Chapter 06anualid flujo caja

Chapter 6 Annual Equivalence Method Identifying Cash Inflows and Outflows

6.1 (10%) $5,000( / ,10%,6)

[$2,000( / ,10%,1)$2,500( / ,10%,6)]( / ,10%,6)$1,103.50

AE A PP F

P F A P

= −+ ++=

6.2 (12%) $20,000( / ,12%,6) $5,000

$3,000( / ,12%,5)( / ,12%,1)( / ,12%,6)$4,303.13

AE A PP G P F A P

= −−=

6.3 (10%) [ $3,000 $3,000( / ,10%,2)

$3,000( / ,10%,4)( / ,10%,2)$1,000( / ,10%,4)( / ,10%,2)]( / ,10%,6)$751.01

AE P AP A P FP G P F A P

= − −++=

6.4 (13%) [ $8,000 $2,000( / ,13%,6)

$1,000( / ,13%,6) $4,000( / ,13%,2)$2,000( / ,13%,4) $1,000( / ,13%,6)]( / ,13%,6)$934.92

AE P AP G P FP F P F A P

= − ++ −− −=

6.5 (8%) $5,000( / ,8%,6) $2,000

[($500 $1,000( / ,8%,2))( / ,8%,2)$500( / ,8%,5)]( / ,8%,6)$421.37

AE A PP A P F

P F A P

= − +− ++=

Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be

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2

6.6 (10%) $2,500( / ,10%,5) $400

$100( / ,10%,5)$78.47 (Reject)

(10%) $4,500( / ,10%,5) $500[$2,500( / ,10%,1) $1,500( / ,10%,2)$500( / ,10%,3)]( / ,10%,5)$338.57 (Accept)

(10%) [ $8,000 $2,

A

B

C

AE A PA G

AE A PP F P F

P F A P

AE

= − ++= −= − ++ ++== − − 000( / ,10%,1)

$2,000( / ,10%,5)]( / ,10%,5)$162.77 (Accept)

(10%) [ $12,000 $2,000( / ,10%,1)$4,000( / ,10%,5)]( / ,10%,5)

$1,868.31 (Accept)

D

P FP F A P

AE P FP F A P

+== − ++ +=

6.7 (12%) [ $500,000 $600,000( / ,12%,1) $400,000( / ,12%, 2)

$300,000( / ,12%,3) $200,000( / ,12%, 4)]( / ,12%, 4)$228,894

AE P F P FP F P F A P

= − + ++ +=

6.8 (13%) $7,500( / ,13%,3) $15,500( / ,13%,3)

$1,373.10 (Accept)(13%) $4,000( / ,13%,3) $1,500

$300( / ,13%,3)$81.53 (Accept)

(13%) $5,000( / ,13%,3) $4,000$1,000( / ,13%,3)$963.62 (Accept)

A

B

C

AE A P A F

AE A PA G

AE A PA G

= − +== − ++== − +−=

(13%) $6,600( / ,13%,3) $3,800$1,004.70 (Accept)

DAE A P= − +=

6.9 Since the project has the same cash flow cycle during the project life, you just can consider the first cycle.




3

(10%) [ $800 $900( / ,10%,1) $700( / ,10%,2)$500( / ,10%,3)]( / ,10%,3)$390.98

AE P F P FP F A P

= − − ++=

∴ Accept the project.

6.10 $10,000( ) $10,000( / ,8%,10) ( / ,8%,10)

0.06$77,199 $67,101$144,300

CE i P A P F= +

= +=

∴ The amount of additional funds should be $44,300.

Capital (Recovery) Cost / Annual Equivalent Cost

6.11 Given: years,$55,000, $6,000, 10I S N= = = 12%i = (a)

(12%) ($55,000 $6,000)( / ,12%,10)$6,000(0.12)$9,392

CR A P= −+=

(b) Operating Revenue(12%) $5,000 $2,500( / ,12%,10)

$13,962

AE A G= +

=

(c) (12%) $13,962 $9,392

$4,570AE = −

=

∴ This is a good investment.

6.12

O&M

(12%) ($45,000 $9,000)( / ,12%, 4) (.12)($9,000)$11,067

(12%) $15,000 $2,000( / ,12%,5)$18,549

(12%) $11,067 $18,549$29,616

CR A P

AE A G

AEC

= − +== +

== +=




4

6.13 Given: years,$250,000, $40,000, 5I S= = N = 18%i =

(18%) ($250,000 $40,000)( / ,18%,5)

$40,000(0.18)$74,353

CR A P= −+=

6.14

n Option 1 Option 2 0 -$600-$5,760 1 0 -$1,560 2 0 -$1,680 3 0 -$1,800 4 +$100 -$1,920

Option 1

Option 2

(8%) $6,360( / ,8%,4) $100( / ,8%,4)

$1,897.89(8%) $1,560 $120( / ,8%,4)

$1,728.36

AEC A P A F

AEC A G

= −

== +

=

∴ Select Option 2.

6.15 Given compounded annually, N =12 years, the effective monthly rate is 6%i =

120.06 (1 ) 10.487% per month

ii= + −=

• Conventional System:

Conventional(6%) $471 $576 $1,047AEC = + = The equivalent monthly cost = $1,047( / ,0.487%,12) $84.94A F =

• IONETIC System:

IONETICS(6%) $185 $1,200( / ,6%,12) $323.13AEC A P= + = The equivalent monthly cost = $328.13( / ,0.487%,12) $26.62A F =




5

6.16 (a)

(13%) $4,000( / ,13%, 4) $1,000( $1,000)( / ,13%,2)( / ,13%, 4)0

(13%) $608.06 0.26328 0$2,309.55

AE A PX P F A P

AE XX

= − ++ −== − + ==

(b)

(13%) $5,500( / ,15%,4) $1,400$526.46 0

AE A P= −= >

∴Accept project B.

6.17 • Option 1: Purchase-annual installment option:

1

$45,000( / ,7%,5) $10,975(10%) $5,000( / ,10%,5) $10,975

$12, 294

A A PAEC A P

= == +=

• Option 2: Cash payment option:

2(10%) $46,000( / ,10%,5)$12,135

AEC A P==

∴ Option 2 is a better choice.

6.18 The total investment consists of the sum of the initial equipment cost and the installation cost, which is . Let$135,000 R denote the break-even annual revenue.

(10%) $135,000( / ,10%,10) $30,000

$5,000 $10,0000

AE A PR

= − −− + +=

Solving for R yields

$46,971R =




6

6.19 • Capital cost:

(15%) ($105,000 $6,000)( / ,15%,5) $6,000(0.15)$30, 432

CR A P= − +=

• Annual operating costs: $4 5,000

(15%) $30, 432 $45,000 $75, 432AEC = + =

Unit-Cost Profit Calculation

6.20 • Capital recovery cost

(10%) ($20,000 $10,000)( / ,10%, 2) $10,000(0.10)$6,792

CR A P= − +=

•

Revenue 2

Net Savings

$30,000 $40,000(10%) [ ]( / ,10%,2)1.1 1.1

$34,762(10%) =$34,762 $6,792

$27,970

AE A P

AE

= +

=−

=

•

Savings 2

(6,000) (8,000)(10%) [ ]( / ,10%,2)1.1 1.1

$6,952$27,970 6,952

$4.02 per hour

C CAE A P

CC

C

= +

===

6.21 Given data: Total cost of building: $110 16 20 $35, 200× × = , Salvage value = $3,520, Annual taxes, insurance, maintenance = $2,112, Other operating cost = $1,600, Number of engineer assigned = 3

• Equivalent annual cost of operating the new building:




7

(12%) ($35, 200 $3,520)( / ,12%, 25) (0.12)($3,520)$2,112 $1,600$8,173.6

AEC A P= − ++ +=

• Required annual increase in productivity per engineer:

$8,173.6 / 3 $2,724.5 per engineer=

6.22

• Equivalent annual cost of owning and operating the vehicle:

(6%) [$4,680( / ,6%,1) $3,624( / ,6%, 2)$3, 421( / ,6%,3)]( / ,6%,3)$3,933

AEC P F P FP F A P

= ++=

• Annualized cost of owning and operating the vehicle as a function of

mileage.

2 3

(14,500) (13,000) (11,500)(6%) [ ]( / ,6%,3)1.06 1.06 1.06

13,058

C C CAEC A P

C

= + +

=

• So,

13,058 $3,933$0.3012 per mile

CC==

6.23 Let T denote the total operating hours in full load.

• Motor I (Expensive)

Annual power cost: 150 (0.746) (0.05) $6.7410.83

T T× × × =

Equivalent annual cost of operating the motor:

(6%) $4,500( / ,6%,10) $675 6.741$1, 286.41 $6.741

IAEC A P TT

= += +

+




8

• Motor II (Less expensive):

Annual power cost: 150 (0.746) (0.05) $6.99380.80

T T× × × =

Equivalent annual cost of operating the motor:

(6%) $3,600( / ,6%,10) $540 6.9938$1,026.11 6.9938

IIAEC A P TT

= += +

+

II

• Let and solve forT . (6%) (6%)IAEC AEC=

$1, 286.41 6.741 $1,029.11 6.9938T T+ = +

1,017.8T = hours per year

6.24 • Option 1: Purchase units from John Holland

Unit cost $25 ($35,000) / 20,000 $23.25= − =

• Option 2: Make units in house

dm 1

dl 1

vo 1

(15%) $63,000( / ,5%,15%,5) $230,241(15%) $190,800( / ,6%,15%,5) $709, 491(15%) $139,050( / ,3%,15%,5) $490,888

(15%) ($230,241 $709,491 $490,888)( / ,15%,5) $70,000$496,776

PW P APW P APW P AAEC A P

= == =

= == + + +=

Unit cost $496,776 / 20,000= $24.84=


6.25 Determine the unit profit of air sample test by the TEM (in-house).

• Sub-contract Option:

Unit profit

(a)

$400 $300 $0.50 $1,500 /1,000 $98= − − − =




9

• TEM Purchase Option:

Unit cost

(b) Let X denote the break-even number of air samples per year.

(15%) ($415,000 $9,500)( / ,15%,8) ($50,000

$6,000 $18,000 $20,000)$188,600

AEC A P= + ++ + +=

$188,600 /1,000 $188.60= = Unit profit $300 $188.60 $111.40= − =

$1,500 $188,600$400 ($300 $0.50 ) $300

X X− + + = −

Solving for X yields

air samples per year

6.26 Option 1: Pay employee $0.38 per mile (Annual cost: $8360)

933.17 934X = ≈

• • Option 2: Provide a car to employee:

operating cost

total cost

(10%) ($25,000 $8,000)( / ,10%,3) (0.10)($8,000)$7,636

(10%) $900 ($0.22)(22,000) $5,740

(10%) $7,636 $5,740 $13,376Operating cost per mile $13,376 / 22,000 $0.61

CR A P

AEC

AEC

= − +== + =

= + == =


6.27

Capital costs:

• Annual battery replacement cost:

•

(7%) ($25,000 $2,000)( / ,7%,12) (0.07)($2,000)$3,036

CR A P= − +=

(7%) [$3,000( / ,7%,3) ( / ,7%,6)

( / ,7%,9)]( / ,7%,12)$765.41

AEC P F P FP F A P

= ++=




10

• Annual recharging cost:

(7%) ($0.015)(20,000) $300AEC = = • Total annual costs:

(7%) $3,036 $765.41 $300 $700$4,801.41

AEC = + + +=

• Costs per mile:

cost/mile $4,801.41/ 20,000 $0.2401= =

6.28• Minimum operating hours:

($2,000) $500$4,153.65

AEC A P

(9%) ($30,000 $2,000)( / ,9%,15)(0.09)

= −+

=

Let denote the annual operating hours. Then the total kilowatt-hours

generated would be . Since the value of the energy generated is considered to be per kilowatt-hour, the annual energy cost is

+

T40T

$0.08

$0.08 40 $4,153.65T× = Solving forT yiel

1, 298T ds

= hours •

5 $3,846.35=

Annual worth of the generator at full load operation:

(9%) ($0.08AE = −)(100,000) $4,153.6

Discounted payback period at full load of operation: •

n Investment Revenue Maintenance Net cost Cash flow

0 -$30,000 -$30,000 1 500 $8,000 -$500 7,

15 +$2,000 8,000 -500 9,500 $30,000 $7,500( / ,9%, )n P A=

Solving for n yields

5.185n = years




11

6.29 • Capital recovery cost:

,000)( / ,6%,12) (0.06)($3,000)$17,714

CR A P= − +=

Annual operating costs:

(6%) ($150,000 $3

•

O&M(6%) $50,000 $10,000 $3,000$63,000

AEC = + +

=

• Total annual system costs:

(6%) $17,714 $63,000 $80,714AEC = + =

• Number of rides required per year:

Number of rides rides

6.30 ent cost

$80,714 /($0.10) 807,140= =

Given: Investm $7= million, plant capacity 200,000= 1bs/hour, plant oper hours per year, O&M costating hours 3,60= $4= m0 illion per year, useful lif years, salvage valuee 15= =$700,000, and MARR = 15%.

/2

0

P A

=

Solving fo

(a) (15%) 0,000 ( $4,000,000)( ,15%,6)

3.7844 $ 2,137,900PW R

R= − + −= −

$7,00

r R yields

per year

(b) Mi es per 1b (after-tax):

$5,849,700R =

nimum proc sing fee

$5,849,700 $0.0081 per(200,000)(3,600)

= 1b

Comments: The minimum processing fee per 1b should be higher before-tax basis.




12

6.31 Given: Investment million, plant capacity$3= 0.4= acre, useful plant life years, salvage value20= = negligible, O&M cost $250,000= per year,

nthly r

$602,379C

= Monthly water bill for each household:

MARR 10%= compounded annually (or effective mo ate of 0.7974%)

) $3,000,000( / ,10%, 20) $250,000AE A P= +

(10%

$602,379( / ,0.7974%,12) $A F

= 162.83295

6.32 • Annual total operating hours:

per year

• The amount of electrici

• Equivalent annual cost:

A P +=

• Cost per kilowatt-hour:

(0.70)(8,760) 6,132= hours

ty generated annually:

50,000 6,132 306,600,000× = kilowatt-hours

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

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

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

ap

A P

• C ital costs:

(1CR 5%) ($12,000,000 $2,000,000)( / ,15%,15)(0.15)($2,000,000)$2,010,171

= −+=

• Annual crew costs: $225,000

s:

• Annual fuel costs for round trip




13

($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 $403,500 $75X X+ + = +

• Total equivalent annual costs:

XX

++ + +=

Solving fo

(15%) $2,010,171 $2AEC = + 25,000 $1,125,696$78,000 $403,500 $75$3, 400

r X yields

1,156X = Passengers per year

Or

1,156 /(52)(3) 7.41 8= ≈ passengers per round trip

omparing Mutually Exclusive Alternatives by the AE Method

(a)

(b) Process A:

C

6.34

(15%) $12,000( / ,15%, 4) [$9,120 $2, 280( / ,15%, 4)]$1,892.95

(15%) $12,000( / ,15%, 4) $6,350$2,146.82

A

B

AE A P A G

AE A P

= − + −== − +=

$1,892.95 / 2,000 $0.9465 /= hour Process B: $2,146.82 / 2,000 $1.0734 /= hour

a better choice.

6.35 • Capital recovery cost for both motors:

(c) Process B is




14

CV(13%) $13,000( / ,1CR A P

PE

3%,20) $1,851(13%) $15,600( / ,13%,20) $2,221CR A P

= == =

• Annual operating cost for both motors:

18.650 kW $0.07 3,120 hrsCV : $4,551

0.895 kWh year18.650 kW $0.07 3,120 hrsPE : $4,380

0.93 kWh year

× × =

× × =

(a) Savings per kWh:

Comments: At 3,120 annual operating hours, it will cost the company an

CV

PE

(13%) $1,851 $4,551 $6,402(13%) $2,221 $4,380 $6,601

AECAEC

= + == + =

additional $370, but energy savings are only $171, which results in a $199loss from each motor. The total output power is 58,188kWh per year (25 HP × 0.746 kW/HP × 3120 hrs/year). Therefore, the savings (losses) per operating hour from switching from conventional motor to the PE is

($199 / yr) ($0.034) / kWh58,188 kWh/yr

=

(b)

6.36 New lighting system cost:

+

• Old lighting system cost:

nnual savings from installing the new lighting system

$1,851 1.4587 $2, 221 1.40380.0549 370

6,737 hours

T TTT

+ = +==

•

(12%) $50,000( / ,12%, 20) ($8,000 $3,000)$17,694

AEC A P= +=

(12%) $20,000AEC =

A $2,306=




15

6.37 Given: interest compounded monthly, the effective annual

=

• Option 1: Buying a bond

• Option 2: Buying and holding a growth stock for 3 years

• Option 3: Receiving $150 interest per year for 3 years

Making the personal loan is the best option.

6.38 Equivalent annual cost:

• Processing cost per ton:

per ton

∴ Incinerator B is a better choice.

6%i =interest 125) 1 6.17%− = per year, effective semiannual interest per semiannual

(1.006(1.005) 1 3.04%= − =

1(3.04%) $2,000( / ,3.04%,6) $100 $2,000( / ,3.04%,6)

$39.20 per semiannual(6.17%) $39.20( / ,3.04%,2)

$79.59 per year

AE A P A F

AE F A

= − + +===

2(6.17%) $2,000( / ,6.17%,3) $2,735.26( / ,6.17%,3)

$107.17AE A P A F= − +

=

3(6.17%) $2,000( / ,6.17%,3) $150 $2,000( / ,6.17%,3)

$126.60AE A P A F= − + +

=

∴

•

(13%) ($1, 200,000 $60,000)( / ,13%, 20)(0.13)($60,000) $50,000 $40,000$260,083

(13%) ($750,000 $30,000)( / ,13%,10)(0.13)($30,000) $80,000 $30,000$216,395

A

B

AEC A P

AEC A P

= −+ + +== −+ + +=

$260,083/(20)(365) $35.63AC = =$216,395 /(20)(365) $29.64BC = = per ton




16

6.39

∴ Project A is a better choice.

(b)

P A A P= − − +

=

∴ Project C is a better choice.

ife-Cycle Cost Analysis

6.40 Assumption: jet fuel cost = $2.10/gallon

• System A : Equivalent annual fuel cost: A1 = ($2.10/gal)(40,000gals/1,000

(a) (15%) [ $2,500 $1,000( / ,15%,1)

$400( / ,15%,4)]( / ,15%,4)$216.06

AAE P FP F A P

= − ++ +

=

(15%) [ $4,000 $100( / ,15%,1)]( / ,15%,4) $1,500$129.40

BAE P F A P= − + +=

(15%) $129.40BAE =

(15%) [ $5,000 $200( / ,15%,2)]( / ,15%,4) $2,000$134.79

CAE

L

hours)(2,000 hours) $168,000= (assuming an end of-year convention)

• System B : Equivalent annual fuel cost: A1 = ($2.10/gal)(32,000gals/1,000

fuel 1(10%) [$168,000( / ,6%,10%,3)]( / ,10%,3)$177,623

(10%) ($100,000 $10,000)( / ,10%,3)(0.10)($10,000) $177,623$$214,813

A

AEC P A A P

AEC A P

=

== −+ +=

hours)(2,000 hours = $134,400

fuel 1(10%) [$134,400( / ,6%,10%,3)]( / ,10%,3)$142,099

(10%) ($200,000 $20,000)( / ,10%,3)(0.10)($20,000) $142,099$216,480

B

AEC P A A P

AEC A P

=

== −+ +=




17

• Cost of owning and operating:

ystem per hour

System per hour

∴ System A is a better choice.

6.41 Since the required service period is 12 year uture replacement cost

for ck remains unchanged, we can easily determine the equivalent annual cost over a 12-year period by simply finding the annual equivalent cost of the first r

• Truck A: Four replacements are required

S $214,813 / 2,000 $107.41A = =$216, 480 / 2,000 $108.24B = =

s and the feach tru

eplacement cycle for each truck.

(12%) ($15,000 $5,000)( / ,12%,3)

(0.12)($5,000) $3,000$7,763.50

AAEC A P= −+ +=

Truck B: Three replacements are required •

(12%) ($20,000 $8,000)( / ,12%,4)

(0.12)($8,000) $2,000$6,910.80

BAEC A P= −+ +=

∴ Truck B is a more economical choice.

6.42

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

Alternative Description

A1 Buy Machine A and use it for 4 years. Then lease the machine for one year.

A2 Buy Machine B and use it for 5 years.

A3 Lease a machine for 5 years.




18

A4 Then buy another Machine A and use it Buy Machine A and use it for 4 years.

for just one year.

A5 ear. Buy Machine A and use it for 4 years. Then buy Machine B and use it for one y

Both A4 and A5 are feasible but may not be practical alternatives. To consider these alternativmach after on

(b) Wi

• F

$100( / ,10%,2) ($3,000 $100)( / ,10%,5)$10,976

(10%) $10,976( / ,10%,5)$2,895.44

PW P F

P F P F

AEC A P

= − + −

− − += −==

• For A2:

$10,042( / ,10%,5)$2,649

PW P FP A P F

A P

= − +− −= −==

• For A3:

A2 is a better choice.

es, we need to know the salvage values of the e-year use. ines

th lease, the O&M costs will be paid by the leasing company:

or A1:

1

1

(10%) $6,500 ($600 $100)( / ,10%, 4)$800( / ,10%, 4) $200( / ,10%,3)P A P F− −

2

2

(10%) $8,500 $1,000( / ,10%,5)$520( / ,10%,5) $280( / ,10%,4)

$10,042(10%)AEC

3(10%) [$3,000 $3,000( / ,10%, 4)]( / ,10%,5)$3,300

AEC P A A P= +

=

∴




19

6.43 • Option 1:

A PA F−

+ +===

• Option 2:

6.44 Given: Required service period

1(18%) $200,000AEC = (180)( / ,18%,20)$(0.08)(200,000)(180)( / ,18%,20)($0.005 0.215)(180,000,000)$46,305,878

cost/lb $46,305,878 /180,000,000$0.2573 per 1b

2(18%) ($0.05 $0.215)(180,000,000)$47,700,000

cost/lb $47,700,000 /180,000,000$0.2650 per 1b

AEC = +===


= indefinite, analysis period= indefinite

Plan A: Incremental investment strategy: • Ca

AEC P F A P= + ∞

P FA P

+× ∞=

ote that the effective interest rate for 15-year period is

pital investment :

1(10%) [$400,000 $400,000( / ,10%,15)]( / ,10%, )$49,576=

• Supporting equipment:

2(10%) [($75AEC = ,000 $75,000 / 3.1772)( / ,10%,30)]( / ,10%, )$565

N

15(1 0.1) 1 3.1772+ − =




20

• Operating cost:

3(10%) [$31,000( / ,10%,15)$62,000( / ,10%,5)( / ,10%,15)]

,000 $1,000( / ,10%, )]0.10/ ,10%, 20)]( / ,10%, )

$40,056

P AP A P F

P G

A P

=+

+ + ∞

× ∞=

Note that or

$63[

(P F

AEC

2( / , , ) 1/P G i i∞ = ( / ,10%, ) 100P G ∞ =

• Total equivalent annual worth:

= $90,197 (10%) $49,576 $565 $40,056AAEC = + +

Plan B: One time investment strategy: • Capit nt:

A P ∞=

•

al investme

1(10%) $550,000(AEC = / ,10%, )$55,000

Supporting equipment:

2(10%)AEC =$150,000 ( / ,10%, )16.4494

$912

A P ∞

=

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

.1) 1 16.4494− =

• Operating cost:

$39,788

AEC P A=

=

annual worth:

30(1 0+

3(10%) [$35,000( / ,10%,15)$55,000( / ,10%, )( / ,10%,15)]( / ,10%, )

P A P FA P

+ ∞× ∞

• Total equivalent




21

(10%) $55,000 $912 $39,788BAEC = + + = $95,700 ∴ Plan A is a better choice.

Minim m

6.45 (a)

• Energy loss in kilowatt-hour:

u Cost Analysis

6.516 $4,709.11(24 365)($0.0825)A A

× =

• Material weight in pounds:

3(200)(12)(555)12

(770.83)A A=

• Total material costs:

(770.83) ($6) $4,625A A=

• Capital recovery cost:

Total equivalent annual cost:

(11%) [$4,625 $1 770.83 ]( / ,11%, 25) $1 770.83 0.11542.44

CR A A A P AA

= − × + × ×=

•

4,709.11(AEC 11%) 542.44AA

= +

• Optimal cross-sectional area:

2

2

(11%) 4,709.11542.44 0

2.9464 inches

dAECdA A

A

= − =

=

(b) inimum annual equivalent total cost:

M

4,709.11(11%) 542.44(2.9464) $3,196.512.9464

AEC = + =




22

(c) Graph is not shown

Short

ST 6.1 This case problem appears to be a trivial decision problem as one alternative (laser blanking method) dominates the other (conventional method). The problem of this nature (from an engineer’s point of view) involves more

s than comparing the accounting data. We will first under each production method. Since all operating costs

are already given in dollars per part, we need to convert the capital expenditure apital recovery cost per unit.

• Conventional method:

Case Studies

strategic planning issuecalculate the unit cost

into the required c

(16%) $106, 480( / ,16%,10)CR A P=

$22,031= per year

Unit capital recovery cost $22,0313,000

=

$7.34= per year • Laser blanking method:

per year

Unit capital recovery cos

(16%) $83,000( / ,16%,10)CR A P=

$17,173=

t $1=

7,1733,000

$5.72= per part

Blanking Method Description Conventional Laser

Steel cost/part 14.98 $8.19 $Transportation cost/part $0.67 $0.42 Blanking cost/part $0.50 $0.40 Capi 7.34 $5.72 tal cost/part $Total unit cost $23.49 $14.73

It appears that the window frawould sav 76 f d. If For ides to make the wi e, the part e betw and $23.49, depending upon the blanking method ad If Ford rel an outside su work should this cost ra Ford were producing the window frames by the conven method, the die investment had alr In this case, one of the tant issues is to address if it is

me production by the laser blanking technique or each part prode about $8.

ame in housuce

cost would rand deceen $1ndow fr g

opted. 4.73

ies on pplier, the subcontracting be in nge. If

tional eady been made. impor




23

worth switching to the laser blanking this time or later. If Ford decides to go with

ST

ST

the laser blanking, it will take 6 months to reach the required production volume. What option should Ford exercise to meet the production need during this start-up period?

6.2 Not given

6.3 Given: annual energy requirement 145,000,000,000= BTUs,1-metric ton 2, 204.6= l bs ( an approximation figure of 2,000 lbs was mentioned in the case

problem), net proceeds from demo

nu s for each alternative: lterna

Weight of dry coal

lishing the old boiler unit $1,000=

(a) An al fuel cost• A tive 1:

145,000,000,000 BTUs(0.75)(14,300)

13,51 14

=

9,8= lbs 13,519,814

= 2, 204.6

6,132.45= tons

Annual fuel cost 6,132.45 $55.5= × = $340,351

• Alternative 2:

Gas cost 145,000,000,000(0.94)$9.5(0.78)(1,000,000)

⎛ ⎞= ⎜ ⎟

⎝ ⎠

= $1,660,064

Oil cost 145,000,000,0$1.35⎛= ⎜00(0.06)

(0.81)(139,400)⎞⎟

⎝ ⎠

= $104,016

Annual fuel cost = $1,660,064 + $104,016

(b) Unit cost per steam pound:

• Alternative 1: Assuming a zero salvage value of the investment

= $1,764,080




24

(10%) ($1,770,300 $1AEC 00,000$1,000)( / ,10%, 20)

= +

$340,351A P−

+$644,571=

$644,571 Unit cost $0.0044 per steam lb

145,000,000= =

• Alternative 2:

(10%) ($889, 200 $1,000)( / ,10%, 20)$1,764,080$1,868, 408

AEC A P= −+=

$1,868,408 Unit cost $0.0129 per steam lb145,000,000

= =

(c) Select alternative 1.

ST 6.4 Assuming that the cost of your drainage pipe has experienced a 4% annual inflation rate, I could estimate the cost of the pipe 20 years ago as follow:

( / , 4%,20) $4,208(1.04) $1,920.48P F −= =

If the pipe had a 50-year service life with a zero salvage value when it was

wo year. In other words, the owner could invest his $1,920.48 at 5% annual interest, if he

did not purchase the pipe.)

the cost of money. With only a 20-year’s usage, he still has 30 more years to go. So, the unrecovered investment at the current point is

20$4,208

placed in service 20 years ago, the annual capital recovery cost to the owner

uld be as follows: (I assumed the owner’s interest rate would be 5% per

(5%) $1,920, 48( / ,5%,50)CR A P= $105.20= per year

You can view this number as the annual amount he would expect to recover from his investment considering

$105.20( / ,5%,30) $1,617.15.P A =

The owner could claim this number, but the city’s interest rate could be different from the owner’s, so there is some room for negotiation.

Assumed interest rate Claim cost




25

0% $1,152.243% $1,462.984% $1,545.895% $1,617.15




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