(e) Compute the equivalent present worth cost for each option at i = 6% /12 = 0.5% per month: • Conventional mortgage: = P $664( = P / A, 0.5%,360) $110, 749.63 • FHA mortgage including mortgage insurance: P = $522.95(P / A,0.5%,12) +$548.21( P / A,0.5%,12)(P / F,0.5%,12) +$574.62(P / A,0.5%,12)(P / F,0.5%,24) +$602.23( P / A,0.5%,12)(P / F,0.5%,36) +$631.09(P / A,0.5%,12)(P / F,0.5%,48) +$661.24(P / A,0.5%,300)(P / F,0.5%,60) = $105,703.95 The FHA option is more desirable (least cost).

Chapter 4 Equivalence Calculations under Inflation 4.1)

= F $3.50(1.04)(1.06)(1.08) = $4.13 = $4.13 $3.50(1 + f )3 f = 5.65%

4.2)

Correction: average price index → average inflation rate (a) 428.76(1 + f ) 4 = 477.91 f = 2.75% (b)

477.91(1 + 0.0275)3 = 518.43

4.3) 100(1 + 0.05)(1 + 0.08) = 113.40 100( F / P, f , 2) = 113.40 f = 6.4894%

4.4) 538,400 − 504,000 = 6.825% 504,000 577,000 − 538,400 = 7.169% f2 = 538,400 629,500 − 577,000 = 9.099% f3 = 577,000 f1 =

 629,500  f =  504,000 

1/ 3

− 1 = 7.69%

4.5) Given : f = 8%

1(1 + 0.08) − n = 0.5 1.08− n = 0.5 (−n) log1.08 = log 0.5 n = − log 0.5 / log1.08 = 9 years

Comments: If you use the Rule of 72, you may find

72 = 9 years which is equal 8

to the actual value. 4.6)

243.73 = 26.87(1 + f )62 f = 3.62% $72,500(1 + 0.0362) −62 = $7,993 4.7) Given: i = 13%, f = 5%, Maintenance costs are given in constant dollars.

i' =

i − f 0.13 − 0.05 = = 7.62% 1 + 0.05 1+ f

= P $25, 000( P / F , 7.62%,1) + $26, 000( P / F , 7.62%, 2) +$28, 000( P / F , 7.62%,3) + $30, 000( P / F , 7.62%, 4) +$32, 000( P / F , 7.62%,5) = $112, 672 A = $112, 672( A / P,13%,5) = $32, 034

4.8)

Given: i = 14%, f = 5%

Actual dollars

Constant Dollars

$25,000

$25,000(P/F,5%,0) = $25,000

35,000

35,000(P/F,5%,4) = 28,795

45,000

45,000(P/F,5%,5) = 35,259

55,000

55,000(P/F,5%,7) = 39,087

4.9) Given: = P $25, = 000, i 0.75% per month, f = 0.33% per month • 20th payment in actual dollars:

A20 = $25, 000( A / P, 0.75%, 48) = $622.13 • 20th payment in constant dollars:

= A '20 $622.13( = P / F , 0.33%, 20) $582.07

4.10) Given: = i 9%, = f 3.8% , we find the inflation-free interest rate as follows: = i'

i − f 0.09 − 0.038 = = 5.01% 1 + 0.038 1+ f

First compute the equivalent present worth of the constant dollar series at i ' : P = $1, 000( P / A,5.01%, 4) = $3,545.13

Then, we compute the equivalent annual payment in actual dollars using i: A = $3,545.13( A / P,9%, 4) = $1, 094.27

4.11)

Given: = i 12%, = f 6% , bond interest rate = 9% compounded semiannually, face

value = $1,000

• The 15th interest payment in actual dollars:

I15 = $1, 000(0.045) = $45 • The 15th interest payment (7.5th year) in constant dollars:

I '15 = $45( P / F , 6%, 7.5) = $29.07

4.12)

0.02 12 (1 + ) −1 ia = 12 = 2.0184% $10, 000[( F / P, 2.0184%,5) − ( F / P,3%,5)] = −$541.97 I will lose $541.97 when the general inflation rate is 3%. 4.13) = i ' 4%, = f 6% i = 0.04 + 0.06 + (0.04)(0.06) = 10.24%

No, I would not be interested in an investment opportunity with 10% of interest rate.

4.14)

Given: = i 8%, = f 4%, 10 annuity payments in actual dollars

P = $72, 000( P / A,8%,10) = $483,126 Comments: Since the annuity payments are made in actual dollars, we use the market interest rate to find its equivalent lump sum amount in today’s dollars.

4.15)

Given: = i 15%, = f 7%

(a) Constant-dollar analysis: we need to find the inflation-free interest rate.

i' =

i − f 0.15 − 0.07 = = 7.48% 1 + 0.07 1+ f

Then, find the equivalent present worth of this geometric series at i ' . = = P $10, 000( P / A1 , g , i ', N ) $10, 000( P / A1 ,8%, 7.48%, 4) 1 − (1 + 0.08) 4 (1 + 0.0748) −4  = $10, = 000   $10, 000(3.7487) − 0.0748 0.08   = $37, 487.20

(b) Actual-dollar analysis Period

Net Cash Flow in Constant $

Conversion factor

Net Cash Flow in Actual $

$10,000.00

(1 + 0.07)1

$10,700.00

$10,800.00

(1 + 0.07) 2

$12,364.92

$11,664.00

(1 + 0.07)3

$14,288.90

$12,597.12

(1 + 0.07) 4

$16,512.25

= P $10, 700.00( P / F ,15%,1) + $12,364.92( P / F ,15%, 2) + $14, 288.90( P / F ,15%,3) + $16,512.25( P / F ,15%, 4) = $37, 490.12 Comments: As an alternative way of finding the equivalent cash flows in actual dollars, we may use the compound growth rate (geometric growth and inflation): g= (1 + 0.08)(1 + 0.07) − 1 = 15.56% P = $10, 000(1.07)( P / A1 ,15.56%,15%, 4) = $37, 490

Slight differences are due to rounding errors.

4.16) (a) i =4% semiannually $50( P / A, 4%,10) + $1, 000( P / F ,8%,5) = $405.54 + $680.58

= $1, 086.12 (b) i− f = 4.85% 1+ f $50( P / A, 2.425%,10) + $1, 000( P / F , 4.85%,5) = $439.30 + $789.15

= i'

= $1, 228.45 4.17) Given: = i′ 4%, = f 6%

i = 0.04 + 0.06 + (0.04)(0.06) = 0.1024 $50,000(1+ f )5 ( P / F ,10.24%,5) = $50,000(1+0.06)5 (0.6142) = $41, 096.91

4.18) Given: i = 1% per month, f = 0.5% per month, P = $25,000, N = 60 months

0.01 − 0.005 1 + 0.005 = 0.4975% A ' = $25, 000( A / P, 0.4975%, 60) = $482.97 i' =

4.19) Given: i ' = 6%, f = 5%, N = 5 years, A = $2.2 million in constant dollars • Market interest rate: i = 0.06 + 0.05 + (0.06)(0.05) = 11.3%

• Actual dollar analysis:

Period

Net Cash Flow in Constant $

Net Cash Flow in Actual $

Equivalent Present Worth

1 2 3 4 5

$2,200,000 2,200,000 2,200,000 2,200,000 2,200,000

$2,310,000 2,425,500 2,546,775 2,674,114 2,807,819

$2,075,472 1,957,992 1,847,162 1,742,606 1,643,968 $9, 267, 200

P = $2,310, 000( P / F ,11.3%,1) +  + $2,807,819( P / F ,11.3%,5) = $9, 267, 200

4.20)

Given: i = 0.75% per

month

(APR

9%),

f = 0.5% per

month,

P = $8, 000, N = 24 months

(a) Inflation-free interest rate:

= im '

0.0075 − 0.005 = 0.2488% per month 1 + 0.005

i ' = (1 + 0.002488 ) − 1 = 3.0263% per year 12

(b) Equal monthly payment in constant dollars:

A ' = $8, 000( A / P, 0.2488%, 24) = $343.80

4.21)

Given: i = 6% compounded monthly, f = 4% compounded annually, number of months to deposit = 240 months, number of annual withdrawals = 15, first withdrawal = 6 months after retirement.

•

Effective inflation rate per semiannual: Since the first withdrawals is made after 6 months from retirement, it is necessary to calculate the effective inflation rate per semiannual.

per semiannual •

Annual withdrawals in actual dollars: On semiannual basis, the first withdrawal will be made after 41 semiannual periods. Then, we can calculate the equivalent amount of this withdrawals in actual dollars using this formula Actual dollar = Constant dollar (F/P, f , N )

•

Then, the conversion of constant dollar to actual dollar is as follows:

Actual dollars

$134,050

$139,411

$144,986

$150,784

$156,815

$163,086

$169,608

$176,391

$183,445

59 61 63 65 67 69

$190,782 $198,427 $206,364 $214,618 $223,203 $232,131

Equivalence calculation: To find the required equal monthly deposit amount (A), we establish the following equivalence relationship:  0.06  = im =  0.5%  12  12

 0.06  ia = 1 +  − 1 = 6.168% 12  

A( F / A,0.5%, 240)( F / P,0.5%,6) = $134,050 + $139, 411( P / F ,6.168%,1) +$144,986( P / F ,6.168%, 2) + ⋅⋅⋅ +$232,131( P / F ,6.168%,14) A = $1,747, 271 / 476.08 = $3,670 per month.

$232 131

$134,050 (240 months)

Years

Monthly Deposits (A)

4.22)

Given : i = 0.5% per month, f = 4% per year (a) • Actual dollar analysis: A( F / A, 0.5%, 480) = $1, 000, 000( F / P, 4%, 40) (1,991.4907) A = $4,801, 021 A = $2, 410.77

(b) • Effective annual interest rate: ia = (1 + 0.06 /12)12 − 1 = 6.1678%

• Equivalent purchasing power of $1,000,000 today at the end of 65th birthday: $1, 000, 000( F / P, 4%, 40) = $4,801, 021 (in actual dollars) • Conversion of gradient series to equivalent uniform series: A = G ( A / G, 6.1678%, 40) = $1, 000(12.1962) = $12,196.20

• Amount of the first deposit ( A1 ) : ( A1 + $12,196.20)( F / A, 6.1678%, 40) = $4,801, 021 ( A1 + $12,196.20)(161.4438) = $4,801, 021 A1 + $12,196.20 = $29, 738.03 A1 = $17,541.83 4.23) (a) i =i′ + f + i′f = 0.05 + 0.06 + 0.05(0.06) = 0.113

(b) = A( F / A,11.3%,8)

[ 40, 000( P / A,11.3%, 4) + 1, 000( P / G,11.3%, 4)] ( F / P,11.3%,1)

11.9897 A = $141,929.68 A = $11,837.63

4.24)

Given: i = 8% per year, f = 6% per year (a) Freshman-year expense in actual dollars:

$40,000(F / P,6%,10) = $71,632 (b) Equivalent single-sum amount at n = 0

i' =

i− f

1+ f = (0.08 − 0.06) / (1 + 0.06)

= 0.01887 P = [$40,000(P / A,1.887%,3) +$40,000](P / F,1.887%,10) = $129,076.84

A = $129,076.84( A / P,8%,10) = $19,236.2 4.25) Consider the following project’s after-tax cash flow and the expected annual general inflation rate during the project period: End of year

Cash flow in actual dollars

Expected general inflation rate

0 1 2 3

-$45,000 32,000 32,000 32,000

3.5% 4.2% 5.5%

(a) The average annual general inflation rate:

(1 + 0.035)(1 + 0.042)(1 + 0.055) = 1.1378 (1 + f )3 = 1.1378 f = 4.40%

(b) Constant dollars:

Actual dollars

Constant dollars

0 1

-$45,000 32,000

-$45,000 32,000(0.9662) = 30,918

2 3

32,000 32,000

32,000(0.9272) = 29,670 32,000(0.8789) = 28,125

Conversion factors: ( P / F ,3.5%,1) = 0.9662 ( P / F , 4.2%,1)( P / F ,3.5%,1) = 0.9272 ( P / F ,5.5%,1)( P / F , 4.2%,1)( P / F ,3.5%,1) = 0.8789

(c) The project is still profitable in an inflationary economy. P= −$45, 000 + $30,918( P / F ,5%,1) +$29, 670( P / F ,5%, 2) + $28,125( P / F ,5%,3) = $35, 653 > 0

4.26) •

Saving $2,000 per month (actual dollars) for 10 years, i=0.8333% per month, 10 years = 120 months

$2, 000( F / A, 0.8333%,120) = $409, 680.57 •

Retirement income $10,000 per month (today dollar), i =0.5% per month, 20 years = 240 months, inflation rate: 4% per year = 0.3333% per month $10, 000( P / A1 , 0.3333%, 0.5%, 240)( P / F ,10%,15) = $471,919.80

•

Contribution to college with $1,000,000 (actual dollars)

$1, 000, 000( P / F , 6%, 20)( P / F ,10%,15) = $74, 643.57 •

A vacation home in Sedona Buying at year 10: $500, 000(1 + 0.04)10 = $740,122.14 in actual dollars House value with increasing rate 5%: $740,122.14(1 + 0.05)35 ( P / F , 6%, 20)( P / F ,10%,15) = $304, 734.26 Therefore, the required savings (A) in each month in years 11 through 25 :

A( P / A, 0.8333%,180) = −$409, 680.57 + $471,919.80 + $74, 643.57 +$740,122.14 − $304, 734.5 = $572.270.68 A = $6,149.51

4.27)

Let i be the effective interest rate per month. Then, (1 + i )12 − 1 =0.0677 (1 + i )12 = 1 + 0.0677 i= (1 + 0.0677 )

1/12

−1

= 0.5474%

P = $10,000 +$100(P / A,0.5474%,480) = $26,938.67 4.28) (a) Real after-tax yield on bond investment: • Nontaxable municipal bond:

0.09 − 0.03 = 5.825% 1 + 0.03

= i 'municipal

• Taxable corporate bond: 0.12(1 − 0.3) − 0.03 = 5.243% 1 + 0.03 The municipal bond provides a greater return on investment. i'corporate =

(b)

Given : i = 6%, and f = 3% , i 'savings = 2.91%

Since i 'municipal >2.91% and i 'corporate >2.91%, 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. After-tax Cash Flow n

Municipal

Corporate

Incremental

-$10,000

$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.

Chapter 5 Present-Worth Analysis 5.1)

$450, 000 (No. of engineer:5) Revenue = $45 × (0.40)(5, 000) × 5 = n

Inflow

Outflow

Net flow

$500,000

-$500,000

$450,000

175,000

275,000

$450,000

175,000

275,000

…

$450,000

175,000

275,000

5.2) (a) Cash inflows: (1) savings in labor, $45,000 per year, (2) salvage value, $3,000 at year 5. (b) Cash outflows: (1) capital expenditure = $30,000 at year 0, (2) operating costs = $5,000 per year. (c) Estimating project cash flows: n

Inflow

Outflow

Net flow

$30,000

-$30,000

$45,000

5,000

40,000

45,000

5,000

40,000

45,000

5,000

40,000

45,000

5,000

40,000

45,000+3,000

5,000

43,000

5.3) Project cash flows over the project life Cost of n

Cmax

Demand

0 1

6,000,000

3,000,000

Revenue

Expense

Bldg.

NCF

1,527,776

-$1,527,776

16,256,976

6,462,108

9,794,868

6,000,000

3,300,000

17,882,673

7,096,319

10,786,354

6,000,000

3,630,000

19,670,941

7,793,951

11,876,990

6,000,000

3,993,000

21,638,035

8,561,346

13,076,689

6,000,000

4,392,300

23,801,838

9,405,481

14,396,358

6,000,000

4,831,530

26,182,022

10,334,029

12,000,000

5,314,683

28,800,224

11,355,432

17,444,793

12,000,000

5,846,151

31,680,247

12,478,975

19,201,272

12,000,000

6,430,766

34,848,271

13,714,872

21,133,399

12,000,000

7,073,843

38,333,099

15,074,359

23,258,739

12,000,000

7,781,227

42,166,408

16,569,795

25,596,613

12,000,000

8,559,350

46,383,049

18,214,775

28,168,274

12,000,000

9,415,285

51,021,354

20,024,252

30,997,102

12,000,000

10,356,814

56,123,490

22,014,678

24,000,000

11,392,495

61,735,839

24,204,145

1,545,123

♠

1,573,302

♠

14,302,870

32,535,510 37,531,693

♠: The cost of building is given as if Cmax is being built from scratch. No “credit” is given for the capacity already in place. This assumption could be rather unrealistic. In that case, what we need to do is to identify the incremental cost of adding the additional capacity above the existing capacity. 5.4) (a) Conventional payback period: $100,000/$25,000 = 4 years (b) Discounted payback period at i = 15%: n

Net Cash Flow

-$100,000

Cost of Funds (15%)

Cumulative Cash Flow -$100,000

$25,000 -$100,000(0.15) =-$15,000

-$90,000

$25,000

-$90,000(0.15) =-$13,500

-$78,500

$25,000

-$11,775

-$65,275

$25,000

-$9,791

-$50,066

$25,000

-$7,510

-$32,576

$25,000

-$4,886

-$12,463

$25,000

-$1,869

$10,668

Payback period = 6.54 years 5.5) (a) Conventional payback period: 0.75 years (b) Discounted payback period = 0.85 years, assuming continuous payments:

Net Cash Flow

-$30,000

+$40,000

Cost of Funds (15%)

Cumulative Cash Flow -$30,000

-$30,000(0.15) = -$4,500

+$5,500

5.6) (a) It will take 3 years to recover the total investment. n

Inflow

Outflow

Net Cash Flow

Cumulative CF

$35,500

-$35,500

$12,000

-$23,500

$12,000

-$11,500

$12,000

$500

$12,000

$12,500

$17,000

$29,500

(b) It will take 5 years to recover the total investment. Cost of funds n

Cash Flow

17%

Cumulative CF

-$35,500

$12,000

-$6,035

-$29,535

$12,000

-$5,021

-$22,556

$12,000

-$3,835

-$14,391

$12,000

-$2,446

-$4,837

$17,000

-$822

$11,341

5.7) (a) Payback Period

Project

(b) Discounted Payback Period

3.40

3.70

1.67

1.99

3.18

3.66

2.60

3.25

5.8) PW(12%) = −$250, 000 − $20, 000 + $90, 000( P / A,12%,5) + $75, 000( P / F ,12%,5)

= $96,986.87 Since PW(12%) > 0, this purchase should be justified. 5.9) (a) PW(10%) = −$28,500 + $25,500( P / F ,10%,1) + $27,980( P / F ,10%, 2) + $32, 660( P / F ,10%,3) + $40, 230( P / F ,10%, 4) = $69,821.36

(b) Draw the graphs using Excel or other software (not provided). 5.10) Given: Estimated remaining service life = 25 years, current rental income = $250,000 per year, O&M costs = $85,000 for the first year increasing by $5,000 thereafter, salvage value = $50,000, and MARR = 12% . Let A0 be the maximum investment required to break even. − A0 + [$250, 000( F / A,12%, 25) + $25, 000( F / A,12%, 20) PW(12%) = +$27,500( F / A,12%,15) + $30, 250( F / A,12%,10) +$33, 275( F / A,12%,5) + $50, 000]( P / F ,12%, 25) −$85, 000( P / A,12%, 25) − $5, 000( P / G,12%, 25) =0

∴ Solving for A0 = yields A0 = $1, 241, 461

5.11) (a) −$800 + $3, 000( P / F ,10%,3) PW(10%) A = = $1, 453.9 −$1,800 + $600( P / F ,10%,1) PW(10%) B = +$900( P / F ,10%, 2) + $1, 700( P / F ,10%,3) = $766.49 −$1, 000 − $1, 200( P / F ,10%,1) PW(10%)C = +$900( P / F ,10%, 2) + $3,500( P / F ,10%,3) = $1282.3 PW(10%) D = −$6, 000 + $1,900( P / A,10%, 2) +$2,800( P / F ,10%,3) = −$598.91

(b) Not provided. 5.12)

PW(15%) = − X + $120, 000( P / A,15%, 4) + 0.1X ( P / F ,15%, 4) = 0 $342,597= X − 0.0572 X X * = $363,382 5.13) PW(15%) = − A0 + $60, 000( P / A,15%,10) =0 A0 = $301,126

5.14)

PW(9%) = −$4, 000 + $3, 400( P / F ,9%,1) +$3, 400( P / F ,12%,1)( P / F ,9%,1) +$1,500( P / F ,10%,1)( P / F ,12%,1)( P / F ,9%,1) +$3,500( P / F ,13%,1)( P / F ,10%,1)( P / F ,12%,1)( P / F ,9%,1) +$4,300( P / F ,12%,1)( P / F ,13%,1)( P / F ,10%,1)( P / F ,12%,1)( P / F ,9%,1) = $7,858.34

5.15) Given: Estimated remaining service life = 25 years , current rental income = $250,000 per year, O&M costs = $65,000 for the first year increasing by $6,000 thereafter, salvage value = $200,000 , and MARR = 15% . Let A0 be the maximum investment required to break even.

= PW(15%) $250, 000( P / A,15%,5) + $275, 000( P / A,15%,5)( P / F ,15%,5) +$302,500( P / A,15%,5)( P / F ,15%,10) +$332, 750( P / A,15%,5)( P / F ,15%,15) +$366, 025( P / A,15%,5)( P / F ,15%, 20) −$65, 000( P / A,15%, 25) − $6, 000( P / G,15%, 25) + $200, 000( P / F ,15%, 25) = $1,116, 775

5.16) (a)

FW(15%) A = −$12,500( F / P,15%,3) + $6, 400( F / P,15%, 2) +$14, 400( F / P,15%,1) + $7, 200 = $13, 213 FW(15%) B = −$12,500( F / P,15%,3) − $3,000( F / P,15%, 2) +$23,000( F / P,15%,1) + $13,000 = $16, 472

= FW(15%)C $22,500( F / P,15%,3) − $7,000( F / P,15%, 2) −$2,500( F / P,15%,1) + $4,000 = $26,087 FW(15%) D = −$14,000( F / P,15%,3) + $7,500( F / P,15%, 2) +$4,500( F / P,15%,1) + $8,500 = $2,302

∴ All projects are acceptable.

(b) Sample calculation for Project A: PB(15%)0 = −$12,500 −$12,500(1 + 0.15) + $6, 400 = −$7,975 PB(15%)1 = −$7,975(1 + 0.15) + $14, 400 = PB(15%) 2 = $5, 228.75 = PB(15%) $5, 228.75(1 + 0.15) + $7,= 200 $13, 213.06 3 The terminal project balance is the same as the net future worth of the project. 5.17) (a)

PB(i )3 =−$7, 000(1 + i ) + $1,840 =−$6, 000 i = 12% (b) The original cash flows of the project are as follows. n

-$10,000

2,200

-9,000

3,080

-7,000

1,840

-6,000

6,820

100

Project Balance

5.18) Given: Initial cost = $3, 000, 000 , annual savings = $1, 200, 000 , Annual O&M costs = $250, 000 , annual income taxes = $150, 000 , Salvage value = $200, 000 , useful life = 10 years, MARR = 18%

PW(18%) = −$3, 000, 000 +[$1, 200, 000 − $250, 000 −$150, 000]( P / A,18%,10) +$200, 000( P / F ,18%,10) = $633, 482 The project is a profitable one. 5.19) (a) No graphs are given. Project Balance, PB(i)n

Period (n)

-$2,500.00

-$3,500.00

-$2,800.00

-$2,300.00

-2,750.00

-2,250.00

-4,880.00

-3,530.00

-3,025.00

325.00

-6,268.00

-1,983.00

-3,327.50

3,857.50

-2,394.80

118.70

+2,539.75

6,443.25

1,865.72

1,630.57

-$3,025

$325

-$6,268

-$1,983

(b) Project PB2

Project B would be preferred because it has recovered all of its investment and some surplus at the end of year 2. 5.20) (a) From the project balance diagram, note that PW(24%)1 = 0 for project 1 and PW(23%) 2 = 0 for project 2.

PW(24%)1 = −$1, 000 + $400( P / F , 24%,1) + $800( P / F , 24%, 2) + X ( P / F , 24%,3) = 0

PW(23%) 2 = −$1, 000 + $300( P / F , 23%,1) + Y ( P / F , 23%, 2) + $800( P / F , 23%,3) = 0

X = $299.58,

Y = $493.49

(b) Since PW(24%)1 = 0 , FW(24%)1 = 0.

ⓑ = $499.58,

5.21) (a) The original cash flows of the project are as follows n

Project Balance

-$3,000

($600)

-$2,700

$1,470

-$1,500

($1,650)

(-$300)

-$300

$600

($270)

(b) • PB(i ) 2 = −$2, 700(1 + i ) + $1, 470 = −$1,500 ∴ i = 10% • PW(10%) $270( = = P / F ,10%,5) $167.65

5.22) 5.1 (a)

Period (n)

Project Balance, PB(i)n A

-$1,000.00

-1,100.00

-506.00

-413.00

-600.00

-1,210.00

37.40

133.70

-60.00

634.00

635.14

634.07

634.00

FW(10%) = FW(10%) = FW(10%) = FW(10%) = $634 A B C D

Note that the terminal project balances are the same (rounding errors), which are the net future worth of the projects. (b) Discounted payback period: A: 2.67 years, B: 1.93 years, C: 1.77 years, D: 2.09 years (c) & (d) Period (n)

Area of Negative Project Balance, PB(i)n A

-$1,000.00

-1,100.00

-506.00

-413.00

-600.00

-1,210.00

Total Area

3,310.00

-60.00 1,506.00

1,413.00

1,660.00

It appears that Project C recovers its investment earliest among the four projects, so we can say that it has the minimum exposure to risk if other things are equal. 5.23) (a)

−$200 + $50( P / A,10%,3) PW(10%) A = −$100( P / F ,10%, 4) +$400( P / A,10%, 2)( P / F ,10%, 4) = $330.20 −$100 + $40( P / A,10%,3) PW(10%) B = +$10( P / A,10%, 2)( P / F ,10%,3) = $12.51 = PW(10%) $120 − $40( P / A,10%,3) C = $20.53

All projects are acceptable. (b)

FW(10%) A = $330.20( F / P,10%, 6) = $584.97 FW(10%) B = $12.51( F / P,10%,5) = $20.15 FW(10%)C = $20.53( F / P,10%,3) = $27.32 All projects are acceptable. (c) FW(i ) A = [−$200( F / P,10%,3) + $50( F / A,10%,3)]( F / P,15%,3) +[−$100( F / P,15%, 2) + $400( F / A,15%, 2)] = 574.60 FW(i ) B = [−$100( F / P,10%,3) + $40( F / A,10%,3)]( F / P,15%,3) +$10( F / P,15%, 2) + $10( F / P,15%,1) = $23.66 FW(i )C = $120( F / P,10%,3)( F / P,15%,3) −$40( F / A,10%,3)( F / P,15%,3) = $41.55

5.24) (a)

PW(0%) A = 0 PW(18%) B $575( = = P / F ,18%,5) $251.34 PW(12%)C = 0 (b) Assume that A2 = $500.

PB(12%) 2 = −$530(1.12) + $500 = X X = −$93.60 . (c) The net cash flows for each project are as follows: Net Cash Flow n

-$1,000

$200

$500

$590

$200

$500

$200

$300

-$106

$200

$300

$147

$200

$300

$100

Sample calculation for Project C: PW(12%)0 = −$1, 000 PW(12%)1 = −$1, 000(1.12) + A1 = $530 Solving for A1 yields A1 = $590. (d)

FW(0%) A = 0 FW(18%) B = $575 FW(12%)C = 0

5.25)

= PW(5%) $1, 000, 000( P / A,5%,5) + $1,300, 000( P / A,5%,5)( P / F ,5%,5) + ($1,500, 000 / 0.05)( P / F ,5%,10) = $8, 739, 412 + $18, 417,398 = $27,156,810

5.26)  Find the equivalent annual series for the first cycle:

A = [$100(P / A,15%, 2) + $40(P / A,15%, 2)(P / F,15%, 2) + $20(P / F,15%, 5)](A / P,15%, 5) = $66.13  Capitalized equivalent amount: CE(15%) =

$66.13 = $440.88 0.15

5.27) Given: r =5% compounded monthly, maintenance cost = $80,000 per year ia = (1 +

0.05 12 ) − 1 = 5.116% 12

CE(5.116%) =

$80, 000 = $1,563, 664 0.05116

5.28) Given: Construction cost = $15,000,000, renovation cost = $3,000,000 every 15 years, annual O & M costs = $1,000,000 and i = 5% per year (a)

P1 = $15, 000, 000 $3, 000, 000( A / F ,5%,15) 0.05 = $2, 780,537 P3 = $1, 000, 000 / 0.05

P2 =

= $20, 000, 000 CE(5%) = P1 + P2 + P3 = $37, 780,537

(b) P1 = $15, 000, 000 $3, 000, 000( A / F ,5%, 20) 0.05 = $1,814,555 P3 = $1, 000, 000 / 0.05

P2 =

= $20, 000, 000 CE(5%) = P1 + P2 + P3 = $36,814,555

(c) • 15-year cycle with 10% of interest: P1 = $15, 000, 000 $3, 000, 000( A / F ,10%,15) 0.1 = $944, 213 P3 = $1, 000, 000 / 0.1

P2 =

= $10, 000, 000 CE(10%) = P1 + P2 + P3 = $25,944, 213

• 20-year cycle with 10% of interest:

P1 = $15, 000, 000 $3, 000, 000( A / F ,10%, 20) 0.1 = $523, 789 P3 = $1, 000, 000 / 0.1

P2 =

= $10, 000, 000 CE(10%) = P1 + P2 + P3 = $25,523, 789

As interest rate increases, CE value decreases. 5.29) Given: Cost to design and build = $830,000 , rework cost = $120,000 every 10 years, new type of gear = $80,000 at the end of 5th year, annual operating costs = $70,000 for the first 15 years and $100,000 thereafter $120,000( A / F,7%,10) 0.07 +$80,000(P / F,7%,5)

CE(7%) = $830,000 +

+$70,000(P / A,7%,15) $100,000 (P / F,7%,15) 0.07 = $2,166,448.62 +

5.30) (a)

PW(0.5%) = $2, 460( P / A, 0.5%, 240) = $343,368.70 (b)

PW(0.5%) = $2, 460( P / A, 0.5%, 480) = $447, 099.06 (c)

A i $2, 460 A= 0.005 = $492, 000

PW(0.5%) =

Comments: Longer life means greater total benefit, but most of the benefit is collected in the first 20 years.

5.31) (a)

−$1, 000 + $912( P / F , 25%,1) PW(25%) A = + $684( P / F , 25%, 2) + $456( P / F , 25%,3) + $228( P / F , 25%, 4) = $494.22 −$1, 000 + $284( P / F , 25%,1) PW(25%) B = + $568( P / F , 25%, 2) + $852( P / F , 25%,3) + $1,136( P / F , 25%, 4) = $492.25 Select project A. (b) Project A

Cost of n

Cash Flow

funds

Project Balance

-$1000

-$1,000

$912

-$250

-$338

$684

-$85

$262

$456

$65

$783

$228

196

$1,207

Project B

Cost of n

Cash Flow

funds

Project Balance

-$1,000

$284

-$250

-$966

$568

-$242

-$640

$852

-$160

$53

$1,136

$13

$1,202

Project B is exposed to higher risk of loss if either project terminates at the end of the year 2, according to the results below.

5.32) PW(12%) A = −$22,500 + $18, 610( P / F ,12%,1) + $15,930( P / F ,12%, 2) + $16,300( P / F ,12%,3) = $18, 417 PW(12%) B = −$16,900 + $15, 210( P / F ,12%,1) + $16, 720( P / F ,12%, 2) + $12,500( P / F ,12%,3) = $18,907

∴ Select project B.

5.33) −$2, 000 + $1,300( P / F , i,1) + $1,500( P / F , i, 2) > 0 PW(i ) B-A = ∆ = −$2, 000 +

$1,300 $1,500 + >0 1+ i (1 + i ) 2

1 = −$2, 000 + $1,300 X − $1,500 X 2 , where X = . 1+ i X= 0.8 → i= 25%

∴ Select project B, if MARR < 25%.

5.34) (a) PW(15%) A = −$15, 000 + $9,500( P / F ,15%,1) +$12,500( P / F ,15%, 2) + $7,500( P / F ,15%,3) = $7, 644.03

(b) PW(15%) B = −$25, 000 + X ( P / A,15%, 2)( P / F ,15%,1) = $9,300 X = $24, 262.57

(c) Note that the net future worth of the project is equivalent to its terminal project balance. PB(15%)3 = $7, 643.7( F / P,15%,3)

= $11, 625.30 (d) Select B, which has the greater PW. 5.35) (a) Project balances as a function of time are as follows: Project Balances n

-$2,500

-$5,000

-$2,100

-$6,000

-$1,660

-$7,100

-$1,176

-$3,810

-$694

-$1,191

-$163

$1,690

$421

$3,859

$763

$7,245

$1,139

All figures above are rounded to nearest dollars. (b) Knowing the relationship FW(i ) = PB(i ) N ,

FW(10%) A = $1,139 FW(10%) D = $7, 245 (c) Assuming a required service period of 8 years −$7, 000 − $1,500( P / A,10%,8) PW(10%) B = −$1, 000( P / F ,10%,1) − $500( P / F ,10%, 2) −$1,500( P / F ,10%, 7) − $1,500( P / F ,10%,8) = −$17, 794 −$5, 000 − $2, 000( P / A,10%, 7) PW(10%)C = −$3, 000( P / F ,10%,8) = −$16,136

Select Project C. 5.36) Given: Required service period = infinite, analysis period = least common multiple service periods (6 years) • Model A:

PW(16%)cycle = −$22, 000 + $17,500( P / F ,16%,1) + $17, 000( P / F ,16%, 2) $15,330 + $15, 000( P / F ,16%,3) = PW(16%) = $15,330[1 + ( P / F ,16%,3)] total = $25,151

• Model B:

PW(12%)cycle = −$27, 000 + $20,500( P / F ,16%,1) + $28, 000( P / F ,16%, 2) = $11, 481 PW(16%)total = $11, 481[1 + ( P / F ,16%, 2) + ( P / F ,16%, 4)] = $26,354 ∴ Model B is preferred.

5.37)

(a) Without knowing the future replacement opportunities, we may assume that both alternatives will be available in the future with the same investment and expenses. We further assume that the required service period is 6 years. (b) With the common service period of 6 years, • Project A1: PW(10%)cycle = −$900 − $400( P / A,10%,3) +$200( P / F ,10%,3) = −$1, 744.48 PW(10%)total = −$1, 744.48[1 + ( P / F ,10%,3)] = −$3, 055.13

• Project A2:

−$1,800 − $300( P / A,10%, 6) PW(10%)cycle = +$500( P / F ,10%, 6) = −$2,824.34 ∴ Project A2 is preferred.

PW(10%) A 2 = −$1,800 − $300( P / A,10%,3) + S ( P / F ,10%,3) = −$2,546.06 + 0.7513S Let PW(10%) A1 = PW(10%) A2 , then S = $1, 067

5.38) (a) Assuming a common service period of 15 years

• Project B1:

PW(12%)cycle = −$20, 000 − $2, 000( P / A,12%,5) + $2, 000( P / F ,12%,5) = −$26, 074.7 PW(12%)total = −$26, 074.7[1 + ( P / A, 76.23%, 2)] = −$49, 266.29 Note: (1.12) − 1 = 76.23% 5

• Project B2:

−$17, 000 − $2,500( P / A,12%,3) + $3, 000( P / F ,12%,3) PW(12%)cycle = = −$20,869.24 −$20,869.24[1 + ( P / A, 40.49%, 4)] PW(12%)total = = −$59,180.43 Note: (1.12) − 1 = 40.49% 3

∴ Select project B1. (b) • Project B1 with 2 replacement cycles:

PW(12%) = −$26, 074.7 − $26, 074.7( P / F ,12%,5) = −$40,870.19 • Project B2 with 4 replacement cycles where the 4th replacement ends at the end of first operating year:: PW(12%) = −$20,869.24[1 + ( P / F ,12%,3) + ( P / F ,12%, 6)] − [$17, 000 − ($2,500 − $9, 000)( P / F ,12%,1)]( P / F ,12%,9) = −$50,334.10

∴ Project B1 is still a better choice.

5.39) • Model A: $25, 000( A / F ,12%,5) CE(12%) A = $60, 000 + $92, 794 = 0.12

• Model B: $180, 000( A / F ,12%,50) CE(12%) B = $150, 000 + = $150, 625 0.12

∴ Project A is more economical.

5.40) • Standard Lease Option:

PW(0.5%)SL = −$5,500 − $1,150 − $1,150( P / A, 0.5%, 23) +$1, 000( P / F , 0.5%, 24) = −$30, 690 • Single Up-Front Option:

PW(0.5%)SU = −$29,500 + $1, 000( P / F , 0.5%, 24) = −$28, 613

∴ Select the single up-front lease option.

5.41) • Machine A: PW(13%) A = −$75, 200 − ($6,800 + $2, 400)( P / A,13%, 6) +$21, 000( P / F ,13%, 6) = −$101,891 • Machine B: PW(13%) B = −$44, 000 − $11,500( P / A,13%, 6)

= −$89,972

∴ Machine B is a better choice. 5.42) (a) • Required HP to produce 10 HP: -Motor= A: X 1 10 = / 0.85 11.765 HP -Motor = B: X 2 10 = / 0.90 11.111 HP • Annual energy cost: -Motor A: 11.765(0.7457)(2,000)(0.09) = $1,579.17 -Motor B: 11.111(0.7457)(2,000)(0.09) = $1,491.39 • Equivalent cost:

−$1, 200 − $1,579.17( P / A,8%,15) PW(8%) A = +$50( P / F ,8%,15) = −$14, 710.11 −$1, 600 − $1, 491.39( P / A,8%,15) PW(8%) B = +$100( P / F ,8%,15) = −$14,334

∴ Motor B is preferred.

(b) With 1,000 operating hours: Annual energy cost: -Motor A: 11.765(0.7457)(1, 000)(0.09) = $789.58 -Motor B: 11.111(0.7457)(1, 000)(0.09) = $745.69 Equivalent cost:

−$1, 200 − $789.58( P / A,8%,15) PW(8%) A = +$50( P / F ,8%,15) = −$7,942.62 −$1, 600 − $745.69( P / A,8%,15) PW(8%) B = +$100( P / F ,8%,15) = −$7,951.19

∴ Motor A is now preferred.

5.43) Since only Model B is repeated in the future, we may have the following sequence of replacement cycles: • Option 1: Purchase Model A now and repeat Model A forever. • Option 2: Purchase Model B now and replace it at the end of year 2 by Model A. Then repeat Model A forever.

−$8, 000 + $3,500( P / A,12%,3) PW(12%) A = = $406.41 −$15, 000 + $10, 000( P / A,12%, 2) PW(12%) B = = $1,900.51 (a) • Option 1: $406.41( A / P,12%,3) 0.12 = $1, 410.08

PW(12%) AAA =

• Option 2: PW(12%) $1,900.51 + = BAA = $3, 024.62

∴ Option 2 is a better choice.

$406.41( A / P,12%,3) ( P / F ,12%, 2) 0.12

(b) Let S be the salvage value of Model A at the end of year 2.

−$8, 000 + $3,500( P / A,12%, 2) + S ( P / F ,12%, 2) = $1,900.51 −$2, 084.82 + S (0.7972) = $1,900.51 Solving for S yields

S = $4,999.16

5.44) • Since either tower will have zero salvage value after 20 years, we may select the analysis period of 35 years:

PW(11%) Bid A = −$137, 000 − $2, 000( P / A,11%,35) = − $154, 710 PW(11%) Bid B = −$128, 000 − $3,300( P / A,11%,35) = − $157, 222 ∴ Bid A is a better choice. • If you assume an infinite analysis period, the present worth of each bid will be:

[−$137, 000 − $2, 000( P / A,11%, 40)]( A / P,11%, 40) 0.11 = − $157,322

PW(11%) Bid A =

[−$128, 000 − $3,300( P / A,11%,35)]( A / P,11%,35) 0.11 = − $161, 407

PW(11%) Bid B =

∴ Bid A is still preferred.

5.45)

• Option 1: Non-deferred plan

PW(12%) = −$300, 000 − $49, 000( P / A,12%,8) = − $543, 414.35 • Option 2: Deferred plan −$140, 000( P / F ,12%, 2) − $14, 000( P / A,12%, 6)( P / F ,12%, 2) PW(12%) = − $160, 000( P / F ,12%,5) − $21, 000( P / A,12%,3)( P / F ,12%,5) − $180, 000( P / F ,12%,8) = − $349, 600.72

∴ Option 2 is a better choice.

5.46) •

Option 1: Tank/tower installation PW(12%)1 = −$164, 000

•

Option 2: Tank/hill installation with the pumping equipment replaced at the end of 20 years at the same cost PW(12%) = −($120, 000 + $12, 000)

−($12, 000 − $1, 000)( P / F ,12%, 20) +$1, 000( P / F ,12%, 40) − $1, 000( P / A,12%, 40) = −$141,374 Option 2 is a better choice. 5.47) •

Project A with repeating 2 times,

PW(12%) A = −1, 000 − 400( P / F ,12%,1) − 400( P / F ,12%, 2) −1, 200( P / F ,12%,3) − 400( P / F ,12%, 4) −400( P / F ,12%,5) − 200( P / F ,12%, 6) = −$3,112.66 •

Project B with repeating 3 times,

PW(12%) B = −800 − 200( P / F ,12%,1) − 1, 000( P / F ,12%, 2) −200( P / F ,12%,3) − 1, 000( P / F ,12%, 4) −200( P / F ,12%,5) − 200( P / F ,12%, 6). = −$2, 768.45 Since $3,112.66 − $2, 768.45 = $344.21 , select (a). 5.48) •

Alternative A: Once-for-all expansion PW(15%) A = −$30 M − $0.40 M ( P / A,15%, 25) +$0.85M ( P / F ,15%, 25) = −$32,559,839

•

Alternative B: Incremental expansion

−$10 M − $18M ( P / F ,15%,10) PW(15%) B = −$12 M ( P / F ,15%,15) + $1.5M ( P / F ,15%, 25) −$0.25M ( P / A,15%, 25) −$0.10 M ( P / A,15%,15)( P / F ,15%,10) −$0.10 M ( P / A,15%,10)( P / F ,15%,15) = −$17, 700, 745 Select alternative B. 5.49) •

Option 1: Process device A lasts only 4 years. You have a required service period of 6 years. If you take this option, you must consider how you will satisfy the rest of the required service period at the end of the project life. One option would subcontract the remaining work for the duration of the required service period. If you select this subcontracting option along with the device A, the equivalent net present worth would be

PW(12%)1 = −$100, 000 − $60, 000( P / A,12%, 4) +$10, 000( P / F ,12%, 4) −$100, 000( P / A,12%, 2)( P / F ,12%, 4) = −$383, 292 •

Option 2: This option creates no problem because its service life coincides with the required service period. PW(12%) 2 = −$150, 000 − $50, 000( P / A,12%, 6) +$30, 000( P / F ,12%, 6) = −$340,371

•

Option 3: With the assumption that the subcontracting option would be available over the next 6 years at the same cost, the equivalent present worth would be PW(12%)3 = −$100, 000( P / A,12%, 6)

= −$411,141 With the restricted assumptions above, option 2 appears to be best alternative. 5.50) (a) Year

Outflow

2011

$14,500,000

2012

$3,500,000

2013

$26,000,000

= FW(15%) $14,500, 000( F / P,15%, 2) + $3,500, 000( F / P,15%,1) +$26, 000, 000 = $49, 201, 250

(b)

A( P / A,15%,10) = FW(15%) = $49, 201, 250 A(5.0188) = $49, 201, 250 A = $9,803, 450

Chapter 6 Annual Equivalence Method 6.1)

6.2)

6.3)

AE(9%) = $300, 000( A / P,9%,5) = $77,127.74

= AE(12%) A= ( P / A,12%,3) $250, 000 A = $104, 087.25

AE(10%) = [−$3, 000 − $3, 000( P / A,10%, 2) +$3, 000( P / A,10%, 4)( P / F ,10%, 2) +$1, 000( P / G,10%, 4)( P / F ,10%, 2)]( A / P,10%, 6) = $751

AE(8%) = −$2,154( A / P,8%,6) $400( P / F ,8%,1) + X ( P / A,8%, 2)( P / F ,8%,1)  +  ( A / P,8%,6)  +$400( P / F ,8%, 4) + X ( P / A,8%, 2)( P / F ,8%, 4)  = $200 6.4)

−465.94 + (370.36 + 1.65 X + 294 + 1.31X )(0.2163) 200 = −1, 489.78 + 2.96 X 924.64 = X = $815.68 6.5)

AE(15%) A = −$3,500( A / P,15%, 4) + $600 $900( P / F ,15%,1) + $1, 200( P / F ,15%, 2)  +  ( A / P,15%, 4)  +$400( P / F ,15%,3)  = $58.13 > 0 (Accept) AE(15%) B = −$4,000( A / P,15%, 4) + $1,500 +$100( P / F ,15%,1)( A / P,15%, 4) = $129.39 > 0 (Accept) −$6,500( A / P,15%, 4) + $2,000 + $500( F / A,15%, 2)( A / F ,15%, 4) AE(15%)C = = −$61.43 < 0 (Reject)

6.6) $2, 000 AE(15%) = $3, 000 + [ ( P / F ,15%, 6)]( A / P,15%, ∞) 0.15 = $3, 000 + $864.66 = $3,864.66

6.7)

PW(10%) = −$300 + $100( P / F ,10%,1) − $160( P / F ,10%, 2) +  + $130( P / F ,10%, 6) = $105.54 AE(10%) = $105.44( A / P,10%, 6) = $24.23 6.8) −$4, 400( A / P,12%,3) + $12,500( A / F ,12%,3) AE(12%) A = = $1,872.42 > 0 (Accept) −$3, 400( A / P,12%,3) + $1, 000 AE(12%) B = +$500( A / G,12%,3) = $46.71 > 0 (Accept) −$3,300( A / P,12%,3) + $3, 000 AE(12%)C = −$1, 000( A / G,12%,3) = $701.43 (Accept) AE(12%) D = −$6, 200( A / P,12%,3) + $3,800 = $1, 218.63 (Accept) 6.9) Since the project has the same cash flow cycle during the project life, you just can consider the first cycle.

AE(10%) = [−$2,500 + $1,800( P / F ,10%,1) + $900( P / F ,10%, 2) +$500( P / F ,10%,3)]( A / P,10%,3) = $102.87 ∴ Accept the project. 6.10)

$1.5 $5 $9 $12 $10 + + + + = $17,835,518.55 2 3 4 1.15 1.15 1.15 1.15 1.155 = AE(15%) $17,835,518.55( = A / P,15%,5) $5,320,335.18 −$5 + PW(15%) =

Yes, the project is justified. 6.11)

3,880 = (25, 000 − S )( A / P,10%,10) + 0.1S S = $3, 006.35

6.12) CR(6%)3 years = ($21, 635 − $6, 057.80)( A / P, 6%,3) + (0.06)($6, 057.80) = $6,191.05 CR(6%)5 years = ($21, 635 − $3, 677.95)( A / P, 6%,5) + (0.06)($3, 677.95) = $4, 483.68 6.13) a) Capital recovery cost: CR(10%) = ($40, 000 − $15, 000)( A / P,10%, 2) + $15, 000(0.10) = $15,905

b) Annual savings: = PW(10%) $28, 000( P / F ,10%,1) + $40, 000( P / F ,10%, 2) = $58,512.40 AE(10%)1 = $58,512.40( A / P,10%, 2)

= $33, 714.84

AE(10%)= $33,714.84 − $15,905 = $17,809.84

c) = AE(10%) C[4, 000( P / F ,10%,1) + 6, 000( P / F ,10%, 2)]( A / P,10%, 2) = 4,952.46C C = 17,809.84 / 4,952.46 = $3.596 / hour 6.14) CR(12%) = ($45, 000 − $9, 000)( A / P,12%,5) + $9, 000(0.12) = $11, 066.4

OC(12%)= $15,000+$2,000(A / G,12%, 5) = $18,549.2

AE(12%) = $11, 066.4 + $18,549.2 = $29, 615.6

6.15)

•

Capital recovery cost: CR(15%) =− ($135, 000 $6, 000)( A / P,15%,5) + $6, 000(0.15) = $39,383

•

Annual operating costs: $75, 000

∴ Equivalent annual cost AEC(15%) = $39, 383 + $75, 000 = $114, 383

6.16) (a)

AE(15%) = −$4,500( A / P,15%, 4) + $1,000 +( X − $1,000)( P / F ,15%, 2)( A / P,15%, 4) =0 AE(15%) = −$841.21 + 0.26468 X = 0 X = $3,175.33

(b) = AE(12%) B $6,500( A / P,12%, 4) − $1, 400 = $740.02 > 0 AE(12%) A = −$744.01 + 0.262464 X > $740.02

X * > $5,654.30 6.17) CE(10%) = $100, 000 +

$10, 000 $20, 000 + ( A / F ,10%, 4) 0.1 0.1

= $243,094.2 6.18) = Given: I $65, = 000, S $5, = 000, N 10 years, i = 9% (a)

AE(9%) = ($65, 000 − $5, 000)( A / P,9%,10) 1 +$5, 000(0.09) = $9, 799 (b)

AE(9%) = $18, 000 + $2,500( A / G,9%,10) 2 = $27, 495

(c)

AE(9%) = $27, 495 − $9,799 = $17,696

∴ This is a good investment. 6.19)

CR(10%) = ($500, 000 − 100, 000)( A / P,10%,15) + 100, 000(0.1) = $62, 600 = AE(10%) $40, 000 X − $30, 000 X CR(10%) = AE(10%) $62, 600 = $10, 000 X X = 6.26 (or rounds up to 7)

6.20) (a) AE(15%) = −$4,500( A / P,15%, 4) + $1,000 +( X − $1,000)( P / F ,15%, 2)( A / P,15%, 4) =0 AE(15%) = −$841.21 + 0.26468 X = 0 X = $3,175.33

(b) = AE(12%) B $6,500( A / P,12%, 4) − $1, 400

= $740.02 > 0 AE(12%) A = −$744.01 + 0.262464 X > $740.02 X * > $5,654.30

6.21) The total investment consists of the sum of the initial equipment cost and the installation cost, which is $145,000. Let R denote the break-even annual revenue.

AE(12%) = −$145, 000( A / P,12%,10) − $50, 000 −$5, 000 + $10, 000 + R =0 Solving for R yields R = $70, 663

6.22) •

New lighting system cost: = AEC(12%) $50, 000( A / P,12%, 20) + ($8, 000 + $3, 000) = $17, 694

•

Old lighting system cost: AEC(12%) = $20, 000

Annual savings from installing the new lighting system = $2,306 6.23) •

Capital recovery cost CR(12%) = ($30, 000 − $18, 000)( A / P,12%, 2) + $18, 000(0.12) = $9, 260

• $35, 000 $42, 000 + ]( A / P,12%, 2) 1.12 1.122 = $38,302 Net annual savings =$38,302 − $9, 260 AE savings = (12%) [

= $29, 042

•

C (6, 000) C (8, 000) + ]( A / P,12%, 2) 1.12 1.122 = $6,943.3C $29, 042 = $6,943.3C C = $4.18 per hour

= AE hours (12%) [

6.24) Capital recovery cost: CR(15%) = ($56, 000 − $5, 000)( A / P,15%,5) + $5, 000(0.15) + $6, 000 = $21,963.3

The machine cost per hour would be $8.78(=21,963.3/2,500)

6.25)

AE(14%) = [−$100, 000 + $35, 000( P / A1 , −3%,14%,5)]( A / P,14%,5) = $4, 095.13

C = $4,095.13 / 3,000 = $1.365 / hr 6.26) •

Option 1: Pay employee $0.56 per mile

AEC(10%) = $0.56(30, = 000) $16,800 total cost •

Option 2: Provide a car to employee: CR(10%) = ($25, 000 − $8, 000)( A / P,10%,3) + (0.10)($8, 000) = $7, 636 O& M(10%) =+ $1, 200 ($0.3)(30, 000) = $10, 200 AEC(10%) =$7, 636 + $10, 200 =$17,836 = Cost per mile $17,836 = / 30, 000 $0.5945

∴ Option 1 is a better choice. 6.27) •

Option 1: Purchase units from Tompkins Unit cost = $25 + ($70, 000 − $35,000) / 20,000 − $3.50 = $23.25

•

Option 2: Make units in house

PW(15%) dm = $63, 000( P / A1 ,5%,15%,5) = $230, 241 PW(15%) dl = $190,800( P / A1 , 6%,15%,5) = $709, 491 PW(15%)vo = $139, 050( P / A1 ,3%,15%,5) = $490,888 = ($230, 241 + $709, 491 AEC(15%) +$490,888)( A / P,15%,5) + $70, 000 = $496, 776 Unit cost = $496, 776 / 20, 000 = $24.84 Option 1 is a better choice. 6.28) •

Capital costs:

CR(7%) = ($25, 000 − $2, 000)( A / P, 7%,12) 1 +(0.07)($2, 000) = $3, 036 Annual battery replacement cost:

•

= AEC(7%) 2 $3, 000[( P / F , 7%,3) + ( P / F , 7%, 6)

•

+( P / F , 7%,9)]( A / P, 7%,12) = $763.14 Annual recharging cost:

•

= AEC(7%)3 ($0.015)(20, = 000) $300 Total annual equivalent costs: AEC(7%)= $3, 036 + $763.14 + $300 + $700 = $4, 798.84

•

Cost per mile: cost/mile = $4,798.84 / 20,000 = $0.2399

6.29) •

Annual total operating hours: (0.70)(8, 760) = 6,132 hours per year

•

Annual electricity generated: 50, 000 × 6,132 = 306, 600, 000 kilowatt-hours

•

Equivalent annual cost:

= AEC(14%) $85, 000, 000( A / P,14%, 25) + $6, 000, 000 = $18,367,364

•

Cost per kilowatt-hour: $18, 367, 364 / 306, 600, 000 = $0.06 per kilowatt-hour

6.30)

•

Annual equivalent revenue:

AE = $32, 000 + 40, 000X Revenue •

Annual equivalent cost:

= AEC(8%)Cost $800, 000( A / P,8%, ∞) + $133, 000 + $50, 000( A / F ,8%,5) =$64, 000 + $133, 000 + $8,525 = $205,525 AE Revenue = AECCost $32, 000 + 40, 000 X = $205,525 X = $4.34 6.31)

Salvage Value: $1, 200, 000( F / P,5%, 25) = $4, 063, 680 CR(12%) = ($6, 000, 000 − $4, 063, 680)( A / P,12%, 25) + (0.12)$4, 063, 680 = $734,522.4 AECO&M= $100 ×12 × 40 + $400, 000= $448, 000 = CR(12%) + AE= AEC(12%) $1,182,522.4 per year ma AEC(0.9489%) Monthly = $1,182,522.4( A / F , 0.9489%,12) = $93,506 per month

6.32) •

Minimum operating hours: AEC(10%) = ($32, 000 − $2, 000)( A / P,10%,15) +(0.10)($2, 000) + $800 = $4,944.21 Let T denote the annual operating hours. Then the total kilowatt-hours generated would be 40T . Since the value of the energy generated is considered to be $0.09 per kilowatt-hour, we can establish the following relationship:

$0.09 × 40T = $4,944.21 Solving for T yields •

T = 1,373.4 hours

Annual worth of the generator at full load operation: AE(10%) = ($0.09)(100, 000) − $4,944.21 = $4, 055.79

•

Discounted payback period at full load of operation:

Investment

0 1  15

-$32,000

Revenue

 +$2,000

$9,000  $9,000

Maintenance cost

Net Cash flow

-$800  -$800

-$32,000 8,200  10,200

$32,000 = $8,200( P / A,10%, n)

Solving for n yields n = 5.19 years

6.33)

•

Capital recovery cost: CR(6%) = ($170, 000 − $12, 000)( A / P, 6%,12) + (0.06)($12, 000) = $19,565.77

•

Annual operating costs:

AEC(6%)O = $70, 000 + $15, 000 + $5, 000 = $90, 000

•

Total annual system costs: AEC(6%)= $19, 565.77 + $90, 000= $109, 565.77

•

Number of rides required per year: Number of rides = $109,565.77 / $0.10 = 1,095,658 rides

6.34) •

Pump I: (

180 )(0.746)(0.06)T = $9.368T 0.86 = AEC(8%) I $6, 000( A / P,8%,12) + $500 + $9.368T = $1, 296.2 + $9.368T

•

Pump II:

180 ( )(0.746)(0.06)T = $10.071T 0.8 = AEC(8%) II $4, 000( A / P,8%,12) + $440 + $10.071T = $970.8 + $10.071T •

6.35)

$1,296.2 + 9.368T = $970.8 + 10.071T T = 463 hours

Given: Investment cost = $10 million, plant capacity = 300,000 lbs/hour, plant operating hours = 4,000 hours per year, O&M cost = $4 million per year, useful life = 15 years, salvage value = $800,000, and MARR = 15%. (a)

PW(15%) = −$10, 000, 000 + ( R − $4, 000, 000)( P / A,15%, 6) = 3.7845 R − $25,137,930.78 =0 Solving for R yields R = $6,642,370 per year

(b) Minimum processing fee per lb (after-tax):

$6,642,370 = $0.00554 per lb. (300,000)(4,000) Comment: The minimum processing fee per lb should be higher on a before-tax basis.

6.36)

Let C denote the green fee per round during the first year. •

Capital cost: = CR(15%) ($20, 000, 000 − $25, 000, 000( A / P,15%,10) + (0.15)($25, 000, 000) = $2, 753, 740

•

Operating and maintenance cost: O&M(15%) = $650, 000 + $50, 000( A / G,15%,10) = $819,160

•

Equivalent annual revenue: AE(15%) Revenue = $15 × 40, 000 +40, 000(1.15)C ( P / A1 ,5%,15%,10)( A / P,15%,10) = $600, 000

•

1 − (1 + 0.05)10 (1 + 0.15) −10  +$46, 000C   ( A / P,15%,10) 0.15 − 0.05   = $600, 000 + 54, 752C Breakeven green fee:

$600, 000 + 54, 752C= $2, 753, 740 + $819,160 54, 752C = $2,972,900 C = $54.30 6.37)

Let X denote the average number of round-trip passengers per year. •

Capital costs: = CR(15%) ($12, 000, 000 − $2, 000, 000)( A / P,15%,15) +(0.15)($2, 000, 000)

•

= $2, 010,171 Annual crew costs: $225,000

•

Annual fuel costs for round trips:

•

($1.10)(3, 280)(2)(3)(52) = $1,125, 696 Annual landing fees: ($250)(3)(52)(2) = $78, 000

•

Annual maintenance, insurance, and catering costs: $237,500 + $166, 000 + $75 X = $403,500 + $75 X

•

Total equivalent annual costs: AEC(15%) = $2, 010,171 + $225, 000 + $1,125, 696 +$78, 000 + $403,500 + $75 X = $3, 400 X Solving for X yields X = 1,156 Passenger-round-trips per year or 1,156 /(52)(3) = 7.41 ≈ 8 Passengers per round trip

6.38) •

Option 1: Purchase-annual installment option:

= A $65, = 000( A / P,9%,5) $16, 711 AE(10%)1 = −$5, 000( A / P,10%,5) − $16, 711 = −$18, 030

•

Option 2: Cash payment option:

AE(10%) 2 = −$66, 000( A / P,10%,5) = −$17, 411

∴ Option 2 is a better choice.

6.39) AEC(10%)1 = ($95, 000 − $12, 000)( A / P,10%,3) + (0.1)$12, 000 +$3, 000 = $37,575.53 AEC(10%) 2 = ($120, 000 − $25, 000)( A / P,10%, 6) + (0.1)$25, 000 +$9, 000 = $33,312.70 AEC(10%)1 − AEC(10%) 2 =$37,575.53 − $33,312.70 = $4, 262.83

6.40) a) We suppose that the current projects are expected to continue for an indefinite period. These two projects will be available in the future, without significant changes in revenue expectations. b) For A,

PW(10%) A-3year-cycle = −$10, 000 + $6, 000( P / F ,10%,1) + $5, 000( P / F ,10%, 2) +$4, 000( P / F ,10%,3) = $2,592 AE(10%) A-3-year-cycle = $2,592( A / P,10%,3) = $1, 042.24

= $2,529[1 + ( P / F ,10%,3)] PW(10%) A-6-year-cycle = $4, 429.08 AE(10%) A-6-year-cycle = $4, 429.08( A / P,10%, 6) = $1, 016.92

For B,

−$12, 000 + $7, 000( P / F ,10%,1) + $8, 000( P / F ,10%, 2) PW(10%) B-2-year-cycle = = $975.21 AE(10%) B-2-year-cycle = $975.21( A / P,10%, 2) = $561.89 PW(10%) B-6-year-cycle = $975.21[1 + ( P / F ,10%, 2) + ( P / F ,10%, 4)] = $2, 447.25 AE(10%) B-6-year-cycle = $2, 447.25( A / P,10%, 6) = $561.89

Choose the project A, which has a higher annual equivalent worth.

6.41) (a)

−$40, 000( A / P,12%, 4) + [$19,120 − $1, 280( A / G,12%, 4)] AE(12%) A = = $4, 211.59 −$40, 000( A / P,12%, 4) + $17,350 AE(12%) B = = $4,180.62

(b) Process A: $4, 211.29 / 3, 000 = $1.40 /hour Process B: $4,180.62 / 3, 000 = $1.39 /hour (c) Process A is a better choice. 6.42) Let T denote the total operating hours in full load. •

Motor I (Expensive): Annual power cost: 180 × (0.746) × (0.05) × T = $8.089T 0.83 Equivalent annual cost of operating the motor: = AEC(6%) I $5,800( A / P, 6%,10) + $870 + $8.089T = $1, 658.03 + $8.089T

•

Motor II (Less expensive): Annual power cost:

180 × (0.746) × (0.05) × T = $8.392T 0.80

Equivalent annual cost of operating the motor: = AEC(6%) II $4, 600( A / P, 6%,10) + $690 + $8.392T = $1,314.99 + $8.392T

•

Let AEC(6%) I =AEC(6%) II and solve for T . $1, 658.03 + $8.089T = $1,314.99 + $8.392T T = 1,132.15 hours per year

6.43) •

Equivalent annual cost: AEC(13%) ($1,300, 000 − $60, 000)( A / P,13%, 20) = A +(0.13)($60, 000) + $70, 000 + $40, 000 = $294,318.70 AEC(13%) ($850, 000 − $30, 000)( A / P,13%,10) = B +(0.13)($30, 000) + $100, 000 + $30, 000 = $285, 017.44

•

Processing cost per ton: C1 = $294,318.70 / (20)(365) = $40.32 per ton C2 = $285,017.44 / (20)(365) = $39.04 per ton

∴ Incinerator B is a better choice. 6.44) Assumption: jet fuel cost = $4.80 /gallon • System A : Equivalent annual fuel cost: A1 = ($4.80/gal)(40,000gals/1,000 hours)(2,000 hours) = $384, 000 (assuming an end of-year convention)

AEC(10%) fuel = [$384, 000( P / A1 , 6%,10%,3)]( A / P,10%,3) = $405,981 AEC(10%) ($100, 000 − $10, 000)( A / P,10%,3) = sys . A +(0.10)($10, 000) + $405,981 = $443,170 • System B : Equivalent annual fuel cost: A1 = ($4.80/gal)(32,000gals/1,000 hours)(2,000 hours) = $307, 200 AEC(10%) fuel = [$307, 200( P / A1 , 6%,10%,3)]( A / P,10%,3) = $324, 785 AEC(10%) ($200, 000 − $20, 000)( A / P,10%,3) = sys . B

•

+ (0.10)($20, 000) + $324, 785 = $399,163 Equivalent operating cost ( including capital cost ) per hour:

System A $443,170 = = / 2, 000 $221.6 per hour System B $399,163 = = / 2, 000 $199.6 per hour System B is a better choice.

6.45) Since the required service period is 12 years and the future replacement cost for each truck remains unchanged, we can easily find the equivalent annual cost over a 12-year period by simply finding the annual equivalent cost of the first replacement cycle for each truck. •

Truck A: Four replacement cycles are required

AEC(12%) = ($15, 000 − $5, 000)( A / P,12%,3) A

•

+(0.12)($5, 000) + $3, 000 = $7, 763.50 Truck B: Three replacement cycles are required = AEC(12%) ($20, 000 − $8, 000)( A / P,12%, 4) B +(0.12)($8, 000) + $2, 000 = $6,910.80 Truck B is a more economical choice.

6.46) (a) Number of decision alternatives (required service period = 5 years):

Alternative

Description

Buy Machine A and use it for 4 years. Then lease a machine for one year.

Buy Machine B and use it for 5 years.

Lease a machine for 5 years.

Buy Machine A and use it for 4 years. Then buy another Machine A and use it for one year.

Buy Machine A and use it for 4 years. Then buy Machine B and use it for one year.

There are five alternatives. Both A4 and A5 are feasible but we do not consider the alternatives because we need to know the salvage values of the machines after one-year use. (b) With lease, the O&M costs will be paid by the leasing company: • For A1: −$6,500 + $600( P / F ,10%, 4) PW(10%)1 = −$800( P / A,10%, 4) − $200( P / F ,10%,3) −$100( P / F ,10%, 2) − ($3, 000 + $100)( P / F ,10%, 4) = −$10,976.33 AEC(10%)1 = $10,976.33( A / P,10%,5) = $2,895.53

• For A2: −$8,500 + $1, 000( P / F ,10%,5) PW(10%) 2 = −$520( P / A,10%,5) − $280( P / F ,10%, 4) = −$10, 042 AEC(10%) 2 = $10, 042( A / P,10%,5) = $2, 649

• For A3:

AEC(10%) = [$3, 000 + $3, 000( P / A,10%, 4)]( A / P,10%,5) 3 = $3,300

∴ A2 is a better choice. 6.47) •

Option 1: AEC(18%)1 = $200, 000(180)( A / P,18%, 20) −(0.08)($200, 000)(180)( A / F ,18%, 20) +($0.005 + 0.215)(180, 000, 000) = $46,305,878

•

cos t/lb = $46,305,878 /180, 000, 000 = $0.2573 per 1b Option 2: AEC(18%) = ($0.05 + $0.215)(180, 000, 000) 2 = $47, 700, 000 cos t/lb = $47, 700, 000 /180, 000, 000 = $0.2650 per 1b Option 1 is a better choice.

6.48) Given: Required service period = indefinite, analysis period = indefinite Plan A: Incremental investment strategy: •

Capital investment :

CR(10%)1 = [$1,500, 000 +$1,500, 000( P / F ,10%,15)]( A / P,10%, ∞) •

= $185,910 Supporting equipment: CR(10%) = [($200, 000 + $200, 000 / 3.1772)( P / F ,10%,30)] 2 ×( A / P,10%, ∞) = $1,507 Note that the effective interest rate for 15-year period is

(1 + 0.1)15 − 1 =3.1772 •

Operating cost:

•

 [$91, 000( P / A,10%,15)     +$182, 000( P / A,10%,5)( P / F ,10%,15)]  OC(10%)3  $185, 000  ( A / P,10%, ∞) + $3, 000( P / G,10%, ∞)]  +[  0.10   × ( P / F ,10%, 20)]   = $117, 681.33 Note that ( P / G, i, ∞) =1/ i 2 or ( P / G,10%, ∞) =100 Total equivalent annual worth: AEC(10%) A= $185,910 + $1,507 + $117, 681 = $305, 098

Plan B: One time investment strategy: •

Capital investment: = CR(10%)1 $1,950, 000( A / P,10%, ∞)

•

= $195, 000 Supporting equipment:

= CR(10%) 2

$350, 000 ( A / P,10%, ∞) 16.4494 = $2,128

Note that the effective interest rate for 30-year period is

(1 + 0.1)30 − 1 = 16.4494 •

Operating cost:

•

OC(10%) = [$105, 000( P / A,10%,15) +$155, 000( P / A,10%, ∞)( P / F ,10%,15)] ×( A / P,10%, ∞) = $80, 235 Total equivalent annual worth:

AEC(10%)= $195, 000 + $2,128 + $80, 235 = $277,363

Plan B is a better choice.

6.49) •

$84,000 = $1,400 per kW. But if you consider the 60 time value of money, say 10% annual interest, the capital cost per kW without considering any salvage value at the end of its service life is as follows:

Installed cost per kilowatt =

$84, 000( A / P,10%,10) $13, 671 = = $227.84 per kW 60 60 or $13, 671 = $0.026 per kWh 60 × 24 × 365 •

Operating cost per kilowatt-hour: $19, 000 = $0.036 (60)(24)(365)

6.50) •

Make option: AEC(14%) Make = $4,582, 254 or $4,582, 254 / (48 × 79,815) = $1.196 / unit

•

Buy option: AEC(14%) = CR (14%) + $4,331,127 Buy = ($405, 000 − $45, 000)( A / P,14%, 7) + (0.14)($45, 000) + $4,331,127 = $90, 249 + $4,331,127 = $4, 421,376 or $4, 421,376 / (48 × 79,815) = $1.154 / unit

6.51) Given: annual energy requirement = 145, 000, 000, 000 BTUs, 1-metric ton = 2, 204.6 lbs (an approximation figure of 2,000 lbs was mentioned in the case problem), net proceeds from demolishing the old boiler unit = $1, 000 . (a) Annual fuel costs for each alternative: •

Proposal 1: 145, 000, 000, 000 BTUs (0.75)(14,300) = 13,519,814 lbs 13,519,814 = 2, 204.6 = 6,132.45 tons Annual fuel cost 6,132.45 × $125 = = $766,566.25

Weight of dry coal =

•

Proposal 2:

145, 000, 000, 000(0.94) (0.78)(1, 000, 000) = $2,533, 782 145, 000, 000, 000(0.06) Oil cost = $2.93 (0.81)(139, 400) = $225, 755.93 Annual= fuel cost $2,533, 782 + $225, 755.93 = $2, 759,537.93 (b) Unit cost per steam pound:

Gas cost = $14.5

•

Proposal 1: Assuming a zero salvage value of the investment AEC(10%)= ($2,570,300 + $145, 000 − $1, 000)( A / P,10%, 20) +$766,568.75 = $1, 085,389.41

Unit cost = $1, 085,389.41/145, 000, 000 = $0.007485 per steam lb •

Proposal 2:

= AEC(10%) ($1, 289,340 − $1, 000)( A / P,10%, 20) +$2, 759,537.93 = $2,910,865.86

unit cost = $2,910,865.86 /145, 000, 000 = $0.020075 per steam lb (c) Select Proposal 1.

Chapter 7 Rate of Return Analysis Note: Symbol convention---The symbol i* represents the breakeven interest rate that makes the PW of the project equal to zero. The symbol IRR represents the internal rate of return of the investment. For a simple (or pure) investment, IRR = i*. For a nonsimple investment, generally i* is not equal to IRR.

7.1) = F 2= P P(1 + i ) 4 log 2 = 4 log (1+i ) log(1 + i ) = 0.07526 ∴ i ∗ = 18.92% Or $2,000 = $1000( F / P, i %, 4) i ∗ = 18.29%

7.2) = $2529.24 $1000( F / P, i %,5) + $1, 000( F / P, i %,3) i ∗ = 6%

7.3)

$24, 500 = $514.55( P / A, i, 60)

i = 0.7917% per month r = 0.7917% ×12 = 9.5%

ia = (1 + 0.007917)12 − 1 = 9.93% per year 7.4)

$950 = $35( P / A, 4%,8) + F ( P / F , 4%,8) 0.7307 F = $714.35 F = $977.64

7.5) $110,500, = 000 $19, 000(1 + i )32 i = 31.11%

7.6)

PW(i ) = −$50, 000 + $100, 000( P / F , i,1) − $40, 000( P / F , i, 2) = 0

i = 44.72%

7.7) PW(i ) = $6,500 + $4,500( P / F , i,1) − $13, 000( P / F , i, 2) = 0 i* = 10.9807%

7.8) (a)

Simple investment: Project A, D

(b) Non-simple investment: Project B (c)

• Project A: PW(i ) = −$32, 000 + $30, 000( P / A, i,3) − $10, 000( P / G, i,3) =0 i* = 49.52% • Project B: PW(i ) = −$38, 000 + $32, 000( P / A, i, 2) − $22, 000( P / F , i,3) =0 = i* 13.45% or − 41.69% • Project C: = PW(i ) $45, 000 − $18, 000( P / A, i, 3) =0 = i* 9.70% → Borrowing rate of return

• Project D:

PW(i ) = −$46, 500 + $2, 500( P / F , i,1) + $6, 459( P / F , i, 2) + $78, 345( P / F , i, 3) =0 i* = 24.94%

7.9) PW(10%) = −$2, 000 + $1, 000( P / F ,10%,1) +$1, 200( P / F ,10%, 2) + $ X ( P / F ,10%,3)

X = $132

7.10)

PW(25%) = −$12,000 + $2,500( P / F , 25%,1) + $5,500( P / F , 25%, 2) + X ( P / A, 25%, 2)( P / F , 25%, 2) =0 $6, 480 = 0.9216 X X = $7,031.25

7.11) Use Excel or Cash Flow Analyzer to find the rate of return:

PW(i ) = −$1, 000 + [$50( F / A, i,12) + $50( F / A, i,5) + $4, 000]( P / F , i,15) =0 Solving for i yields

i* = 12.08% 7.12) (a) Classification of investment projects: • Simple projects: B and E • Non-simple projects: A, C and D (b)

−$250 +

Let X =

$450 $120 − = 0 1 + i (1 + i ) 2

1 , then, (1 + i )

−$250 + $450 X − $120 X 2 = 0 = X 1 0.6782, = X 2 3.071 ∴ i* = 47.44% or -67.44%

A B C D E

47.44%, -67.44%% 20.01% 12.89% -39.46% 23.53%

7.13) (a) Classification of investment projects: • Simple projects: A,B, and D • Non-simple projects: C (b) • Project A: i* = −1.08% • Project B: i * = 11.84% • Project C: i* = 18.99% • Project D: i* = 34.76% . 7.14) (a)

−$90,000 + ($27,000 − $8,000)( P / A,i,6) + $10,000( P / F ,i,6) = 0

Solving for i yields i* = 9.4208% . (b) With the geometric expense series −$90, 000 + $27, 000( P / A, i, 6) − $8, 000( P / A1 , 7%, i, 6) +$10, 000( P / F , i, 6) = 0 Solving for i* yields i* = 7.1745% .

(c) To maintain i* = 9.4208% PW(i ) = −$90, 000 + $27, 000( P / A1 , g ,9.4208%, 6) −$8, 000( P / A1 , 7%,9.4208%, 6) + $10, 000( P / F ,9.4208%, 6) =0 Solving for g yields

7.15)

g = 2.20%

PW(15%) = −$150, 000 + $120, 000( P / A,15%,5) + $25, 000( P / F ,15%,5) = $264, 694

7.16) The present worth of the project cash flow is PW(i ) = −$35, 000 + $15, 000( P / F , i,1) + $14, 520( P / F , i, 2) +$13, 990( P / F , i, 3) =0

∴ Since= IRR 11.88% > 10% = MARR, the project is profitable.

7.17)

(a) Since IRR = 10% and PW(10%) = 0, we have,

PW(10%) = −$12,500 + $8, 000( P / F ,10%,1) + $4, 000( P / F ,10%, 2) + X ( P / F ,10%,3) = 0 ∴ X = $2557.50 (b) Since IRR > 8%, the project is acceptable.

7.18) Let X be the annual rent per apartment unit. Then the expected net cash flows are: N 0

Capital Investment 14,500,000

Revenue

Maintenance

Manager

Net Cash Flow -14,500,000

50 X

-350,000

-85,000

50X - 435,000

50 X

-400,000

-85,000

50X - 485,000

50 X

-450,000

-85,000

50X - 535,000

50 X

-500,000

-85,000

50X - 585,000

50 X

-550,000

-85,000

50X +15,365,000

16,000,000

Through the Excel Solver function by setting PW(15%) = 0, ∴ X = $49,473.35 (or $4,122.78 per month) 7.19) Let X be the annual savings in labor PW(12%) = −$35, 000 + ( X − $4, 000)( P / A,12%, 6) +$5, 000( P / F ,12%, 6) =0 X = $11,896.82

7.20) PW(18%) = −$150, 000 + ( X − 50, 000)( P / A,18%,10) +15, 000( P / F ,18%,10) =0 X = $82, 739.26

7.21) •

Net cash flow table:

N Land Building Equipment Revenue Expenses Net Cash Flow 0 -$1.50 -$3 -$4.50 1 -$4 -$4.00 2 $3.50 -$1.40 $2.10 3 $3.68 -$1.47 $2.21 4 $3.86 -$1.54 $2.32 5 $4.05 -$1.62 $2.43 6 $4.25 -$1.70 $2.55 7 $4.47 -$1.79 $2.68 8 $4.69 -$1.88 $2.81 9 $4.92 -$1.97 $2.95 10 $5.17 -$2.07 $3.10 11 $5.43 -$2.17 $3.26 12 $5.43 -$2.17 $3.26 13 $5.43 -$2.17 $3.26 14 $2 $1.40 $0.50 $5.43 -$2.17 $7.16

•

Rate of return calculation:

PW(i ) = −$4.5 − $4( P / F , i,1) + $2.1( P / F , i, 2) +  + $7.16( P / F , i,14) = 0 ∴ i* = 24.85% •

Since this is a simple investment, IRR = 24.85%. At MARR = 15%, the project is economically attractive.

7.22)

(a)

PW(i ) = −$30 + $9( P / F , i,1) + $18( P / F , i, 2) + $20( P / F , i,3) + $18( P / F , i, 4) + $10( P / F , i,5) + $5( P / F , i, 6) =0

This is a simple investment. Therefore, IRR= i*= 40.20% . ∴ Since IRR is higher than MARR, the project is acceptable. (b) IRR = 45.47% $220, 000 + $72, 000( P / F ,12%,3) = $217, 248 $94, 000( F / P,15%, 2) + $144, 000( F / P,15%,1) = $289,915 $217, 248(1 + MIRR)3 = $289,915 MIRR = 10.10% $9( P / F , i,1) + $18( P / F , i, 2) + $20( P / F , i,3)  PW(i ) = −$30 + 1.1    +$18( P / F , i, 4) + $10( P / F , i,5) + $5( P / F , i, 6)  =0

$9( P / F , i,1) + $18( P / F , i, 2) + $20( P / F , i,3)  PW(i ) = −$35 + 0.9    +$18( P / F , i, 4) + $10( P / F , i,5) + $5( P / F , i, 6)  =0 7.23) (a) Nonsimple investment as there are more than one sign change in cash flows. (b)

$220, 000 + $72, 000( P / F ,12%,3) = $271, 248 $94, 000( F / P,15%, 2) + $144, 000( P / F ,15%,1) = $289,915 $271, 248(1 + MIRR)3 = $289,915 MIRR = 2.24%

(c) Since MIRR(2.24%) < 15%, Reject the project! 7.24) (a) Rate of return calculation: • Project A: i* = 19.50% • Project B: i* = 9.18% (b)

PW Plot $40,000.00 $30,000.00

$20,000.00

Project A

$10,000.00 $0.00 ($10,000.00)

($20,000.00)

8 10 12 14 16 18 20 22 24 26 28 30

Interest Rate (%)

The interest rate that makes two projects equivalent is 40%.

7.25) (a) Project A: IRR = 16.01% Project B: IRR = 18.18% (b) Both projects are simple investments, so MIRR would be the same as the IRR. Project A and B are acceptable

(c) Since the incremental cash flow displays a nonsimple investment, we attempt to find the rate of return on incremental investment: PW(i)A−B = −$15,000 + $10,000(P / F , i ,1) + $10,000(P / F , i ,2) −$10,000(P / F , i ,3) =0 A mixed investment → Need to calculate MIRR on incremental investment. $15,000+$10,000(P/F,12%,3)=$22,117 $10,000(F/P,15%,2)(F/P,15%,1)=$24,725 $22,117(1+MIRR)3 = $24,725 11.79% < 15% → accept B MIRRA−= B

7.26) (a) Project A: IRR = 18.33% Project B: IRR = 23.77% (b) Both projects are acceptable (c) n 0 1 2 3

Project A

Project B

-$100,000.00 -$100,000.00 $10,500.00 $70,000.00 $60,000.00 $50,000.00 $80,000.00 $20,500.00

A-B $0.00 -$59,500.00 $10,000.00 $59,500.00

IRR A-B = 8.76% < 10%( MARR) Project B is better choice.

7.27) Option 1: Buy a certificate. Option 2: Purchase a bond, and assume that MARR = 5% .

n 0 1 2 3 4 5

Option 1 -$10,000 0 0 0 0 15,386.24

Net Cash Flow Option 2 Option 1 – Option 2 -$10,000 $0 650 -650 650 -650 -650 650 650 -650 10,650 4736.24

The rate of return on incremental investment is

i*1−2 = 25.49% > 5% ∴ Option 1 is a better choice.

7.28) Determine the cash flow on incremental investment: Net Cash Flow

Project A1

Project A2

A2 – A1

-$4,000

-$5,000

-$1,000

2,600

3,600

1,000

2,800

3,200

400

i* A2− A=1 30.62% > 15% → accept A2. 7.29) (a) IRR on the incremental investment: Net Cash Flow n Project A1 Project A2 0 -12,000 -14,000 1 5,000 6,200 2 5,000 6,200 3 5,000 6,200

A2 – A1 -$2,000 1,200 1,200 1,200

i* A2− A1 = 36.31% (b) Since it is an incremental simple investment, IRR A2-A1 = 36.31% > 10% . Therefore, select project A2.

7.30) (a) n 0 1 2

A1 -$16,000 $7,500 $7,500

A2 -$20,000 $5,000 $15,000

A2 – A1 -$4,000 -$2,500 $7,500

$7,500

$8,000

$500

IRR A 2− A1 = 13.08% (b) Select Project A2.

7.31) (a)

IRR B = 25.99% NPWA = $2,557.74 (b)

$5,500( P / A= ,15%,3) $10,000(1 + MIRR)3 MIRR = 24.07% (c) Incremental analysis: n 0 1 2 3

Net Cash Flow Project A Project B -$10,000 -$20,000 5,500 0 5,500 0 5,500 40,000

B–A -$10,000 -5,500 -5,500 34,500

IRR B −= 24.24% > 15% → Select B. A 7.32) Incremental cash flows (Model A – Model B):

n 0 1 2 3 4

A–B -$2,376 0 0 0 2,500

IRR A− B = 1.28% ∴ If MARR < 1.28%, Model A is preferred.

7.33)

PW(i ) A = −$8, 000 + ($900 − $150)( P / A, i, 20) + $500( P / F , i, 20) PW(i ) B = −$12, 000 + ($1,100 − $100)( P / A, i, 20) + $600( P / F , i, 20) PW(i ) B − A = −$4, 000 + $250( P / A, i, 20) + $100( P / F , i, 20) =0 i = 7.11% * A

iB* = 5.65% iB* − A= 2.39% < 12% → Select A, if no do-nothing is allowed.

7.34) Let A0 = current practice, A1 = just-in-time system, A2 = stock-less supply system. •

Comparison between A0 and A1: n A0 A1 0 0 -$3,000,000 1-8 -9,000,000 -3,800,000

A1 – A0 -$3,000,000 5,200,000

i* A1−= IRR A1−= A0 A 0 173.28% > 10% → Select A1 •

Comparison between A1 and A2:

n 0 1-8

A2 -$6,000,000 -1,900,000

A1 -$3,000,000 -3,800,000

A2 – A1 -$3,000,000 1,900,000

i* A2− A1 = IRR A2− A1 = 62% > 10% → Select A2.

7.35) (a)

i1* = 54.52%, i2* = 57.61%, and i3* = 38.41% (b)

• Project 1 versus Project 2:

n 0 1 2

Project 1 -$1,500 700 2,500

Project 2 -$5,000 7,500 600

2–1 -$3,500 6,800 -1,900

$6,800(1.15) = [$3,500 + $1,900(1.15)−2 ](1 + MIRR)2 MIRR2−= 25.86% > 15% → Select Project 2. . 1 RIC2−= 47.08% > 15% → Slect Project 2. 1 • Project 2 versus Project 3:

n 0 1 2

Project 2 -$5,000 7,500 600

Project 3 -$2,200 1,600 2,000

2–3 -$2,800 5,900 -1,400

$5,900(1.15) = ($2,800 + $1,400(P / F ,15%,2))(1 + MIRR)2 MIRR = 32.61% > 15% → Select Project 2. RIC2−= 67.23% > 15% → Select Project 2. 3 Comments: If you want to apply the IRR decision rule to the non-simple investments, you should apply the net investment test and make the selection by calculating the return on invested capital (or true internal rate of return).

7.36) (c) Comments: Statement (a): Incorrect. The rate of return is still 21.46%. Statement (b): Incorrect. The rate of return is 16.016% Statement (c): Correct. F $238, = = 080( F / P,16.016%, 27) $13,143,344 Statement (d): Incorrect 7.37) All projects would be acceptable because individual ROR exceed the MARR. Based on the incremental analysis, we observe the following relationships:

IRR A2− A1 = 10% < 15% (Select A1) IRR A3− A1 = 18% > 15% (Select A3) IRR A3− A2 = 23% > 15% (Select A3) Therefore, A3 is the best alternative. 7.38)

From the incremental rate of return table, we can deduce the following relationships:

IRR A2− A1 = 8.9% < 15% (Select A1) IRR A3− A2 = 42.7% > 15% (Select A3)

IRR A4− A3 = 0% < 15% (Select A3) IRR A5− A4 = 20.2% > 15% (Select A5) IRR A6− A5 = 36.3% > 15% (Select A6) It is necessary to determine the preference relationship among A1, A3, and A6.

IRR A3− A1 = 16.66% > 15% (Select A3) IRR A6− A3 = 20.18% > 15% (Select A6) IRR A6− A1 = 18.24% > 15% (Select A6) A6 is the best alternative. 7.39) All projects are acceptable on individual basis as IRR > 29%. iB*− A = 85% > 29% → Select B. ← Investment

iB*−C = 30% > 29% → Select C. ← Borrowing iD* −C = 25% < 29% → Select C. ← Investment iA* −D = 50% > 29% → Select D. ← Borrowing B  A,C  D,C  B,D  A → Best project is C. 7.40) Let A0 = current practice, A1 = just-in-time system, A2 = stock less supply system. •

Comparison between A0 and A1: n 0 1-8

A0 0 -$5,000,000

A1 -$2,500,000 -$2,900,000

A1 – A0 -$2,500,000 $2,100,000

= i* A1− A0 IRR = 83.34% > 10% A1− A 0 A1 is a better choice.

•

Comparison between A1 and A2: n 0

A2 -$5,000,000

A1 -$2,500,000

A2 – A1 -$2,500,000

1-8

-$1,400,000

-$2,900,000

$1,500,000

58.49% > 10% i* A 2− A1 IRR = = A 2 − A1 A2 is a better choice. That means that the stockless supply system is the final choice.

7.41)

n 0 1 2

Current Pump(A) Larger Pump(B) $0 -$1,600,000 $10,000,000 $20,000,000 $10,000,000 $0

B-A -$1,600,000 $10,000,000 -$10,000,000

i* =

25% 400%

The incremental cash flows result in multiple rates of return (25% and 400%), so we may abandon the rate of return analysis. Using the PW analysis, PW(20%) = −$1.6M + $10M( P / F , 20%,1) − $10M( P / F , 20%, 2) = −$0.21M < 0 Reject the larger pump.

Comments: If we follow the procedure outlined in Appendix 7A, we will find the return on invested capital to be 4.16% at MARR of 20%, so we will reject the larger pump. 7.42) With the least common multiple of 6 project years,

n 0 1 2 3 4 5 6

Project A -$5,000 3,000 4,000 4,000 -5,000 3,000 4,000 4,000

Net Cash Flow Project B -$10,000 8,000 8,000 – 10,000 8,000 8,000 – 10,000 8,000 8,000

B–A -$5,000 5,000 -6,000 9,000 -5,000 4,000 4,000

Since the incremental cash flow series is a mixed investment, we may need to calculate the RIC (or MIRR) on incremental investment or abandon the IRR

analysis and use the PW decision rule. •

RIC: Use the Cash Flow Analyzer to find the RIC :

RICB−= 25.67% > 15% → Select B. A •

MIRR: Equi. cash outlay at n = 0: $5,000 + 6,000(1.15)−2 + $5,000(1.15)−4 = $12,395.62 Equi. cash revenue at n = 6: $5,000(1.15)5 + $9,000(1.15)3 + $4,000(1.15) + $4,000 = $32,344.66

•

$29,302.90 = $9,395.62(1 + MIRR)6 MIRRB−= 17.34% > 15% → Select B. A

PW analysis: −$5,000 + $5,000( P / F ,15%,1) PW(15%) B − A = +  + $4,000( P / F ,15%,6) = $1,587.86 > 0 → Select B.

7.43) Assumptions are required that required service period is indefinite and both projects can be repeated at the same cost in the future.

n 0 1 2 3 4 5 6 7 8 9 10 11 12

Net Cash Flow Machine A Machine B -$40,000 -$55,000 -$15,000 -$10,000 -$15,000 -$10,000 -$15,000-$40,000 -$10,000 -$15,000 -$10,000-$55,000 -$15,000 -$10,000 -$15,000-$40,000 -$10,000 -$15,000 -$10,000 -$15,000 -$10,000-$55,000 -$15,000-$40,000 -$10,000 -$15,000 -$10,000 -$15,000 -$10,000 -$15,000-$40,000 -$10,000-$55,000

A–B $15,000 -$5,000 -$5,000 -$45,000 $50,000 -$5000 -$45000 -$5,000 $50,000 -$45,000 -$5,000 -$5,000 $10,000

= IRR 42.72% > 10% A− B (Machine A should be purchased.)

7.44)

(a) Since there is not much information given regarding the future replacement options and required service period, we may assume that the required service period is 3years and project A2 can be repeated at the same cost in the future. (b) The analysis period may be chosen as the least common multiple of project lives, which is 3 years.

n 0 1 2 3

A2 – A1 -$5,000 0 0 15,000

IRR A2− A1 = 44.22% The MARR must be less than 44.22% for Project A1 to be preferred.

7.45) (a) PW(i ) = −$1, 250, 000 + $731,500( P / A, i,15) + $80, 000( P / F , i,15) =0 i* = 58.47% (a) IRR exceeds Marco’s MARR, the project is attractive and should be accepted.

i* =58.47% > MARR (=18%) 7.46) (a) Analysis period of 40 years (unit: thousand $): •

Without “mothballing” cost:

−$1,500, 000 + ($207, 000 − $69, 000)( P / A1 , 0.05%, i, 40) PW(i ) = =0 i* = 8.95%

•

With “mothballing” cost of $0.75 billion: −$1,500, 000 + ($207, 000 − $69, 000)( P / A1 , 0.05%, i, 40) PW(i ) = − $750, 000( P / F , i, 40) =0 i = 8.77% *

For a 40-year analysis period, the drop of IRR with the mothballing cost is only 0.18%, which is relatively insignificant. (b) Analysis period of 25 years (unit: thousand $): • Without “mothballing” cost: −$1,500, 000 + ($207, 000 − $69, 000)( P / A1 , 0.05%, i, 25) PW(i ) = =0 i = 7.84% *

• With “mothballing” cost of $0.75 billion: −$1,500, 000 + ($207, 000 − $69, 000)( P / A1 , 0.05%, i, 25) PW(i ) = − $750, 000( P / F , i, 25) =0 i = 6.80% *

For a 25-year analysis period, the drop of IRR with the mothballing cost is about 1.04%, which is relatively significant.

7.47) Let’s assume that your college expenses for four years would be $120,000 ($30,000 per year) and you only consider about your future asset in 30years. Then, you can think about the cash flows in both cases.

n 0

Net Cash Flow A : High-school graduates B : College graduates $0

B-A $0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

$24,315.84 $25,047.31 $25,741.26 $26,401.34 $27,030.71 $27,632.10 $28,207.89 $28,760.17 $29,290.79 $29,801.38 $30,293.39 $30,768.15 $31,226.80 $31,670.41 $32,099.94 $32,516.25 $32,920.13 $33,312.30 $33,693.43 $34,064.11 $34,424.90 $34,776.33 $35,118.85 $35,452.92 $35,778.93 $36,097.28 $36,408.30 $36,712.34 $37,009.69 $37,300.65

-$30,000.00 -$30,000.00 -$30,000.00 -$30,000.00 $48,891.69 $50,515.47 $52,070.13 $53,561.32 $54,994.01 $56,372.62 $57,701.10 $58,982.95 $60,221.34 $61,419.12 $62,578.87 $63,702.92 $64,793.42 $65,852.31 $66,881.36 $67,882.22 $68,856.38 $69,805.24 $70,730.08 $71,632.07 $72,512.33 $73,371.88 $74,211.66 $75,032.57 $75,835.44 $76,621.04

-$54,315.84 -$55,047.31 -$55,741.26 -$56,401.34 $21,860.98 $22,883.37 $23,862.24 $24,801.14 $25,703.22 $26,571.25 $27,407.70 $28,214.80 $28,994.54 $29,748.71 $30,478.93 $31,186.67 $31,873.29 $32,540.01 $33,187.94 $33,818.11 $34,431.48 $35,028.92 $35,611.23 $36,179.16 $36,733.40 $37,274.60 $37,803.36 $38,320.24 $38,825.75 $39,320.39

IRR B − A = 10.06% In this example, if MARR is larger than 10.06%, then your college degree is rather an expensive proposition in economic terms.

Chapter 8 Benefit-Cost Analysis 8.1) B = $117, 400( P / A, 6%,5) = $494,535.76 = C $5, 000 + $48,830( P / A, 6%,5) = $210, 691.49

$494,535.76 $210, 691.49 = 2.35 > 1

BC(6%) =

This project is justifiable based on the benefit-cost analysis.

8.2) = B $250, 000( P / A, 6%, 25) + $50, 000( P / F , 6%, 25) = $3, 207,500 = C $1, 200, 000 + $100, 000( P / A, 6%, 25) = $2, 478,340 $3, 207,500 $2, 478,340 = 1.29 > 1

BC(6%) =

8.3) (a)

analysis:

• Design A:

• Design B:

• Incremental analysis: Fee collections in the amount of $85,000 will be the same for ratio. If this both alternatives. Therefore, we will not be able to compute the happens, we may select the best alternative based on either the least cost criterion or the incremental PI(i ) criterion. Using the incremental PI(i ) criterion,

PI(8%)= A− B

∆B − ∆C ' 0 − ($427,974 − $684, 758) = = 2.57 > 1 ∆I $100, 000

∴Select design A.

(b) Incremental analysis (A – C):

PI(8%)= A−C

∆B − ∆C ' 0 − ($427,974 − $556,366) = = 2.57 > 1 ∆I $50, 000

∴Select design A. 8.4) • Building X: 000( P / A,10%, 20) $16, 686, 656 BX $1,960, = = C X $8, 000, 000 + $240, 000( P / A,10%, 20) = − $4,800, 000( P / F ,10%, 20) = $9,329,984 $16, 686, 656 BC(10%) = = 1.79 > 1 X $9,329,984

• Building Y: = = 000( P / A,10%, 20) $11, 237,952 BY $1,320, = CY $12, 000, 000 + $180, 000( P / A,10%, 20) − $7, 200, 000( P / F ,10%, 20) = $12, 462,528 = BC(10%) Y

$11, 237,904 = 0.90 < 1 $12, 462,528

∴ Since Building Y is not desirable at the outset, we don’t need an incremental analysis. Building X becomes the better choice. 8.5) • Option 1 – The “long” route:

• Option 2 – Shortcut:

• Incremental analysis (Option 2 - Option 1):

$1, 200, 000 $4, 766, 674 − $2, 287, 448 = 0.48 < 1

BC(10%) 2−1 =

Assuming that there is no do-nothing alternative, select option 1. 8.6) Incremental BC(i ) analysis:

Present Worth I B C’ BC(i )

A1 $100 $400 $100 2

Proposals A2 $300 $700 $200 1.4

A3 $200 $500 $150 1.43

Incremental A3-A1 A2-A1 $100 $200 $100 $300 $50 $100 0.67 1

C $1,600 $5,656 $2,922 1.25

Incremental C-B A-B $720 $1,560 -$1,414 $754 -$472 $471 -5.7 0.37

Select either A1 or A2.

8.7) Incremental BC(i ) analysis:

Present Worth B I C’ BC(i )

A $2,440 $7,824 $3,865 1.24

Design B $880 $7,070 $3,394 1.65

Select design B.

8.8) (a) The benefit-cost ratio for each alternative: • Alternative A: B= ($1, 000, 000 + $250, 000 + $350, 000 + $100, 000)( P / A,10%,50) = $16,855,185 = C $8, 000, 000 + $200, 000( P / A,10%,50) = $9,982,963 = BC(10%) A

$16,855,185 = 1.69 > 1 $9,982,963

• Alternative B: B= ($1, 200, 000 + $350, 000 + $450, 000 + $200, 000)( P / A,10%,50) = $21,812,592 = C $10, 000, 000 + $250, 000( P / A,10%,50) = $12, 478, 704 $21,812,592 = = 1.75 > 1 BC(10%) B $12, 478, 704

• Alternative C: B= ($1,800, 000 + $500, 000 + $600, 000 + $350, 000)( P / A,10%,50) = $32, 223,147 = C $15, 000, 000 + $350, 000( P / A,10%,50) = $18, 470,185 $32, 223,147 = = 1.74 > 1 BC(10%) B $18, 470,185

(b) Select the best alternative based on BC(i ) : $21,812,592 − $16,855,185 $12, 478, 704 − $9,982,963 = 1.99 > 1 (Select B)

BC(10%) B − A =

$32, 223,147 − $21,812,592 $18, 470,185 − $12, 478, 704 = 1.74 > 1 (Select C)

BC(10%)C − B =

8.9) (a) Find the equivalent annual value of each element - I, B, and C’ , as B and C’ are already given on annual basis. Designs

(b)

$5.83

$9.32

$12.82

$20

$40

$60

$18

$25

AE(i)

$6.17

$12.68

$22.18

BC(i)

1.45

1.46

1.59

(c)

8.10) • Multiple alternatives: Projects PW of Benefits A1 $40 A2 $150 A3 $70 A4 $120

PW of Costs $85 $110 $25 $73

Net PW -$45 $40 $45 $47

B/C ratio 0.47 1.36 2.80 1.64

Since the BC ratio for project A1 is less than 1, we delete it from our comparison.

• Incremental analysis - A3 versus A4:

$120 − $70 $73 − $25 = 1.04 > 1

BC(10%) A 4− A3 =

Select A4.

- A2 versus A4:

$150 − $120 $110 − $73 = 0.81 < 1

BC(10%) A 2− A 4 =

Select A4.

8.11)

Project A1

Project A2

Incremental (A2 - A1)

(900,000) $

(1,200,000) $

(300,000)

1 2 3 4 5 6

$ $ $ $ $ $

400,000 340,000 300,000 240,000 200,000 150,000

200,000 300,000 350,000 440,000 400,000 350,000

(200,000) (40,000) 50,000 200,000 200,000 200,000

$ $ $ $ $ $

(a) •

A1:

= B − C ' $400, 000( P / F , 6%,1) + $340, 000( P / F , 6%, 2) +  + $150, 000( P / F , 6%, 6) = $1,377,141 I = $900, 000 = PI

•

$1,377,141 = 1.530 > 1 $900, 000

A2:

= B − C ' $200, 000( P / F , 6%,1) + $300, 000( P / F , 6%, 2) +  + $350, 000( P / F , 6%, 6) = $1, 643, 706 I = $1, 200, 000 $1, 643, 706 = PI = 1.370 > 1 $1, 200, 000

(b) (1) No budget limit: Select both projects as its PI > 1. (2) If A1 and A2 are mutually exclusive, select A1.

∆( B − C ') = −$200, 000( P / F , 6%,1) − $40, 000( P / F , 6%, 2) +  + $200, 000( P / F , 6%, 6) = $266,564 ∆I =$300, 000 $266,564 ∆PI A 2− A1 = =0.889 < 1, Select A1 $300, 000

8.12)

Project A1

Project A2

Incremental (A2 - A1)

(750,000) $

(1,000,000) $

(250,000)

1 2 3 4 5 6 7 8 9 10

$ $ $ $ $ $ $ $ $ $

100,000 100,000 100,000 240,000 200,000 180,000 180,000 180,000 180,000 180,000

200,000 200,000 200,000 300,000 300,000 250,000 250,000 150,000 100,000 50,000

100,000 100,000 100,000 60,000 100,000 70,000 70,000 (30,000) (80,000) (130,000)

$ $ $ $ $ $ $ $ $ $

(a) •

A1:

= B − C ' $100, 000( P / F , 7%,1) + $100, 000( P / F , 7%, 2) +  + $180, 000( P / F , 7%,10) = $1,114,333 I = $750, 000 $1,114,333 = PI = 1.486 > 1 $750, 000

•

A2:

= B − C ' $200, 000( P / F , 7%,1) + $200, 000( P / F , 7%, 2) +  + $50, 000( P / F , 7%,10) = $1, 457, 013 I = $1, 000, 000 = PI

$1, 457, 013 = 1.457 > 1 $1, 000, 000

Project A1 has a higher PI. (b) Since A1 and A2 are mutually exclusive, select A2. A2-A1: = ∆( B − C ') $100, 000( P / F , 6%,1) + $100, 000( P / F , 7%, 2) +  − $130, 000( P / F , 7%,10) = $342, 680 ∆I =$250, 000 $342, 680 ∆PI A 2− A1 = =1.371 > 1, Select A2 $2500, 000

8.13) Given i = 8%, g = 10%, garbage amount/day = 300 tons (a) The operating cost of the current system in terms of $/ton of solid waste: •

Annual garbage collection required (assuming 365 days): Total amount of garbage = 300 tons ×365 days = 109,500 tons/year

•

Equivalent annual operating and maintenance cost:

PW(8%) = −$905, 400( P / A1 ,10%,8%, 20) = −$20, 071,500 AEC(8%) = $20, 071,500( A / P,8%, 20)

= $2, 044,300

•

Operating cost per ton: $2, 044,300 cost per ton = 109,500

= $18.67 ton

(b) The economics of each solid-waste disposal alternative in terms of $/ton of solid waste: • Site 1:

AEC(8%)1 = $4, 053, 000( A / P,8%, 20) +$342, 000( P / A1 ,10%,8%, 20)( A / P,8%, 20) −($13, 200 + $87, 600) = $1, 084, 773.719 Cost per ton = $1, 084, 773.719 /109,500

= $9.91 per ton

• Site 2:

AEC(8%) 2 = $4,384, 000( A / P,8%, 20) +$480, 000( P / A1 ,10%,8%, 20)( A / P,8%, 20) −($14, 700 + $99,300) = $1, 417, 042.61 Cost per ton = $1, 417, 042.61/109,500

= $12.94 per ton • Site 3:

AEC(8%)3 = $4, 764, 000( A / P,8%, 20) +$414, 000( P / A1 ,10%,8%, 20)( A / P,8%, 20) −($15,300 + $103,500) = $1,301,871.57 Cost per ton = $1,301,871.57 /109,500

= $11.89 per ton • Site 4:

AEC(8%) 4 = $5, 454, 000( A / P,8%, 20) +$408, 000( P / A1 ,10%,8%, 20)( A / P,8%, 20) −($17,100 + $119, 400) = $1,340,928.66 Cost per ton = $1,340,928.66 /109,500

= $12.25 per ton Site 1 is the most economical choice.

Site2

Site3

Site4

System B

$989.67

$1,119.26

$1,166.39

$1,340.17

$4,053.0

$4,384.0

$4,764.0

$5,454.0

$20,071.48

$7,581.68

$10,640.95

$9,177.82

$9,044.81

$12,489.80

$9,430.53

$10,893.66 $11,026.67

Reduction in C’ over the present system

•

Site 1 vs. Site 2:

($1,119.26 + $9, 430.53) − ($989.67 + $12, 489.80) ∆BC(8%) 2−1 = $4,384 − $4, 053 −$2,929.68 = 331 = −8.85 < 1 Select Site 1. •

Site 1 vs. Site 3: ($1,166.39 + $10,893.66) − ($989.67 + $12, 489.80) ∆BC(8%)3−1 = $4, 764 − $4, 053 −$1, 419.42 = 711 = −1.996 < 1

Select Site 1. •

Site 1 vs. Site 4 ($1,340.17 + $11, 026.67) − ($989.67 + $12, 489.80) ∆BC(8%) 4−1 = $5, 454 − $4, 053 −$1,112.63 = $1, 401 = −0.7942 < 1 Select Site 1. The ultimate choice: Site 1.

Chapter 9 Accounting for Depreciation and Income Taxes Note: For the most up-to-date depreciation and income tax information, consult the IRS website at https://www.irs.gov/ 9.1)

(a), (b), (e), (f), (h) (amortization, rather than depreciation)

9.2) The loss of value is defined as the purchase price of an asset less its market value, also known as economic depreciation. Economic depreciation = $30,000 - $12,000 = $18, 000 9.3)  Total property value with the house: Original cost Add: New building Demolition expenses Property value

Land $155,000

$155,000

Building $245,000 $1,250,000 $15,000 $1,510,000

Total property value = $155,000+ $1,510,000= $1,665,000  Cost basis for depreciation = $15,000+ $1,250,000= $1,265,000 (Note: The demolition expense is treated as a site preparation expense)

9.4)  Unrecognized profit Old drill press (Book value) Trade-in allowance Unrecognized loss

$46,220 $40,000 $6,220

 Cost basis Cost of new drill Plus: unrecognized loss Cost basis of new drill

$148,000 $6,220 $154,220

Comments: If the old drill were sold on the market (instead of trade-in), there would be no unrecognized loss. In that situation, the cost basis for the new drill would be $148,000.

9.5)  Unrecognized profit Old lift truck (Book value) Trade-in allowance Unrecognized gains

$7,808 $9,000 $1,192

 Cost basis Cost of new truck Minus: unrecognized gains Cost basis of new truck

$38,000 $1,192 $36,808

Comments: If the old truck were sold on the market (instead of trade-in), there would be no unrecognized gains. In that situation, the cost basis for the new drill would be $38,000.

9.6) Cost basis for flexible manufacturing cells: Flexible Manufacturing cells ($400,000 x 3) Freight charges Handling fee Site preparation costs Start-up and testing costs Special wiring and material costs Cost basis

$1,200,000 $20,000 $15,000 $45,000 $24,000 $3,500 $1,307,500

(Note: start-up and testing costs = $20 x 40 x 6 x 5 = $24,000)

9.7) Depreciation allowances and book values: (a) Depreciation rate = 1/6 for SL, (b) Depreciation rate = 1.5/6 = 1/4 for 150% DB SL n

0 1 2

150% DB Bn

$32,000 $4,500 $4,500

$27,500 $23,000

Bn $32,000

$8,000 $6,000

$24,000 $18,000

9.8)

(a) SL

(b) DDB

$35,000

$200,000

$94,000

$141,000

$35,000

$165,000

$56,400

$84,600

$35,000

$130,000

$24,600

$60,000

$35,000

$95,000

$60,000

$35,000

$60,000

9.9) Given: I = $200,000, S = $32,000, N = 8 years n

1 2 3 4 5 6 7 8

$50,000 $37,500 $28,125 $21,094 $15,820 $11,865 $3,596 0

$150,000 $112,500 $84,375 $63,281 $47,461 $35,596 $32,000 $32,000

9.10) Given: I = $125,000, n = 3 years, N = 8.

1 ) 0.25 = α 2(= 8 D3= 0.25 ×125, 000 × (1 − 0.25)3−1 = $17,578

9.11) DDB switching to SL: n

1 2 3 4 5 6 7

$35,141 $25,104 $17,933 $12,805 $10,676 $8,341 $0

$87,859 $62,755 $44,822 $32,017 $21,341 $13,000 $13,000

9.12) Given: I = $88,000, S = $13,000, N = 6 years. (a) D1 = $29,333 , D2 = $19,556 , D3 = $13, 037 (b) DDB Switching to SL Dn n

1 2 3 4

$29,333 $19,556 $13,037 $8,691

$58,667 $39,111 $26,074 $17,383

$4,383

$13,000

Comments: If the regular DDB deduction were taken during the fifth year, B5 would be less than the salvage value. Therefore, it is necessary to adjust D5 . The number in the box represents the adjusted value. No switching is common for this type of situation whenever the salvage value is high. 9.13) 1 (a) = α =  2 0.3333 6 (b) D2 (0.3333)(0.6667)(42, = = 000) $9,333

9.14) Given: I = $55,000, N = 5 years, S = $5,000 n

(a) SL

(b) DDB

1 2 3 4 5

$10,000 $10,000 $10,000 $10,000 $10,000

$22,000 $13,200 $7,920 $4,752 $2,128

9.15) Given: I = $56,000, S = $6,500, N = 12 years (a) = D (b)

$56, 000 − $6,500 = $4,125 12

D3 = $6, 481

9.16) Depreciation amount for each year of Limo #1 = D1

24,000 ($32,000 −= $3,000) $3, 480 200,000

28,000 ($32,000 −= $3,000) $4,060 200,000 The book value of Limo #1 at the end of year 2: = D2

B2 = 32,000 − 3, 480 − 4,060 = $24, 460

9.17) = D

$90, 000 − $5, 000 = (30, 000) $10, 200 250, 000    Depreciation per mile

9.18)

$68, 000 − $7,500 (5,500) 50, 000 = $6, 655

D5,000 hours =

9.19) 

Truck A: D=

25, 000 ($50, 000 − $5, 000) = $5, 625 200, 000

12, 000 ($25, 000 − $2,500) = $2, 250 120, 000

15, 000 ($18,500 − $1,500) = $2,550 100, 000

20, 000 ($35, 600 − $3,500) = $3, 210 200, 000

 Truck B:

 Truck C:

 Truck D:

9.20)

(a) Book depreciation: 

Truck 22, 000 D1 = ($25, 000 − $2, 000) = $2,530 200, 000 25, 000 D2 = ($25, 000 − $2, 000) = $2,875 200, 000

 Lathe and building:

Building

Lathe DDB n

$45,000

$800,000

$7,500

$37,500

$14,000

$786,000

$6,250

$31,250

$14,000

$772,000

$5,208

$26,042

$14,000

$758,000

$4,340

$21,701

$14,000

$744,000

$3,617

$18,084

$14,000

$730,000

$3,014

$15,070

$14,000

$716,000

$2,512

$12,559

$14,000

$702,000

$2,093

$10,466

$14,000

$688,000

$1,744

$8,721

$14,000

$674,000

$1,454

$7,268

$14,000

$660,000

$1,211

$6,056

…… .

…..

…….

$1,009

$5,047

$14,000

$100,000

(b) Allowed annual depreciation:

n 0 1 2 3 4 5 6 7 8 9 10 11 12

With switching From DDB to SL Dn Bn $7,500 $6,250 $5,208 $4,340 $3,617 $3,014 $2,512 $2,093 $1,866 $1,866 $1,866 $1,866

The switching occurs at the 9th year.

$45,000 $37,500 $31,250 $26,042 $21,701 $18,084 $15,070 $12,559 $10,466 $8,599 $6,733 $4,866 $3,000

9.21)

(a) Straight line: D1 = UP: D1 (b) =

257, 000 − 32, 000 = $22,500; = B1 $234,500 10

257, 000 − 32, 000 (23, 450) $21,105; = = B1 $235,895 250, 000

257, 000 − 32, 000 (2, 450) $18,375; = = B1 $238, 625 30, 000

(d) DDB(without conversion to SL): = D1 0.2($257, = 000) $51, = 400; B1 $205, 600 (e) DDB(with conversion to SL): D1 0.2(257, = = 000) $51, = 400; B1 $205, 600 Note that: (d) and (e) are indifferent in year 1.

9.22) Given: I = $35,000, S = $6,000, N = 8 years, and 5-year MACRS n

Book Depreciation

MACRS Depreciation

1 2 3 4 5 6 7 8

$3,625 $3,625 $3,625 $3,625 $3,625 $3,625 $3,625 $3,625

$7,000 $11,200 $6,720 $4,032 $4,032 $2,016 -

9.23) (a) Cost basis: $190, 000 + $25, 000 = $215, 000

(b) = D1 $30, = 723.5, D2 $52, = 653.5, D3 $37, = 603.5, D4 $26,853.5

D= D= $19,199.5, D= $19,178, D= $19,199.5, D= $9,589 5 7 6 7 8 9.24) Let I denote the cost basis for the equipment.

B3 = I − ( D1 + D2 + D3 ) I = I − (0.1429 + 0.2449 + 0.1749) I = I − 0.5627 I = 0.4373($185, 000) = $80,901 9.25) Given: I = $92,000, S = $12,000, N = 5 years, 7-year MACRS depreciation class D1 = $13,147 D2 = $22,531 D3 = $16, 091 D4 = $11, 491 D5 = $8, 216

9.26) Given: = I machine tool $5,000, = I CNC machine $125,000, = and I warehouse $335,000  Machine tool n 0 1 2 3 4

Dep. Rate 0.3333 0.4444 0.1481 0.0741

Dn $1,667 $2,222 $741 $370

Bn $5,000 $3,333 $1,111 $370 $0

 CNC machine n 0 1 2 3 4 5

Dep. Rate 0.1429 0.2449 0.1749 0.1249 0.0892

Dn $17,857 $30,612 $21,866 $15,618 $11,156

Bn $125,000 $107,143 $76,531 $54,665 $39,046 $27,890

6 7 8

0.0892 0.0892 0.0446

$11,156 $11,156 $5,578

$16,734 $5,578 $0

Bn $335,000 $330,347 $321,757 $313,168

Dep. Rate

…….

39 40

0.0256 0.0118

$4,653 $8,590 $8,590 …….

0.0139 0.0256 0.0256

…..

n 0 1 2 3 …….

 Warehouse

$8,590 $3,937

$3,937 $0

9.27) Since the land is not depreciable, just consider the building depreciations. Given: I = $250,000, tax depreciation method = 27.5-year MACRS property

n 1 2 3 4 5

Depreciation Allowed rate depreciation 2.5758% $6,439 3.6364% $9,091 3.6364% $9,091 3.6364% $9,091 3.1818% $7,955

9.28) Given: Residential real property (27.5-year), I = $270,000 (a)  100%  2.5 D1 =    27.5  12 = (0.00758)($270, 000) = $2, 045 (b) Total amount of depreciation over the 4-year ownership, assuming that the asset is sold at the end of 4th calendar year:

Rate

1 0.7576% $2,045 2 3.6364% $9,818 3 3.6364% $9,818 4 3.4848% $9, 409 Total amount of depreciation allowed = $31,091. Note that the 4th year depreciation reflects the mid-month convention (11.5 months). B4 = $270, 000 − $31, 091 + $80, 000(land) = $318,909

9.29) Given: I = $1,000,000, 39 years-MACRS real property n 0 1 2

Dep. Rate

0.016026 0.025641

$16,026 $25,641

Bn $1,000,000 $983,974 $969,017

9.30) Types of depreciation method (a). B (b). A (c). D (d). C (e). None

9.31) (a) Book depreciation methods:  Straight-line method: n

Cum. Dn

1 2 3 4 5

$15,800

$73,200

$15,800

$15,800 $15,800

$57,400 $41,600

$31,600 $47,400

$15,800

$25,800

$63,200

$15,800

$10,000

$79,000

 DDB method: n

Cum. Dn

1 2 3 4 5

$35,600

$53,400

$35,600

$21,360 $12,816 $7,690 $1,534

$32,040 $19,224 $11,534 $10,000

$56,960 $69,776 $77,466 $79,000

(b) Tax depreciation: 7-year MACRS n

Cum. Dn

1 2 3 4 5 6 7 8

$12,718

$76,282

$12,718

$21,796 $15,566

$54,486 $38,920

$34,514 $50,080

$11,116 $7,948 $7,939 $7,948

$27,804 $19,856 $11,917 $3,969

$61,196 $69,144 $77,083 $85,031

$3,969

$89,000

$38,920

Less: Trade-in allowance

$20,000

Unrecognized loss

($18,920)

Cost of new equipment Plus: Unrecognized loss on trade-in

$92,000 $18,920

Cost basis of new lathe

$110,920

Comments: If the old equipment were sold on the market (instead of trade-in), there would be no unrecognized loss. In that situation, the cost basis for the new equipment would be just $92,000. No half-year convention.

9.32) (a) and (b):

n 0 1 2 3 4 5 6 7 8

MACRS rate

0.1429 $ 0.2449 $ 0.1749 $ 0.1249 $ 0.0893 $ 0.0892 $ 0.0893 $ 0.0446 $

Bn $ $ $ $ $ $ $ $ $

500,150 857,150 612,150 437,150 312,550 312,200 312,550 156,100

3,500,000 2,999,850 2,142,700 1,530,550 1,093,400 780,850 468,650 156,100 -

Property taxes

B n-1

$ 3,500,000 $ 2,999,850 $ 2,142,700 $ 1,530,550 $ 1,093,400 $ 780,850 $ 468,650 $ 156,100

$ $ $ $ $ $ $ $

9.33) Net income calculation: Gross income Expenses: Sarlaries Wages Depreciation Loan interest

34,000,000

$ $ $ $

5,000,000 4,000,000 1,000,000 210,000

Taxable income Income Taxes(21%)

$ $

23,790,000 4,995,900

Net income

18,794,100

a) The marginal tax rate: 21% b) The average (effective) tax rate: 21% c) The net income: $18,794,100

9.34)

(a) Taxable income = $8,500,000 - $2,280,000 - $456,000 = $5,764,000 (b) Income tax calculation using tax formula

42,000 35,998 25,712 18,367 13,121 9,370 5,624 1,873

Income taxes = $5,764,000(0.21) = $1,210,440 9.35) (a) Depreciation expenses:  Building (39-year class, placed in service in February):  100%   10.5  Dbuilding = $400, 000     39   12  = $400, 000(2.2436%) = $8,974

 Equipment (5-year MACRS):

Dequipment = $200, 000 (20% ) = $40, 000  Total depreciation allowed in year 2019: D = $8,974 + $40, 000 = $48,974

(b) Tax liability: Sales revenue Expenses: Cost of goods sold Bond interest Depreciation Taxable income Income taxes Net income

$2,500,000 $800,000 $50,000 $48,974 $1,601,026 $336,215 $1,264,810

Note: Income taxes = $1,601,026(0.21) = $336,215 9.36) (a) Disposed of in year 3: Allowed depreciation = $80,000(0.20 + 0.32 + 0.192 / 2) = $49, 280 Book = value $80,000 − $49, 280 = $30,720 Gains = $40,000 − $30,720 = $9, 280

(b) Disposed of in year 5: Allowed depreciation = $80, 000(0.20 + 0.32 + 0.192 + 0.1152 + 0.1152 / 2) = $70, 784 Book = value $80, 000 − $70, 784 = $9, 216 Taxable gains = $30, 000 − $9, 216 = $20, 784

9.37) Allowed depreciation = $350, 000(0.1429 + 0.2449 + 0.1749 +0.1249 + 0.0893 / 2) = $256, 288 Book = value $350, 000 − $256, 288 = $93, 713

(a) If sold at $20,000: Loss = $20, 000 − $93, 713 = ($73, 713) = Loss credit $73, = 713(0.21) $15, 480

(b) If sold at $99,000: Gains = $99, 000 − $93, 713 = $5, 287 = Gains tax $5, = 287(0.21) $1,110

9.38) (a) Taxable operating income (Do not include ordinary gains):

Revenues: Gross income Expenses: Labor Materials Depreciation Office supplies Interest Rental

$ 4,250,000

Taxable income Income taxes

$ 1,550,000 $ 785,000 $ 332,500 $ 15,000 $ 42,200 $ $ , 45,000 , $ 1,480,300 $ 310,863

Net income

$$ 1,169,437

Taxable income for 2018: $1,480,300 (b) Taxable gains: $43,000 - $30,000 = $13,000 (c) Total taxes: income taxes = $1, 480,300(0.21) = $310,863 gain taxes = (0.21)($13, 000) = $2, 730 total = taxes $310,863 + $2, 730 = $313,593

9.39) (a) Incremental Operating income:

Revenue Expenses: Mfg. cost O&M costs Depreciation Taxable income Income taxes (25%) Net income

Operating Costs Year 1 Year 2 $15,000,000 $15,000,000 $6,000,000 $1,200,000 $714,500 $7,085,500 $1,771,375 $5,314,125

$6,000,000 $1,200,000 $1,224,500 $6,575,500 $1,643,875 $4,931,625

Year 3 Year 4 Year 5 $15,000,000 $15,000,000 $15,000,000 $6,000,000 $1,200,000 $874,500 $6,925,500 $1,731,375 $5,194,125

$6,000,000 $1,200,000 $624,500 $7,175,500 $1,793,875 $5,381,625

$6,000,000 $1,200,000 $223,250 $7,576,750 $1,894,188 $5,682,562

(b) Gains or losses: Total depreciation = $3, 661, 250 = B5 $5, 000, 000 − $3, 661, 250 = $1,338, 750 Taxable = gains $1, 600, 000 − $1,338, 750 = $261, 250

9.40) (a) Book value: $4, 000 − $0 (3) 6 = $2, 000 = B3 $4, 000 − $2, 000

Total depreciation =

= $2, 000 (b) Cost basis: Depreciation base = $14,000 + $800 + $200 = $15,000 (c) Taxable gains and gains taxes

Taxable= gain $2,500 − $2, 000 = $500 Gains tax = (0.40)($500) = $200 (d) Capital gains:

B3 = $2, 000 ordinary gain = $4, 000 − $2, 000 = $2, 000 gain taxes = $2, 000(0.40) = $800 capital gain = $5, 000 − $4, 000 = $1, 000 capital gain taxes = $1, 000(0.40) = $400 total gains taxes = $800 + $400 = $1, 200

(e) Book value at the end of year 3 under 175% DB: B3 = $1, 422 With switching From DDB to SL Dn Bn

0 1 2 3 4 5 6

$1,167 $826 $585 $474 $474 $474

$4,000 $2,833 $2,007 $1,422 $948 $474 $0

(f) Optimal time to switch: during the 4th year

9.41) Note: Personal income tax brackets and amount of personal exemption are updated yearly, so you need to consult the IRS tax manual for the tax rates as well as the amount of exemption that are applicable to your tax year. In this solution, we assumed the tax rate schedule of year 2018. For 2018, the amount of deduction for joint filers is $24,000. The personal exemptions were eliminated. And we assume that Mr. Zodrow would take a standard deduction instead of itemized deductions. (a) Business form: Corporate or Limited Partnership •

Corporate taxes (at 21%): Gross income Expenses:

Year 1 $200,000

Year 2 $215,000

Year 3 $230,000

Salary Business expenses Taxable income Income taxes

•

$100,000

$110,000

$120,000

$25,000

$30,000

$40,000

$75,000 $15,750

$70,000 $14,700

Personal income taxes (with the assumption of the standard deduction as well as the individual tax rates remaining unchanged over the 3-year period). We also assume that Mr. Zodrow can enjoy a 20% deduction on the pass-through income.)

Gross income Deductions: Pass-through (20%) deduction Standard deduction Taxable income Income taxes

100,000

$ 110,000

120,000

$ $ $ $

20,000 24,000 56,000 6,339

$ $ $ $

22,000 24,000 64,000 7,299

$ $ $ $

24,000 24,000 72,000 8,259

Note that, in Year 2018, the personal income tax rates for married filing jointly are as follows:

Income Tax Rate Income Levels for Those Filing As: 2017 2018-2025 Single Married-Joint 10% 10% $0-$9,525 $0-$19,050 15% 12% $9,525-$38,700 $19,050-$77,400 25% 22% $38,700-$82,500 $77,400-$165,000 28% 24% $82,500-$157,500 $165,000-$315,000 33% 32% $157,500-$200,000 $315,000-$400,000 33%-35% 35% $200,000-$500,000 $400,000-$600,000 39.6% 37% $500,000+ $600,000+ •

Total taxes = corporate taxes + personal taxes:

Year 1 = $15, 750 + $6,339 = $22, 089 Year 2 = $15, 750 + $7, 299 = $23, 049 Year 3 = $14, 700 + $8, 259 = $22,959

(b) Business form: sole ownership Year 1

Year 2

Year 3

Gross income

$200,000

$215,000

$230,000

Standard deduction

$24,000

Business expenses

$25,000

$30,000

$40,000

Taxable income

$151,000

$161.000

$166,000

Income taxes

$25,099

$27,299

$28,419

Expense: Exemptions

The corporate business form is preferred.

9.42) (a) Let ic denote the interest rate for a corporate bond:

9.5% = ic (1 − 0.25) ic = 12.67% (b) Let A denote the annual interest payment from the corporate bond. Since Julie’s opportunity cost rate is 9.5%, we can establish the following equivalence relationship: $50, 000 = (1 − 0.25) A( P / A,9.5%,3) + [$50, 000 +(1 − 0.25)(0.05)($50, 000)]( P / F ,9.5%,3) = 1.8817 A + $39,510.79

Solving for A yields A = $5,574.39

This is equivalent to receiving a bond interest rate of ic = $5,574.39 / $50, 000 = 11.15%

(c)

PW(9.5%) = −$50, 000 + [$75, 000 −($75, 000 − $50, 000)(0.25)]( P / F ,9.5%,3) = $2,363.70 > 0 IRR= 11.20% > 9.5% ⇒ Better than investment in bonds.

Investment in a tract of land is more economically desirable.

Chapter 10 Project Cash Flow Analysis Note to Instructors: Regular MACRS depreciation is assumed in all problems unless otherwise mentioned. However, instructors may adopt the capital expensing option under the 2017 Tax Cuts and Job Acts instead. In that case, the entire capital expenditure would be claimed as depreciation amount at the end of first year. Then, any salvage value at the end of project life will be subject to taxable gains. 10.1) • Storage and material handling costs for raw materials: product cost (indirect costs) • Gains or loss on disposal of factory equipment: period costs • Lubricants for machinery and equipment used in production: product cost (mfg. Overhead) • Depreciation of a factory building: product cost (mfg. Overhead) • Depreciation of manufacturing equipment: product cost (mfg. Overhead) • Depreciation of the company president’s automobile: period cost • Leasehold costs for land on which factory buildings stand: period cost • Inspection costs of finished goods: product cost • Direct labor cost: product cost • Raw materials cost: product cost • Advertising expenses: period cost 10.2) • 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 the plant: Variable cost • Heat and air-conditioning for the 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

10.3) a) Paint shop superintendent’s salary: Fixed cost b) Labor costs in assembling a product: Variable cost c) Rent on a factory building: Fixed cost d) RFID units embedded in final product during shipping: Variable cost e) Depreciation on machinery: Variable cost f) Lubricants used for machines: Variable cost g) CPU chips used in notebook production: Variable cost h) Paint used in automobile production: Variable cost i) Janitorial and custodian salaries: Fixed cost j) Coffee beans used in packaging roasted coffee: Variable cost k) Sugar used in ice cream production: Variable cost l) Electricity for operation of machines: Variable cost m) Electricity for heating and cooling the factory building: Fixed cost n) Glue used in electronic board production: Variable cost 10.4)

10.5) Given: p = selling price per unit = $30, v = variable cost per unit = $10 v a) The contribution margin percentage(rate) =1 − =66.67% p

b) The breakeven point =

F = 2,500 units p−v

10.6) a) The contribution margin percentage = b) 1 −

$250, 000 = 50% $500, 000

v 1 = → The selling price per unit = p = 2v = $30 p 2

10.7) Given: The contribution margin rate =1 −

v =0.7 , F = $60,000 p

a) The cost-volume-profit diagram

b) The break-even point (Sales Volume) = $85,715 c)

v v 1 0.3 = = = 1− 1− 0.6739; p 0.92 0.92 p 0.92 $60, 000 = R = $89, 033 0.6739 Break-even sales volume would increase by $3,319 1−

10.8) (a)

Since belt A has maximum contribution margin, we choose to first pursue belt A at its demand. And we choose to produce belt A at its maximum demand.

Then the fixed cost remains $255, 000 − $3(20, 000) = $195, 000 and we $195, 000 = 97,500 units. Since the calculate breakeven for belt B: Vb = $2 demand for a belt B is 100,000 units, which are greater than breakeven unit. Therefore, we opt to produce belt B for 97,500 units. Then, the policy is to produce 20,000 units of belt A and 97,500 units of belt B, respectively in order to breakeven. (b)

The total contribution margin when 200,000 units are sold is $3(20, 000) + $2(100, 000) + $1(80, 000) = $340, 000 . And the operating income is $340, 000 − $255, 000 = $85, 000 .

(c)

Operating income = $3(20, 000) + $2(80, 000) + $1(100, 000) − $255, 000 = $65, 000 . Once again, we choose to produce belt A at its maximum demand. Then the new fixed cost is $255, 000 − 20, 000($3) = $195, 000 . Then, we choose belt B based on the contribution margin, the breakeven for belt B is $195, 000 Vb = = 97,500 units. However, since the new demand for belt B $2 is 80,000 units, which are less than the breakeven unit, therefore we can only produce belt B at its maximum demand; 80,000 units. The new fixed cost is $195,000 - $2(80,000) = $35,000. Then we proceed to produce belt $35, 000 = 35, 000 units. Thus, the new C. The breakeven for belt C is Vb = $1 breakeven is 20,000 units of belt A, 80,000 units of belt B, and 35,000 units of belt C.

10.9)

Good

Economic condition Fair

Poor

Taxable income Before expansion Due to expansion After expansion

$ $ $

2,500,000 $ 2,000,000 $ 4,500,000 $

2,500,000 $ 500,000 $ 3,000,000 $

2,500,000 (100,000) 2,400,000

Income Taxes

945,000

630,000 $

504,000

(a) Marginal tax rate

21%

(b) Average tax rate

21%

Note: With the 2017 Tax Cuts and Jobs Act, the marginal and average tax rates are the same.

10.10) (a) Economic depreciation for the milling machine $200, 000 − $50, 000 = $150, 000

(b) Marginal tax rates with the project: n

Revenue

$80,000

$28,580

Taxable income $51,420

Combined Marginal income rate $476,420 21%

$80,000

$48,980

$31,020

$456,020

21%

$80,000

$34,980

$45,020

$470,020

21%

$80,000

$24,980

$55,020

$480,020

21%

$80,000

$17,860

$62,140

$487,140

21%

$80,000

$8,920

$71,080

$496,080

21%

Combined income taxes $100,048 $95,764 $98,704 $100,804 $102,299 $104,177

Net income

Depreciation

Net cash flow

$376,372 $360,256 $371,316 $379,216 $384,841 $391,903

$28,580 $48,980 $34,980 $24,980 $17,860 $8,920

Year 1

Year 2

$220,000 $150,000 $12,000 $58,000

$220,000 $150,000 $19,200 $50,800

10.11) Incremental tax rate calculation:

Revenue Operating costs Depreciation Taxable income

$404,952 $409,236 $406,296 $404,196 $402,701 $400,823

Taxable income without project Income taxes

Year 1 $650,000 $136,500

Year 2 $650,000 $136,500

Taxable income with project Income taxes

$708,000 $148,680

$700,800 $147,168

Incremental taxable income Incremental income taxes Incremental tax rate (%) Net cash flow (net income + dep)

$58,000 $12,180 0.21 57,822

$50,800 $10,668 0.21 59,332

Comment: Note that the marginal tax rates over the project life remain unchanged with the flat corporate tax rate of 21%. 10.12) Taxable income from the project during year 1: = D1 0.20($100, = 000) $20, 000 Taxable income = $80, 000 − $20, 000 = $60, 000 a) Increment in income tax due to the project during year 1:

Taxable income without project Income taxes

Year 1 $195,000 $40.950

Taxable income with project Income taxes

$255,000 $53,550

Incremental taxable income Incremental income taxes Incremental tax rate

$60,000 $12,600 21%

b) Incremental net cash flow in year 1: •

$60,000 - $12,600 + $20,000 = $67,400

10.13) Incremental tax calculations: (a) Additional taxable income due to project:

Year 1

Year 2

Year 3

Annual revenue

$80,000

Operating cost

$20,000

Depreciation

$16,665

$22,225

$3,703

Taxable income

$43,335

$37,775

$56,298

(b) Additional income tax calculation: Year 1

Year 2

Year 3

$350,000

$73,500

Taxable income with project Income taxes

$393,335 $82,600

$387,775 $81,433

$406,298 $85,323

Incremental taxable income Incremental income taxes Incremental tax rate

$43,335 $9,100 21%

$37,775 $7,933 21%

$56,298 $11,823 21%

Taxable income without project Income taxes

(c) Gains tax: Total depreciation = $42,593 Book value = $50, 000 − $42,593 = $7, 408 Taxable gains = $10, 000 − $7, 408 = $2,593 = Gains tax (0.21)($2,593) = $545

10.14) (a) Taxable gain:

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

Gross income Interest income Bond interest income Expenses: Labor Materials Depreciation Interest Rental

$ $ $

2,250,000 6,000 4,000

$ $ $ $ $

550,000 385,000 132,500 22,200 , 45,000 ,

Taxable income Income taxes (21%)

$ $

1,125,300 236,313

Net income

888,987

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 = 21% Average (effective) tax rate = 21%

(d) Net cash flow: Net income Adjustments: Add depreciation Proceeds from sale Gains tax

888,987

$ $ $

132,500 23,000 630

Net cash flow

$ 1,043,857

10.15) (a) Income tax liability: Gross revenues Expenses: Manufacturing Operating Interest

$ 3,500,000 $ $ $

Taxable operating income Adjustment: (loss)

$ 2,490,000 $ (15,000)

Taxable income Income taxes (21%)

$ 2,475,000 $ 519,750

Net income

$ 1,955,250

650,000 320,000 40,000

Note: book loss = $60,000 - $75,000 = ($15,000)

(b) Operating income: Taxable operating income Income taxes (21%)

$ 2,490,000 $ 522,900

Net operating income

$ 1,967,100

(c) Since the depreciation figure is included in the manufacturing expenses, the first task is to determine the amount of depreciation. In the problem statement in the first printing, the depreciation information is missing, so we assume $20,000. • • • • •

Cash flow from operation: $1,955,250 + $20,000 = $1,975,250 Bank loan = $90,000 Cash dividends paid = ($80,000) Equipment sold = $60,000 Net cash flow = $1,975,250 + $90,000 - $80,000 + $60,000 = $2,045,250

Note: With disposal of the equipment at $60,000, there is a loss in the amount of $15,000, such that the firm would save $15,000(0.21) = $3,150. This amount is already reflected in the net income calculation.

10.16) (a) 0

$320,000

$130,000 $30,000

$130,000 $48,000

$130,000 $28,800

$130,000 $17,280

$130,000 $8,640

Taxable Income Income Taxes (25%)

$160,000 $40,000

$142,000 $35,500

$161,200 $40,300

$172,720 $43,180

$181,360 $45,340

Net Income

$120,000

$106,500

$120,900

$129,540

$136,020

$120,000 $30,000

$106,500 $48,000

$120,900 $28,800

$129,540 $17,280

$136,020 $8,640

Income Statement Revenues Expenses Labor & Mat'l Costs Depreciation

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(12%)

($150,000) $0 $0 ($150,000) $463,554.16

$150,000

(b) PW(12%) = $463,554 > 0 , Accept the investment.

$154,500

$149,700

$146,820

$144,660

10.17)

Investment in industrial robot:

$125,000

35,725

61,225

43,725

31,225

11,163

Taxable Income Income Taxes (25%)

$89,275 22,319

$63,775 15,944

$81,275 20,319

$93,775 23,444

$113,838 28,459

Net Income

$66,956

$47,831

$60,956

$70,331

$85,378

$66,956 $35,725

$47,831 $61,225

$60,956 $43,725

$70,331 $31,225

$85,378 $11,163

Income Statement Revenues (savings) Expenses: Depreciation

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

($250,000) $50,000 $4,234.38 ($250,000)

$102,681

$109,056

PW(15%)= IRR=

$123,607 33% >15%(MARR)

$104,681

$101,556

$150,775

Accept the investment

10.18)

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

10.19) Investment in energy management system: N = 7 years 0

$14,000

$0 $18,332

$0 $24,448

$0 $8,146

$0 $4,076

$0 $0

Taxable Income Income Taxes (25%)

($4,332) ($1,083)

($10,448) ($2,612)

$5,855 $1,464

$9,925 $2,481

$14,000 $3,500

Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax

($3,249)

($7,836)

$4,391

$7,443

$10,500

($3,249) $18,332

($7,836) $24,448

$4,391 $8,146

$7,443 $4,076

$10,500 $0

$0 $0

$10,500

Income Statement Revenues Expenses O&M Depreciation

Net Cash Flow PW(12%) = IRR =

($55,000)

($55,000) $3,980.61 14.47%

$15,083

$16,612

$12,536

$11,519

With N = 6 years, IRR = 11.46%. With N = 7 years, IRR = 14.47%, indicating that it needs to operate at least for 7 years.

10.20) Maximum amount to invest = $373,116 0

$150,000

$150,000 $150,000

$150,000

$30,000 $124,360

$30,000 $165,850

$30,000 $55,258

$30,000 $27,648

$30,000 $0

Taxable Income Income Taxes (25%)

($4,360) ($1,090)

($45,850) ($11,463)

$64,742 $16,185

$92,352 $23,088

$120,000 $30,000

Net Income

($3,270)

($34,388)

$48,556

$69,264

$90,000

($3,270) $124,360

($34,388) $165,850

$48,556 $55,258

$69,264 $27,648

$90,000 $0

Income Statement Revenues Expenses O&M Depreciation

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow PW(15 %)

($373,116) $0 $0 ($373,116) ($0)

$121,090

$131,463

$103,815

$96,912

$90,000

10.21) .We will assume that these lease payments are received at year end. 0 Income Statement Lease Revenues Expenses O&M Depreciation

$15,698

$0 $10,600

$0 $16,960

$0 $10,176

Taxable Income Income Taxes (25%)

$5,098 $1,275

($1,262) ($315)

$5,522 $1,381

Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Security Deposit

$3,824

($946)

$4,142

$3,824 $10,600

($946) $16,960

$4,142 $10,176

Net Cash Flow PW(10 %)

($53,000)

$ 22,000 $ (346) $ 1,500

(1,500) ($54,500) $0

$14,424

$16,014

$37,472

total depreciation amount = $32,648 book value = $53,000 - $32,648 = $20,352 taxable gain = $22,000 - $20,352 = $1,648 gains tax = (0.25)($1,648) = $346 net proceeds from sale = $22,000 - $346= $21,654

10.22)

0 Income Statement Revenues (savings) Expenses: Required annual digging (ft) Number of hours to operate Operating cost (@$10/hr) Depreciation

$50,000

8,000 500 $5,000 $60,000

8,000 500 $5,000 $96,000

8,000 500 $5,000 $57,600

8,000 500 $5,000 $34,560

8,000 500 $5,000 $17,280

Taxable Income Income Taxes (25%)

($15,000) ($3,750)

($51,000) ($12,750)

($12,600) ($3,150)

$10,440 $2,610

$27,720 $6,930

Net Income

($11,250)

($38,250)

($9,450)

$7,830

$20,790

($11,250) $60,000

($38,250) $96,000

($9,450) $57,600

$7,830 $34,560

$20,790 $17,280

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

($300,000) $100,000 ($16,360) ($300,000)

$48,750

IRR= PV(15%)= $

1.8% (97,534) < 0

$57,750

$48,150

Not Acceptable

$42,390

$121,710

BV=

$34,560

10.23) Tucson Solar Company: (a) (a) 0

Income Statement Revenues (savings) Expenses: Operating Expenses Depreciation

$66,000

$70,000

$74,000

$80,000

$64,000

$50,000

29,000 10,800

28,400 17,280

32,000 10,368

38,800 6,221

31,000 6,221

25,000 3,110

Taxable Income Income Taxes (25%)

$26,200 6,550

$24,320 6,080

$31,632 7,908

$34,979 8,745

$26,779 6,695

$21,890 5,472

Net Income

$19,650

$18,240

$23,724

$26,234

$20,084

$16,417

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment $ Salvage Gains Tax

$ $

19,650 10,800

$ $

18,240 17,280

$ $

23,724 10,368

$ $

26,234 6,221

$ $

20,084 6,221

$ $

16,417 3,110

$ $

8,000 (2,000)

Net Cash Flow

(54,000)

($54,000) NPV=

(b)

$30,450 $74,255

$35,520

( $74, 255) × ( A / P /12%, 6)

AE(12%) =

= $18, 061

$34,092 AE(12%)=

$32,455 $18,061

$26,305

$25,528

10.24)

$250,000

$50,000 $46,000

$50,000 $73,600

$50,000 $44,160

$50,000 $26,496

$50,000 $13,248

Taxable Income Income Taxes (25%)

$154,000 $38,500

$126,400 $31,600

$155,840 $38,960

$173,504 $43,376

$186,752 $46,688

Net Income

$115,500

$94,800

$116,880

$130,128

$140,064

$115,500 $46,000

$94,800 $73,600

$116,880 $44,160

$130,128 $26,496

$140,064 $13,248

Income Statement Revenues Expenses O&M Depreciation

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities Investment Salvage Gains Tax Net Cash Flow IRR =

($230,000) $5,000 $5,374 ($230,000) $161,500 65.03%

$168,400

$161,040

$156,624

$163,686

10.25) (a) Investment in Mazda automatic screw machine: Input Tax Rate(%) = MARR(%) =

Output PW(i) = IRR(%) =

25 15

$53,441 40.82%

Income Statement Revenues (savings) Expenses: Depreciation

$38,780

9,817

16,825

12,016

8,581

6,135

3,064

Taxable Income Income Taxes (25%)

$28,963 7,241

$21,955 5,489

$26,764 6,691

$30,199 7,550

$32,645 8,161

$35,716 8,929

Net Income

$21,722

$16,466

$20,073

$22,649

$24,484

$26,787

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

$ $ $

21,722 9,817

$ $

16,466 16,825

$ $

20,073 12,016

$ $

22,649 8,581

$ $

24,484 6,135

$ $

26,787 3,064

(68,701) $ 3,500 $ 2,190.78 ($68,701)

(b) Since PW(15%) > 0, accept the investment. (c) IRR = 40.82%

$31,539

$33,291

$32,089

$31,230

$30,619

$35,542

10.26)

$125,000

35,725

61,225

43,725

31,225

11,163

Taxable Income Income Taxes (25%)

$89,275 22,319

$63,775 15,944

$81,275 20,319

$93,775 23,444

$113,838 28,459

Net Income

$66,956

$47,831

$60,956

$70,331

$85,378

$66,956 $35,725

$47,831 $61,225

$60,956 $43,725

$70,331 $31,225

$85,378 $11,163

Income Statement Revenues (savings) Expenses: Depreciation

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Working Capital Net Cash Flow

($250,000) $50,000 $4,234 $30,000

($30,000) ($280,000)

$102,681

$109,056

PW(15%)= IRR=

$108,522 29% >15%(MARR)

$104,681

$101,556

$180,775

Accept the investment

10.27)

Income Statement Revenues (savings) Expenses: Required annual digging (ft) Number of hours to operate Operating cost (@$10/hr) Depreciation

$50,000

8,000 500 $5,000 $60,000

8,000 500 $5,000 $96,000

8,000 500 $5,000 $57,600

8,000 500 $5,000 $34,560

8,000 500 $5,000 $17,280

Taxable Income Income Taxes (25%)

($15,000) ($3,750)

($51,000) ($12,750)

($12,600) ($3,150)

$10,440 $2,610

$27,720 $6,930

Net Income

($11,250)

($38,250)

($9,450)

$7,830

$20,790

($11,250) $60,000

($38,250) $96,000

($9,450) $57,600

$7,830 $34,560

$20,790 $17,280

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Working Capital Net Cash Flow

($300,000) $100,000 ($16,360) $50,000

($50,000) ($350,000)

$48,750

IRR= PW(15%)= $

1.5% (122,675) < 0

$57,750

$48,150

Not Acceptable

$42,390

$171,710

BV=

$34,560

10.28) a) Equal repayment of the principal: n 0 1 2 3 4 5 6

Loan Interest

Repayment Loan Principal Balance $300,000 $36,000 $50,000 $250,000 $30,000 $50,000 $200,000 $24,000 $50,000 $150,000 $18,000 $50,000 $100,000 $12,000 $50,000 $50,000 $6,000 $50,000 0

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

Loan Interest

Repayment Loan Principal Balance $300,000 $36,000 $300,000 $36,000 $300,000 $36,000 $300,000 $36,000 $300,000 $36,000 $300,000 $36,000 $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

Repayment Loan Principal Balance $300,000 $36,000 $36,968 $263,032 $31,564 $41,404 $221,628 $26,000 $46,373 $175,255 $21,031 $51,937 $123,318 $14,798 $58,170 $65,148 $7,818 $65,148 0

10.29) Note: The net income figures in Table P10.29 are incorrect. The correct net income figures are $13,001, and $15,023, respectively.

Income Statement Savings Expenses: O&M Interest Expense Depreciation Taxable Income Income Taxes (25%)

Net Income 0 Operating Activities Net income Depreciation Investment Activities Investment ($20,000) Salvage Gains Tax(25%) Financing Activities Borrowed funds $10,000 Principal repayment Net Cash Flow ($10,000) PW(15%) $19,938

1 2 $ 30,000 $ 30,000 $ 5,000 $ 1,000 $ 6,666 $ 17,334 $ 4,334

$ 5,000 $ 524 $ 4,445 $ 20,031 $ 5,008

$ 13,001 1

$ 15,023 2

$13,001 $6,666

$15,023 $4,445

$8,000 $222

($4,762) $14,905

($5,238) $22,452

Note: Annual installments for the loan = $10, 000( A / P,10%, 2) = $5, 762

10.30) (a), (b), and (c)

Income Statement 0

$95,000

30,000 10,800

48,000 9,100

28,800 7,196

17,280 5,063

8,640 2,675

Taxable Income Income Taxes (35%)

$54,200 $18,970

$37,900 $13,265

$59,004 $20,651

$72,657 $25,430

$83,685 $29,290

Net Income Cash Flow Statement Cash from operation: Net Income Depreciation Investment / Salvage Gains Tax Loan repayment

$35,230

$24,635

$38,353

$47,227

$54,395

35,230 $ 30,000 $

24,635 $ 48,000 $

38,353 $ 28,800 $

(150,000)

90,000 $ (14,167) $

(15,867) $

(17,771) $

47,227 $ 17,280 $ $ $ (19,903) $

54,395 8,640 10,000 2,548 (22,292)

($60,000)

$51,063

$56,768

$49,382

$44,603

$53,291

PW (20%) = IRR =

$93,479 82.19%

Income Statement Revenue Expenses: Depreciation Interest (12%)

Net Cash Flow (actual)

$ $

10.31) (a) Development of after-tax cash flows (b) IRR = 82.74% (c) Since PW(18%) = $376,060>0, accept the investment. Input Tax Rate(% )= MARR(%) =

25 18

Output PW(i) = IRR(%) =

$376,060 82.74%

(a) Income Statement Revenues: Additional revenue Labor & materials savings Expenses: Depreciation Debt interest Taxable Income Income Taxes

Net Cash Flow

5-7

$85,000 $85,000 $85,000 $85,000 $85,000 $85,000 $85,000 $85,000 $ 65,000 $ 65,000 $ 65,000 $ 65,000 $ 65,000 $ 65,000 $ 65,000 $ 65,000 $ 31,438 $ 10,800 $ 107,762 $ 26,941

Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

$80,822

$ 53,878 $ 38,478 $ 27,478 $ 19,646 $ 9,812 $ 7,200 $ 3,600 $ 88,922 $ 107,922 $ 122,522 $ 130,354 $ 140,188 $ 150,000 $ 150,000 $ 22,231 $ 26,981 $ 30,631 $ 32,589 $ 35,047 $ 37,500 $ 37,500 $66,692

$80,942

$ 80,822 $ 66,692 $ 80,942 $ $ 31,438 $ 53,878 $ 38,478 $ $

$91,892

$97,766

$105,141

$112,500

91,892 $ 97,766 $ 105,141 $ 112,500 $ 112,500 27,478 $ 19,646 $ 9,812 $ $ -

(220,000) $ 20,000 $ (5,000)

120,000 $ (40,000) $ (40,000) $ (40,000) ($100,000)

$72,260

$80,570

$79,420

$119,370

$117,412

$114,953

$112,500

$127,500

10.32) (a) and (b)

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

Output 35 18

PW(i) = AE(i)=

($1,648,462) ($527,142)

Income Statement Revenues (savings) Expenses: Depreciation Debt interest

357,250 100,000

612,250 83,620

437,250 65,603

312,250 45,783

111,625 23,982

Taxable Income Income Taxes (35%)

($457,250) (160,038)

($695,870) (243,555)

($502,853) (175,998)

($358,033) (125,312)

($135,607) (47,462)

Net Income

($297,213)

($452,316)

($326,854)

($232,721)

($88,144)

(297,213) $ 357,250 $

(452,316) $ 612,250 $

(326,854) $ 437,250 $

(232,721) $ 312,250 $

(88,144) 111,625

$ $

250,000 146,781

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment Net Cash Flow

$ $ $

(2,500,000)

1,000,000 $

(163,797) $

(180,177) $

(198,195) $

(218,014) $

(239,816)

($1,500,000) $

(103,760) $

(20,243) $

(87,799) $

(138,486) $

180,446

AEC(18%) = $527,142, an equivalent annual revenue that must be generated each to break even.

10.33) (a) and (b)

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

Output PW(i) = $16,114 IRR(%) = 18.78%

25 14 1

Revenues (savings) Expenses: O&M cost Depreciation Debt interest

$48,000

11000 21435 5000

11000 36735 4563

11000 26235 4082

11000 18735 3553

11000 13395 2971

11000 13380 2331

11000 13395 1627

11000 6690 852

Taxable Income Income Taxes (25%)

$10,565 2,641

($4,298) (1,074)

$6,683 1,671

$14,712 3,678

$20,634 5,159

$21,289 5,322

$21,978 5,495

$29,458 7,364

Net Income

$7,924

($3,223)

$5,012

$11,034

$15,476

$15,967

$16,484

$22,093

Income Statement

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

$ (150,000)

Net Cash Flow

($100,000) $24,987

$ 7,924 $ (3,223) $ 5,012 $ 11,034 $ 15,476 $ 15,967 $ 16,484 $ 22,093 $ 21,435 $ 36,735 $ 26,235 $ 18,735 $ 13,395 $ 13,380 $ 13,395 $ 6,690

$ 10,000 $ (2,500) $ 50,000 $ (4,372) $ (4,809) $ (5,290) $ (5,819) $ (6,401) $ (7,041) $ (7,746) $ (8,520) $28,702

$25,957

$23,950

$22,469

$22,305

$22,133

$27,763

10.34) (a) With 100% equity financing: PW(9%) = $1,962 Input Tax Rate(%) = MARR(%) =

Output 25 9

PW(i) = AE(i)=

$1,962 $504

$2,500

2,666 0

3,556 0

1,185 0

593 0

0 0

Taxable Income Income Taxes (25%)

($166) (42)

($1,056) (264)

$1,315 329

$1,907 477

$2,500 625

Net Income

($125)

($792)

$986

$1,430

$1,875

(125) $ 2,666 $

(792) $ 3,556 $

986 $ 1,185 $

1,430 $ 593 $

1,875 -

$ $

2,000 (500)

2,023 $

3,375

Income Statement Revenues (savings) Expenses: Depreciation Debt interest

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment Net Cash Flow

$ $ $

(8,000)

($8,000) $

2,542 $

2,764 $

2,171 $

(b) With a 100% debt financing: PW(9%) = $2,429

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

Output 25 9

PW(i) = AE(i)=

$2,429 $624

$2,500

2,666 720

3,556 600

1,185 469

593 326

0 170

Taxable Income Income Taxes (25%)

($886) (222)

($1,656) (414)

$847 212

$1,582 395

$2,330 583

Net Income

($665)

($1,242)

$635

$1,186

$1,748

(665) $ 2,666 $

(1,242) $ 3,556 $

635 $ 1,185 $

1,186 $ 593 $

1,748 -

$ $

2,000 (500)

Income Statement Revenues (savings) Expenses: Depreciation Debt interest

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment Net Cash Flow

$ $ $

(8,000)

8,000 $0 $

($1,337)

($1,457)

($1,588)

($1,731)

($1,887)

665 $

857 $

232 $

48 $

1,361

10.35)

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

Output PW(12%) =

25 12

$108,785

Financial Data year Depreciation Book value Salvage value Gains tax Loan payment schedule Interest Principal Revenues O&M costs

1 2 3 4 5 $ 19,292 $ 33,062 $ 23,612 $ 16,862 $ 6,028 135,000 $ 115,709 $ 82,647 $ 59,036 $ 42,174 $ 36,146 $ 50,000 $ (3,463)

$ 135,000 $ $

13,500 $ 11,289 $ 8,856 $ 6,181 $ 3,238 22,113 $ 24,324 $ 26,756 $ 29,432 $ 32,375 68,000 $ 68,000 $ 68,000 $ 68,000 $ 68,000

0 ($135,000)

Cash Flow Statement

Investment Net proceeds from sale Investment in working capital Recovery of working capital (1 - 0.25) (Revenue) -(1 - 0.25) (Expenses) -(1 - 0.25) (Debt interest) + (0.25) (Depreciation) Borrowed funds Principal repayment Net Cash Flow

5 $46,537

$51,000 $51,000 $51,000 $51,000 $51,000 $ (10,125) $ (8,467) $ (6,642) $ (4,636) $ (2,428) $ 4,822.88 $ 8,265.38 $ 5,902.88 $ 4,215.38 $ 1,506.94 $

135,000 $ $0

(22,113) $ (24,324) $ (26,756) $ (29,432) $ (32,375) $23,585

$26,475

$23,504

$21,148

$64,240

10.36)

Cash Flow Statement Investment Net proceeds from sale Investment in working capital Recovery of working capital

0 $

(270)

(1 - 0.25)(Revenue) -(1 - 0.25) (Debt interest) +(0.25)(Depreciation)

$ $ $

Borrowed funds Principal repayment

243

Net cash flow

(27.00) $

Cash Flow Statement

33.75 $ (21.87) $ 9.65 $

33.75 $ (21.87) $ 16.53 $

33.75 $ (21.87) $ 11.81 $

33.75 $ (21.87) $ 8.44 $

33.75 $ (21.87) $ 6.02 $

33.75 $ (21.87) $ 6.03 $

33.75 (21.87) 6.02

21.53

28.41

23.69

20.32

17.90

17.91

17.90

$ $ $

33.75 $ (21.87) $ 3.01

33.75 $ (21.87) $

14.89

11.88

Borrowed funds Principal repayment Net cash flow

PW(18%) =

$ $

33.75 $ (21.87)

Investment Net proceeds from sale (1 - 0.25)(Revenue) -(1 - 0.25) (Debt interest) +(0.25)(Depreciation)

30.38

33.75

64.13

(243)

$ (231.12) $

42 > 0, Accept the investment.

10.37) •

Option 1: Lease PW(12%)lease = −$144, 000(1 − 0.25)(1 + ( P / A,12%, 29))

•

= −$974,354 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) = $15,972.22 D2 to D29 = (12 /12)(1/ 39)($650, 000) = $16, 666.67

Note 2: Property tax calculation: ($800,000)(0.05) = $40,000 Input Tax Rate(% )= MARR(%) =

25 12

Output PW(i) =

($1,000,356)

Income Statement Revenues: Expenses: Depreciation Property tax Taxable Income Income Taxes

$ 15,972 $ 16,667 $ 16,667 $ $ 40,000 $ 40,000 $ 40,000 $ $ (55,972) $ (56,667) $ (56,667) $ $ (13,993) $ (14,167) $ (14,167) $

16,667 40,000 (56,667) (14,167)

$ $ $ $

16,667 $ 15,972 40,000 $ 40,000 (56,667) $ (55,972) (14,167) $ (13,993)

Net Income

$ (41,979) $ (42,500) $ (42,500) $

(42,500)

(42,500) $ (41,979)

$ (41,979) $ (42,500) $ (42,500) $ $ 15,972 $ 16,667 $ 16,667 $

(42,500) 16,667

$ $

(42,500) $ (41,979) 16,667 $ 15,972

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment(land) Investment(structure) Salvage Gains Tax Net Cash Flow

$ $

(150,000) (650,000) $ 215,000 $ 21,597

(800,000) $ (26,007) $ (25,833) $ (25,833) $

PW(12%) purchase = −$1, 000,356

(25,833)

(25,833) $ 210,590

•

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: Property tax calculation: $660,000 (0.05) = $33,000 Cost basis for property tax= Land + building + remodeling cost = $660,000

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

25 12

Income Statement Revenues: Expenses: Depreciation Property tax Lease fee(Parking lot) Taxable Income Income tax Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment(Remodeling)

Output PW(i) =

($559,031)

$ $

7,372 $ 7,692 $ 7,692 $ 7,692 $ 33,000 $ 33,000 $ 33,000 $ 33,000 $ $9,000 $9,500 $10,000 $10,500 $ (49,372) $ (50,192) $ (50,692) $ (51,192) $

7,692 $ 7,372 33,000 $ 33,000 $23,000 $23,500 (63,692) $ (63,872)

$ (12,343) $ (12,548) $ (12,673) $ $ (37,029) $ (37,644) $ (38,019) $

(12,798) $ (38,394) $

(15,923) $ (15,968) (47,769) $ (47,904)

$ (37,029) $ (37,644) $ (38,019) $ $ 7,372 $ 7,692 $ 7,692 $

(38,394) $ 7,692 $

(47,769) $ (47,904) 7,692 $ 7,372

(300,000)

Salvage Gain tax Net Cash Flow

(300,000) $ (29,657) $ (29,952) $ (30,327) $

PW(12%) remodel = −$559,031

Option 3 is the least cost alternative.

(30,702) $

$ $

30,000 9,968

(40,077) $

(564)

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

Book Value (n = 20) Salvage Value Taxable gains Gains tax (25%) Net Proceeds from sale

$380.61 $853.00 $469.39 $117.35 $735.65

$423.80 $470.56 $949.80 $1,054.60 $526 $584.04 $131.50 $146.01 $818.30 $908.59

Plant A • Capital recovery cost with return: A1 = ($8,530 − $735.65)( A / P,12%, 20) + $735.65(0.12) = $1,131.78 •

After-tax O&M cost: A2 = (1 − 0.25)($1,964) = $1, 473

• Depreciation tax shield: = A3 0.39($8,530) [ 0.0375( P / F ,12%,1) + ] ( A / P,12%, 20) •

= $110.40 Total equivalent annual cost:

A= $1,131.78 + $1, 473 − $110.40= $2, 494.38 •

Unit cost: $2, 494,380 = $0.0499 / kWh 50, 000, 000 kWh

Plant B • Capital recovery cost with return: A1 = ($9, 498 − $818.30)( A / P,12%, 20) + $818.30(0.12) = $1, 260.22 •

After-tax O&M cost: A2 = (1 − 0.25)($1, 744) = $1,308

• Depreciation tax shield: = A3 0.25($9, 498) [ 0.0375( P / F ,12%,1) + ] ( A / P,12%, 20) •

= $122.92 Total equivalent annual cost:

A = $1, 260.22 + $1,308 − $122.92 = $2, 445.30 • Unit cost: $2, 445,300 = $0.0489 / kWh 50, 000, 000 kWh

Plant C • Capital recovery cost with return:

A1 = ($10,546 − $908.59)( A / P,12%, 20) + $908.59(0.12) = $1,399.28 •

After-tax O&M cost:

A2 = (1 − 0.25)($1, 632) = $1, 224 •

Depreciation tax shield:

= A3 0.25($10,546) [ 0.0375( P / F ,12%,1) + ] ( A / P,12%, 20) = $136.49 •

Total equivalent annual cost:

A= $1,399.28 + $1, 224 − $136.49= $2, 486.79 •

Unit cost: $2, 486, 790 = $0.0497 / kWh 50, 000, 000 kWh

Plant B is the most economical.

10.39) (a) Prescott Welding’s cost of leasing in present worth:

after-tax lease expense = (1 - 0.25)($12,000) = $9,000 PW(15%)lease = −$9, 000 − $9, 000( P / A,15%,3) = −$29,549 Input Tax Rate(%) = MARR(%) = 0

25 15% 1

Output PW(i) =

($29,549)

Income Statement Revenues (savings) Expenses: O&M cost Leasing cost Debt interest

0 $12,000

$12,000

Taxable Income Income Taxes (25%)

($12,000) (3,000)

$0 0

Net Income

($9,000)

(9,000) $

($9,000)

Cash Flow Statement Operating Activities: Net Income Net Cash Flow

(b) Prescott Welding’s cost of owning in present worth: •

PW of after-tax maintenance expenses:

P1 = −$1, 200(1 − 0.25)( P / A,15%, 4) = −$2,569 •

PW of after-tax loan repayment P2 = −$14,815( P / A,15%, 4) = −$42, 296

•

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

Sum

$9,000

$5,400

$14,400

$3,600

$14,400

$4,270

$18,670

$4,668

$8,640

$3,005

$11,645

$2,911

$2,592

$1,587

$4,179

$1,045

P3=

•

Combined tax

savings

$9,171

PW of net proceeds from sale: Total depreciation amount=

$34,632

Book value=

$10,368

Salvage=

$10,000

Taxable loss=

($368)

Loss credit=

($92)

Net proceeds from sale=

$10,092

P 4=

$5,770

PW(15%) buy = P1 + P2 + P3 + P4 = −$29,926

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

25 15%

Output PW(i) =

($29,926)

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

$ $ $

1,200 9,000 5,400

$ 1,200 $ 14,400 $ 4,270

$ $ $

Taxable Income Income Taxes (25%)

($15,600) (3,900)

($19,870) (4,968)

($12,845) (3,211)

($5,379) (1,345)

Net Income

($11,700)

($14,903)

($9,634)

($4,035)

(9,634) $ 8,640 $

(4,035) 2,592

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment Net Cash Flow

$ (11,700) $ (14,903) $ $ 9,000 $ 14,400 $ $

1,200 8,640 3,005

$ $ $

1,200 2,592 1,587

(45,000) $ 10,000 $ 92

45,000 $ $0

(9,416) $ (10,545) $ (11,811) $ (13,228)

$ (12,116) $ (11,048) $ (12,804) $

(4,579)

(c) Should the truck be leased or purchased? The leasing option is a better choice. 10.40) (a) PW (incremental) cost of owning the equipment: •

PW of after-tax loan repayment: P1 = −$37,857( P / A,15%, 4) = −$108, 080

•

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

n Dn In Combined Tax Savings 1 $24,000 $12,000 $36,000(0.25) = $9,000 2 $38,400 $9,414 $47,814(0.25) = $11,954 3 $23,040 4 $6,912

$6,570 $3,441

$29,610(0.25) = $7,403 $10,353(0.25) = $2,588

= P2 $9, 000( P / F ,15%,1) + $11,954( P / F ,15%, 2) +$7, 403( P / F ,15%,3) + $2,588( P / F ,15%, 4) = $23, 212 •

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 = (0.25)(-$7,648) = -$1,912 net proceeds from sale = $20,000 -(-$1,912) = $21,912 P3 = $21,912(P / F ,15%, 4) = $12,528 PW(15%) buy = P1 + P2 + P3 = −$72,340

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

25 15%

Output PW(i) =

($72,339)

Revenues (savings) Expenses: O&M cost Depreciation Debt interest

24000 12000

38400 9414

0 23040 6570

0 6912 3441

Taxable Income Income Taxes (25%)

($36,000) (9,000)

($47,814) (11,954)

($29,610) (7,403)

($10,353) (2,588)

Net Income

($27,000)

($35,861)

($22,208)

($7,765)

Income Statement

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

$ (27,000) $ (35,861) $ (22,208) $ $ 24,000 $ 38,400 $ 23,040 $

(7,765) 6,912

$ $

20,000 1,912

(120,000)

120,000

Net Cash Flow

$ (25,856) $ (28,442) $ (31,286) $ (34,415) $0

($28,856)

($25,903)

($30,454)

($13,356)

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

PW of after-tax operating cost: common cost for both alternative, so we can ignore this item in incremental analysis.

•

PW of after-tax leasing

P = −$44, 000(1 − 0.25) − $44, 000(1 − 0.25)( P / A,15%,3) = −$108,346

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

25 15% 1

Output PW(i) =

($108,346)

Income Statement Revenues (savings) Expenses: O&M cost Leasing cost Debt interest

0 $44,000

$44,000

Taxable Income Income Taxes (25%)

($44,000) (11,000)

$0 0

Net Income

($33,000)

Cash Flow Statement Operating Activities: Net Income Net Cash Flow

(33,000) $ (33,000) $ (33,000) $ (33,000) $ ($33,000)

($33,000)

10.41)

(a) Boggs’ PW cost of leasing: PW(15%)leasing = (1 − 0.25)[$15, 000 + $15, 000( P / A,15%, 2)] =$36,936

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

25 15%

Output PW(i) =

($36,936)

Income Statement Revenues (savings) Expenses: O&M cost Leasing cost Debt interest

0 $15,000

$15,000

Taxable Income Income Taxes (25%)

($15,000) (3,750)

$0 0

Net Income

($11,250)

Cash Flow Statement Operating Activities: Net Income

Net Cash Flow

(11,250) $ (11,250) $ (11,250) $ (11,250) $ ($11,250)

($11,250)

(b) Boggs’ PW cost of owning: •

PW of after-tax maintenance expenses: P1 = −$5, 000(1 − 0.25)( P / A,15%,3) = −$8,562

•

PW cost of after-tax loan repayment: P2 = −$41, 635( P / A,15%,3) = −$95, 062

•

PW of tax credit (shield) on depreciation and interest: Combined Tax Savings n Dn In 1 $20, 000 $12, 000 $32, 000(0.25) = $8, 000 2 $32, 000 $8, 444 $40, 444(0.25) = $10,111 3 $9, 600 $4, 461 $14, 061(0.25) = $3,515

= P3 $8, 000( P / F ,15%,1) + $10,111( P / F ,15%, 2) +$3,515( P / F ,15%,3) = $16,913 •

PW of net proceeds from sale:

total depreciation amount = $61,600 book value = $38,400 taxable gain = $50,000 - $38,400 = $11,600 gains tax = (0.25)($11,600) = $2,900 net proceeds from sale = $50,000 - $2,900 = $47,100 P4 = $45,360(P / F ,15%,3) = $30,969

PW(15%) buy =P1 + P2 − P3 − P4 =−$55, 742

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

Output PW(i) =

25 15%

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

$ $ $

5,000 20,000 12,000

$ $ $

5,000 32,000 8,444

$ $ $

5,000 9,600 4,461

Taxable Income Income Taxes (25%)

($37,000) (9,250)

($45,444) (11,361)

($19,061) (4,765)

Net Income

($27,750)

($34,083)

($14,296)

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Financing Activities: Borrowed funds Principal repayment

$ (27,750) $ (34,083) $ (14,296) $ 20,000 $ 32,000 $ 9,600 $

(100,000) $ $

Net Cash Flow

50,000 (2,900)

100,000 $ (29,635) $ (33,191) $ (37,174) $0

$ (37,385) $ (35,274) $

5,230

($55,742)

10.42) (a) and (b) Income Statement Revenue Expenses: O&M Depreciation Interest

$ $ $

56,490 11,000 5,000

$ $ $

59,315 8,800 2,619

Taxable Income Income Taxes

$ $

41,510 10,378

$ $

43,266 10,816

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Working Capital Gains Tax Loan Repayment

31,133

32,449

$ $

31,133 11,000

$ $ $ (600) $ $ (23,810) $

32,449 8,800 29,768 12,600 1,358.13 (26,190)

($17,000) ($17,000)

$17,723 $16,879

$58,785 $53,319

PW(18%) = $ IRR (%) =

40,238 133.57%

Net Cash Flow (actual) Net Cash Flow (constant)

$114,000

$ $

(55,000) (12,000) $

50,000

$114,000

Note: Use the market interest rate to find the equivalent present worth of the cash flow series in actual dollars.

10.43) (a) and (b) 0

Income Statement Revenue Expenses: O&M Depreciation Interest

$ 86,100 $ 90,405 $ 94,925 $ 99,672 $ 104,655 $ 109,888 $ 24,000 $ 38,400 $ 23,040 $ 13,824 $ 13,824 $ 6,912 $ 10,800 $ 10,800

Taxable Income Income Taxes (25%)

$31,350 $7,838

$20,258 $5,065

$49,891 $12,473

$62,752 $15,688

$66,582 $16,646

$77,514 $19,379

Net Income

$23,513

$15,194

$37,418

$47,064

$49,937

$58,136

$23,513 $24,000

$15,194 $38,400

$37,418 $23,040

$47,064 $13,824

$49,937 $13,824

$58,136 $6,912

$152,250

Cash Flow Statement Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage $ (120,000) Gains Tax Working Capital Cash from financing activities: Loan repayment $ 120,000 Net Cash Flow (actual)

$159,863 $167,856 $176,248 $185,061 $194,314

$ 20,101 $ (5,025)

$ (120,000) $47,513

($66,407)

$60,458

PW (18%) = $118,325

(c) Present value gain (or loss) due to inflation: 0.18 − 0.05 = 12.38% 1 + 0.05 PW(12.38%) no inflation = $111, 089 = i′

PW(18%) with inflation = $118,325 Present value gains = $118,325 - $111,089 =$7,236

$60,888

$63,761

$80,124

Project Cash Flows without Inflation 0

Income Statement Revenue Expenses: O&M Depreciation Interest

$ 82,000 $ 82,000 $ 82,000 $ 82,000 $ 82,000 $ 82,000 $ 24,000 $ 38,400 $ 23,040 $ 13,824 $ 13,824 $ 6,912 $ 10,800 $ 10,800

Taxable Income Income Taxes

$28,200 $7,050

$13,800 $3,450

$39,960 $9,990

$49,176 $12,294

$56,088 $14,022

Net Income

$21,150

$10,350

$29,970

$36,882

$42,066

$36,882 $13,824

$42,066 $6,912

$145,000

$145,000 $145,000 $145,000 $145,000 $145,000

Cash Flow Statement Cash from operation Net Income $21,150 $10,350 $29,970 Depreciation $24,000 $38,400 $23,040 Cash from investing activities: Investment / Salvage $ (120,000) Gains Tax Working Capital Cash from financing activities: Loan repayment $ 120,000 $ (120,000) Net Cash Flow (actual)

$45,150

($71,250)

$53,010

$ 15,000 $ (3,750)

$50,706

$60,228

PW (12.38%) = $111,089

(d) Present value gain due to borrowing: n 0 1 2

Net Financing Cost Net Principal Interest (A/T) Loan Flow +$120,000 +$120,000 -(1-0.25)(10,800) -$8,100 -$120,000 -(1-0.25)(10,800)-$128.100

PW(18%) Loan = +$120, 000 − $8,100( P / F ,18%,1) −$128,100( P / F ,18%, 2) = $21,136 Comments: Present value gain due to inflation in (c) is largely from the fact that the firm was able to finance the project at 9% interest whereas the market interest rate is 18%. In practice, it is not likely that the firm would be able to access such a cheap money during the inflationary period.

10.44) a), (b), and (c). The project is acceptable Income Statement 0

$84,000

$88,200

$92,610

21,435 $

36,735 $

13,118

Taxable Income Income Taxes (25%)

$62,565 $15,641

$51,465 $12,866

$79,493 $19,873

Net Income

$46,924

$38,599

$59,619

$46,924 $21,435

$38,599 $36,735

$59,619 $13,118

Income Statement Revenue (Labor Savings) Expenses: O&M Depreciation Interest

inflation 5%

Cash Flow Statement Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage Gains Tax Working capital Cash from financing activities: Loan repayment Net Cash Flow (actual) Net Cash Flow (constant)

(150,000)

(10,000) $

$ $ (864) $

(800) $

($160,000) ($160,000)

$67,559 $63,735

PW (20%) = $ PW (13.21%) = $

42,967 42,965

$74,470 $66,278

80,000 (322) 11,664

$164,079 $137,764

Therefore, the project is acceptable.

Note: The general inflation rate is 6% and this rate should be used in converting the actual dollars to its equivalent constant dollars

= i'

0.14 − 0.06 = 0.13208 1 + 0.06

10.45)

0 Income Statement Revenue Expenses: O&M Depreciation

1 $22,000

$22,000

9,500 $ 9,500 $ 9,500 $ 8,860 $ 15,184 $ 10,844 $

9,500 $ 7,744 $

9,500 $ 5,537 $

9,500 $ 5,530 $

9,500 $ 5,537 $

9,500 2,765

Taxable Income Income Taxes (25%)

$3,640 910

($2,684) (671)

$1,656 414

$4,756 1,189

$6,963 1,741

$6,970 1,742

$6,963 1,741

$9,735 2,434

Net Income

$2,730

($2,013)

$1,242

$3,567

$5,223

$5,227

$5,223

$7,301

3,567 $ 7,744 $

5,223 $ 5,537 $

5,227 $ 5,530 $

5,223 $ 7,301 5,537 $ 2,765 $ 5,000 $ (1,250) $ 10,000

Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Net Cash Flow

$ $

$ $ $

(62,000)

(10,000)

2,730 $ (2,013) $ 1,242 $ 8,860 $ 15,184 $ 10,844 $

($72,000)

$11,590

PW (12.38%)=

($10,193.04)

$13,171

$12,086 IRR (%)

$11,311 =

8.40%

$10,759

$10,758

$10,759

$23,816

Income Statement 0

$23,760

$25,661

$27,714

$29,931

$32,325

$34,911

$37,704

$40,720

10,070 8,860

10,674 15,184

11,315 10,844

11,994 7,744

12,713 5,537

13,476 5,530

14,284 5,537

15,142 2,765

Taxable Income Income Taxes (25%)

$4,830 1,208

($197) (49)

$5,555 1,389

$10,193 2,548

$14,075 3,519

$15,905 3,976

$17,883 4,471

$22,814 5,703

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Loan repayment

$3,623

($148)

$4,166

$7,645

$10,557

$11,929

$13,412

$17,110

$3,623 $8,860

($148) $15,184

$4,166 $10,844

$7,645 $7,744

$10,557 $5,537

$11,929 $5,530

$13,412 $5,537

($10,000)

($800)

($864)

($933)

($1,008)

($1,088)

($1,175)

($1,269)

$17,110 $2,765 $7,387 ($1,847) $17,138

($72,000) ($72,000)

$11,682 $11,126

$14,172 $12,854

$14,077 $12,160

$14,381 $11,831

$15,005 $11,757

$16,284 $12,151

$17,679 $12,564

$42,554 $28,802

Income Statement Revenue Expenses: O&M Depreciation

($62,000)

Net Cash Flow (actual dollars) Net Cash Flow (constant dollars

PW (18%)= ($6,474) The project is not acceptable.

IRR (%) = IRR'(%) =

15.50% < 18% 10.00% < 12.38%

10.46) (a) Real after-tax yield on bond investment: • Nontaxable municipal bond: 0.09 − 0.03 = 5.825% 1 + 0.03 • Taxable corporate bond: ′ = imunicipal

′ = icorporate

0.12(1 − 0.3) − 0.03 = 5.243% 1 + 0.03

(b) Given i = 6%, and f = 3% ′ isavings = 2.91% ′ ′ Since imunicipal > 2.91% and icorporate > 2.91% , both bond investments are better than the savings account. The real return is best with the municipal bond.

10.47) (a), (b), and (c) – Select Engine B. Engine A 1

Income Statement Revenue Expenses: O&M Depreciation

$637,200 $12,000

$688,176 $12,000

$743,230 $12,000

$802,688 $12,000

$866,904 $12,000

Taxable Income Income Taxes (25%)

($649,200) ($162,300)

($700,176) ($175,044)

($755,230) ($188,808)

($814,688) ($203,672)

($878,904) ($219,726)

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

($486,900)

($525,132)

($566,423)

($611,016)

($659,178)

($486,900) $12,000

($525,132) $12,000

($566,423) $12,000

($611,016) $12,000

($659,178) $12,000 $40,000

($100,000)

($474,900)

($513,132)

($554,423)

($599,016)

($607,178)

PW (20%)=

($1,705,827)

AE (20%)=

($570,394)

FW (20%)=

($4,244,643)

Net Cash Flow

($100,000)

Engine B 0

Income Statement Revenue Expenses: O&M Depreciation

$407,808 $24,000

$440,433 $24,000

$475,667 $24,000

$513,721 $24,000

$554,818 $24,000

Taxable Income Income Taxes (25%)

($431,808) ($107,952)

($464,433) ($116,108)

($499,667) ($124,917)

($537,721) ($134,430)

($578,818) ($144,705)

Net Income

($323,856)

($348,324)

($374,750)

($403,290)

($434,114)

Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax

($323,856) $24,000

($348,324) $24,000

($374,750) $24,000

($403,290) $24,000

($434,114) $24,000 $80,000

($200,000)

($299,856)

($324,324)

($350,750)

($379,290)

($330,114)

PW (20%)=

($1,193,665)

AE (20%)=

($399,137)

FW (20%)=

($2,970,221)

Net Cash Flow

($200,000)

10.48) (a) & (b) Actual and constant dollar analysis: (b) 0

$126,000

$132,300

$62,400 $12,000

$64,896 $9,600

Taxable Income Income Taxes (25%)

$51,600 $12,900

$57,804 $14,451

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Working capital Gains Tax

$38,700

$43,353

$38,700 $12,000

$43,353 $9,600 $40,000 $5,200 ($400)

Income Statement Revenue Expenses: O&M Depreciation

Net Cash Flow (actual) Net Cash Flow (constant)

($60,000) ($5,000)

($200)

($65,000) ($65,000)

$50,500 $46,759

IRR'(%)

55.08%

(c) Given = f 8%, = i 15% 0.15 − 0.08 = 6.48% (Inflation-free MARR) 1 + 0.08 Since IRR’> 6.48%, the project is a profitable one.

= i′

$97,753 $83,807

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

$84,800

$89,888

$95,281

$100,998

$107,058

$113,482

20,000

32,000

19,200

11,520

5,760

Taxable Income Income Taxes (25%)

$64,800 $16,200

$57,888 $14,472

$76,081 $19,020

$89,478 $22,370

$95,538 $23,885

$107,722 $26,931

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Loan repayment

$48,600

$43,416

$57,061

$67,109

$71,654

$80,792

48,600 20,000

43,416 32,000

57,061 19,200

67,109 11,520

71,654 11,520

80,792 5,760 42,556 (10,639)

$68,600 $55,547

$75,416 $59,392

$76,261 $54,448

$78,629 $51,650

$83,174 $51,443

$118,468 $67,625

Income Statement Revenue Expenses: O&M Depreciation Interest

Net Cash Flow (actual $) Net Cash Flow (constant $)

(100,000)

($100,000) ($100,000) PW (18%) = IRR'(%) =

$179,509 51.53%

$84,800

$89,888

$95,281

$100,998

$107,058

$113,482

20,000 12,000

32,000 10,521

19,200 8,865

11,520 7,010

11,520 4,933

5,760 2,606

Taxable Income Income Taxes (25%)

$52,800 13,200

$47,367 11,842

$67,216 16,804

$82,468 20,617

$90,605 22,651

$105,116 26,279

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Loan repayment

$39,600

$35,525

$50,412

$61,851

$67,954

$78,837

39,600 20,000

35,525 32,000

50,412 19,200

61,851 11,520

67,954 11,520

78,837 5,760 42,556 (10,639)

100,000

(12,323)

(13,801)

(15,457)

(17,312)

(19,390)

(21,717)

$0 $0

$47,277 $37,130

$53,724 $41,491

$54,154 $37,004

$56,058 $34,605

$60,084 $34,743

$94,797 $51,213

(100,000)

Net Cash Flow (actual $) Net Cash Flow (constant $) PW (18%) =

$201,903 IRR'(%) =

No IRR

(d) The present value loss due to inflation: 0

$80,000

20,000

32,000

19,200

11,520

5,760

Taxable Income Income Taxes (25%)

$60,000 15,000

$48,000 12,000

$60,800 15,200

$68,480 17,120

$74,240 18,560

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Investment / Salvage Gains Tax Working capital Loan repayment

$45,000

$36,000

$45,600

$51,360

$55,680

45,000 20,000

36,000 32,000

45,600 19,200

51,360 11,520

55,680 5,760 30,000 (7,500)

($100,000)

$65,000

$68,000

$64,800

$62,880

$83,940

PW (11.32%) =

$182,076

Income Statement Revenue Expenses: O&M Depreciation Interest

Net Cash Flow (actual $)

(100,000)

Present value loss = $179,509 - $182,076 = ($2,567) (e) Required additional before-tax annual revenue in actual dollars (equal amount) to make-up the inflation loss. $2,567( A / P,18%, 6) = $978.58 1 − 0.25

10.50) (a), (b), and (c) (Unit:$000) 0

5-7

10-11

$ 84,000

$ 140,000

$ 224,000

224,000

$ 45,000

75,000

$ 120,000

120,000

$ 529 $ 14,290

$ 1,154 $ 24,490

$ 1,154 $ 17,490

$ 1,154 $ 12,490

$ $

1,154 8,930

$ $

1,154 4,460

1,154

1,106

Taxable Income Income Taxes (25%)

$ 24,181 $ 6,045

$ 13,356 $ 3,339

$ 20,356 $ 5,089

$ 25,356 $ 6,339

$ $

54,916 13,729

$ $

98,386 24,597

$ 102,846 $ 25,712

$ $

102,894 25,724

Net Income

$ 18,136

$ 10,017

$ 15,267

$ 19,017

41,187

73,790

77,135

77,171

$ 10,017 $ 25,644

$ 15,267 $ 18,644

$ 19,017 $ 13,644

$ $

41,187 10,084

$ $

73,790 5,614

$ $

77,135 1,154

$ $

77,135 1,154

$ $

77,171 1,106

$ $ $

8,000 30,000 10,000

Income Statement Revenue Expenses: Production costs Depreciation : Building Equipment

Cash Flow Statement Operating Activities: Net Income $ 18,136 Depreciation $ 14,819 Investment Activities: Land $ (5,000) Building $ (45,000) Equipment $ (100,000) Gains Tax Land (25%) Building (25%) Equipment (25%) Net Cash Flow

($150,000)

$32,955

PW(15%) =

$120,994

1,154

$ (750) $ 456.68 $ (2,502.50) $35,661

$33,911

$32,661

$51,271

IRR =

27.87%

$79,403

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.

$78,288

$123,481

10.51) (a) The net after-tax cash flows for each financing option: •

Option 1: Equity Financing (Retained earnings) Input Tax Rate(%) = MARR(%) = 0

25 18

Output PW(i) = IRR(%) =

$220,799 50.74%

Revenues (savings) Expenses: O&M costs Depreciation Debt interest

$174,000

$22,000 $28,580

$22,000 $48,980

$22,000 $34,980

$22,000 $24,980

$22,000 $17,860

$22,000 $8,930

Taxable Income Income Taxes (25%)

$123,420 $30,855

$103,020 $25,755

$117,020 $29,255

$127,020 $31,755

$134,140 $33,535

$143,070 $35,768

$92,565

$77,265

$87,765

$95,265

$100,605

$107,303

$92,565 $28,580

$77,265 $48,980

$87,765 $34,980

$95,265 $24,980

$100,605 $17,860

$107,303 $8,930

Income Statement

Net Income Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Working capital Financing Activities: Borrowed funds Principal repayment Net Cash Flow

($200,000) $30,000 $1,423 $25,000

($25,000)

($225,000)

$121,145

$126,245

$122,745

$120,245

$118,465

$172,655

•

Option 2: Debt Financing at 12% Input Tax Rate(%) = MARR(%) =

25 18

Output PW(i) = IRR(%) =

$265,586 317.03%

Revenues (savings) Expenses: O&M costs Depreciation Debt interest

$174,000

$22,000 28,580 24,000

$22,000 48,980 21,043

$22,000 34,980 17,730

$22,000 24,980 14,020

$22,000 17,860 9,866

$22,000 8,930 5,212

Taxable Income Income Taxes (25%)

$99,420 24,855

$81,977 20,494

$99,290 24,823

$113,000 28,250

$124,274 31,069

$137,858 34,465

Net Income

$74,565

$61,483

$74,468

$84,750

$93,206

$103,394

93,206 17,860

$ 103,394 $ 8,930

Income Statement

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Working capital Financing Activities: Borrowed funds Principal repayment Net Cash Flow

$ $ $

74,565 28,580

$ $

61,483 48,980

$ $

74,468 34,980

$ $

84,750 24,980

$ $

(200,000) $ 30,000 $ 1,423 $ 25,000

(25,000) 200,000 (24,645)

(25,000) $

78,500

(27,603) $

82,860

(30,915) $

78,533

(34,625) $

75,105

(38,780)

(43,433)

72,286

$ 125,313

•

Option 3: Lease Financing Input Tax Rate(%) = MARR(%) = 0

Output PW(i) = $270,452 IRR(%) = 134.72%

25 18 1

$174,000

Income Statement Revenues (savings) Expenses: Lease Payment O&M costs Debt interest

$55,000

$55,000 0

Taxable Income Income Taxes (25%)

($55,000) $119,000 ($13,750) 29,750

$119,000 29,750

$174,000 43,500

Net Income

($41,250)

$89,250

$130,500

(41,250) $ 89,250

$ 89,250

$ 130,500

(25,000)

(66,250) $ 89,250

$ 25,000

$ 89,250

$ 155,500

(b) Vermont’s PW cost of owning the equipment by borrowing: •

PW of total after-tax revenue:

= P1 $174, 000(1 − 0.25)( P / A,18%, 6) •

= $456, 437 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 − 0.25)( P / A,18%, 6) = $57, 710

•

PW cost of owning by borrowing:

Net cost = P1 − P2 − P3 − $265, 799 = $117,189 (c) Vermont’s PW cost of leasing the equipment: •

PW cost of after-tax leasing

= P $55, 000(1 − 0.25) + $55, 000(1 − 0.25)( P / A,18%,5) = $170, 246 (d) Lease the tipping machine. (Note: we assumed that the entire O&M costs are picked up by the lessor.)

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

2017 -2

2018 -1

2019 0

Revenues: Sales unit Unit price Sales volume Expenses: Fixed costs Variable costs Depreciation : Building Equipment Amortization Taxable Income Income Taxes (25%) Net Income Cash Flow Statement Operating Avtivities: Net Income Depreciation Amortization Investment activities Opportunity cost* Land Building Equipment Gains Taxes Land Building Equipment Working capital

($3,040,000) ($2,500,000)

Net Cash Flow (actual) Net Cash Flow (constant)

($5,540,000) ($5,540,000)

($3,500,000)

PW (15%,n =-2)

($3,500,000) ($3,333,333)

2020

2021

200,000 $400 $80,000,000

200,000 $420 $84,000,000

$8,500,000 $52,000,000

2022 3

2023

2024

2025 6

2026

2027

200,000 $441 $88,200,000

200,000 $463 $92,610,000

200,000 $486 $97,240,500

200,000 200,000 $511 $536 $102,102,525 $107,207,651

200,000 $563 $112,568,034

$8,925,000 $54,600,000

$9,371,250 $57,330,000

$9,839,813 $60,196,500

$10,331,803 $63,206,325

$10,848,393 $66,366,641

$11,390,813 $69,684,973

$11,960,354 $73,169,222

$258,017 $2,715,100 $1,000,000 $15,526,884 $3,881,721

$269,231 $4,653,100 $1,000,000 $14,552,670 $3,638,167

$269,231 $3,323,100 $1,000,000 $16,906,420 $4,226,605

$269,231 $2,373,100 $1,000,000 $18,931,357 $4,732,839

$269,231 $1,696,700 $1,000,000 $20,736,441 $5,184,110

$269,231 $1,694,800 $1,000,000 $21,923,460 $5,480,865

$269,231 $1,696,700 $1,000,000 $23,165,934 $5,791,484

$269,231 $847,400 $1,000,000 $25,321,828 $6,330,457

$11,645,163

$10,914,502

$12,679,815

$14,198,518

$15,552,331

$16,442,595

$17,374,451

$18,991,371

$11,645,163 $2,973,117 $1,000,000

$10,914,502 $4,922,331 $1,000,000

$12,679,815 $3,592,331 $1,000,000

$14,198,518 $2,642,331 $1,000,000

$15,552,331 $1,965,931 $1,000,000

$16,442,595 $1,964,031 $1,000,000

$17,374,451 $1,965,931 $1,000,000

$18,991,371 $1,116,631 $1,000,000

($7,000,000) ($19,000,000)

$4,500,000 $3,000,000 $3,500,000

($9,600,000)

($480,000)

($504,000)

($529,200)

($555,660)

($583,443)

($612,615)

($643,246)

($500,000) $1,339,343 ($875,000) $13,508,164

($35,600,000) ($32,290,249)

$15,138,279 $13,077,015

$16,332,833 $13,437,062

$16,742,945 $13,118,536

$17,285,188 $12,898,474

$17,934,819 $12,745,941

$18,794,010 $12,720,526

$19,697,135 $12,696,949

$45,580,508 $27,982,478

$29,339,006 PW (15%,n = 0) PW(F /P ,15%,2) PW(A /P ,15%,8)= AE(15%)=

$38,800,835.12 $8,646,769.58

IRR' =

25.55%

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 (0.25)($8,000,000) = $2,000,000. If the firm decides to undertake this project, then an opportunity cost of $2,000,000 will be incurred.

(e) Income Statement

2017 -2

2018 -1

2019 0

($3,040,000) ($2,500,000)

Net Cash Flow (actual) Net Cash Flow (constant)

($5,540,000) ($5,540,000)

(f)

2021 2

132,016 $400 $52,806,477

132,016 $420 $55,446,801

$8,500,000 $34,324,210

2022 3

2023

2024

2025

2026

2027

132,016 $441 $58,219,141

132,016 $463 $61,130,098

132,016 $486 $64,186,603

132,016 $511 $67,395,933

132,016 $536 $70,765,729

132,016 $563 $74,304,016

$8,925,000 $36,040,420

$9,371,250 $37,842,441

$9,839,813 $39,734,564

$10,331,803 $41,721,292

$10,848,393 $43,807,356

$11,390,813 $45,997,724

$11,960,354 $48,297,610

$269,231 $2,715,100 $1,000,000 $5,997,936 $1,499,484

$269,231 $4,653,100 $1,000,000 $4,559,050 $1,139,762

$269,231 $3,323,100 $1,000,000 $6,413,119 $1,603,280

$269,231 $2,373,100 $1,000,000 $7,913,391 $1,978,348

$269,231 $1,696,700 $1,000,000 $9,167,577 $2,291,894

$269,231 $1,694,800 $1,000,000 $9,776,153 $2,444,038

$269,231 $1,696,700 $1,000,000 $10,411,262 $2,602,815

$269,231 $847,400 $1,000,000 $11,929,421 $2,982,355

$4,498,452

$3,419,287

$4,809,839

$5,935,043

$6,875,683

$7,332,115

$7,808,446

$8,947,066

$4,498,452 $2,984,331 $1,000,000

$3,419,287 $4,922,331 $1,000,000

$4,809,839 $3,592,331 $1,000,000

$5,935,043 $2,642,331 $1,000,000

$6,875,683 $1,965,931 $1,000,000

$7,332,115 $1,964,031 $1,000,000

$7,808,446 $1,965,931 $1,000,000

$8,947,066 $1,116,631 $1,000,000

($7,000,000) ($19,000,000)

$4,500,000 $3,000,000 $3,500,000

($6,336,777)

($316,839)

($332,681)

($349,315)

($366,781)

($385,120)

($404,376)

($424,594)

($500,000) $1,336,539 ($875,000) $8,916,482

($3,500,000) ($32,336,777) ($3,333,333) ($29,330,410)

$8,165,944 $7,054,049

$9,008,937 $7,411,675

$9,052,855 $7,093,149

$9,210,593 $6,873,087

$9,456,494 $6,720,554

$9,891,769 $6,695,139

$10,349,783 $6,671,562

$30,941,718 $18,995,530

($3,500,000)

PW (15%,n =-2) =

2020 1

$3,347,321 PW (15%,n =0) =PW(F /P ,15%,2)

$4,198,879.81

Income Statement

2017 -2

2018 -1

2019 0

($3,040,000) ($2,500,000)

Net Cash Flow (actual) Net Cash Flow (constant)

($5,540,000) ($5,540,000)

2020

2021

200,000 $400 $80,000,000

200,000 $388 $77,600,000

$8,500,000 $52,000,000

2022 3

2023

2024

2025

2026

2027

200,000 $376 $75,272,000

200,000 $365 $73,013,840

200,000 $354 $70,823,425

200,000 $343 $68,698,722

200,000 $333 $66,637,760

200,000 $323 $64,638,628

$8,925,000 $50,440,000

$9,371,250 $48,926,800

$9,839,813 $47,458,996

$10,331,803 $46,035,226

$10,848,393 $44,654,169

$11,390,813 $43,314,544

$11,960,354 $42,015,108

$269,231 $2,715,100 $1,000,000 $15,515,670 $3,878,917

$269,231 $4,653,100 $1,000,000 $12,312,670 $3,078,167

$269,231 $3,323,100 $1,000,000 $12,381,620 $3,095,405

$269,231 $2,373,100 $1,000,000 $12,072,701 $3,018,175

$269,231 $1,696,700 $1,000,000 $11,490,465 $2,872,616

$269,231 $1,694,800 $1,000,000 $10,232,129 $2,558,032

$269,231 $1,696,700 $1,000,000 $8,966,473 $2,241,618

$269,231 $847,400 $1,000,000 $8,546,536 $2,136,634

$11,636,752

$9,234,502

$9,286,215

$9,054,526

$8,617,849

$7,674,097

$6,724,855

$6,409,902

$11,636,752 $2,984,331 $1,000,000

$9,234,502 $4,922,331 $1,000,000

$9,286,215 $3,592,331 $1,000,000

$9,054,526 $2,642,331 $1,000,000

$8,617,849 $1,965,931 $1,000,000

$7,674,097 $1,964,031 $1,000,000

$6,724,855 $1,965,931 $1,000,000

$6,409,902 $1,116,631 $1,000,000

($7,000,000) ($19,000,000)

$4,500,000 $3,000,000 $3,500,000

($9,600,000)

$288,000

$279,360

$270,979

$262,850

$254,964

$247,315

$239,896

($500,000) $1,336,539 ($875,000) $7,756,635

($3,500,000) ($35,600,000) ($3,333,333) ($32,290,249)

$15,909,083 $13,742,864

$15,436,193 $12,699,394

$14,149,524 $11,086,523

$12,959,706 $9,670,732

$11,838,744 $8,413,574

$10,885,443 $7,367,696

$9,930,681 $6,401,405

$27,244,706 $16,725,886

($3,500,000)

PW (15%,n =-2) = $13,988,120 PW (15%) = AE(15%) =

PW(F /P ,15%,2) PW(A /P ,15%,8)

$18,499,289 $4,122,568

IRR' =

18.99%

10.53) (a) The net cash flow from the cogeneration project with bond financing: (a) The net cash flow from the cogeneration project with bond financing: 0 1 2 Income Statement Revenue Electricity bill $6,120,000 $6,120,000 Excess power $480,000 $480,000 Expenses: O&M $500,000 $500,000 Misc. $1,000,000 $1,000,000 Standby power $6,400 $6,400 Fuel $1,280,000 $1,280,000 Other Overhaul Standby power(overhaul)

$6,120,000 $480,000

$500,000 $1,000,000 $6,400 $1,280,000

$1,500,000

$100,000

Depreciation Unit

$500,000

$950,000

$855,000

$770,000

$693,000

$623,000

Inter Equipment

$100,000

$160,000

$96,000

$57,600

$28,800

Interest (9%)

$945,000

Taxable Income

$2,268,600

$1,758,600

$317,600

$2,041,000

$567,150

$439,650

$79,400

$510,250

$1,701,450

$1,318,950

$238,200

$1,701,450

$1,318,950

Income Taxes (25%)

Net Income

$590,000

$591,000

$590,000

$591,000

$295,000

$945,000

$2,118,000

$616,800

$2,278,600

$677,600

$2,278,600

$2,277,600

$973,600

$529,500

$154,200

$569,650

$169,400

$569,650

$569,400

$243,400

$1,530,750

$1,588,500

$462,600

$1,708,950

$508,200

$1,708,950

$1,708,200

$730,200

$238,200

$1,530,750

$1,588,500

$462,600

$1,708,950

$508,200

$1,708,950

$1,708,200

$730,200

Cash Flow Statement Cash from operation Net Income Depreciation Unit

$500,000

$950,000

$855,000

$770,000

$693,000

$623,000

$590,000

$591,000

$590,000

$591,000

$295,000

Inter Equipment

$100,000

$160,000

$96,000

$57,600

$28,800

Investment / Salvage Unit Inter Equipment

($10,000,000)

$1,000,000

($500,000)

Gains Tax Loan repayment

$340,500 $10,500,000

Net Cash Flow (actual)

$0 PW (27%) =

($10,500,000)

$2,301,450 $6,592,272

$2,428,950

$1,189,200

$2,358,350

$2,339,100

$1,114,400

$2,298,950

$1,099,200

$2,298,950

$2,299,200

($8,134,300)

(b). The maximum annual lease amount that ACC is willing to pay is $1,297,463.

7 - 11

Income Statement Revenue Electricity bill Excess power Expenses: O&M Misc. Standby power Overhead Lease

$6,120,000 $480,000

$500,000 $1,000,000 $6,400 $1,280,000 $1,297,463

Taxable Income Income Taxes (25%)

$2,516,137 $629,034

Net Income Cash Flow Statement Cash from operation Net Income

$1,887,103

PW (27%) =

$6,592,272

Net Cash Flow (actual)

10.54) (a) & (b) The project cash flows and IRR with no inflation: Income Statement Revenue Expenses: O&M Labor Material Energy Depreciation : Building Milling machine Jigs & dies

1 2 $80,000 $80,000

3 $80,000

4 $80,000

5 $80,000

6 $80,000

7 $80,000

8 $80,000

9 $80,000

10 $80,000

$3,000 $3,000 $15,000 $15,000 $9,000 $9,000 $4,500 $4,500

$3,000 $15,000 $9,000 $4,500

$15,719 $26,939 $3,333 $4,445

$19,239 $1,481

$13,739 $741

$9,823 $0

$9,812 $3,333

$9,823 $4,445

$4,906 $1,481

$0 $741

$0 $0

Taxable Income Income Taxes (25%)

$29,448 $17,116 $7,362 $4,279

$27,780 $6,945

$34,020 $8,505

$38,677 $9,669

$35,355 $8,839

$34,232 $8,558

$42,113 $10,528

$47,759 $11,940

$48,500 $12,125

Net Income Cash Flow Statement Cash from operation Net Income Depreciation Building Milling machine Jigs & dies Investment / Salvage Building Milling machine Jigs & dies (Replacement) Gains Taxes: Building Milling machine Jigs & dies

$22,086 $12,837

$20,835

$25,515

$29,008

$26,516

$25,674

$31,585

$35,819

$36,375

$22,086 $12,837

$20,835

$25,515

$29,008

$26,516

$25,674

$31,585

$35,819

$36,375

$15,719 $26,939 $3,333 $4,445

$19,239 $1,481

$13,739 $741

$9,823 $0

$9,812 $3,333

$9,823 $4,445

$4,906 $1,481

$0 $741

$0 $0

Net Cash Flow

($110,000) ($10,000)

($120,000)

$10,000

$41,138 $44,221

PW (11.32%) = $110,619

$41,555

$39,995

IRR (%) =

31.39%

$300 ($10,000)

$300

($75)

($2,500) ($75)

$29,056

$39,661

$39,942

$37,972

$36,560

$44,100

(c) & (d) (c) and (d) Income Statement (with inflation) 0 1 Revenue $85,600 Expenses: O&M $3,090 Labor $15,750 Material $9,360 Energy $4,635 Depreciation : Building Milling machine $15,719 Jigs & dies $3,333

2 $91,592

3 $98,003

4 $104,864

5 $112,204

6 $120,058

7 $128,463

8 $137,455

9 $147,077

10 $157,372

$3,183 $16,538 $9,734 $4,774

$3,278 $17,364 $10,124 $4,917

$3,377 $18,233 $10,529 $5,065

$3,478 $19,144 $10,950 $5,217

$3,582 $20,101 $11,388 $5,373

$3,690 $21,107 $11,843 $5,534

$3,800 $22,162 $12,317 $5,700

$3,914 $23,270 $12,810 $5,871

$4,032 $24,433 $13,322 $6,048

$26,939 $4,445

$19,239 $1,481

$13,739 $741

$9,823 $0

$9,812 $3,333

$9,823 $4,445

$4,906 $1,481

$0 $741

$0 $0

Taxable Income Income Taxes (25%)

$33,713 $8,428

$25,979 $6,495

$41,599 $10,400

$53,181 $13,295

$63,592 $15,898

$66,468 $16,617

$72,021 $18,005

$87,088 $21,772

$100,470 $25,118

$109,537 $27,384

$25,285

$19,485

$31,200

$39,886

$47,694

$49,851

$54,016

$65,316

$75,353

$82,153

$25,285

$19,485

$31,200

$39,886

$47,694

$49,851

$54,016

$65,316

$75,353

$82,153

$15,719 $3,333

$26,939 $4,445

$19,239 $1,481

$13,739 $741

$9,823 $0

$9,812 $3,333

$9,823 $4,445

$4,906 $1,481

$0 $741

$0 $0

Net Cash Flow (actual) Net Cash Flow (constant)

($110,000) ($10,000)

($120,000) ($120,000)

$10,000

$44,337 $38,647

PW (11.32%) = $108,411

$50,869 $42,960

$51,920 $40,100 IRR' (%) =

$54,366 $38,850

$300 ($10,000)

$300

($75)

($2,500) ($75)

$47,742 $30,899

$62,996 $39,724

30.53% >11.32%

(e). The economic loss in present worth due to inflation = $108,411- $110,619 = ($2,208).

$68,284 $40,620

$71,703 $39,523

$76,094 $39,093

$89,878 $43,496

Chapter 11 Handling Project Uncertainty Note to Instructors: Regular MACRS depreciation is assumed in all problems unless otherwise mentioned. However, instructors may adopt the capital expensing option under the 2017 Tax Cuts and Job Acts instead. In that case, the entire capital expenditure would be claimed as depreciation amount at the end of first year. Then, any salvage value at the end of project life will be subject to taxable gains. 11.1) (a)

(b)

(c)

AEC(10%) = (25, 000 − 5, 000)( A / P,10%, 6) + 0.1(5, 000) + 3, 000 = $8, 092

AEC(10%) = (25, 000 − 5, 000)( A / P,10%,5) + 0.1(5, 000) + 3, 000 = $8, 776 AEC(10%) = (25, 000 − 5, 000)( A / P,10%, 6) + 0.1(5, 000) + 3,300 = $8,392

11.2)

(a) Project cash flows based on most-likely estimates: without working capital 0 Income Statement Labor Savings Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gains Tax Net Cash Flow PW (10%) =

$35,000 21,600 $13,400 3,350 $10,050

$35,000 34,560 $440 110 $330

$35,000 20,736 $14,264 3,566 $10,698

$35,000 6,221 $28,779 7,195 $21,584

10,050 21,600

330 34,560

10,698 20,736

21,584 6,221 30,000 (1,279)

31,650

34,890

31,434

56,526

(108,000) (108,000) $11,832

The project is acceptable.

(b) Project cash flows based on most-likely estimates: with working capital 0 Income Statement Labor Savings Depreciation Taxable Income Income Tax (25%) Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gains Tax Working Capital Net Cash Flow PW (10%) =

$35,000 21,600 $13,400 3,350

$35,000 34,560 $440 110

$35,000 20,736 $14,264 3,566

$35,000 6,221 $28,779 7,195

$10,050

$330

$10,698

$21,584

10,050 21,600

330 34,560

10,698 20,736

21,584 6,221 30,000 (1,279) 5,000

31,650

34,890

31,434

61,526

(108,000) (5,000) (113,000) $10,247

The project is still acceptable.

Labor Savings Depreciation

$ 39,079 $ $ 21,600 $

39,079 $ 34,560 $

39,079 $ 20,736 $

39,079 6,221

Taxable Income Income Tax (25%)

$ 17,479 $ $ 4,370 $

4,519 $ 1,130 $

18,343 $ 4,586 $

32,858 8,214

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment &Salvage Gains Tax

$ 13,109 $

3,389 $

13,757 $

24,643

$ 13,109 $ $ 21,600 $

3,389 $ 34,560 $

13,757 $ 20,736 $ $ $

24,643 6,221 30,000 (2,047)

$ (108,000) $ 34,709 $

37,949 $

34,493 $

58,817

Net Cash Flow

$ (108,000)

PW(18%) =

11.3) • Project’s IRR if the investment is made now:

PW(i ) = −$500, 000 + $200, 000( P / A, i,5) = 0

i = 28.65% • Let X denote the new after-tax annual cash flow: PW(28.65%) = −$500, 000 + X ( P / A, 28.65%, 4)( P / F , 28.65%,1) = 0 X = $290, 248

The needed additional flow is $290,248-$200,000 = $90,248.

11.4) Income statement Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$40 $40 $40 $40 $40 100,000 100,000 100,000 100,000 100,000 $4,000,000 $4,000,000 $4,000,000 $4,000,000 $4,000,000 $18 $18 $18 $18 $18 $1,800,000 $1,800,000 $1,800,000 $1,800,000 $1,800,000 $230,000 $230,000 $230,000 $230,000 $230,000 $90,000 $90,000 $90,000 $90,000 $90,000

Taxable income Income taxes (35%)

$1,880,000 $1,880,000 $1,880,000 $1,880,000 $1,880,000 $658,000 $658,000 $658,000 $658,000 $658,000

Net income

$1,222,000 $1,222,000 $1,222,000 $1,222,000 $1,222,000

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

$1,222,000 $1,222,000 $1,222,000 $1,222,000 $1,222,000 $90,000 $90,000 $90,000 $90,000 $90,000 ($500,000)

$50,000 0

($500,000) $1,312,000 $1,312,000 $1,312,000 $1,312,000 $1,362,000 $3,443,777

The breakeven value of unit price is $22.28.

As long as the unit price is greater than

$22.28, the project rate of return will be positive.

11.5)

(a) Economic building height • 5% < i < 20% : The optimal building height is 5 floors. • 20% ≤ i < 30% : The optimal building height is 2 floors.

Net Cash Flows n 0 1 2 3 4 5

2 Floors ($500,000) $199,100 $199,100 $199,100 $199,100 $799,100

3 Floors ($750,000) $169,200 $169,200 $169,200 $169,200 $1,069,200

4 Floors ($1,250,000) $149,200 $149,200 $149,200 $149,200 $2,149,200

5 Floors ($2,000,000) $378,150 $378,150 $378,150 $378,150 $3,378,150

Sensitivity Analysis PW(i) as a Function of Interest Rate i (%) 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

2 Floors $832,115 $787,037 $744,141 $703,298 $664,388 $627,298 $591,924 $558,167 $525,937 $495,148 $465,720 $437,580 $410,657 $384,885 $360,205 $336,557 $313,889 $292,150 $271,292 $251,271 $232,044 $213,572 $195,817 $178,745 $162,323 $146,519

3 Floors $687,721 $635,264 $585,441 $538,091 $493,067 $450,230 $409,452 $370,612 $333,599 $298,309 $264,644 $232,512 $201,829 $172,516 $144,496 $117,701 $92,066 $67,527 $44,029 $21,516 ($62) ($20,753) ($40,601) ($59,650) ($77,939) ($95,505)

4 Floors $963,010 $873,011 $787,722 $706,879 $630,199 $557,428 $488,330 $422,686 $360,291 $300,953 $244,495 $190,751 $139,565 $90,792 $44,298 ($46) ($42,357) ($82,746) ($121,319) ($158,173) ($193,399) ($227,084) ($259,308) ($290,148) ($319,674) ($347,955)

5 Floors $1,987,770 $1,834,680 $1,689,448 $1,551,593 $1,420,666 $1,296,250 $1,177,957 $1,065,427 $958,321 $856,326 $759,148 $666,513 $578,166 $493,867 $413,393 $336,533 $263,091 $192,883 $125,737 $61,490 ($9) ($58,903) ($115,327) ($169,407) ($221,261) ($271,002)

Best Floor Plan 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 2 2 2 2 2 2 2 2 2 2 2

(b) Effects of overestimation on resale value: Present Worth as a Function of Number of Floors

Resale value

2 Floors $465, 720 $435,890 $29,831

Base 10% error Difference

3 Floors $264, 644 $219,898 $44, 746

4 Floors $244, 495 $145, 060 $99, 435

5 Floors $759,148 $609,995 $149,153

11.6) Sensitivity graph $2,000,000 V = 6000

$1,800,000 $1,600,000

NPW ($)

$1,400,000

V = 5000

$1,200,000 $1,000,000

V = 4000

$800,000 V = 3000

$600,000 $400,000

V = 2000

$200,000

V = 1000

$0 20

Sales Price (X)

11.7) (a) Project cash flows based on most-likely estimates: 1

Income Statement Revenue: Bill savings Mile Savings Expenses: Depreciation

$3,000,000 1,250,000

$3,000,000 $1,250,000

2,000,000

3,200,000

1,920,000

1,152,000

576,000

Taxable Income Income Tax (25%)

2,250,000 562,500

1,050,000 262,500

2,330,000 582,500

3,098,000 774,500

3,674,000 918,500

4,250,000 1,062,500

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage

$1,687,500

$787,500

$1,747,500

$2,323,500

$2,755,500

$3,187,500

1,687,500 2,000,000

787,500 3,200,000

1,747,500 1,920,000

2,323,500 1,152,000

2,755,500 576,000

3,187,500 0

(10,000,000)

Net Cash Flow

(10,000,000)

3,687,500

3,987,500

3,667,500

3,475,500

3,331,500

3,187,500

PW (18%) =

$4,615,437

(b) Sensitivity analysis: Percentage deviation

Savings In T.B

-30% -20% -10% 0 (base) +10% +20% +30%

$2,100,000 2,400,000 2,700,000 3,000,000 3,300,000 3,600,000 3,900,000

PW(18%)

Savings In D.M

PW(18%)

$1,863,080 2,780,532 3,697,985 4,615.437 5,532,889 6,450,341 7,367,794

$875,000 1,000,000 1,125,000 1,250,000 1,375,000 1,500,000 1,625,000

$3,468,821 3,850,893 4,233,165 4,615,437 4,997,709 5,379,980 5,762,252

11.8) • PW of net investment:

P0 = −$2, 200, 000 − $600, 000 − $400, 000 = −$3, 200, 000 • PW of after-tax revenue: P1 = −$4, 000(365) X (1 − 0.31)( P / A,10%, 25) = $9,144, 210 X

• PW of after-tax operating costs: P2 = −($230, 000 + $170, 000 X )(1 − 0.31)( P / A,10%, 25) = −$1, 440,526 − 1, 064, 737 X

• PW of tax credit (shield) on depreciation:

Depreciation Building Furniture $54,060 $57,160

Combined Tax savings $111,220(0.31) = $34,478

56,410

97,960

154,370(0.31) = 47,855

56,410

69,960

126,370(0.31) = 39,175

56,410

49,960

106,370(0.31) = 32,975

56,410

35,720

92,130(0.31) = 28,560

56,410

33,680

92,090(0.31) = 28,548

56,410

35,720

92,130(0.31) = 28,560

56,410

17,840

74,250(0.31) = 23,018

9-24

56,410

56,410(0.31) = 17,487

54,060

54,060(0.31) = 16,759

P3 = $34, 478( P / F ,10%,1) + $47,855( P / F ,10%, 2) +  + $16, 759( P / F ,10%, 25) = $247, 461

• PW of net proceeds from sale:

Property (asset) Furniture

Cost basis

Building

2,200,000

600,000

2,031,813

Land

$400,000

Salvage Book value value $0 $0

Gains (losses) $0

Gains Taxes

794,450

(794,450)

(246,280)

600,000

1,431,813

443,862

Net proceeds from sale = $2, 031,813 + $246, 280 − $443,862 = $1,834, 231 P4 = $1,834, 231( P / F ,10%, 25) = $169, 292

PW(10%) = P0 + P1 + P2 + P3 + P4 = −$4, 223, 772 + 8, 079, 473 X =0 X = 52.28%

11.9) (a) With an assumption of the unit price of $100, the PW(12%) for options A and B would be the same when unit cost is set at $9.35. Here we also assumed that the old machine is sold off and the current net after-tax salvage value in the amount of $18,750 is credited for Option B. If the firm retains the old machines for other use, then the unit cost would be $8.93.

Option A

100 0

$2,500,000

Expenses: Operating cost Depreciation

225,000 0

Taxable Income Income Tax (25%)

2,275,000 568,750

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes

$1,706,250

1,706,250 0

1,706,250 0 6,000 (1,500)

1,706,250

1,710,750

Income Statement Revenue:

Net Cash Flow

0 PW (12%) =

$4,101,328

Option B

100 0

unit cost= 2

9.345227724 3

Income Statement Revenue: $2,500,000

$2,500,000

Expenses: Operating cost Depreciation

233,631 0

Taxable Income Income Tax (25%)

2,266,369 566,592

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes

$1,699,777

1,699,777 0

25,000 (6,250)

Net Cash Flow

18,750 PW (12%) =

$4,101,328

0 1,699,777

1,699,777

(b) If we assume the unit price of $100, the PW(12%) for options B and C would be $4,181,646, when unit cost is set at $7.14.

Option C 0

Income Statement Revenue: $2,500,000

$2,500,000

Expenses: Operating cost Depreciation

168,750 7,860

168,750 13,470

168,750 4,810

Taxable Income Income Tax (25%)

2,323,391 580,848

2,317,781 579,445

2,326,440 581,610

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes Net Sale of old machine

$1,742,543

$1,738,335

$1,744,830

1,742,543 7,860

1,738,335 13,470

1,744,830 4,810 15,000 3,465

1,750,402

1,751,805

1,768,105

unit cost= 2

7.145517276 3

(55,000) 18,750

Net Cash Flow

(36,250) PW (12%) =

$4,181,640

Option B

100 0

Income Statement Revenue: $2,500,000

$2,500,000

Expenses: Operating cost Depreciation

178,638 0

Taxable Income Income Tax (25%)

2,321,362 580,341

Net Income Cash Flow Statement Cash From Operation: Net Income Depreciation Investment&Salvage Gain taxes

$1,741,022

1,741,022 0

1,741,022 0 0

Net Cash Flow

0 PW (12%) =

$4,181,640

(c). Option C is the most economical.

1,741,022

11.10) Useful life of the old bulb: 14, 600 /(19 × 365) = 2.1 years

For computational simplicity, let’s assume a useful life of 2 years for the old bulb. Then, the new bulb will last 4 years. Let X denote the price for the new light bulb. With an analysis period of 4 years, we can compute the equivalent present worth cost for each option as follows:

PW(15%)old = (1 − 0.25)[$45.90 + $45.90( P / F ,15%, 2)] = $60.45 PW(15%) new = (1 − 0.25)( X + $16) The break-even price for the new bulb will be

0.75 X + 12 = $60.45 X = $64.6 Since the new light bulb costs only $60, it is worth switching to the new light bulb

11.11) • PW of net investment:

P0 = −$250, 000 • PW of after-tax rental revenue: P1 = X (1 − 0.30)( P / A,15%, 20) = $4.3815 X

• PW of after-tax operation costs: P2 = −(1 − 0.30)$12, 000( P / A,15%, 20) = −$52,578

• PW of tax credit (shield) on depreciation: (In this problem, we assume that the purchasing cost of $250,000 does not include any land value. Therefore, the entire purchasing cost will be the cost basis for depreciation purpose.)

Depreciation n Building 1 $6,143 2-19 6,410 20 6,143

Combined Tax savings $6,143(0.30) = $1,843 6,410(0.30) = 1,923 6,143(0.30) = 1,843

P3 = $1,843( P / F ,15%,1) + $1,923( P / A,15%,18)( P / F ,15%,1) +$1,843( P / F ,15%, 20) = $11,962 • PW of net proceeds from sale: Total depreciation = $127, 666 Book value = $250, 000 − $127, 666 = $122,334 Salvage value = $250, 000(1.05) 20 = $663,324 Taxable gain = $663,324 − $122,334 = $540,990 Gains tax = $540,990(0.30) = $162, 297 Net proceeds from sale = $663,324 − $162, 297 = $501, 027 P4 = $501, 027( P / F ,15%, 20) = $30, 613

• The break-even rental: PW(15%) = P0 + P1 + P2 + P3 + P4 = −$260, 003 + 4.3815 X =0 X = $59,341

11.12) Model A: 0

Income Statement Revenues (savings) Expenses: Operating Expenses Depreciation

22,000 16,000

22,000 25,600

22,000 15,360

22,000 9,216

22,000 4,608

Taxable Income Income Taxes (25%)

($38,000) (9,500)

($47,600) (11,900)

($37,360) (9,340)

($31,216) (7,804)

($26,608) (6,652)

Net Income

($28,500)

($35,700)

($28,020)

($23,412)

($19,956)

(28,500) $ 16,000 $

(35,700) $ 25,600 $

(28,020) $ 15,360 $

(23,412) $ 9,216 $

(19,956) 4,608

$ $

20,000 (5,000)

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment $ Salvage Gains Tax Net Cash Flow

$ $ (80,000)

($80,000) PW(20%)=

($12,500) ($117,425)

($10,100)

($12,660) AE(20%)=

($14,196) -$35,310

($14,196)

Net annual cost for Model A: $35,310 Net before-tax annual revenue required to breakeven: $35,310 = (1 – 0.25)XA or XA = $47,080

($348)

Model B: 0

Income Statement Revenues (savings) Expenses: Operating Expenses Depreciation

17,000 10,400

17,000 16,640

17,000 9,984

17,000 5,990

17,000 2,995

Taxable Income Income Taxes (25%)

($27,400) (6,850)

($33,640) (8,410)

($26,984) (6,746)

($22,990) (5,748)

($19,995) (4,999)

Net Income

($20,550)

($25,230)

($20,238)

($17,243)

($14,996)

(20,550) $ 10,400 $

(25,230) $ 16,640 $

(20,238) $ 9,984 $

(17,243) $ 5,990 $

(14,996) 2,995

$ $

15,000 (3,750)

Cash Flow Statement Operating Activities: Net Income Depreciation Investment Activities: Investment $ Salvage Gains Tax Net Cash Flow

$ $ (52,000)

($52,000) PW(20%)=

($10,150) ($82,558)

($8,590)

($10,254) AE(20%)=

($11,252) -$24,826

($11,252)

($751)

Net annual cost for Model B: $24,826 Net before-tax annual revenue required to breakeven: $24,826 = (1 – 0.25)XB, or XB = $33,101 Additional annual revenue required for Model A to be breakeven with Model B: $47,080 - $33,101 = $13,979

11.13) Let X denote the number of copies to break-even. • A/T annual revenue (0.75)[$0.05 + $0.25]X = 0.225X = • A/T O&M cost = −(0.75)[$300, 000(12) + $0.15 X ] = −$2, 700, 000 − 0.1125X = Depreciation tax credit (0.25)[$85, 714( P / F ,13%,1) +  +$26, 775( P / F ,13%,8)]( A / P,13%,10) = $18,303 CR(13%) = −$600, 000( A / P,13%,10) + $100, 000( A / F ,13%,10) = −$105,145 AE(13%) = 0.225 X − $2, 700, 000 − 0.1125 X + $18,303 − $105,145 = 0.1125 X − $2, 613,158 = 0

∴ X = 23, 228, 071 copies per year or 23,228,071/300 =77,427 copies per day

11.14) (a) With infinite planning horizon: We assume that both machines will be available in the future with the same cost. Model A Financial Data n Depreciation Book value Market value Gain/Loss Operation Cost

0 $6,000

Cash Flow Statement Investment +(.30)*(Depreciation) -(1-0.30)*(Operation cost) Net proceeds from sale

1 $857 $5,143

2 $1,469 $3,673

3 $1,049 $2,624

4 $749 $1,874

5-7 $536 $1,339

8 $268 $0 $500 $500 $700

$700

$257 ($490)

$441 ($490)

$315 ($490)

$225 ($490)

$161 ($490)

$80 ($490) $350

($233)

($49)

($175)

($265)

($329)

($60)

($6,000)

Net Cash Flow

($6,000) PW (10%) = ($7,152)

AE (10%) = ($1,341)

Model B Financial Data n Depreciation Book value Market value Gain/Loss Operation Cost

0 $8,500

Cash Flow Statement Investment +(.30)*(Depreciation) -(1-0.30)*(Operation cost) Net proceeds from sale Net Cash Flow

Model A is preferred.

2 $2,082 $5,204

3 $1,487 $3,717

4 $1,062 $2,655

5-7 $759 $1,896

8 $379 ($0)

$520

$364 ($364)

$624 ($364)

$446 ($364)

$318 ($364)

$228 ($364)

$114 ($364)

$0 ($364)

$0 ($364) $700

$260

$82

($46)

($136)

($250)

($364)

$336

($0)

10 ($0) $1,000 $1,000 $520

($8,500)

($8,500) PW (10%) = ($8,627)

1 $1,215 $7,285

AE (10%) = ($1,404)

(b)Break-even annual O&M costs for machine A: Let X denotes a before-tax annual operating cost for model. −$6, 000 + ($257 − 0.7 X )( P / F ,10%,1) +  PW(10%) A = +($430 − 0.7 X )( P / F ,10%,8) = −$4,538 − 3.734 X −$851 − 0.7 X AE(10%) A = Let AE(10%) A = AE(10%) B , and solve for X. −$851 − 0.7 X = −$1, 404 X = $791 per year

Model A Financial Data n Depreciation Book value Market value Gain/Loss Operation Cost

1 2 3 4 5-7 $857 $1,469 $1,049 $749 $536 $6,000 $5,143 $3,673 $2,624 $1,874 $1,339

$791

Cash Flow Statement Investment +(.30)*(Depreciation) -(1-0.30)*(Operation cost) Net proceeds from sale Net Cash Flow

$791

8 $268 $0 $500 $500 $791

($6,000) $257 $441 $315 $225 $161 $80 ($553) ($553) ($553) ($553) ($553) ($553) $350 ($6,000) ($296) ($113) ($239) ($329) ($393) ($123)

PW (10%) = ($7,490)

AE (10%) = ($1,404)

Net Cash Flow Model A Model B -$6,000 -$8,500 -233 0 -49 260 -175 82 -265 -46 2,172 2,883 -$5,216 -$6,464

Model A is still preferred over Model B.

11.15) (a) Base case scenario: Income statement

Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$500 $500 $500 $500 $500 $500 $500 $500 20,000 20,000 20,000 20,000 20,000 20,000 20,000 20,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $10,000,000 $120 $120 $120 $120 $120 $120 $120 $120 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,400,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $2,500,000 $1,143,200 $1,959,200 $1,399,200 $999,200 $714,400 $713,600 $714,400 $356,800

Taxable income Income taxes (35%)

$3,956,800 $3,140,800 $3,700,800 $4,100,800 $4,385,600 $4,386,400 $4,385,600 $4,743,200 $1,384,880 $1,099,280 $1,295,280 $1,435,280 $1,534,960 $1,535,240 $1,534,960 $1,660,120

Net income

$2,571,920 $2,041,520 $2,405,520 $2,665,520 $2,850,640 $2,851,160 $2,850,640 $3,083,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(12%)

$2,571,920 $2,041,520 $2,405,520 $2,665,520 $2,850,640 $2,851,160 $2,850,640 $3,083,080 $1,143,200 $1,959,200 $1,399,200 $999,200 $714,400 $713,600 $714,400 $356,800 ($8,000,000)

$500,000 0

($8,000,000) $3,715,120 $4,000,720 $3,804,720 $3,664,720 $3,565,040 $3,564,760 $3,565,040 $3,939,880 $10,576,354

(b) Best case scenario: Income statement Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$600 $600 $600 $600 $600 $600 $600 $600 24,000 24,000 24,000 24,000 24,000 24,000 24,000 24,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $14,400,000 $96 $96 $96 $96 $96 $96 $96 $96 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $2,000,000 $914,560 $1,567,360 $1,119,360 $799,360 $571,520 $570,880 $571,520 $285,440

Taxable income Income taxes (35%)

$9,181,440 $8,528,640 $8,976,640 $9,296,640 $9,524,480 $9,525,120 $9,524,480 $9,810,560 $3,213,504 $2,985,024 $3,141,824 $3,253,824 $3,333,568 $3,333,792 $3,333,568 $3,433,696

Net income

$5,967,936 $5,543,616 $5,834,816 $6,042,816 $6,190,912 $6,191,328 $6,190,912 $6,376,864

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(12%)

$5,967,936 $5,543,616 $5,834,816 $6,042,816 $6,190,912 $6,191,328 $6,190,912 $6,376,864 $914,560 $1,567,360 $1,119,360 $799,360 $571,520 $570,880 $571,520 $285,440 ($6,400,000)

$600,000 0

($6,400,000) $6,882,496 $7,110,976 $6,954,176 $6,842,176 $6,762,432 $6,762,208 $6,762,432 $7,262,304 $27,967,318

(c) Worst case scenario: Income statement Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$400 $400 $400 $400 $400 $400 $400 $400 16,000 16,000 16,000 16,000 16,000 16,000 16,000 16,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $6,400,000 $144 $144 $144 $144 $144 $144 $144 $144 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $2,304,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $1,371,840 $2,351,040 $1,679,040 $1,199,040 $857,280 $856,320 $857,280 $428,160

Taxable income Income taxes (35%)

-$275,840 -$1,255,040 -$96,544 -$439,264

-$583,040 -$204,064

-$103,040 -$36,064

$238,720 $83,552

$239,680 $83,888

$238,720 $83,552

$667,840 $233,744

Net income

-$179,296

-$378,976

-$66,976

$155,168

$155,792

$155,168

$434,096

-$179,296 -$815,776 -$378,976 -$66,976 $1,371,840 $2,351,040 $1,679,040 $1,199,040

$155,168 $857,280

$155,792 $856,320

$155,168 $857,280

$434,096 $428,160

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(12%)

-$815,776

($9,600,000)

$400,000 0

($9,600,000) $1,192,544 $1,535,264 $1,300,064 $1,132,064 $1,012,448 $1,012,112 $1,012,448 $1,262,256 ($3,611,477)

(d) We see that the base case produces a positive PW, the worst case produces a negative PW, and the best case produces a large positive PW.

11.16)

(a) The breakeven sales volume

PW (20%) = 0 ⇒ Demand (units ) = 38,388 ⇒ Sales _ revenue = $2, 418, 440 Income statement Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$63 $63 $63 $63 $63 $63 38,388 38,388 38,388 38,388 38,388 38,388 $2,418,440 $2,418,440 $2,418,440 $2,418,440 $2,418,440 $2,418,440 $42 $42 $42 $42 $42 $42 $1,612,293 $1,612,293 $1,612,293 $1,612,293 $1,612,293 $1,612,293 $532,000 $532,000 $532,000 $532,000 $532,000 $532,000 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

$114,147 $39,951

$18,147 $6,351

$120,547 $42,191

$181,987 $63,695

$228,067 $79,823

Net income

$74,195

$11,795

$78,355

$118,291

$148,243

$74,195 $160,000

$11,795 $256,000

$78,355 $153,600

$118,291 $92,160

$148,243 $46,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

($800,000)

($800,000) $0

$100,000

$234,195

$267,795

$231,955

$210,451

$294,323

(b) The base case: Income statement Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$63 $63 $63 $63 $63 $63 65,000 65,000 65,000 65,000 65,000 65,000 $4,095,000 $4,095,000 $4,095,000 $4,095,000 $4,095,000 $4,095,000 $42 $42 $42 $42 $42 $42 $2,730,000 $2,730,000 $2,730,000 $2,730,000 $2,730,000 $2,730,000 $532,000 $532,000 $532,000 $532,000 $532,000 $532,000 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

$673,000 $235,550

$577,000 $201,950

$679,400 $237,790

$740,840 $259,294

$786,920 $275,422

Net income

$437,450

$375,050

$441,610

$481,546

$511,498

$437,450 $160,000

$375,050 $256,000

$441,610 $153,600

$481,546 $92,160

$511,498 $46,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

($800,000)

($800,000) $1,208,007

$100,000

$597,450

PW(20%) = $1,208,007

$631,050

$595,210

$573,706

$657,578

(b) The sales price per unit increases to $400 Income statement

$400 2,252 $900,723

$42 $94,576 $532,000 $160,000

$42 $94,576 $532,000 $256,000

$42 $94,576 $532,000 $153,600

$42 $94,576 $532,000 $92,160

$42 $94,576 $532,000 $46,080

Taxable income Income taxes (35%)

$114,147 $39,951

$18,147 $6,351

$120,547 $42,191

$181,987 $63,695

$228,067 $79,823

Net income

$74,195

$11,795

$78,355

$118,291

$148,243

$74,195 $160,000

$11,795 $256,000

$78,355 $153,600

$118,291 $92,160

$148,243 $46,080

Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

($800,000)

($800,000) $0

$100,000

$234,195

$267,795

The required break-even volume is 2,252.

$231,955

$210,451

$294,323

(d) •

Best case

Best Case Income statement Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$72.45 $72.45 $72.45 $72.45 $72.45 $72.45 74,750 74,750 74,750 74,750 74,750 74,750 $5,415,638 $5,415,638 $5,415,638 $5,415,638 $5,415,638 $5,415,638 $35.70 $35.70 $35.70 $35.70 $35.70 $35.70 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $452,200 $452,200 $452,200 $452,200 $452,200 $452,200 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

$2,134,863 $2,038,863 $2,141,263 $2,202,703 $2,202,703 $2,248,783 $747,202 $713,602 $749,442 $770,946 $770,946 $787,074

Net income

$1,387,661 $1,325,261 $1,391,821 $1,431,757 $1,431,757 $1,461,709

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

$1,387,661 $1,325,261 $1,391,821 $1,431,757 $1,431,757 $1,461,709 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080 ($800,000)

$100,000

($800,000) $1,547,661 $1,581,261 $1,545,421 $1,523,917 $1,523,917 $1,607,789 $4,367,942

•

Worse case

Worse Case Income statement

Revenues: Unit price Demand (units) Sales revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$53.55 $53.55 $53.55 $53.55 $53.55 $53.55 55,250 55,250 55,250 55,250 55,250 55,250 $2,958,638 $2,958,638 $2,958,638 $2,958,638 $2,958,638 $2,958,638 $48.30 $48.30 $48.30 $48.30 $48.30 $48.30 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $2,668,575 $611,800 $611,800 $611,800 $611,800 $611,800 $611,800 $160,000 $256,000 $153,600 $92,160 $92,160 $46,080

Taxable income Income taxes (35%)

-$481,738 -$168,608

-$577,738 -$202,208

-$475,338 -$166,368

-$413,898 -$144,864

-$367,818 -$128,736

Net income

-$313,129

-$375,529

-$308,969

-$269,033

-$239,081

-$313,129 $160,000

-$375,529 $256,000

-$308,969 $153,600

-$269,033 $92,160

-$239,081 $46,080

Cash flow statement Operating activities Net income Depreciation Investment activities Investment Salvage Gains tax Net cash flow PW(20%)

11.17)

($800,000)

($800,000) ($153,129) ($119,529) ($155,369) ($176,873) ($176,873) ($1,288,052)

E[return] = (0.15 × 5%) + (0.25 ×15%) + (0.35 × 22%) + (0.15 × 30%) + (0.1× 40%) = 20.7%

σ 2 = (0.15 × (5 − 20.7) 2 ) + (0.25 × (15 − 20.7) 2 ) + (0.35 × (22 − 20.7) 2 ) + (0.15 × (30 − 20.7) 2 ) + (0.1× (40 − 20.7) 2 ) = 95.91 σ = 9.79%

$100,000

($93,001)

11.18) We can calculate the mean and variance for each periods with the three- point estimate. Period (n) 0 1 2

Pessimistic ($10,000) $5,000 $4,000

Most Likely ($8,000) $12,000 $10,000

E[PW(10%)] = −$8,167 +

Optimistic ($7,000) $15,000 $13,000

E[An]

Var[An]

($8,167) $11,333 $9,500

250,000 2,777,778 2,250,000

$11,333 $9,500 + 1.1 1.12

= $9,986.97 2, 777, 778 2, 250, 000 + 1.12 1.14 = 4, 082, 464.57 0 − 9,986.97 z= = −4.9428 4, 082, 464 NORMDIST(-4.9428,0,1,1)= 0.0000385% V [PW(10%)] = 250, 000 +

11.19)

−$8, 000, 000 + $1,300, 000( P / A,12%,3) PW(12%)light = = −$4,877, 619 −$8, 000, 000 + $2,500, 000( P / A,12%, 4) PW(12%) moderate = = −$406, 627 −$8, 000, 000 + $4, 000, 000( P / A,12%, 4) PW(12%) high = = $4,149,397 E[PW(12%)] = −$4,877, 619(0.20) − $406, 627(0.40) +$4,149,397(0.40) = $521,584

Since E[PW] is positive, it is good to invest.

11.20)

(a) 1.

A1 and A2 are mutually independent E[PW(10%)] = −500 +

200

(1 + 0.1)

500

(1 + 0.1)

= $95.04

502 502 Var[PW(10%)] = 0+ + = 3773.65 2 4 (1 + 0.1) (1 + 0.1) A1 and A2 are partially correlated with ρ12 = 0.3 200 500 E[PW(10%)] = −500 + + = $95.04 1 2 (1 + 0.1) (1 + 0.1)  0.3(50)(50)  502 502 Var[PW(10%)] = 0+ + + 2 = 4900.62 2 4 3  (1 + 0.1) (1 + 0.1)  (1 + 0.1) 

(b)

σ [PW(10%)] = Var[PW(10%)] σ [PW(10%)] = 3773.65 z=

X −µ

= 61.43 0 − 95.04 = 61.43 = −1.547

P( z < −1.547) = 0.06093 or 6.093%

11.21) (a) −$300 + E[PW(10%)] =

$120 $150 $150 $110 $100 + + + + 1.1 1.12 1.13 1.14 1.15

= $182.98 102 152 202 252 302 Var[PW(10%)] = 20 + 2 + 4 + 6 + 8 + 10 1.1 1.1 1.1 1.1 1.1 = 1,501 2

(b)

Pr(PW(10%) ≥ $200) = 1 − Pr(PW(10%) ≤ $200)  200 − 182.98  = 1− Φ   = 1 − Φ (0.4393) 1,501   = 0.67

11.22) (a) $5 $8 $12 $10 $5 + + + + 2 3 4 1.12 1.12 1.12 1.12 1.125 = $10.58M

E[PW(12%)] = −$18 +

82 92 102 52 32 + + + + 1.122 1.124 1.126 1.128 1.1210 = 166.155

Var[PW(12%)] =+ 02

(b)

0 − 10.58 = −0.8208 166.155 NORMDIST(-0.8208,0,1,1) = 20.59% z=

$5 $8 $12 $10 $5 + + + + 2 3 4 (1 + i ) (1 + i ) (1 + i ) (1 + i ) (1 + i )5

=0 i = 32.38%

11.23) (a) 1.

Mutually independent: E[PW(10%)] = −500 +

200

(1 + 0.1)

1002 + Var[PW(10%)] =

502

(1 + 0.1)

500

(1 + 0.1)

+ 2

= $95.04

502

(1 + 0.1)

13, 773.65 =

Partial correlations: E[PW(10%)] = −500 +

200

(1 + 0.1)

500

(1 + 0.1)

= $95.04

Var[PW(10%)] = 100 + 2

502

(1 + 0.1)

502

(1 + 0.1)

 0.2(100)(50)   0.2(100)(50)  +2  + 2   1 2  (1 + 0.1)   (1 + 0.1)   0.5(50)(50)  +2  3   (1 + 0.1)  = 13, 773.65 + 5,349.36 = 19,123.01

Perfect positive correlations: E[PW(10%)] = −500 +

200

(1 + 0.1)

Var[PW(10%)] = 1002 +

502

(1 + 0.1)

500

(1 + 0.1)

+ 2

= $95.04

502

(1 + 0.1)

1(100)(50)  1(100)(50)  +2  + 2 1  2   (1 + 0.1)   (1 + 0.1)  1(50)(50)  +2  3   (1 + 0.1)  = 13, 773.65 + 21,111.95 = 34,885.60

(b)

σ [PW(10%)] = Var[PW(10%)] σ [PW(10%)] = 13, 773.65 = $117.36 X − µ 100 − 95.04 z = = = 0.0423 117.36 σ P ( z > 0.0423) = 1 − P ( z < 0.0423) = 1 − 0.51687 =0.4831 (or 48.31%)

11.24) (a) The mean return for projects E[return]A = (0.1× −20%) + (0.2 × 0%) + (0.25 ×10%) + (0.3 ×15%) + (0.1× 20%) + (0.05 × 40%) = 9% E[return]B = (0.1× −35%) + (0.2 × −10%) + (0.25 ×15%) + (0.3 × 25%) + (0.1× 40%) + (0.05 × 50%) = 12.25%

(b) The variance of return for projects (0.1× (−20 − 9) 2 ) + (0.2 × (0 − 9) 2 ) σ 2= A + (0.25 × (10 − 9) 2 ) + (0.3 × (15 − 9) 2 ) + (0.1× (20 − 9) 2 ) + (0.05 × (40 − 9) 2 ) = 171.50 (0.1× (−35 − 12.25) 2 ) + (0.2 × (−10 − 12.25) 2 ) σ 2= B + (0.25 × (15 − 12.25) 2 ) + (0.3 × (25 − 12.25) 2 ) + (0.1× (40 − 12.25) 2 ) + (0.05 × (50 − 12.25) 2 ) = 521.19

(c) It is not a clear case, because E[return]B > E[return]A but also VarB > VarA . If you make decision solely based on the principle of maximization of expected return, you may prefer project B.

11.25) (a) (0.3) [ −$150, 000 + $35, 000( P / A,12%,5) ] E[PW(12%)]A = + (0.5) [ −$150, 000 + $40, 000( P / A,12%,5) ] + (0.2) [ −$150, 000 + $50, 000( P / A,12%,5) ] = −$4, 006.6 (0.3) [ −$180, 000 + $45, 000( P / A,12%,5) ] E[PW(12%)]B = + (0.5) [ −$180, 000 + $55, 000( P / A,12%, 5) ] + (0.2) [ −$180, 000 + $67, 000( P / A,12%,5) ] = $16,100

(0.3 × (−23,832 + 4, 006.6) 2 ) + (0.5 × (−5,808 + 4, 006.6) 2 ) σ 2= A + (0.2 × (30, 240 + 4, 006.6) 2 ) = 354, 097, 713 (0.3 × (−17, 785 − 16,100) 2 ) + (0.5 × (18, 263 − 16,100) 2 ) σ 2= B + (0.2 × (61,520 − 16,100) 2 ) = 759,392,532

(b).Project A has a higher probability to lose money. 0 + 4, 006.6 = z A = 0.2129 354, 097, 713 NORMDIST(0.2129,0,1,1)=58.43% zB =

0 − 16,100 = −0.58428 759,392,532

NORMDIST(-0.58428,0,1,1)= 27.95%

11.26) (a) The PW distribution for project 1: Event (x,y)

Joint probability

($20,$10) ($20,$20) ($40,$10) ($40,$20)

0.18 0.12 0.42 0.28

PW(10%) $400 0 2,400 1,600

(b) The mean and variance of the PW for Project 1:

E[PW(10%)]1= $400(0.18) + $0(0.12) + $2, 400(0.42) +$1, 600(0.28) = $1,528 Var[PW(10%)]1= (400 − 1,528) 2 (0.18) + (0 − 1,528) 2 (0.12) +(2, 400 − 1,528) 2 (0.42) +(1, 600 − 1,528) 2 (0.28) = 830, 016 (c) The mean and variance of the PW for Project 2: E[PW(10%)]2 = $0(0.24) + $400(0.20) + $1, 600(0.36) +$2, 400(0.20) = $1,136 Var[PW(10%)]2 = (0 − 1,136) 2 (0.24) + (400 − 1,136) 2 (0.20) +(1, 600 − 1,136) 2 (0.36) +(2, 400 − 1,136) 2 (0.20) = 815,104 (d) It is not a clear case, because E1 > E2 but also Var1 > Var2 . If Juan makes the decision solely based on the principle of maximization of expected value, she may prefer contract A. 11.27) (a) • Option 1:

E[ R ]1 = $2, 450(0.25) + $2, 000(0.45) + $1, 675(0.30) −$150( F / P, 7.5%,1) = $1,854 • Option 2:

E[ R]2 = $25, 000(0.075) = $1,875 ∴ Option 2 is the better choice based on the principle of expected value maximization. (b)

Potential return

Prob.

Op.1

Op.2

Optimal choice

High

0.25

$2,288.75

$1,875

Op.1

$413.75

Medium

0.45

1,838.75

1,875

Op.2

Low

0.3

1,513.75

1,875

Op.2

1,853.75

1,875

Expected value

Opp. Loss

103.44

EPPI 0.25 = ($2, 288.75) + 0.45($1,875) + 0.3($1,875) $1,978.44 EV = $1,875 EVPI = EPPI - EV = $103.44

11.28) (a)

E[NPW]1 = ($2, 000)(0.20) + ($3, 000)(0.60) + ($3,500)(0.20) − $1, 000 = $1,900 E[NPW]2 = ($1, 000)(0.30) + ($2,500)(0.40) + ($4,500)(0.30) − $800 = $1,850 Project 1 is preferred over Project 2.

(b)

Var[NPW]1 = (2, 000 − 2,900) 2 (0.20) + (3, 000 − 2,900) 2 (0.60) +(3,500 − 2,900) 2 (0.20) = 240, 000 Var[NPW]2 = (1, 000 − 2, 650) 2 (0.30) + (2,500 − 2, 650) 2 (0.40) +(4,500 − 2, 650) 2 (0.30) = 1,852,500 ∴ Select Project 1.

11.29) 12.1 (a) Mean and variance calculations:

E[PW]1 = ($100, 000)(0.20) + ($50, 000)(0.40) + (0)(0.40) = $40, 000 = E[PW]2 ($40, 000)(0.30) + ($10, 000)(0.40) + (−$10, 000)(0.30) = $13, 000 Var[PW]1 = (100, 000 − 40, 000) 2 (0.20) + (50, 000 − 40, 000) 2 (0.40) +(0 − 40, 000) 2 (0.40) = 1, 400, 000, 000 Var[PW]2 = (40, 000 − 13, 000) 2 (0.30) + (10, 000 − 13, 000) 2 (0.40) +(−10, 000 − 13, 000) 2 (0.30) = 381, 000, 000

It is not a clear case, because E1 > E2 but also Var1 > Var2 . If she makes decision solely based on the principle of maximization of expected value, she may prefer contract A. If she is a risk-averse decision maker, then she has to ask herself whether or not $27,000 increase in expected value is worth taking an increase in risk of 1,019,000,000. (b) Assuming that both contracts are statistically independent from each other, Joint event (PWA >PWB )

Joint Probability

($100,000,$40,000) ($100,000,$10,000) ($100,000,-$10,000) ($50,000,$40,000) ($50,000,$10,000) ($50,000,-$10,000) ($0,-$10,000)

(0.20)(0.30) = 0.06 (0.20)(0.40) = 0.08 (0.20)(0.30) = 0.06 (0.40)(0.30) = 0.12 (0.40)(0.40) = 0.16 (0.40)(0.30) = 0.12 (0.40)(0.30) = 0.12

P ( PWA >PWB ) = 0.72

11.30) (a) • Machine A:

Σ = 0.72

CR(10%) A = ($60, 000 − $22, 000)( A / P,10%, 6) + (0.10)($22, 000) = $10,924 E = [AE(10%)]A ($5, 000)(0.20) + ($8, 000)(0.30) +($10, 000)(0.30) + ($12, 000)(0.20) + $10,924 = $19, 725 Var[AE(10%)]A = (15,924 − 19, 725) 2 (0.20) + (18,924 − 19, 725) 2 (0.30) +(20,924 − 19, 725) 2 (0.30) + (22,924 − 19, 725) 2 (0.20) = 5,560, 000 • Machine B:

CR(10%) B = $35, 000( A / P,10%, 4) = $11, 042 E = [AE(10%)]B ($8, 000)(0.10) + ($10, 000)(0.30) +($12, 000)(0.40) + ($14, 000)(0.20) + $11, 042 = $22, 441 Var[AE(10%)]B = (19, 042 − 22, 442) 2 (0.10) + (21, 042 − 22, 442) 2 (0.30) +(23, 042 − 22, 442) 2 (0.40) + (25, 042 − 22, 442) 2 (0.20) = 3, 240, 000 (b) Prob[ AE (10%) A > AE (10%) B ] : Joint event (O&M A ,O&M B ) (AE A >AE B )

Joint Probability

($10,000, $8,000)

($20,924, $19,042)

(0.30)(0.10) = 0.03

($12,000, $8,000)

($22,924, $19,042)

(0.20)(0.10) = 0.02

($12,000, $10,000)

($22,924, $21,042)

(0.20)(0.30) = 0.06

Σ = 0.11

11.31)

(a) Mean and variance calculation (Note: For a random variable Y, which can be expressed as a linear function of another random variable X (say, Y = aX , where a is a constant) the variance of Y can be calculated as a function of variance of X, Var[Y ] = a 2Var[ X ] .

E[PW]A = −$5, 000 + $4, 000( P / A,15%, 2) = $1,502.84 E[PW]B = −$10, 000 + $6, 000( P / F ,15%,1) + $8, 000( P / F ,15%, 2) = $1, 266.54 1, 0002 + ( P / F ,15%,1) 21, 0002 + ( P / F ,15%, 2) 21,5002 V [PW]A = = 3, 042,588 2, 0002 + ( P / F ,15%,1) 21,5002 + ( P / F ,15%, 2) 2 2, 0002 V [PW]B = = 7,988,336 (b) Comparing risky projects.

E[PW]

Project A $1,503

Project B $1,267

Var[PW]

3,042,588

7,988,336

Project A is preferred because of higher E[PW] and lower Var[PW] .

11.32) After-tax cost of debt: a) (0.12)(1 – 0.25) = 0.09 or (9%) b) (0.14)(1 – 0.34) = 0.924 or (9.24%) c) (0.15)(1 – 0.22) = 0.117 or (11.7%)

11.33) 0.07 + 1.7(0.14 – 0.07) = 18.9%

11.34)

ie = 0.22 id= (0.13)(1 − 0.25)= 0.0975 = k (0.0975)(0.45) + (0.22)(0.55) = 0.1649

11.35)

Cost of equity:

ie =+ rf β (rM − rf )

= rf 1.32%, = rM 10.5% where

11.36)

Verizon

0.56

6.46%

Amazon

1.38

13.99%

General Electric

1.20

12.34%

Google

1.00

10.50%

0.2 = 0.05 + β (0.15 − 0.05) β = 1.5

11.37) −$1, 000 + PW(25%)certainty equivalent =

$500 $1,500 $800 + + 1.25 1.252 1.253

= $769.6 > 0 Yes, it would be justified.

11.38)

Certainty equivalent value: −$300 + PW(18%)certainty equivalent =

$120 $150 $150 $110 $100 + + + + 1.18 1.182 1.183 1.184 1.185

= $101 > 0

Yes, it would be justified.

11.39) (a)

(b)

VaR = −1.647 × 0.6916% + 0.1769% = −0.9622% $1, 000, 000 × (−0.9622 /100) =−$9, 622

11.40) (a) 0 − 120   P ( NPW ≤ 0) = 0.0082 Φ z = = −2.4  = 50  

(b)

VaR =−1.647 × $50 + $120 =+$37.65 > 0 ⇒ a very safe project

11.41) Not given

11.42)  Project cash flow and PW calculation at MARR of 15%

Income Statement Revenue: Unit price Demand(units) Sales Revenue Expenses: Unit variable cost Variable cost Fixed cost Depreciation

$50

2,000

$100,000

$15

$30,000

$10,000

$17,863

$30,613

$21,863

$15,613

$5,581

Taxable Income Income Tax (25%)

$42,138

$29,388

$38,138

$44,388

$54,419

$10,534

$7,347

$9,534

$11,097

$13,605

Net Income Cash Flow Statement Cash From Operation: Net Income

$31,603

$22,041

$28,603

$33,291

$40,814

$31,603

$22,041

$28,603

$33,291

$40,814

$17,863

$30,613

$21,863

$15,613

$5,581

Depreciation Investment Salvage Gains tax

($125,000) $40,000 ($1,633)

Net Cash Flow

($125,000)

$49,466

PW(15%)=

$61,111

$52,653

$50,466

$48,903

$84,763

 Sensitivity analysis for three key input variables

Variable Unit price Demand Unit variable cost

-20%

-15%

$ 10,829 $ 23,400 $ 25,914 $ 34,713 $ 76,196 $ 72,435

-10%

Percent Deviation from Base Base -5% 0% 5%

35970 $ 48,541 $ 43513 $ 52,313 $ 68654 $ 64,883 $

10%

20%

61,111 $ 73,682 $ 86,253 $ 98,823 $ 111,394 61,111 $ 69,911 $ 78,710 $ 87,510 $ 96,309 61,111 $ 57,340 $ 53,569 $ 49,798 $ 46,027

 Sensitivity graph for the BMC’s transmission-housings project Not provided.

15%

11.43) (a), (b) & (c) f = 0.25(3%) + 0.5(5%) + 0.25(7%) = 5% i = i′ + f + i′f = 0.1 + 0.05 + 0.1(0.05) = 0.155 Salvage_year 2=$6,000(1.05) 2 =$6,615 Gain tax_year 2=$902 (tax credit) Working capital_year 2=$3,308 PW(15.5%) = −$26, 000 + (0.75 X + 0.25($7, 666))( P / F ,15.5%,1)) + (0.75 X + 0.25($5,112) + $6, 615 + $902 + $3, 308)( P / F ,15.5%, 2)) = 1.211559 X − 81, 331.85

 X = $15,000, PW(15.5%) = $3,326 Income Statement

Revenue (Savings) Expenses: O&M Depreciation Interest

inflation 5%

2 $15,750

1 $15,000

7,666

5,112

Taxable Income Income Taxes (25%)

$7,334 $1,834

$10,638 $2,660

Net Income

$5,501

$7,979

$5,501 $7,666

$7,979 $5,112

Cash Flow Statement

Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage Gains Tax Working capital Cash from financing activities: Loan repayment

(23,000)

(3,000)

Net Cash Flow (actual) Net Cash Flow (constant)

6,615 902 3,308

($26,000)

$13,166

$23,915

(26,000)

12,540

21,691

PW (15.5%) =

 X = $25,000, PW(15.5%) = $15, 723

$ $ $

3,326

Revenue (Savings) Expenses: O&M Depreciation Interest

$25,000

7,666

$26,250

5,112

Taxable Income Income Taxes (25%)

$17,334 $4,334

$21,138 $5,285

Net Income

$13,001

$15,854

$13,001 $7,666

$15,854 $5,112

Cash Flow Statement

Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage Gains Tax Working capital Cash from financing activities: Loan repayment

(23,000)

(3,000)

Net Cash Flow (actual) Net Cash Flow (constant)

6,615 902 3,308

($26,000)

$20,666

$31,790

(26,000)

19,682

28,834

PW (15.5%) =

 X = $35,000, PW(15.5%) = $28,120

$ $ $

15,723

inflation 5%

Income Statement

Revenue (Savings) Expenses: O&M Depreciation Interest

1 $35,000

7,666

2 $36,750

5,112

Taxable Income Income Taxes (25%)

$27,334 $6,834

$31,638 $7,910

Net Income

$20,501

$23,729

$20,501 $7,666

$23,729 $5,112

Cash Flow Statement

Cash from operation Net Income Depreciation Cash from investing activities: Investment / Salvage Gains Tax Working capital Cash from financing activities: Loan repayment

(23,000)

(3,000)

Net Cash Flow (actual) Net Cash Flow (constant)

$ $ $

($26,000)

$28,166

$39,665

(26,000)

26,825

35,977

PW (15.5%) =

28,120

X = $15,000, PW(15.5%) = $3,326 X = $25,000, PW(15.5%) = $15, 723 X = $35,000, PW(15.5%) = $28,120

E[PW(15.5%)] =$3,326(0.2) + $15, 723(0.5) + $28,120(0.3) = $16,962.7 Var[PW(15.5%)] =(3,326 − 16,962.7) 2 (0.2) + (15, 723 − 16,962.7) 2 (0.5) +(28,120 − 16,962.7) 2 (0.3) = 75,305,948

6,615 902 3,308

11.44) Since the amount of annual labor savings is the same for both alternatives, this labor savings factor is not considered in the following analysis. (a) After-tax cash flows: After-Tax Cash Flows n Lectra System Tex System 0 -$136,150 -$195,500 1 117,927 149,075 2 124,462 158,459 3 117,491 148,449 4 113,308 142,443 5 113,308 142,443 6 122,171 146,939 PW(12%) $350,189 $415,383 AE(12%) $85,175 $101,032 Based on the most-likely estimates, the Tex system is the better choice. (b) Let X and Y denote the annual material savings for the Lectra system and Tex system, respectively.

n 0 1 2 3 4 5 6

After-Tax Cash Flows Lectra System Tex System -$136,150 -$195,500 0.6X-20,073 0.6Y-15,325 0.6X-13,538 0.6Y-5,941 0.6X-20,508 0.6Y-15,951 0.6X-24,691 0.6Y-21,956 0.6X-24,691 0.6Y-21,956 0.6X-15,828 0.6Y-17,461

 Lectra System:

−$52,824 + 0.6 X AE(12%) Lectra = E[ X ] = $224, 000 Var[ X ] = 2,124, 000, 000 E[AE(12%)]Lectra = −$52,824 + 0.6 E[ X ] = $81,576 Var[AE(12%)]Lectra = (0.6) 2 Var[ X ] = 764, 640, 000

 Tex System: −$63,368 + 0.6Y AE(12%)Tex =

E[Y ] = $259, 400 Var[Y ] = 1, 718, 440, 000 E[AE(12%)]Tex = −$63,368 + 0.6 E[Y ] = $92, 272 Var[AE(12%)]Tex = (0.6) 2 Var[Y ] = 618, 638, 400

11.45) (a) Incremental project cash flows (FMS - CMT): No. of part types No. of pieces per year Year

3,000

544,000

Income Statement

Revenue (Savings): Labor Material Overhead Tooling Inventory Expenses: Depreciation

$462,400

233,920

1,200,000

170,000

109,500

928,850

1,591,850

1,136,850

811,850

580,450

579,800

580,450

289,900

Taxable Income Income Tax (40%)

1,246,970

583,970

1,038,970

1,363,970

1,595,370

1,596,020

1,595,370

1,885,920

2,175,820

498,788

233,588

415,588

545,588

638,148

638,408

638,148

754,368

870,328

Net Income

$748,182

$350,382

$623,382

$818,382

$957,222

$957,612

$957,222

$1,131,552

$1,305,492

1,305,492

Cash Flow Statement

Cash From Operation: Net Income Depreciation Investment&Salvage Gains Tax (40%) Net Cash Flow PW (15%) =

(6,500,000)

748,182

350,382

623,382

818,382

957,222

957,612

957,222

1,131,552

928,850

1,591,850

1,136,850

811,850

580,450

579,800

580,450

289,900 500,000 (200,000)

$ (6,500,000) $ 1,677,032 $ 1,942,232 $ 1,760,232 $ 1,630,232 $ 1,537,672 $ 1,537,412 $ 1,537,672 $ 1,421,452 $ 1,305,492 $ 1,605,492

$1,756,225

(b)&(c) Sensitivity analysis: AOC = annual overhead cost VLC = variable labor cost / part AIC = annual inventory cost ATC = annual tooling cost VMC = variable material cost / part Deviation AOC

-30%

-20%

-10%

10%

20%

30%

$3,517,813

$2,930,617

$2,343,421

$1,756,225

$1,169,030

$581,834

-$5,362

VLC

$2,395,095

$2,182,138

$1,969,182

$1,756,225

$1,543,268 $1,330,313 $1,117,356

AIC

$1,784,682

$1,775,196

$1,765,711

$1,756,225

$1,746,740 $1,737,225 $1,727,769

ATC

$2,027,239

$1,936,901

$1,846,563

$1,756,225

$1,665,888 $1,575,550 $1,485,212

VMC

$2,296,807

$2,116,613

$1,936,419

$1,756,225

$1,576,032 $1,395,838 $1,215,644

(d) Best and worst scenarios: • Best case: Material cost = $1.00 per part, annual inventory cost = $25,000 PW (15%) FMS −CMT = $1,939, 611 • Worst case: Material cost = $1.40 per part, annual inventory cost = $100,000 PW (15%) FMS −CMT = $1, 058,516 (e) Mean and variance: • E[ PW (15%) FMS −CMT ] : $1,595,123 • Var[ PW (15%) FMS −CMT ] : 46,073,274,329 (f) In no situation, the FMS would be a more expensive investment option than the

Chapter 12 Replacement Decisions Note to Instructors: Regular MACRS depreciation is assumed in all problems unless otherwise mentioned. However, instructors may adopt the capital expensing option under the 2017 Tax Cuts and Job Acts instead. In that case, the entire capital expenditure would be claimed as depreciation amount at the end of first year. Then, any salvage value at the end of project life will be subject to taxable gains. 12.1) Tax Rate(%) = MARR(%) =

0.00% 10.00%

PW(i) = AE(%) =

($20,065) ($6,329.8)

Revenues (savings) Expenses: O&M Depreciation

$2,500 $0

$3,000 $0

$3,500 $0

$4,000 $0

Taxable Income Income Taxes (%)

($2,500) 0

($3,000) 0

($3,500) 0

($4,000) 0

Net Income

($2,500)

($3,000)

($3,500)

($4,000)

(2,500) $ $0

(3,000) $ $0

(3,500) $ $0

(4,000) $0

3,000 $0.00

Income Statement

Cash Flow Statement

Operating Activities: Net Income Depreciation Investment Activities: $ Investment Salvage Gains Tax Net Cash Flow

(12,000)

($12,000)

($2,500)

($3,000)

($3,500)

($1,000)

PW(10%) = −$12, 000 − $2,500( P / F ,10%,1) ⋅⋅⋅⋅⋅ −$1, 000( P / A,10%, 4) = −$20, 065 AEC(10%) = $20, 065( A / P,10%, 4) = $6,329.8

12.2) A. Original cost: The printing machine was purchased for $20,000 B. Market value: The old machine’s market value is estimated at $10,000. C. Book value: If the machine sold now its book value is $14,693. D. Trade in allowance: This amount is the same as the market value. The market value is the most relevant information, but the defender’s current book value is also relevant as this will be the basis to determine the gains or losses related to disposal of the defender.

12.3) Option 1: Keep the defender −$8, 000( P / A,12%,3) + $2,500( P / F ,12%,3) PW(12%) D = = −$17, 434.9 AEC(12%) D = $17, 434.9( A / P,12%,3) = $7, 259.1

Option 2: Replace the defender with the challenger PW(12%)C = −$5, 000 − $6, 000( P / A,12%,3) + $5,500( P / F ,12%,3) = −$15, 495.9 AEC(12%)C = $15, 495.9( A / P,12%,3) = $6, 451.9

The replacement should be made now. 12.4) (a) Purchase cost = $15,000, market value = $6,000, sunk cost = $15,000 - $6,000 = $9,000 (b) Opportunity cost = $6,000 (c)

PW(15%) = −$6, 000 − $1,500 − $3, 000( P / F ,15%,1) −($3,500 − $3, 000)( P / F ,15%, 2) = $10, 486.77 AEC(15%) = $10, 486.77( A / P,15%, 2) = $6, 450.58

(d)

PW(15%) = −$7,500 − $3, 000( P / F ,15%,1) − $3,500( P / F ,15%, 2) −$3,800( P / F ,15%,3) − $4,500( P / F ,15%, 4) −$9,800( P / F ,15%,5) = −$22, 698.98 AEC(15%) = $22, 698.98( A / P,15%,5) = $6, 771.46

12.5) (a) Opportunity cost = $30,000 (b) Assume that the old machine’s operating cost is $30,000 per year. Then the n ew machine’s operating cost is zero per year. The cash flows associated with r etaining the defender for two more years are n 0 1 2 Cash Flows: -$30,000 -$30,000 -$18,000

AEC D = ($30, 000 − $12, 000)( A / P,12%, 2) + $12, 000(0.12) +$30, 000 = $42, 090.57 (c) Cash flows for the challenger: Year 0: -$165,000; Years 1-7: 0; Year 8: $5,000 AECC =− ($165, 000 $5, 000)( A / P,12%,8) + $5, 000(0.12) = $32,808.45

(d) Since AEC D > AECC , we should replace the defender now. 12.6) (a) Initial cash outlay for the new machine = $120,000 (b) Cash flows for the defender: Year 0: -$10,000 Years 1-5: 0 (c)

AE(15%) D = −$10, 000( A / P,15%,5) = −$2,983

AE(15%)C = − [ ($120, 000 − $30, 000)( A / P,15%, 7) + $30, 000(0.15) ] +$50, 000 = $23,868

We should purchase the new machine because it has a higher annual equivalent cash flow.

12.7) (a) Cash flows Year: 0 Defender -$10K Challenger -$75K

1 0 $30K

2 0 $30K

3 0 $30K

4 0 $30K

5 $5K $30K

(b) PW(15%) D = −$10 K + $5 K ( P / F ,15%,5) = −$7,514 PW(15%)C = $25,565 −$75 K + $30 K ( P / A,15%,5) = Replace the defender.

12.8) (a) and (b) Cash flows: Year: 0 1 2 3 4 5 Defender -$5,500 $21,000 $21,000 $22,200 Challenger -$36,500 $24,000 $24,000 $24,000 $24,000 $30,300 • • (c)

Revenue for defender = ($19 - $12) ´ 3,000 = $21,000 per year Revenue for challenger = ($19 - $11) ´ 3,000 = $24,000 per year

− [ ($5,500 − $1, 200)( A / P,12%,3) + $1, 200(0.12) ] + $21, 000 AE(12%) D = = $19, 066 − [ ($36,500 − $6,300)( A / P,12%,5) + $6,300(0.12) ] + $24, 000 AE(12%)C = = $14,866 Keep the defender for now.

12.9)

Annual changes in MV Interest rate

20% 12%

Market Value

O&M Costs CR(12%)

n 0 1 2 3 4 5 6 7 8

$200,000 $130,000 $104,000 $83,200 $66,560 $53,248 $42,598 $34,079 $27,263

$20,000 $20,000 $20,000 $20,000 $20,000 $22,000 $25,000 $28,000

$94,000 $69,283 $58,614 $51,920 $47,100 $43,396 $40,446 $38,044

OC(12%) AEC(12%)

$20,000 $20,000 $20,000 $20,000 $20,000 $20,246 $20,718 $21,310

The economic life of the asset is eight years; that is, N * = 8 years. Thus,

12.10) (a) Interest i = 10% n

AEOC

CR(10%)

AEC(10%)

0 $15,000 1 $2,500 $12,800 $2,500 2 $3,200 $8,100 $2,833 3 $5,300 $5,200 $3,579 4 $6,500 $3,500 $4,208

$3,700

$6,200

$4,786 $4,461 $3,978

$7,619 $8,039 $8,186

5 $7,800

$4,796

$3,957

$8,753

AE OC

CR(10%)

AEC(10%)

$15,000 $12,800 $2,500

$4,450

$6,950

$8,100 $5,200 $3,500 0

$5,459 $5,072 $4,553 $4,475

$8,285 $8,610 $8,684 $9,150

(b) Interest i = 15% n 0 1 2

OC $2,500 $3,200

3 $5,300 4 $6,500 5 $7,800

$2,826 $3,538 $4,131 $4,675

∴ In both cases, the economic service life is 1 year.

$114,000 $89,283 $78,614 $71,920 $67,100 $63,642 $61,163 $59,354

12.11)

Year (n) 0 1 2 3 4

OC 3,200 3,700 4,800 5,850

MV 7,700 4,300 3,300 1,100 0

(a) Interest i = 12% Defender: Year 0 1 2 3 4

MV $7, 700

$3, 200 $4, 300 $3, 700 $3, 300 $4, 800 $1,100 $5, 850 0

CR (12%) Total AEC (12%)

$3, 200 $3, 436 $3, 840 $4, 261

$4, 324 $2, 999 $2, 880 $2, 535

$7, 524 $6, 435 $6, 720 $6, 796

The defender’s remaining useful (economic) life is 2 more years with an AEC value of $6,435, i.e., . (b) N C = 10 years

Since

don’t purchase the challenger now.

12.13) •

Defender: Economic service year is 2 years

Interest rate

n 0 1 2 3 4 5

•

15%

Market Value O&M Costs CR(15%) OC(15%) AEC(15%) $5,000 $4,000 $3,000 $2,000 $1,000 $0

$1,200 $2,000 $3,500 $5,000 $6,500 $8,000

$1,750 $1,680 $1,614 $1,551 $1,492

$3,380 $3,436 $3,886 $4,410 $4,942

$5,130 $5,116 $5,500 $5,961 $6,434

Challenger: Economic service year is 4 years

Interest rate

15%

Market Value O&M Costs CR(15%) OC(15%) AEC(15%) 0 1 2 3 4 5

$10,000 $6,000 $5,100 $4,335 $3,685 $3,132

$2,000 $2,800 $3,600 $4,400 $5,200

$5,500 $3,779 $3,131 $2,765 $2,519

$2,000 $2,372 $2,726 $3,061 $3,378

$7,500 $6,151 $5,857 $5,826 $5,897

Since AEC D < AECC , we should not replace the defender now. If no technological advances are expected in the next few years, the defender should be used for at least 2 more years. However, it is not necessarily best to replace the defender at the end of its economic year either.

•

Marginal analysis: 1. Opportunity cost at the end of year two, which is equal to the market value then, or $3,000 2. Operating cost for the third year: $5,000 3. Salvage value of the defender at the end of year three: $2,000 The cost of using the defender for one more year from the end of its economic service life is = F3 $3, 000( F / P,15%,1) + $5, 000 − $2, 000 = $6, 450 Compare this cost with AECC∗ = $5,826 of the challenger.

Since keeping the defender for the 3rd year is more expensive than replacing it with the challenger, do not keep the defender beyond its economic service life.

12.14)

It is assumed that the required service period is very long. AECD = $6, 000( A / P,12%, 6) + $2, 000 − $1,500( A / F ,12%, 6) = $3, 274.52 AECC = $21, 000( A / P,12%,12) + $1, 000 − $500( A / F ,12%,12) = $4,369.46 We should continue to use the present machine. The economic advantage is $4,369.46-$3,274.52 = $1,094.94 per year.

12.15) (a) and (b) n Defender Challanger 0 -$4,000 -$6,000 1 -$3,000 -$2,000 2 3

-$4,500 -$5,000

-$3,000 -$2,000

Now is the time to replace the defender. 12.16) (a) Opportunity cost = $0 (b) The cash flows are:

Year: 0 1 2 3 4 5 Defender $0 -$3K -$3K -$3K -$3K -$3K Challenger -$10K 0 0 0 0 0 C-D -$10K $3K $3K $3K $3K $3K

(c)

∴ We find i* = 15.24% . Since Challenger.

the firm should buy the

12.17) (a)

Yes, the new machine should be purchased now.

(b) Let P as the current market value of the old machine − P( A / P,15%,5) + $10,000 − $7,000 = $3,216.84

We find P = −$726.9 . Let P as the cost of new machine − P( A / P,15%,5) + $11,500 − $5,000 + $2,000( A / F,15%,5) = $2,701.68

We find P = $13,727 .

12.18) Assume that the old system has a current market value of P.

Let and solve for P. We find that P = $76,941.73. If the resale value of the defender is higher than $76,941.73, the installation of the system is justified.

12.19)

Since

, do not replace the defender for now.

12.20) For the challenger, we have:

For the defender, since salvage value at year 10 is $1,000 and the problem is stated that if sold at the end of first year, it will bring $1,500. We assume the market values will be declined same amount (($1,500-$1,000)/4 = $125) for the next years.

Year 0 1 2 3 4 5

CR(14%) AEC(14%)

AEOC

$3,800 $3,800 $3,800 $3,800 $3,800

$2,000 $1,500 $1,375 $1,250 $1,125 $1,000

$3,800 $3,800 $3,800 $3,800 $3,800

$780 $572 $498 $458 $431

$4,580 $4,372 $4,298 $4,258 $4,231

with i = 14%. Since

the new machine should not be purchased.

12.21) AEC(15%)Option 1 = $15,000 + ($6,000 − 0)( A / P ,15%,10) +$12,000 + ($48,000 − $5,000)( A / P ,15%,10) + $5,000(0.15) = $37,513.35 AEC(15%)Option 2 =$24,000 + ($84,000 − $9,000)( A / P ,15%,10) + $9,000(0.15) = $40,293.90 Since

, Option 1 should be selected.

12.22)

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

25.00% 8.00%

PW(i) = AE(%) =

($15,829)

($4,779)

Revenues (savings) Expenses: O&M Depreciation

$2,500 $2,400

$3,000 $3,840

$3,500 $2,304

$4,000 $691

Taxable Income Income Taxes (25%)

($4,900) (1,225)

($6,840) (1,710)

($5,804) (1,451)

($4,691) (1,173)

Net Income

($3,675)

($5,130)

($4,353)

($3,518)

Income Statement

Cash Flow Statement

Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax

Net Cash Flow

(3,518) $691

3,000 ($59)

(12,000)

($12,000)

AEC(8%) = $15,829( A / P,8%, 4) = $4, 779

$ (3,675) $ (5,130) $ (4,353) $ $2,400 $3,840 $2,304

($1,275)

($1,290)

($2,049)

$114

12.23)

(a) & (b) Yes, replace the defender. •

Defender

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

Output PW(i) = AE(%) =

25.00% 12.00%

($10,988) ($3,048.2)

Revenues (savings) Expenses: O&M Depreciation

$0 $10,000

Taxable Income Income Taxes (%)

($10,000) (2,500)

Net Income

($7,500)

Income Statement

Cash Flow Statement

Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

$ (7,500) $ (7,500) $ (7,500) $ (7,500) $ (7,500) $10,000 $10,000 $10,000 $10,000 $10,000 $ (20,000)

($20,000)

$2,500

Opportunity cost associated with retaining the defender. • Current market value: $10,000 • Tax credit on loss: ($10,000 - $50,000)(0.25) = ($10,000) • Total opportunity cost = $10,000 + $10,000 = $20,000

$2,500

•

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

25.00% 12.00%

Output PW(i) = AE(%) =

$85,205 $18,670

Income Statement

Revenues (savings) Expenses: Depreciation

$50,000

$50,000 $50,000

$50,000

$17,148

$29,388

$20,988

$14,988

$10,716 $10,704

$5,358

Taxable Income Income Taxes (%)

$32,852 $20,612 $8,213 $5,153

$29,012 $7,253

$35,012 $39,284 $39,296 $8,753 $9,821 $9,824

$44,642 $11,161

Net Income

$24,639

$15,459

$21,759

$26,259

$29,463 $29,472

$33,482

$24,639 $15,459 $17,148 $29,388

$21,759 $20,988

$26,259 $29,463 $29,472 $14,988 $10,716 $10,704

$33,482 $5,358

Cash Flow Statement

Operating Activities: Net Income Depreciation Investment Activities: $(120,000) Investment Salvage Gains Tax Net Cash Flow

$(120,000) $ 41,787 $ 44,847

$30,000 $2,678 $ 42,747 $ 41,247 $ 40,179 $40,176 $ 71,517

12.24) (a) & (b): Decision - Replace the defender now.

Defender: 0

$9,600.00 $14,400 $10,000 $11,100

$5,760.00 $8,640

$5,760.00 $2,880

$2,880.00 $0

$0.00

Financial Data

Depreciation Book value Market value Opportunity cost

5000

Cash Flow Statement Depreciation

$5,760

$2,880

Taxable Income Income Taxes (%)

($5,760) (1,440)

($2,880) (720)

$0 0

Net Income

($4,320)

($2,160)

Cash Flow Statement

Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

(4,320) $ (4,320) $ (2,160) $ $5,760 $5,760 $2,880

$ $0

$ (11,100)

($11,100) $

1,440 $

720 $

$ $

5,000 (1,250)

3,750

Challenger: Input Tax Rate(%) = MARR(%) =

Output PW(i) = AE(%) =

25.00% 15.00%

$13,434 $4,007.6

$75,000

$15,000 $60,000

$24,000 $36,000

$14,400 $21,600

$8,640 $12,960

$4,320 $8,640

Cash Flow Statement Savings Depreciation

$30,000 $15,000

$30,000 $24,000

$30,000 $14,400

$30,000 $8,640

$30,000 $4,320

Taxable Income Income Taxes (%)

$15,000 3,750

$6,000 1,500

$15,600 3,900

$21,360 5,340

$25,680 6,420

Net Income

$11,250

$4,500

$11,700

$16,020

$19,260

16,020 $ $8,640

19,260 $4,320

$ $

2,160

24,660 $

25,740

Financial Data

Depreciation Book value

Cash Flow Statement

Operating Activities: Net Income Depreciation Investment Activities: Investment Salvage Gains Tax Net Cash Flow

$ 11,250 $ 4,500 $ 11,700 $ $15,000 $24,000 $14,400 $ (75,000)

($75,000) $ 26,250 $ 28,500 $ 26,100 $

12.25) Economic service life With i = 12% and tax rate = 40%: Economic service life = 1 year

Tax Rat e MARR Hol di ng Per i od 0 1 2 3 4 5

$15, 000 I nvest ment Book val ue $15, 000 Per mi t t ed Annual Depr eci at i on Amount s over t he Hol di ng Per i od 3 4 5 6 7

25% 12%

$3,000 $3,000 $3,000 $3,000 $3,000

$2,400 $4,800 $4,800 $4,800

$1,440 $2,880 $2,880

$864 $1,728

$864

Tot al Depr eci at i on

Book Val ue

$3, 000 $5, 400 $9, 240 $11, 544 $13, 272

$12, 000 $9, 600 $5, 760 $3, 456 $1, 728

Tot al PW of O&M Cost s

Tot al PW of A/T O&M Cost s

$2, 232 $4, 783 $8, 556 $12, 686 $17, 112

$1, 674 $3, 587 $6, 417 $9, 515 $12, 834

2880 Annual O&M Cost s over t he Hol di ng Per i od Hol di ng Per i od 0 1 2 3 4 5

Hol di ng Per i od 0 1 2 3 4 5

8 `

$2, 500 $2, 500 $2, 500 $2, 500 $2, 500 Expect ed Mar ket Val ue $12, 800 $8, 100 $5, 200 $3, 500 $0

$3, 200 $3, 200 $3, 200 $3, 200

Taxabl e Gai ns $800 ($1, 500) ($560) $44 ($1, 728)

$5, 300 $5, 300 $5, 300

$6, 500 $6, 500

Gai ns Tax

Net A/T Mar ket Val ue

A/T Oper at i ng Cost s (i n PW) over t he Hol di ng Per i od O&M Cost s Tax Shi el d Tot al OC

OC(12%)

CR(12%)

Tot al AEC(12%)

$200 ($375) ($140) $11 ($432)

$12, 600 $8, 475 $5, 340 $3, 489 $432

$1, 674 $3, 587 $6, 417 $9, 515 $12, 834

$1, 125 $1, 443 $1, 888 $2, 383 $2, 857

$4, 200 $4, 878 $4, 663 $4, 208 $4, 093

$5, 325 $6, 321 $6, 551 $6, 592 $6, 950

$7, 800

$670 $1, 148 $1, 883 $2, 276 $2, 536

$1, 004 $2, 439 $4, 534 $7, 239 $10, 298

12.26) (a) Interest i = 12% Defender: The depreciation schedule when the defender was placed in service: D1 = $3,573, D2 = $6,123, D3 = $4,373, D4 = $3,125, D5 = $2,231, D6 = $2,231, D7 = $2,231, D8 = $1,116. Tax Rate MARR Holding Period 0 1 2 3 4

25% 12%

Permitted Annual Depreciation Amounts over the Holding Period 2 3 4 5 6

$2,231 $2,231 $2,231 $2,231

$2,231 $2,231 $2,231

$2,231 $2,231

$1,116

Annual O&M Costs over the Holding Period Holding Period 0 1 2 3 4

$3,200 $3,200 $3,200 $3,200

Holding Period 0 1 2 3 4

Expected Market Value $7,700 $4,300 $3,300 $1,100 $0

Total Book Depreciation Value $7,810 $2,231 $5,578 $4,462 $3,347 $6,693 $1,116 $7,809 $0 Total PW Total PW of of A/T O&M Costs O&M Costs

` $3,700 $3,700 $3,700

Taxable Gains ($1,278) ($47) ($16) $0

$4,800 $4,800

$5,850

$2,857 $5,807 $9,223 $12,941

$2,143 $4,355 $6,917 $9,706

Net A/T A/T Operating Costs (in PW) Total Market over the Holding Period OC(12%) CR(12%) AEC(12%) Value O&M Costs Tax Shield Total OC $8,118 ($320) $4,620 $2,143 $498 $1,645 $1,842 $4,128 $5,970 ($12) $3,312 $4,355 $943 $3,412 $2,019 $3,059 $5,078 $6,917 $1,340 $5,578 $2,322 $2,925 ($4) $1,104 $5,247 $9,706 $1,517 $8,189 $2,696 $2,571 $0 $0 $5,267

Gains Tax

Note that the cost of retaining the defender on after-tax basis is $8,118, instead of $7,700. The scheduled depreciation amount during the fourth year of ownership is $3,125. Since the asset will be disposed of during the recovery period, the allowed depreciation amount will be (0.5) ($3,125) = $1,561. Then, the book value becomes $9,370, instead of $7,810. With the market value of $7,700, there will be a loss of $1,670. The tax credit on this loss will be $1,670(0.25) = $417.50. Finally, the net proceeds from sale of old asset will be $8,118 (= $7,700 + $417.50). The defender’s remaining useful (economic) life is 2 more years with an AEC value of $5,078, i.e., = N *D 2,= AEC D $5, 078 .

(b) N C = 10 years AECC = $5,197 Input Tax Rate(%) = MARR(%) =

25.00% 12.00%

Revenues (savings) Expenses: O&M Depreciation

$1,000 $4,430

$1,000 $7,592

$1,000 $5,422

$1,000 $3,872

$1,000 $2,768

$1,000 $2,765

$1,000 $2,768

$1,000 $1,000 $1,383 $0

Taxable Income Income Taxes (%)

($5,430) ($8,592) (1,357) (2,148)

($6,422) (1,605)

($4,872) ($3,768) ($3,765) ($3,768) ($2,383) ($1,000) ($1,000) (1,218) (942) (941) (942) (596) (250) (250)

Net Income

($4,072) ($6,444)

($4,816)

($3,654) ($2,826) ($2,824) ($2,826) ($1,787)

Output PW(i) = ($29,357) AE(12%) = ($5,195.6) 9

Income Statement

($750)

$1,000 $0

($750)

Cash Flow Statement

Operating Activities: $ (4,072) $ (6,444) $ (4,816) $ (3,654) $ (2,826) $ (2,824) $ (2,826) $ (1,787) $ (750) $ (750) Net Income $4,430 $7,592 $5,422 $3,872 $2,768 $2,765 $2,768 $1,383 $0 $0 Depreciation Investment Activities: $ (31,000) Investment $ 2,500 Salvage ($625) Gains Tax Net Cash Flow

$ (31,000)

$357

$1,148

$605

$218

($58)

($59)

($58)

($404)

($750) $ 1,125

From n = 2 to n = 3: $3,312(1.12) − $1,104 + $5,578 = $8,183 > $5, 078

Keep the defender for two years, which happens to be the same as the economic service life as calculated before. (In general, you should not expect this to happen all the time.)

12.27)

(a) Economic service life = 5 years with MARR of 15% Tax Rat e MARR Hol di ng Per i od 0 1 2 3 4 5 6

I nvest ment $10, 000 Book val ue $10, 000 Per mi t t ed Annual Depr eci at i on Amount s over t he Hol di ng Per i od 3 4 5 6 7

25% 15%

$2, 000 $2, 000 $2, 000 $2, 000 $2, 000 $2, 000

$3, 200 $3, 200 $3, 200 $3, 200 $3, 200

$1, 920 $1, 920 $1, 920 $1, 920

$1, 152 $1, 152 $1, 152

$576

Tot al Book Depr eci at i on Val ue $10, 000 $2, 000 $8, 000 $5, 200 $4, 800 $7, 120 $2, 880 $8, 272 $1, 728 $9, 424 $576 $10, 000 $0

Tot al PW Tot al PW of of A/T O&M Cost s O&M Cost s

$1, 152 $1, 152

Annual O&M Cost s over t he Hol di ng Per i od Hol di ng Per i od 0 1 2 3 4 5 6

Hol di ng Per i od 0 1 2 3 4 5 6

8 `

$1, 500 $1, 500 $1, 500 $1, 500 $1, 500 $1, 500

$2, 100 $2, 100 $2, 100 $2, 100 $2, 100

$2, 700 $2, 700 $2, 700 $2, 700

$3, 400 $3, 400 $3, 400

Expect ed Mar ket Val ue

Taxabl e Gai ns

Gai ns Tax

Net A/T A/T Oper at i ng Cost s (i n PW) Mar ket over t he Hol di ng Per i od OC(15%) Val ue O&M Cost s Tax Shi el d Tot al OC

($2, 700) ($900) ($80) $72 $824 $600

($675) ($225) ($20) $18 $206 $150

$5, 975 $4, 125 $2, 820 $1, 782 $1, 194 $450

$5, 300 $3, 900 $2, 800 $1, 800 $1, 400 $600

$4, 200 $4, 200

$978 $2, 169 $3, 501 $4, 959 $6, 525 $8, 114

$4, 900

$435 $1, 040 $1, 355 $1, 520 $1, 663 $1, 725

$543 $625 $1, 129 $695 $2, 145 $940 $3, 439 $1, 204 $4, 862 $1, 450 $6, 388 $1, 688

$1, 304 $2, 892 $4, 668 $6, 612 $8, 700 $10, 818

$978 $2, 169 $3, 501 $4, 959 $6, 525 $8, 114

CR(15%)

Tot al AEC(15%)

$5, 525 $4, 233 $3, 568 $3, 146 $2, 806 $2, 591

$6, 150 $4, 927 $4, 507 $4, 350 $4, 256 $4, 279

(b)Economic service life = 5 years with MARR of 10%

Tax Rat e MARR Hol di ng Per i od 0 1 2 3 4 5 6

I nvest ment $10, 000 Book val ue $10, 000 Per mi t t ed Annual Depr eci at i on Amount s over t he Hol di ng Per i od 3 4 5 6 7

25% 10%

$2, 000 $2, 000 $2, 000 $2, 000 $2, 000 $2, 000

$3, 200 $3, 200 $3, 200 $3, 200 $3, 200

$1, 920 $1, 920 $1, 920 $1, 920

$1, 152 $1, 152 $1, 152

$1, 152 $1, 152

$2, 000 $5, 200 $7, 120 $8, 272 $9, 424 $10, 000

$576

Annual O&M Cost s over t he Hol di ng Per i od Hol di ng Per i od 0 1 2 3 4 5 6

Hol di ng Per i od 0 1 2 3 4 5 6

Tot al Dep.

Book Val ue $10, 000 $8, 000 $4, 800 $2, 880 $1, 728 $576 $0

Tot al PW Tot al PW of of A/T O&M Cost s O&M Cost s

` $1, 500 $1, 500 $1, 500 $1, 500 $1, 500 $1, 500

$2, 100 $2, 100 $2, 100 $2, 100 $2, 100

$2, 700 $2, 700 $2, 700 $2, 700

$3, 400 $3, 400 $3, 400

Expect ed Mar ket Taxabl e Val ue Gai ns

Gai ns Tax

Net A/T A/T Oper at i ng Cost s ( i n PW) Mar ket over t he Hol di ng Per i od Val ue O&M Cost s Tax Shi el d Tot al OC

( $675) ( $225) ( $20) $18 $206 $150

$5, 975 $4, 125 $2, 820 $1, 782 $1, 194 $450

$5, 300 $3, 900 $2, 800 $1, 800 $1, 400 $600

( $2, 700) ( $900) ( $80) $72 $824 $600

$4, 200 $4, 200

$1, 023 $2, 324 $3, 846 $5, 587 $7, 543 $9, 618

$1, 364 $3, 099 $5, 128 $7, 450 $10, 058 $12, 824

$1, 023 $2, 324 $3, 846 $5, 587 $7, 543 $9, 618

OC( 10%)

CR( 10%)

Tot al AEC( 10%)

$625 $696 $953 $1, 235 $1, 501 $1, 764

$5, 025 $3, 798 $3, 169 $2, 771 $2, 442 $2, 238

$5, 650 $4, 494 $4, 122 $4, 006 $3, 944 $4, 002

$4, 900

$455 $1, 116 $1, 476 $1, 673 $1, 852 $1, 933

$568 $1, 209 $2, 369 $3, 914 $5, 692 $7, 685

12.28)

(a) At i = 12%, the economic service life = 7 years: Tax Rat e MARR Hol di ng Per i od 0 1 2 3 4 5 6 7 8 9 10

$30, 000 I nvest ment $30, 000 Book val ue Per mi t t ed Annual Depr eci at i on Amount s over t he Hol di ng Per i od 3 4 5 6 7

25% 12%

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000

$3,000 $3,000

$3,000

Annual O&M Cost s over t he Hol di ng Per i od Hol di ng Per i od 0 1 2 3 4 5 6 7 8 9 10

Hol di ng Per i od 0 1 2 3 4 5 6 7 8 9 10

Tot al Depr eci at i on

Book Val ue

$3, 000 $6, 000 $9, 000 $12, 000 $15, 000 $18, 000 $21, 000 $24, 000 $27, 000 $30, 000

$27, 000 $24, 000 $21, 000 $18, 000 $15, 000 $12, 000 $9, 000 $6, 000 $3, 000 $0

Tot al PW Tot al PW of of A/T O&M Cost s O&M Cost s

8 `

$3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 Expect ed Mar ket Val ue $20, 000 $18, 000 $16, 000 $14, 000 $12, 000 $10, 000 $8, 000 $6, 000 $4, 000 $2, 000

$3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450

Taxabl e Gai ns ( $7, 000) ( $6, 000) ( $5, 000) ( $4, 000) ( $3, 000) ( $2, 000) ( $1, 000) $0 $1, 000 $2, 000

$3, 968 $3, 968 $3, 968 $3, 968 $3, 968 $3, 968 $3, 968 $3, 968

$4, 563 $4, 563 $4, 563 $4, 563 $4, 563 $4, 563 $4, 563

Gai ns Tax

Net A/T Mar ket Val ue

( $1, 750) ( $1, 500) ( $1, 250) ( $1, 000) ( $750) ( $500) ( $250) $0 $250 $500

$21, 750 $19, 500 $17, 250 $15, 000 $12, 750 $10, 500 $8, 250 $6, 000 $3, 750 $1, 500

$5, 247 $5, 247 $5, 247 $5, 247 $5, 247 $5, 247

$6, 034 $6, 034 $6, 034 $6, 034 $6, 034

$6, 939 $6, 939 $6, 939 $6, 939

A/T Oper at i ng Cost s ( i n PW) over t he Hol di ng Per i od O&M Cost s Tax Shi el d Tot al OC $2, 009 $4, 072 $6, 190 $8, 365 $10, 598 $12, 891 $15, 245 $17, 656 $20, 138 $22, 686

$670 $1, 268 $1, 801 $2, 278 $2, 704 $3, 084 $3, 423 $3, 726 $3, 996 $4, 238

$1, 339 $2, 804 $4, 389 $6, 087 $7, 894 $9, 807 $11, 822 $13, 930 $16, 142 $18, 449

$7, 960 $7, 960 $7, 960

$9, 177 $9, 177

$10, 554

$2, 679 $5, 429 $8, 253 $11, 153 $14, 130 $17, 187 $20, 326 $23, 541 $26, 850 $30, 249

$2, 009 $4, 072 $6, 190 $8, 365 $10, 598 $12, 891 $15, 245 $17, 656 $20, 138 $22, 686

OC( 12%)

CR( 12%)

Tot al AEC( 12%)

$1, 500 $1, 659 $1, 827 $2, 004 $2, 190 $2, 385 $2, 590 $2, 804 $3, 029 $3, 265

$11, 850 $8, 553 $7, 378 $6, 739 $6, 315 $6, 003 $5, 756 $5, 551 $5, 377 $5, 224

$13, 350 $10, 212 $9, 206 $8, 743 $8, 505 $8, 388 $8, 346 $8, 355 $8, 406 $8, 489

(b)At i = 20%, the economic service life = 9 years I nvest ment $30, 000 Book val ue $30, 000 Per mi t t ed Annual Depr eci at i on Amount s over t he Hol di ng Per i od 3 4 5 6 7

25% 20%

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000 $3,000

$3,000 $3,000 $3,000

$3,000 $3,000

Tot al Depr eci at i on

Book Val ue

$3,000

$3, 000 $6, 000 $9, 000 $12, 000 $15, 000 $18, 000 $21, 000 $24, 000 $27, 000 $30, 000

$27, 000 $24, 000 $21, 000 $18, 000 $15, 000 $12, 000 $9, 000 $6, 000 $3, 000 $0

Annual O&M Cost s over t he Hol di ng Per i od 1

Tot al PW Tot al PW of of A/T O&M Cost s O&M Cost s

8 `

$3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 $3, 000 Expect ed Mar ket Val ue $20, 000 $18, 000 $16, 000 $14, 000 $12, 000 $10, 000 $8, 000 $6, 000 $4, 000 $2, 000

$3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450 $3, 450

Taxabl e Gai ns ($7, 000) ($6, 000) ($5, 000) ($4, 000) ($3, 000) ($2, 000) ($1, 000) $0 $1, 000 $2, 000

$9, 177 $9, 177 $10, 554

$2, 500 $4, 896 $7, 192 $9, 393 $11, 501 $13, 522 $15, 459 $17, 310 $19, 088 $20, 793

$1, 875 $3, 672 $5, 394 $7, 044 $8, 626 $10, 142 $11, 594 $12, 982 $14, 316 $15, 595

$3, 968 $3, 968 $3, 968 $3, 968 $3, 968 $3, 968 $3, 968 $3, 968

$4, 563 $4, 563 $4, 563 $4, 563 $4, 563 $4, 563 $4, 563

Gai ns Tax

Net A/T A/T Oper at i ng Cost s (i n PW) Mar ket over t he Hol di ng Per i od Val ue O&M Cost s Tax Shi el d Tot al OC

OC(20%)

CR(20%)

Tot al AEC(20%)

($1, 750) ($1, 500) ($1, 250) ($1, 000) ($750) ($500) ($250) $0 $250 $500

$21, 750 $19, 500 $17, 250 $15, 000 $12, 750 $10, 500 $8, 250 $6, 000 $3, 750 $1, 500

$1, 500 $1, 653 $1, 811 $1, 971 $2, 134 $2, 300 $2, 466 $2, 633 $2, 802 $2, 970

$14, 250 $10, 773 $9, 503 $8, 794 $8, 318 $7, 964 $7, 684 $7, 455 $7, 262 $7, 098

$15, 750 $12, 426 $11, 313 $10, 766 $10, 452 $10, 263 $10, 150 $10, 088 $10, 064 $10, 068

$5, 247 $5, 247 $5, 247 $5, 247 $5, 247 $5, 247

$1, 875 $3, 672 $5, 394 $7, 044 $8, 626 $10, 142 $11, 594 $12, 982 $14, 316 $15, 595

$6, 034 $6, 034 $6, 034 $6, 034 $6, 034

$625 $1, 146 $1, 580 $1, 942 $2, 243 $2, 494 $2, 703 $2, 878 $3, 023 $3, 144

$6, 939 $6, 939 $6, 939 $6, 939

$1, 250 $2, 526 $3, 814 $5, 103 $6, 383 $7, 647 $8, 891 $10, 105 $11, 293 $12, 450

$7, 960 $7, 960 $7, 960

Capital recovery cost: = S n − Bn gain gain= tax tm ( S n − Bn ) net proceeds = (1 − tm ) S n + tm Bn = (1 − 0.25)(22, 000 − 2000n) + 0.25(30, 000 − 3, 000n) I − (1 − tm ) S n − tm Bn CR = n 6, 000 + 2, 250n = n 6, 000 = + 2, 250 n

•

Equivalent annual O&M cost: n −1 A/T O&M= (1 − tm ) 3, 000 (1.15 )   

= 2, 250 (1.15 )

∑ =

AE O & M

n n =1

n −1

2, 250 (1.15 )

n −1

∑ (1.15) = 2, 250 n

n −1

n =1

n 2, 250 (1.15 − 1) (1.15n − 1) = = 15, 000 1.15 − 1 n n n

•

Depreciation tax credit:

(t × D ) ∑ = AE = , where D $3, 000 n

n =1

= $750

•

Minimum total annual equivalent cost: 6, 000 (1.15n − 1) + 2, 250 + 15, 000 − 750 AEC(0%) = n n 6, 000 (1.15n − 1) = + 15, 000 + 1,500 n n 1.15n 9, 000 = 15, 000 − + 1,500 n n

n 1 2 3 4 5 6 7 8 9 10

1st 2nd 3rd term term term AEC(0%) 17250 -9000 1500 9750 9919 -4500 1500 6919 7604 -3000 1500 6104 6559 -2250 1500 5809 6034 -1800 1500 5734 5783 -1500 1500 5783 5700 -1286 1500 5914 5736 -1125 1500 6111 5863 -1000 1500 6363 6068 -900 1500 6668

12.29) (a) Keep the defender Financial Data Depreciation Book value Current market value O&M cost

-4 $20,000

-3 $2,858 $17,142

-2 $4,898 $12,244

-1 $3,498 $8,746

Cash Flow Statement (-0.7)*(O&M cost) +(.3)*(Depreciation) Investment Net proceeds from sale

1 $1,786 $4,462

2 $1,784 $2,678

3 $1,786 $892

4 $892 $0

$2,000

$0 $1,500 $2,000

(1,400) 536

(1,400) 535

(1,400) 536

(1,400) 268

(1,400) 0

(6,074) 1,050

Net Cash Flow PW (10%) =

0 $2,498 $6,248 $6,000

$0 ($10,064)

AEC(10%) =

$2,311

($6,074)

($864)

($865)

($864)

($1,132)

($1,400)

($350)

(b) Replace the defender Financial Data Depreciation Book value O&M cost Cash Flow Statement +(.3)*(Depreciation) (-0.7)*(O&M cost) Investment Net proceeds from sale Net Cash Flow PW (10%) =

n $21,000

1 $3,001 $17,999 $1,000

2 $5,143 $12,856 $1,000

3 $3,673 $9,183 $1,000

4 $2,623 $6,560 $1,000

5 $1,875 $4,685 $1,000

6 $1,873 $2,812 $1,000

7 $1,875 $937 $1,000

8 $937 ($0) $1,000

$0 ($0) $1,000

10-11 $0 ($0) $1,000

900 (700)

1,543 (700)

1,102 (700)

787 (700)

563 (700)

562 (700)

563 (700)

281 (700)

0 (700)

$0 ($0) 1000

0 (700)

(21,000) 350 ($21,000) ($21,113)

$200

$843

$402

AEC(10%) =

$3,099

$87

($137)

($138)

($137)

($419)

($700)

($350)

12.30) (a), (b), and (c):

(a) , (b) and ( c ) Option 1 : Keep the defender Financial Data

Depreciation Book value Expected Market value O&M cost

0 $4,000 $0

1 $800 $3,200 $0 $0

2 $800 $2,400 $0 $0

3 $800 $1,600 $0 $0

4 $800 $800 $0 $0

5 $800 $0 $0 $0

0 200

$200

AEC(10%) =

$222

1 $1,429 $8,571 $0 $3,000

2 $2,449 $6,122 $0 $3,000

3 $1,749 $4,373 $0 $3,000

4 $1,249 $3,124 $0 $3,000

5 $446 $2,678 $0 $3,000

$357 $2,250

$612 $2,250

$437 $2,250

$312 $2,250

$112 $2,250

Cash Flow Statement

(-0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

(1,600)

Net Cash Flow PW (10%) =

($1,600) ($842)

Option 2 : Replace the defender Financial Data

Depreciation Book value Expected Market value Savings in O&M cost

0 $10,000 $0

Cash Flow Statement

+(.25)*(Depreciation) (0.75)*(Savings in O&M cost) Investment Net proceeds from sale Net Cash Flow PW (10%) =

($10,000) $670 ($10,000)

$387

$2,607

$2,862

AEC(10%) =

($102)

$2,687

$2,562

$3,031

12.31) Replacement analysis: Let X denote the current market value of the old callswitching system: AEC(14%) $20, 000(0.75) + 0.75 X ( A / P,14%,5) = defender = $15, 000 + (0.75 X )(0.29128) = $15, 000 + 0.2185 X AEC(14%)challenger = $180, 642( A / P,14%,10) = $34, 631

To justify the new call-switching system now, we must have

AEC(14%) defender >AEC(14%)challenger $15, 000 + 0.2185 X > $34, 631 X > $89,844 •

Challenger:

Challenger Financial Data

Depreciation Book value Salvage value O&M cost

0 $200,000

$40,000 $160,000

$64,000 $96,000

$38,400 $57,600

$23,040 $34,560

$23,040 $11,520

$5,000

($3,750) $10,000

($3,750) $16,000

($3,750) $9,600

($3,750) $5,760

$6,250

$12,250

$5,850

$2,010

Cash Flow Statement

-(0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

($200,000)

Net Cash Flow

Financial Data

($200,000)

Depreciation Book value Salvage value O&M cost Cash Flow Statement -(0.75)*(Savings in O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale Net Cash Flow PW (14%) = AEC (14%) =

6 $11,520 $0

$0 $0

$5,000

$18,000 $5,000

($3,750) $2,880

($3,750) $0

($3,750) $0 $13,500

($870) ($180,642) $34,631

($3,750)

$9,750

12.32) Defender analysis (a) and (b): Keep the defender for three years. 

Defender: Tax Rate

25%

MARR

18%

$12,375

Investment

$15,000 Book value Permitted Annual Depreciation Amounts over the

Holding

Holding Period

Period

Total

Book

Depreciation

Value

0 1 2

$4,000 $4,000

$4,000

Annual O&M Costs over the Holding Period Holding

$4,000

$15,000 $11,000

$8,000

$7,000

$12,000

$3,000

Total PW of A/T O&M Costs O&M Costs

Total PW of

Period

8 `

$4,500

$5,300

$4,500

$5,300

$6,100

Expected

Net A/T

$3,814

$2,860

$7,620

$5,715

$11,333

$8,499

A/T Operating Costs (in PW)

Holding

Market

Taxable

Gains

Market

over the Holding Period

Period

Value

Gains

Tax

Value

O&M Costs Tax Shield Total OC

($1,450) ($875)

$6,650 $4,375

$2,860

$847

$5,715

$1,566

($450)

$1,650

$8,499

$2,174

Total OC(18%)

CR(18%)

AEC(18%)

$2,013

$2,375

$7,953

$10,328

$4,149

$2,650

$5,897

$8,547

$6,325

$2,909

$5,230

$8,139

0 1 2

$5,200 $3,500

($5,800) ($3,500)

$1,200

($1,800)

Note: The opportunity cost of retaining the defender is as follows: • • • • •

Current market value = $11,500 Current book value = $15,000 Losses = ($11,500 - $15,000) = ($3,500) Loss tax credit = $3,500(0.25) = $875 Cost of retaining the defender = $11,500+ $875 = $12,375



Challenger: Financial Data

Depreciation Book value

1 $6,216

2 $10,653

3 $7,608

4 $5,433

5 $3,885

$43,500

$37,284

$26,631

$19,023

$13,589

$9,705

$1,500

($1,125)

$1,554

$2,663

$1,902

$1,358

$971

$429

$1,538

$777

$233

($154)

Salvage value O&M cost Cash Flow Statement

-(0.75)*(O&M cost) +(.25)*(Depreciation) Investment

($43,500)

Net proceeds from sale ($43,500)

Net Cash Flow

Financial Data

Depreciation

$3,880

$3,885

$1,940

Book value

$5,825

$1,940

($0)

Salvage value O&M cost

$1,500

$3,500 $1,500

($1,125) $970

($1,125) $971

($1,125) $485

($1,125) $0

Cash Flow Statement

-(0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

$2,625 $0

Net Cash Flow PW (18%) = AEC(18%) =

($155)

($154)

($640)

($1,125)

($41,749) $9,290

Optimal time to replace: Since the remaining useful life for the defender is 3 years, which is the same as the physical life, keep the defender for 3 years.

$1,500

12.33) Decision: Do not replace the defender for now.

Financial Data

Depreciation Book value Market value

0 $0

$8,700

($6,525)

AEC(10%) =

$8,207

1 $7,645 $45,855

2 $13,102 $32,753

3 $9,357 $23,396

4 $6,682 $16,713

5 $2,386 $14,327

$5,700

$12,000 $6,200

$8,500

Operation Cost Cash Flow Statement

+(.25)*(Depreciation) Opportunity cost -(1-0.25)*(Operation cost)

($6,375)

Net Cash Flow

($6,375) PW (10%) = ($31,110)

Replace the defender n

Financial Data

Depreciation Book value

0 $53,500

Market value Operation Cost

$53,500 $4,200

$4,700

$5,200

Cash Flow Statement

Investment Net proceeds from sale +(.25)*(Depreciation) -(1-0.25)*(Operation cost)

($53,500)

Net Cash Flow

($53,500) PW (10%) = ($52,488)

$1,911 ($3,150)

$3,276 ($3,525)

$2,339 ($3,900)

$1,671 ($4,275)

$12,582 $597 ($4,650)

($1,239)

($249)

($1,561)

($2,604)

$8,528

AEC(10%) =

$13,846

11.34) Option 1:

Option 1 Financial Data

Depreciation Book value Current Market value O&M cost

0 $48,000 $6,000

$6,859 $41,141

$11,755 $29,386

$8,395 $20,990

$5,995 $14,995

$4,286 $10,709

$27,000

($20,250) $1,715

($20,250) $2,939

($20,250) $2,099

($20,250) $1,499

($20,250) $1,072

($18,535)

($17,311)

($18,151)

($18,751)

($19,178)

Cash Flow Statement

-(0.75)*(O&M cost) +(.25)*(Depreciation) Opportunity cost Investment Net proceeds from sale

($3,600) ($48,000)

Net Cash Flow

($51,600)

$4,282 $6,427

$4,286 $2,141

$2,141 $0

$0 $0

$27,000

$5,000 $27,000

-(0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale

($20,250) $1,070

($20,250) $1,072

($20,250) $535

($20,250) $0

Net Cash Flow

($19,180)

($19,178)

AEC(12%) =

$27,722

Financial Data

Depreciation Book value Salvage value O&M cost Cash Flow Statement

PW (12%) =

3,750

($156,638)

($19,715)

($20,250)

($16,500)

Option 2:

Option 2 0 Depreciation Book value O&M cost

$84,000

Cash Flow Statement -(0.75)*(O&M cost) +(.25)*(Depreciation) Investment ($84,000) Net proceeds from sale Net Cash Flow

Depreciation Book value Salvage value O&M cost Cash Flow Statement -(0.75)*(O&M cost) +(.25)*(Depreciation) Investment Net proceeds from sale Net Cash Flow

Select Option 1.

($84,000)

$12,004 $71,996

$20,572 $51,425

$14,692 $36,733

$10,492 $26,242

$7,501 $18,740

$24,000

($18,000) $3,001

($18,000) $5,143

($18,000) $3,673

($18,000) $2,623

($18,000) $1,875

($14,999)

($12,857)

($14,327)

($15,377)

($16,125)

$7,493 $11,248

$7,501 $3,746

$3,746 ($0)

$0 ($0)

$24,000

$9,000 $24,000

($18,000) $1,873

($18,000) $1,875

($18,000) $937

($18,000) $0

($18,000) $0 $6,750

($16,127)

($16,125)

PW (12%) = AEC(12%) =

($169,231) $29,951

($17,063)

($18,000)

($11,250)

10.35) Replacement analysis The remaining useful life of the defender is 1 year. Its annual equivalent cost is $1,666. When the defender is replaced now by the challenger, its equivalent annual cost is $2,191, indicating that the defender should be kept for now. (a) Economic service life = one year Tax Rat e MARR

30% 12%

Hol di ng Per i od 0 1 2 3 4 5

I nvest ment $1, 050 Book val ue $0 Per mi t t ed Annual Depr eci at i on Amount s over t he Hol di ng Per i od 3 4 5 6 7

Tot al Depr eci at i on

$0 $0 $0 $0 $0

Tot al PW of O&M Cost s

Tot al PW of A/ T O&M Cost s

$1, 696 $3, 530 $5, 452 $7, 422 $9, 351

$1, 188 $2, 471 $3, 816 $5, 195 $6, 546

OC( 12%)

CR( 12%)

Tot al AEC( 12%)

$1, 330 $1, 462 $1, 589 $1, 710 $1, 816

$336 $291 $333 $346 $291

$1, 666 $1, 753 $1, 922 $2, 056 $2, 107

Annual O&M Cost s over t he Hol di ng Per i od Hol di ng Per i od 0 1 2 3 4 5

Hol di ng Per i od 0 1 2 3 4 5

Book Val ue

8 `

$1, 900 $1, 900 $1, 900 $1, 900 $1, 900 Expect ed Mar ket Val ue $1, 200 $1, 000 $500 $0 $0

$2, 300 $2, 300 $2, 300 $2, 300

Taxabl e Gai ns $1, 200 $1, 000 $500 $0 $0

$2, 700 $2, 700 $2, 700

$3, 100 $3, 100

Gai ns Tax

Net A/ T Mar ket Val ue

$360 $300 $150 $0 $0

$840 $700 $350 $0 $0

$3, 400 A/ T Oper at i ng Cost s ( i n PW) over t he Hol di ng Per i od O&M Cost s Tax Shi el d Tot al OC $1, 188 $2, 471 $3, 816 $5, 195 $6, 546

$0 $0 $0 $0 $0

$1, 188 $2, 471 $3, 816 $5, 195 $6, 546

(b) Replace the defender Financial Data Depreciation Book value Salvage value Operation Cost

$6,000

Cash Flow Statement Investment Net proceeds from sale +(.30)*(Depreciation) -(1-0.30)*(Operation cost)

1 $1,200 $4,800

2 $1,920 $2,880

3 $1,152 $1,728

$691 $1,037

$1,100

$1,300

$1,500

$1,700

$346 $691 $1,000 $1,800

360 (770)

576 (910)

346 (1,050)

207 (1,190)

907 104 (1,260)

($410)

($334)

($704)

($983)

($249)

AEC(12%) =

$2,191

(6,000)

Net Cash Flow

($6,000) PW (12%) =

($7,899)

12.36) (a) and (b): Quintana should purchase the new equipment.

Financial Data

Depreciation Book value Market value Savings

0 $150,000 $150,000

1 $21,435 $128,565

2 $36,735 $91,830

3 $26,235 $65,595

4 $18,735 $46,860

5 $13,395 $33,465

6 $13,380 $20,085

7 $13,395 $6,690

8 $6,690 $0

9 $0 $0

10 $0 $0

$30,000

$22,500 $5,359

$22,500 $9,184

$22,500 $6,559

$22,500 $4,684

$22,500 $3,349

$22,500 $3,345

$22,500 $3,349

$22,500 $1,673

$22,500 $0

$31,684

$29,059

$27,184

$25,849

$25,845

$25,849

$24,173

$22,500

AE (10%) =

$2,491

Cash Flow Statement

+(1-0.25)*(Savings) +(.25)*(Depreciation) Investment

($150,000)

Net Cash Flow

($150,000)

$27,859

PW (10%) =

$15,307

1 $12,000 $60,000

2 $12,000 $48,000

3 $12,000 $36,000

4 $12,000 $24,000

5 $12,000 $12,000

6 $12,000 $0

$3,000

AEC (10%) =

$2,561

(b) Defender Financial Data

Depreciation Book value Current market value

$72,000

$3,000

Cash Flow Statement

+(.25)*(Depreciation) Investment

($28,800)

Net Cash Flow

($28,800)

$3,000

PW (10%) =

($15,734)

(c): Purchase the new equipment; (d): Retain the old machine (c) Defender with a current market value of $ 45,000 n 0 1 $12,000 Depreciation $60,000 $72,000 Book value $45,000 Current market value

$12,000 $48,000

$12,000 $36,000

$12,000 $24,000

$12,000 $12,000

$12,000 $0

$3,000

AEC(10%) =

$6,955

Financial Data

9 - 10

Cash Flow Statement

+(.25)*(Depreciation) Investment

($55,800)

Net Cash Flow

($55,800)

$3,000

PW (10%) =

($42,734)

(d) Challenger with an extended service life of 12 years n 0 1 $21,435 Depreciation $150,000 $128,565 Book value $15,000 Savings

$36,735 $91,830

$26,235 $65,595

$18,735 $46,860

$13,395 $33,465

$13,380 $20,085

$13,395 $6,690

$6,690 $0

$0 $0

$15,000

$5,359 $11,250

$9,184 $11,250

$6,559 $11,250

$4,684 $11,250

$3,349 $11,250

$3,345 $11,250

$3,349 $11,250

$1,673 $11,250

$0 $11,250

$20,434

$17,809

$15,934

$14,599

$14,595

$14,599

$12,923

$11,250

AEC(10%) =

$6,794

Financial Data

9-12

Cash Flow Statement

+(.25)*(Depreciation) +(0.75)*(Savings) Investment

($150,000)

Net Cash Flow

($150,000)

$16,609

PW (10%) =

($46,292)

(e) and (f): (e) Defender (Model A) Original Investment = $150,000 n

Depreciation Book value Current market value

Decision: Replace Model A with Model B 0

1 $26,235

2 $18,735

3 $13,395

4 $13,380

5 $13,395

6 $6,690

$91,830

$65,595

$46,860

$33,465

$20,085

$6,690

(f) It is rather difficult to predict

8 $0

$0 what technological advances would

$0 be made on a typical equipment in the

future. If the industrial engineer had

Cash Flow Statement

+(.25)*(Depreciation) Investment

$6,559

$4,684

$3,349

$3,345

$3,349

$1,673

all the information available in one or two years, he $0 could defer the replacement

($36,732)

decision. Since Model A was already placed in service, the

Net Cash Flow

($36,732)

$6,559

$4,684

AEC(10%) =

$3,575

1 $42,870

2 $73,470

$3,349

$3,345

$3,349

$1,673

$0 amount of $ 150,000 expended is a sunk cost, and it should not be

PW (10%) = ($19,075)

considered in future replacement decisions.

Challenger (Model B) n

Depreciation Book value Savings

3 $52,470

$300,000 $257,130 $183,660 $131,190 $75,000

$75,000

4 $37,470

5 $26,790

6 $26,760

7 $26,790

8 $13,380

$93,720

$66,930

$40,170

$13,380

$75,000

Cash Flow Statement

+(.25)*(Depreciation) +(0.75)*(Savings) Investment

$10,718

$18,368

$13,118

$9,368

$6,698

$6,690

$6,698

$3,345

$56,250

($300,000)

Net Cash Flow

($300,000) $66,968

$74,618

$69,368

$65,618

$62,948

$62,940

$62,948

$59,595

$56,250

PW (10%) = $99,741

AE (10%) = $16,232

12.37) 

Option 1: Keep the defender Option 1 : Keep the defender

Financial Data

Depreciation Book value Market value Setup cost Operating cost

0 $8,930 $13,387 $40,000

1 $8,920 $4,460

2 $4,460

$16,000 $15,986

$16,000 $16,785

$16,000 $17,663

$16,000 $18,630

$16,000 $19,692

$16,000 $20,861

$16,000 $22,147

$16,000 $23,562

$2,230

$1,115

($12,000) ($11,990)

($12,000) ($12,589)

($12,000) ($13,247)

($12,000) ($13,973)

($12,000) ($14,769)

($12,000) ($15,646)

($12,000) ($16,610)

($12,000) ($17,672)

($21,760)

($23,474)

($25,247)

($25,973)

($26,769)

($27,646)

($28,610)

($29,672)

Cash Flow Statement

+(.25)*(Depreciation) Opportunity cost -(1-0.25)*(Setup) -(1-0.25)*(Operating cost)

($34,463)

Net Cash Flow

($34,463) PW (12%) = AEC(12%) =

($161,202) $32,450

Note: Opportunity cost (Investment required to keep the defender) $40,000 – 0.25($40,000 – ($13,387+0.5($8,930))) = $34,463



Option 2: Purchase a used machine Option 2 : Purchase a used machine Financial Data Depreciation Book value Salvage value Setup cost Operating cost Savings

0 $148,200

Cash Flow Statement +(1-0.25)*(Savings) +(.25)*(Depreciation) Investment -(1-0.25)*(Setup) -(1-0.25)*(Operating cost) Net Cash Flow

2 $36,294 $90,728

3 $25,920 $64,808

4 $18,510 $46,298

5 $13,234 $33,063

6 $13,219 $19,844 $15,000 $14,245 36,000

7 $13,234 $6,610 $0 $15,000 $14,950 36,000

8 $6,610 $0 $0 $15,000 $15,745 36,000

$15,000 $11,500 36,000

$15,000 $11,950 36,000

$15,000 $12,445 36,000

$15,000 $12,990 36,000

$15,000 $13,590 36,000

$27,000.00 $5,294.45

$27,000.00 $9,073.55

$27,000.00 $6,480.05

$27,000.00 $4,627.55

$27,000.00 $3,308.57

$27,000.00 $3,304.86

$27,000.00 $3,308.57

$27,000.00 $1,652.43

($148,200.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($8,625.00) ($8,962.50) ($9,333.75) ($9,742.50) ($10,192.50) ($10,683.75) ($11,212.50) ($11,808.75) ($148,200)

PW (12%) = AEC(12%) =

1 $21,178 $127,022

($93,449) $18,811

$12,419

$15,861

$12,896

$10,635

$8,866

$8,371

$7,846

$5,594

Option 3: Keep the defender one year and switch to a brand new machine Option 3 : Keep the old machine for one more year (Assume that the market value will be $ 30,000.) n 0 1 2 3 Financial Data $8,930 $4,460 Depreciation $13,138 $8,927 Book value $40,000 $30,000 Market value $16,000 Setup cost $16,785 Operating cost Cash Flow Statement $1,115 +(.25)*(Depreciation) ($34,401) Opportunity cost ($12,000) -(1-0.25)*(Setup) ($12,589) -(1-0.25)*(Operating cost) $21,570 Net proceeds from sale Net Cash Flow

($34,401)

Option 3 : Purchase a new machine after 1 year n Financial Data Depreciation Book value Market value Setup cost Operating cost Savings Cash Flow Statement +(1-0.25)*(Savings) +(.25)*(Depreciation) Investment -(1-0.25)*(Setup) -(1-0.25)*(Operating cost) Net Cash Flow Combined cash flow

($1,904)

$15,000 $12,821 $36,000

8 $17,900 $8,940 $0 $15,000 $13,455 $36,000

9 $8,940 $0 $0 $15,000 $14,171 $36,000

$27,000.00 $4,470.04

$27,000.00 $4,475.05

$27,000.00 $2,235.02

($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($11,250.00) ($7,762.50) ($8,066.25) ($8,400.00) ($8,768.25) ($9,173.25) ($9,615.75) ($10,091.25)

($11,250.00) ($10,628.25)

$200,450 $200,450

2 $28,644 $171,806

3 $49,090 $122,715

4 $35,059 $87,657

5 $25,036 $62,621

6 $17,900 $44,720

7 $17,880 $26,840

$15,000 $10,350 $36,000

$15,000 $10,755 $36,000

$15,000 $11,200 $36,000

$15,000 $11,691 $36,000

$15,000 $12,231 $36,000

$27,000.00 $7,161.08

$27,000.00 $12,272.55

$27,000.00 $8,764.68

$27,000.00 $6,259.05

$27,000.00 $4,475.05

($200,450.00)

($34,401)

($200,450) ($202,354)

PW (12%) =

($153,897)

Conclusion: Option 2 is the least cost option.

$15,149 $15,149

$19,956 $19,956

$16,115 $16,115

AEC(12%) =

$28,883

$13,241 $13,241

$11,052 $11,052

$10,604 $10,604

$10,134 $10,134

$7,357 $7,357

(b) Answer is not provided. (Note: Analysis similar to Part(a) except that all cash flows will be truncated over 5 years.)

Chapter 13 Understanding Financial Statements 13.1) (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital) 13.2) (7), (8), (1), (11), (3), (9) 13.3) (a) • • • •

Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 Working capital = $580,000 - $230,000 = $350,000 Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000

(b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15; Market price = $15 + $15 = $30 per share

13.4)

(a) Shareholder’s equity in 2011 = $700 - $510 = $190(M) Shareholder’s equity in 2012 = $900 - $640 = $260(M) (b) Net working capital in 2011 = $100 - $60 = $40(M) Net working capital in 2012 = $200 - $90 = $110(M) (c) The income taxes in year 2012 = ($2,350 - $1,130-$420-$210)*0.35 = $206.5(M) (d) $383.50 + $420=$803.50 (M) (Cash from Operating activities = Net income + Depreciation)

13.5)

(a) ROE (= Net income/Equity) ROA (= Net income + interest expense(1-tax rate) /average total assets)

Apple 26.03%

Oracle 22.29%

17.34%

12.59%

(b) Apple has performed better in terms of profitability.

(c) If two companies were merged, the impact on the results of ROE could be positive under the situation where the Apple leads the acquisition using stock swap instead of issuing new stocks for M&A cost. If Apple uses stock swap, the stock value wouldn’t be decreased in terms of scarcity.

13.6) Inventory turnover ratio (2011) = Sales/Average inventory balance = $3,776,395 / $(202,794 + 231,313)*0.5 = 17.4 times Inventory turnover ratio (2012) = 15.6 times This ratio shows how many times the inventory of a firm is sold and replaced over a specific period. From the data, Broadcom Corporation was holding more stocks of inventory than last year; having more inventories on stock is unproductive.

13.7) (b) 13.8) (b) 13.9) (d)

13.10) Given Olson’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets Net Income = $8 X Where X = the number of outstanding shares

• EPS =

• •

•

Total shareholders' equity = $80 X Retained earnings = Net income – Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X – 4X = $24 million, or X = 6 million shares Book value =

From the book value per share, we know that total shareholders’ equity = 80X, or $480 million; Total assets = Total liabilities + Total shareholders’ equity = $240 million + $480 million = $720 million

•

Debt ratio = $240 million/$720 million = 0.33

13.11) (a) Debt ratio (= Total debt/Total assets) = $19,483,000/$38,599,000 = 50.48% (b) Times-interest-earned ratio (=EBIT/Interest expense) = Not defined (c) Current ratio (= Current assets/Current liabilities) = 29,021,000/19,483,000 = 1.49 (d) Quick (acid test) ratio (= (Current assets-Inventories)/Current liabilities)) = (29,021,000-1,301,000)/19,483,000 = 1.42 (e) Inventory-turnover ratio(= Sales/Avg. inventory balance) = 61,494,000/((1,301,000+1,051,000)*0.5) = 52.29 (f) Days-sales-outstanding ratio (= Receivables/(Annual sales/365)) =10,136,000/(61,494,000/365) = 60.16 (g) Total-assets-turnover ratio (= Sales/Total assets) = 61,494,000/38,599,000 = 1.59 (h) Profit margin on sales (= Net income available to common stockholders/Sales) = 2,635,000/61,494,000 = 4.28% (i) Return on total assets (= (Net income + interest expense(1-tax rate))/Avg. total assets) = 2,635,000/((38,599,000 + 33,652,000)*0.5) = 7.29% (j) Return on common equity (= (Net income available to common stockholders)/Avg. common equity) = 2,635,000/((7,766,000 + 5,641,000)*0.5) = 39.31%

(k) Price/earnings ratio (= Price per share/Earnings per share) =13.47/(3,350,000/1,944,000) = 7.82 (l) Book value per share (= (Total stockholders’ equity-Preferred stock)/Shares outstanding) = 7,766,000/1,944,000 = $3.99 To make an informed analysis of Dell’s financial health, we need to calculate the various financial ratios of Dell’s competitors as well as the S&P 500. 13.12) Income Statement: A

$900,000

$585,000

$315,000

$270,000

$108,000

$162,000

Balance Sheet: 











$160,000

$120,000

$320,000

$600,000

$900,000

$1,500,000











$450,000

$700,000

$100,000

$700,000

$800,000

•

From Current ratio Total current assets = 2.4 × $250,000 = $600,000 ----------------------------------- ③ Plant and equipment, net = $1,500,000-$600,000=$900,000----------------------- 

•

From Quick ratio Inventory = $600,000 - (1.12 × $250,000) = $320,000 -----------------------------②

•

From Inventory Turnover Net Revenue = (($320,000 +$280,000)/2) × 6.0 =$1,800,000 Cost of goods sold = $1,800,000- $900,000= $900,000 ------- A

•

From DSO Accounts receivable = 24.3333 × ($1,800,000 ÷365) = $120,000 ------------------ ①

Cash = ③-(②+①) = $160,000 -----------------------------------------------------------ⓞ •

From interest expense of income statement Bond = $450,000 ----------------------------------- ⑥ 250,000 + ⑥ = $700,000 --------------------------⑦

•

From Debt to Equity ratio Total Equity ⑩ = $700,000 ÷ 0.875 =$ 800,000 ------------- ⑩ Total assets or Total liabilities and equity = ⑦ + ⑩ = $1,500,000 ------⑤

•

From Return on total assets Net income F = 14%× ($1,350,000) - ($45,000)(0.6)=$162,000

•

From F, D =F ÷ 0.6 = $270,000, E = D× (0.4) =$108,000 C = D+45,000 = $315,000 B=$900,000-C = $585,000

•

From EPS Stock Outstanding = F ÷ 4.05 = 40,000 shares Common stock = $2.50 × 40,000 = $100,000 -------------------------------⑧ Retained Earnings = ⑩ - ⑧ = $700,000 ------------------------------------

13.13) Select (b)

13.14) Not provided

13.15) (a) Working capital = Current assets – Current liabilities; Working capital requirements = Changes in current assets – Changes in current liabilities; WC req. = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000, indicating that additional financing is needed to fund the increase in current assets.

(b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000 (d) Net cash flow: • Operating activities = net income + depreciation – WC = $408,000 + $200,000 - $90,000 = $518,000 • Investing activities = equipment purchase = ($400,000) • Financing activities = borrowed fund = $200,000 • Net cash flow = $518,000 - $400,000 + $200,000 = $318,000

13.16) Not provided (Visit the websites and get the most recent financial statements available)

ERRATA Park, Fundamentals of Engineering Economics, 4th edition 4/20/18 Pg. 130, Problem 3.18: 6% → 9% | (c) → Eliminated 6/27/18 Pg. 93: Cash flow for Problem 2.69

Pg. 168, Problem 4.2, Part (a): average (geometric) price index → inflation rate Pg. 240, Figure 6.3 caption: Computing the required the annual savings to cover the capital and operating costs. → Eliminate second “the” Pg. 325, Problem 7.39, Lines 4-5: year 0 only one. → year 0 only. Pg. 404, Problem 9s.3, Line 2: 200% method → 200% DB method Pg. 450, Problem 10s.13, Line 5: year MACRS property → 7-year MACRS property Pg. 455, Continued from Problem 10.15 at top of page, Part (c): Question should read as “What is the net cash flow with the depreciation expense of $20,000?” Pg. 623, Problem 3s.4: 9.45% → 9.381% | $91,260 → $90,764 | $99,884 → $99,279 Pg. 634, Problem 6s.15: (c) → (b) Pg. 639, Problem 9s.5: $63,725 → $74,970 Pg. 649, Problem 11s.10: (d) → (c) | Insert Y = 2X – 11, E[Y] = 2E[X] – 11 = 49 above “Var[Y] =…” 11/14/18 Pg. 692, Equal Payment Series, Future Worth formula numerator: (1 + 𝑖)𝑁−1 → (1 + 𝑖)𝑁 − 1