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Contemporary Engineering Economics, Fifth Edition, by Chan S. Park.ISBN: 0-13-611848-8 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Chapter 6 Annual Equivalence Method Identifying Cash Inflows and Outflows 6.1 AE(10%) = −$10,000( A / P,10%,6) +[$3,000( P / F,10%,1) + L +$4,500(P / F,10%,6)]( A / P,10%,6) = $1,085.10 6.2 AE(12%) = $20,000( A / P,12%,6) − $5,000 −$3,000(P / G,12%,5)( P / F,12%,1)( A / P,12%,6) = −$4,303.13 6.3 AE(10%) = [$100(P / F,10%,1) + $150( P / F,10%,2) + ⋅ ⋅ ⋅ +$200( P / F,10%,6)]( A / P,10%,6) = $161.01 6.4 AE (8%) = [−$3,000 − $3,000( P / A,8%,2) +$3,000( P / A,8%,4)( P / F ,8%,2) +$1,000( P / G ,8%,4)( P / F ,8%,2)]( A / P,8%,6) = $898.95 6.5 AE (12%) = [−$8, 000 + $2, 000( P / A,12%, 6) +$1, 000( P / G,12%, 6) − $4, 000( P / F ,12%, 2) −$4, 000( P / F ,12%, 4) − $1, 000( P / F ,12%, 6)]( A / P,12%, 6) = $709.13 Page | 1 Contemporary Engineering Economics, Fifth Edition, by Chan S. Park. ISBN: 0-13-611848-8 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 6.6 AE (10%) = −$5, 000( A / P,10%, 6) + $2, 000 −[($500 + $1, 000( P / A,10%, 2))( P / F ,10%, 2) +$500( P / F ,10%,5)]( A / P,10%, 6) = $356.47 6.7 AE (10%) A = −$1,500( A / P,10%,5) + $400 +$100( A / G,10%,5) = $185.32 (Accept) AE (10%) B = −$3,500( A / P,10%,5) + $500 +[$2,500( P / F ,10%,1) + $1,500( P / F ,10%, 2) +$500( P / F ,10%,3)]( A / P,10%,5) = $602.37 (Accept) AE (10%)C = [−$6, 000 − $3, 000( P / F ,10%,1) + $2, 000( P / F ,10%,5)]( A / P,10%,5) = $450.55 (Accept) AE (10%) D = −$15, 000( A / P,10%,5) + $2, 000 +$2, 000( A / G,10%,5) = $1, 663.29 (Accept) 6.8 AE (10%) = [−$600, 000 + $400, 000( P / F ,10%,1) + $300, 000( P / F ,10%, 2) +$200, 000( P / F ,10%,3) + $100, 000( P / F ,10%, 4)]( A / P,10%, 4) = $72, 600.73 Page | 2 10%.12%.12%.3) + $16. 634. or transmission in any form or by any means.. by Chan S. 266.197 = $426.04 = $147.76 (Accept) AE (12%) B = −$3.500( P / F . 000( P / A.12%. 202 + $279.1S S = $3.800( A / P.3) + $1.3) = $2.3) + $3. 2) +$500( P / F . 000 ( P / F .10) + 0. 6. Upper Saddle River.10 Since the project has the same cash flow cycle during the project life. N = 10 years. AE (10%) = [−$2. All rights reserved. recording.99 (Accept) 6.12%. i = 9% Page | 3 .10%. 200( A / P. 6..10%.500( A / F .18 (Accept) AE (12%)C = −$2.12%. 6%.12%. storage in a retrieval system. 006. mechanical.3) + $3.399.3) = $909. Upper Saddle River.11 PW (i ) = $20. photocopying.10%. Fifth Edition.300( A / P.12 3. Inc. or likewise.3)]( A / P. you just can consider the first cycle.399 ∴ The amount of additional funds should be $226. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.3) = $127. S = $5. 000. 000 − $1.10%.13 Given: I = $65. 6%. ISBN: 0-13-611848-8 © 2011 Pearson Education. NJ 07458. electronic.800 = $1.61 (Accept) AE (12%) D = −$5.3) = $339. Capital (Recovery) Cost / Annual Equivalent Cost 6.79 ∴ Accept the project. 000. 000 − S )( A / P. 200 +$600( A / G. Inc. For information regarding permission(s). 000( A / G.35 6.10) + $20. Park.1) + $700( P / F .Contemporary Engineering Economics. write to: Rights and Permissions Department. NJ. 400( A / P. Pearson Education.880 = (25.10) 0.12%. 000 −$1.9 AE (12%) A = −$6. 16 Given: I = $350. 495 − $9. 000 − $5.000 − $12.549. electronic.62 AE(12%)cost = [$18.000( A / G. 000. 000)( A / P. S = $60.368. or likewise.. recording. mechanical. 6) + (0.549.15) = $95.500 = $23.500( A / G.917. photocopying. NJ. by Chan S. storage in a retrieval system. 000(0. i = 15% CR (15%) = ($350.680 2 0 -$2.10) = $27. Upper Saddle River. 000)( A / P.12%.5) = $21.14 6. 799 = $17. 495 (c) AE (12%) = $27.12%.000)( A / P..240 1 0 -$2. (a) AE (9%)1 = ($65. 000) + $13.5) +$60. 696 ∴ This is a good investment. 000(0.9%.12)($12.920 4 +$100 -$3. All rights reserved.000) = $13. 799 (b) AE (9%) 2 = $18.2 AEC (12%) = $13. ISBN: 0-13-611848-8 © 2011 Pearson Education.000 + $2. 000 + $2.040 Option 2 (Buy KS every year) Page | 4 .09) = $9. Inc. Park.06)($23.800 3 0 -$2. 421. write to: Rights and Permissions Department.17 n Option 1 (Buy storage tank) 0 -$600-$10. 6%.512 6. 000 − $23.62 + $21. Fifth Edition.368.33 CR(12%) = ($55.Contemporary Engineering Economics. Upper Saddle River.15%.9%. or transmission in any form or by any means. NJ 07458. 000.4) + (0. N = 5 years.82 6. 6.2 = $34.10) +$5. For information regarding permission(s).15 AE (6%) = ($65. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. 000 − $60. 000)( A / P. Inc. Pearson Education. 26468 X = 0 X = $3. 4) − $1.10%.02 > 0 ∴Accept project B.13per year 6. 000 + ( A / F .200( A / P. 4) 0.851. 6. ISBN: 0-13-611848-8 © 2011 Pearson Education. 711 = −$18. mechanical. 4) = −$2. 2)( A / P. 4) AE (6%)Option 2 = −$3. Inc.. For information regarding permission(s). photocopying. 680 − $120( A / G.500( A / P.04 (b) AE (12%) = $6. or transmission in any form or by any means.. 4) + $100( A / F .12) = $328. 6. by Chan S.26 ∴ Select Option 2. 6%. N=12 years.10%. storage in a retrieval system.21 + 0. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. 000 + $10. 400 = $740. or likewise.500( A / P.18 Given i = 6% compounded annually.105. recording. 6%. Fifth Edition.12%. 4) + $1.19 CE (10%) = $100. 000 + ( X − $1. 000( A / P.21 • Option 1: Purchase-annual installment option: A = $65. NJ. 000( A / P.20 (a) AE (15%) = −$4. 6%.1 = $243. Park.1 0.Contemporary Engineering Economics. NJ 07458. electronic.5) = $16.15%. AE (6%)Option 1 = −$10. write to: Rights and Permissions Department.176.15%.5) − $16.6%.15%. Upper Saddle River. 4) =0 AE (15%) = −$841. All rights reserved. Inc. Upper Saddle River.094.9%. 030 • Option 2: Cash payment option: Page | 5 . 000 $20. Pearson Education. 711 AE (10%)1 = −$5.47 = −$2.840( A / P. 000)( P / F . • IONETIC System: AE(6%) IONETICS = $185 + $1.2 6. 12%. 000(0. 000 $42.15%.12 1. 000)( A / P.302 Net annual savings =$38..12%.24 • Capital recovery cost CR(12%) = ($30. or transmission in any form or by any means. For information regarding permission(s). 000 −$5.12) = $9. Park. 2) + 1. AE (10%)2 = −$66.15) = $39. Pearson Education. All rights reserved. Inc. or likewise. Upper Saddle River. Inc. 042 AEsavings (12%) = [ Page | 6 . AE (12%) = −$145. 000 + R =0 Solving for R yields R = $70.Contemporary Engineering Economics. 000 + $10. 6. 000)( A / P. 260 • $35. 000 ∴ Equivalent annual cost AE (15%) = $39.5) + $6. 000 = $114. 000(0.383 • Annual operating costs: $75. 000( A / P. 260 = $29. 2) + $18.22 The total investment consists of the sum of the initial equipment cost and the installation cost.5) = −$17.122 = $38.. mechanical.10%.000. by Chan S. Upper Saddle River. NJ 07458.383 Unit-Cost Profit Calculation 6. 000 ]( A / P. 411 ∴ Option 2 is a better choice.10) − $50. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.12%. Fifth Edition. Let R denote the break-even annual revenue. storage in a retrieval system. photocopying. which is $145. 000 − $18.23 • Capital recovery cost: CR(15%) = ($135. recording. 000 − $6. 663 6. electronic.383 + $75.302 − $9. NJ. ISBN: 0-13-611848-8 © 2011 Pearson Education. 000( A / P. write to: Rights and Permissions Department. 095.173. 600 = $8.112.12%. 2) + $1.800)( A / P. 000) + ]( A / P.3) = $1. 000 − $4.26 Given data: Total cost of building: $110 × 16 × 20 = $35.122 = $6. 2) 1.25 AE (14%) = [−$100.Contemporary Engineering Economics.365 / hr 6.600.6 / 3 = $2.13 C = $4. 095. Upper Saddle River.18 per hour AEhours (12%) = [ 6.07) = $2. 698.000 − ($2.879 + $1. All rights reserved. or likewise.776 + $1. storage in a retrieval system. 042 = $6. Other operating cost: $1.3)]( A / P. −3%.5 per engineer 6.. 7%. Number of engineer assigned: 3 • Equivalent annual cost of operating the new building: AE (12%) = ($35. ISBN: 0-13-611848-8 © 2011 Pearson Education.5) = $4. Upper Saddle River. 000 + $35.848( P / F . electronic.3) + $4.173. • C (6.97 Page | 7 .724.3C $29.112 + $1.520) + $2. 200 − $3. NJ. recording. by Chan S.000 = $1.801( P / F .943.1) + $1. 7%. NJ 07458. or transmission in any form or by any means. maintenance: $2. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.520. For information regarding permission(s).12%. Fifth Edition.12 1. Salvage value: $3.14%. 7%. Inc. 7%.800(0.27 • Salvage value: $11. photocopying. 000) C (8. Park.6 • Annual increase in productivity per engineer: $8. 7%.3C C = $4. 000( P / A1 . Annual taxes. Inc..800 • Capital recovery cost: CR(7%) = ($11.5)]( A / P.200. write to: Rights and Permissions Department.13 / 3. mechanical.12)($3.545) = $4.839.876( P / F . 25) + (0.943. insurance.52 • Equivalent annual cost of operating and maintenance: AE (7%) = [$1.520)( A / P. Pearson Education.14%. 03 − $8.97 = $4.132.538. Park. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.314.500) C (13.49 = 13. NJ 07458. or likewise. All rights reserved.314.28 Let T denote the total operating hours in full load. 658.3) 1. Inc. • Motor I (Expensive): Annual power cost: 180 × (0.05) × T = $8.392T T = 1.839.Contemporary Engineering Economics. or transmission in any form or by any means.49 • Annualized driving miles: C (14. electronic. Inc.. ISBN: 0-13-611848-8 © 2011 Pearson Education. 7%.35 per mile 6. storage in a retrieval system.07 1.83 Equivalent annual cost of operating the motor: AE(6%) I = −$5..658.392T 0.07 2 1.80 Equivalent annual cost of operating the motor: AE (6%) II = −$4.089T • Motor II (Less expensive): Annual power cost: 180 × (0. NJ. Upper Saddle River. Pearson Education.089T 0. Upper Saddle River.746) × (0. • Total annual equivalent cost: $2.52 + $1. 698.500) + + ]( A / P.089T = −$1.392T • Let AE (6%) I = AE (6%) II and solve for T .6%.746) × (0.800( A / P. 6%. photocopying.392T = −$1. −$1. write to: Rights and Permissions Department. mechanical.538.99 − $8.089T = −$1. Fifth Edition.03 − $8.05) × T = $8. recording. 600( A / P.15 hours per year Page | 8 . 068C C = $0.10) − $690 − $8.99 − $8. For information regarding permission(s). 000) C (11.073 = 13. by Chan S. 068C AEmiles (7%) = [ • So. $4.10) − $870 − $8. electronic. 491 + $490. • TEM Purchase Option: AE (15%) = ($415.75 • Option 2: Make units in-house PW (15%) dm = $63. 600 /1.40 (b) • Sub-contract Option: Unit profit = $400 − $300 − $0. mechanical. 000 = $24.16 ≈ 933 air samples per year Page | 9 . Fifth Edition. 000 = $496.29 • Option 1: Purchase units from John Holland Unit cost = $25 − ($35. storage in a retrieval system. 050( P / A1 . or likewise.15%. 000 + $9. 000 = $98 Let X denote the break-even number of air samples per year. All rights reserved. Inc.50 + $1.888)( A / P. or transmission in any form or by any means.000 / 20. photocopying. 600 / X Solving for X yields X = 933.5) + $70..84 ∴ Option 1 is a better choice. $400 − ($300 + $0. For information regarding permission(s). 600 Unit cost = $188.5) = $230. recording. write to: Rights and Permissions Department.800( P / A1 . Park.5 = $19.15%.888 ∴ AE (15%) = ($230. Inc. 241 + $709.30 (a) Determine the unit profit of air sample test by the TEM (in-house). NJ 07458. 6.500 / X ) = $300 − $188. ISBN: 0-13-611848-8 © 2011 Pearson Education.60 Unit profit = $300 − $188. Upper Saddle River.000) − $3. 491 PW (15%)vo = $139. 6. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.500 /1..15%. 6%. 000( P / A1 . 000 + $20. 776 Unit cost = $496. Pearson Education. 000 = $188.3%. 776 / 20. 000 + $18.60 = $111. 000) = $188.5%.5) = $490. 000 +$6.8) + ($50. Upper Saddle River.50 − $1.5) = $709.500)( A / P.15%. by Chan S.15%. NJ.Contemporary Engineering Economics. 241 PW (15%) dl = $190. 33 • Minimum operating hours: AEC(10%) = ($32. Upper Saddle River.55 per mile AEC(10%) total cost = $0.10%.000 − $2. or transmission in any form or by any means.801.15) +(0.000)( A / P.12) • = $765. NJ.7%.000 − $2.32 • • Capital costs: AEC(7%)1 = ($25. 200 AEC (10%) total cost = $7.944.6) +(P / F. Park.10)($2. For information regarding permission(s). 6. 000) = $10. ISBN: 0-13-611848-8 © 2011 Pearson Education.3)(30.9)]( A / P. or likewise.41 Annual recharging cost: • AEC(7%)3 = ($0.. 636 + $10. recording.500 • Option 2: Provide a car to employee AEC (10%)capital cost = ($25. mechanical.000[( P / F.10%. storage in a retrieval system.Contemporary Engineering Economics. 6.000) = $3.036 + $765.3) + (0. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.7%.41 Costs per mile: cost/mile = $4. write to: Rights and Permissions Department.801. 000 − $8..7%. All rights reserved.12) + (0. 000 = $0. electronic.000) = $300 Total annual costs (including annual maintenance cost): AEC(7%) = $3.015)(20.7%.41 + $300 + $700 • = $4. by Chan S. Inc.31 • Option 1: Pay employee $0.036 Annual battery replacement cost: AEC(7%)2 = $3.55(30. 000) AEC (10%)operating cost = $7.3) + ( P / F.07)($2.7%. Inc.2401 6. 200 = $17.10)($8.000) + $800 = $4. 636 = $1. Pearson Education. 000)( A / P.836 ∴ Option 1 is a better choice. 200 + ($0.41/ 20.21 Page | 10 .000)( A / P.000) = $16. Fifth Edition. photocopying. Upper Saddle River. NJ 07458. by Chan S. n) Solving for n yields n = 5.79 • Discounted payback period at full load of operation: n Investment 0 1 -$32.19 years (or 6 years) 6.944.000 15 +$2.77 Page | 11 .200 $32. electronic.000) = $19.21 = $4.000 + $5.565.000 = $109.77 • Annual operating costs: AEC(6%)O = $70.34 • Capital recovery cost: CR(6%) = ($170..000 • Total annual system costs: AEC(6%) = $19. we can establish the following relationship: $0. Upper Saddle River. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. NJ 07458.944.06)($12. or likewise. storage in a retrieval system.000 8. Upper Saddle River.000 − $12. Inc.000)( A / P. All rights reserved.Contemporary Engineering Economics.565. Pearson Education. NJ.000 + $15.09 per kilowatt-hour. write to: Rights and Permissions Department.000 -$800 10. photocopying.565. Since the value of the energy generated is considered to be $0. or transmission in any form or by any means.055.09)(100.200 $9.77 + $90. ISBN: 0-13-611848-8 © 2011 Pearson Education.21 Solving forT yields T = 1.4 hours • Annual worth of the generator at full load operation: AE (10%) = ($0.6%.200( P / A. recording. For information regarding permission(s). Then the total kilowatt-hours generated would be 40T .12) + (0.000) − $4. Let T denote the annual operating hours.10%.. Fifth Edition.373. mechanical. Inc.000 Revenue Maintenance cost Net Cash flow $9.09 × 40T = $4. Park.000 = $8.000 = $90.000 -$800 -$32. Contemporary Engineering Economics. and MARR = 15%.35 Given: Investment cost = $10 million. Upper Saddle River.000)(4.70)(8.37 • Annual total operating hours: (0.6) = 3. photocopying. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.12( A / F.77 / $0. 000 per year.20) + $250. MARR = 10% compounded annually (or effective monthly rate of 0. Inc.10%. 600. plant capacity = 0.642.0.642. 760) = 6. Inc. or likewise. or transmission in any form or by any means. NJ 07458.137.. For information regarding permission(s). Upper Saddle River.000 + (R − $4.000( A / P. 6.000)(P / A.78 =0 Solving for R yields R = $6. Park. 000 kilowatt-hours Page | 12 .000 lbs/hour.15%.000. All rights reserved.7845R − $25. Pearson Education. useful plant life = 20 years. storage in a retrieval system. by Chan S. write to: Rights and Permissions Department.12 Monthly water bill for each household: $837. O&M cost = $4 million per year. electronic.000.00554 per lb (300. salvage value = negligible. plant capacity = 300.7974%.658 rides 6. mechanical. salvage value = $800.000) Comment: The minimum processing fee per lb should be higher on a before-tax basis.565.4 acre-foot. recording.10 = 1.132 hours per year • Annual electricity generated: 50.298.000.33 295 6. NJ.000 hours per year.930. • Number of rides required per year: Number of rides = $109.370 = $0. O&M cost = $250. (a) PW (15%) = −$10..000. Fifth Edition. ISBN: 0-13-611848-8 © 2011 Pearson Education.370 per year (b) Minimum processing fee per lb (after-tax): $6.7974%) AEC(10%) = $5. useful life = 15 years.095.12) = $226.298. 000 × 6.36 Given: Investment = $5 million. plant operating hours = 4.132 = 306.000 = $837. 125.41 ≈ 8 Passengers per round trip Page | 13 .. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.364 / 306. 010. or likewise..25) + $6. 000) = $2. 000 + $1. 280)(2)(3)(52) = $1. and catering costs: $237. storage in a retrieval system.38 Let X denote the average number of round-trip passengers per year. 696 +$78.156 Passengers per year 1. insurance. ISBN: 0-13-611848-8 © 2011 Pearson Education. • Capital recovery costs: CR (15%) = ($12. by Chan S. 000)( A / P.000 = $18. NJ 07458.364 • Cost per kilowatt-hour: $18. All rights reserved.15) +(0.500 + $75 X • Total equivalent annual costs: AE (15%) = $2. 000. 400 X Solving for X yields or X = 1.125. For information regarding permission(s).000( A / P. Upper Saddle River. Inc.14%. 600. 010.15)($2.10)(3.171 + $225. • Equivalent annual cost: AEC(14%) = $85.367. Pearson Education. Inc. 000 = $0. 000 − $2. photocopying. 000 + $75 X = $403.Contemporary Engineering Economics. 000. mechanical. 000 + $403. 000 • Annual maintenance.171 • Annual crew costs: $225.500 + $166.06 per kilowatt-hour 6.000. 000. recording. NJ.15%. Upper Saddle River. Park.500 + $75 X = $3. or transmission in any form or by any means.156 /(52)(3) = 7.367.000 • Annual fuel costs for round trips: ($1. write to: Rights and Permissions Department. 696 • Annual landing fees: ($250)(3)(52)(2) = $78. electronic.000. Fifth Edition. 20) = −$2.255 /hour (c) Process A is a better choice.41 • Capital recovery cost for both motors: CR(12%)CV = −$14.4)] = $795.1) + + $600( P / F .69 ∴ Project B is a better choice. storage in a retrieval system. recording. NJ. 2)]( A / P. Upper Saddle River.40 AE (15%)C = [−$6.12%.Contemporary Engineering Economics. by Chan S. Fifth Edition.14 AE(15%) B = [−$4.000( A / P. Inc.500 = $129.000 + $100(P / F. Pearson Education.000( A / P.15%.40 ∴ Project B is a better choice. ISBN: 0-13-611848-8 © 2011 Pearson Education.. electronic.600( A / P. 6.4) + $1.30 CR(12%) PE = −$16. or likewise.500 = $113.4) + $7.15%.222.15%.000( A / P.500( P / F . For information regarding permission(s). mechanical.15%. 6.40 (a) AE (15%) A = [−$3.98 AE(12%) B = −$20.12%..15%. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. or transmission in any form or by any means.000 = $0.280( A / G. 4)]( A / P. write to: Rights and Permissions Department.12%.31 / 3.15%.1)]( A / P.39 Page | 14 .265 /hour Process B: $765.98 / 3.12%. Upper Saddle River.20) = −$1. All rights reserved.500 + $1. Inc.15%.000 = $0.4) + [$9. 000 − $500( P / A. photocopying. 4) = $58.12%.31 (b) Process A: $795. NJ 07458. 4) + $2.39 (a) AE(12%) A = −$20.350 = $765. Park. (b) AE (15%) B = $129.120 − $1. Comparing Mutually Exclusive Alternatives by the AE Method 6.874. 00254 / kWh 0.52 (a) Savings per kWh: Savings = $70.55 − $67.120 h × 20 = $65. Park.918. 696.139. write to: Rights and Permissions Department. ISBN: 0-13-611848-8 © 2011 Pearson Education.000 Annual savings from installing the new lighting system = $2.39 + $65.30 + $68. recording. storage in a retrieval system..650kWh × 20 (b) 18.056T AECV = 0.39 + $21..000) = $17. mechanical.25 0.75 × 3.895 18.000( A / P.12%.75 × × $0. 265.696. Pearson Education. • Annual operating cost for both motors: 18.09 T = 422.650kW × $0. or transmission in any form or by any means.88T = $2. electronic.056T 0.07 / kWh × 3. NJ 07458.75 × × $0.93 $1.55 AEC PE = $2.895 18.93 AECV = 0.918.07 / kWh × T × 20 = $21. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Inc.650kW AEPE = 0.265.694 • Old lighting system cost: AEC(12%) = $20.650kW = 0. Upper Saddle River.874. or likewise. photocopying.650kW × $0.120 h × 20 = $68. Fifth Edition.52 = $0.000 + $3.139.07 / kWh × 3.306 Page | 15 .75 × 0. Upper Saddle River.Contemporary Engineering Economics.20) + ($8.13 = $67. by Chan S.07 / kWh ×T × 20 = $21. All rights reserved.222.42 • New lighting system cost: AEC(12%) = $50.222.824T = 348. Inc.30 + $21.44 hrs (with considering 75% load) or 316.83 hrs (without considering 75% load) 6. NJ.75 × AEPE • Total annual cost for both motors: AECCV = $1.120h × 18.25 = $70. For information regarding permission(s).13 0.874.88T 0. All rights reserved.5)( A / P.6.000 + X ( P / F.17%.15%.15)$15. Park. the effective annual interest = (1. 6.63 = $66.680.000 +$47. or likewise.3.95 6. 2) = $79.04% per semiannual • Option 1: Buying a bond AE (3.3) + $150 + $2.1195X AEC(15%)1 = AEC(15%)2 $69. Upper Saddle River.05 + 0.1195X X = $30.17%)3 = −$2.. mechanical.680.054.000( A / P.7) = $66.15%. 6.000)( A / P. or transmission in any form or by any means. NJ 07458.04%.005)12 − 1 = 6.04%.44 Given: i = 6% interest compounded monthly.17%. recording.17%.17 • Option 3: Receiving $150 interest per year for 3 years AE(6.347.3) = $26.000 = $69. effective semiannual interest = (1.26( A / F . electronic. 6.6.3.7) + (0. 735.7) + $30. Page | 16 . 6) = $39.17%) = $39. Inc.20 per semiannual AE (6.054.3) = $107.60 ∴ Buying the growth stock is the best option.04%.3) + $2. 000( A / F . For information regarding permission(s). This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.43 AEC(15%)1 = ($100.15%.000( A / P.17%.04%)1 = −$2. ISBN: 0-13-611848-8 © 2011 Pearson Education. photocopying.63 AEC(15%)2 = $150.59 per year • Option 2: Buying and holding a growth stock for 3 years AE (6.17% per year. 000( A / P. 6) + $100 + $2. Pearson Education. Inc. Fifth Edition.05 + 0.Contemporary Engineering Economics. by Chan S.15%. Upper Saddle River.17%) 2 = −$2.20( F / A. 000( A / P.. write to: Rights and Permissions Department. storage in a retrieval system.3.005)6 − 1 = 3.000( A / F. NJ.000 − $15. 000 + $30.Contemporary Engineering Economics.318.32 per ton C2 = $285.000 +$30. electronic.26 or 7 machines per year Page | 17 .70 / (20)(365) = $40.51 = $10.589.000 = $294...000 − $60.000)( A / P. 6.10) +(0. write to: Rights and Permissions Department.13)($60. or transmission in any form or by any means. NJ 07458.000)( A / P.70 AEC(13%) B = ($850.000 + $40. All rights reserved. storage in a retrieval system. Fifth Edition. recording.000 X AEC(10%)2 = ($500. mechanical.44 / (20)(365) = $39.10%. Inc.000 − $100.000 − $30.000) + $70.000)( A / P.13%.44 • Processing cost per ton: C1 = $294. Inc.318.45 • Equivalent annual cost: AEC(13%) A = ($1. photocopying. Pearson Education.46 Let X be the number of machines per year AEC(10%)1 = $40. or likewise.000) + $100. 000 X X = 6.1)$100. 6.017. NJ.300.13%. by Chan S. For information regarding permission(s). Park.000 = $285.13)($30.15) + (0.017. Upper Saddle River. ISBN: 0-13-611848-8 © 2011 Pearson Education. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.000 X AEC (10%)1 = AEC (10%) 2 $62.04 per ton ∴ Incinerator B is a better choice. Upper Saddle River.20) +(0. 53 − $33.000 hours) = $134. 6) + (0.000 gals/1.10%. Park..3) + (0. 623.10/gal)(40. All rights reserved. 000 − $20.575.1)$12.41per hour System B = $216.312. recording.10%.53 • System B : Equivalent annual fuel cost: A1 = ($2. electronic. Fifth Edition.10 /gallon • System A : Equivalent annual fuel cost: A1 = ($2.3) = $177.53 / 2. 6%.20 AEC (10%) A = ($100. 000)( A / P. Inc. 400( P / A1 .47 AEC (10%)1 = ($95. by Chan S. storage in a retrieval system. For information regarding permission(s). 000 − $10.000 hours) = $168.3) + (0. ISBN: 0-13-611848-8 © 2011 Pearson Education. 000 +$3. NJ. 098.10)($10. 000) + $142.10)($20. 000)( A / P.000 = $108. Upper Saddle River. photocopying. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. write to: Rights and Permissions Department.813.56 AEC (10%) B = ($200.. Upper Saddle River. Life-Cycle Cost Analysis 6.83 6. 000 − $25.1)$25.10/gal)(32.23 / 2.3) + (0.000 = $107.10%. mechanical. 000 +$9. 000 = $37. or likewise. Inc.3) = $142.000 gals/1. 000)( A / P.24 per hour ∴ System A is a slightly better choice.70 = $4.3)]( A / P.48 Assumption: jet fuel cost = $2.000 hours)(2.262.400 AEC (10%) fuel = [$134. 000)( A / P.20 = $214.10%.56 = $216.813.000 (assuming an end of-year convention) AEC (10%) fuel = [$168.23 • Equivalent operating cost (including capital cost) per hour: System A = $214. Pearson Education.10%.312. 6%.575. Page | 18 .000 hours)(2. 000) + $177.3)]( A / P. 479.10%. 000( P / A1 .53 AEC (10%) 2 = ($120.70 AEC(10%)1 − AEC(10%)2 = $37. 000 − $12.10%. 623. NJ 07458.479.10%. 000 = $33.Contemporary Engineering Economics. 098. or transmission in any form or by any means. by Chan S. Then buy Machine B and use it for one year. Inc. 000)( A / P.12%.12)($8. A5 Buy Machine A and use it for 4 years. 000 − $5. Then lease a machine for one year. 000)( A / P. NJ. write to: Rights and Permissions Department. Inc. 6. recording. Park. A2 Buy Machine B and use it for 5 years. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. ISBN: 0-13-611848-8 © 2011 Pearson Education. Upper Saddle River. Fifth Edition. Pearson Education. or likewise.3) +(0. photocopying. For information regarding permission(s). 000 = $6. 763. electronic. Alternative Description A1 Buy Machine A and use it for 4 years. 6. A3 Lease a machine for 5 years. storage in a retrieval system. All rights reserved. Then buy another Machine A and use it for one year.. Page | 19 . or transmission in any form or by any means. NJ 07458. 4) + (0. 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. A4 Buy Machine A and use it for 4 years. 000 − $8. • Truck A: Four replacements are required AEC (12%) A = ($15. mechanical.12)($5.80 ∴ Truck B is a more economical choice. 000) + $2.50 • Truck B: Three replacements are required AEC (12%) B = ($20.49 Since the required service period is 12 years and the future replacement cost for each truck remains unchanged. 000 = $7.50 (a) Number of decision alternatives (required service period = 5 years): We can think of five alternatives initially..910.12%. 000) + $3. Upper Saddle River.Contemporary Engineering Economics. 4) −$800(P / A. 4) = −$10.3) −$100(P / F. Upper Saddle River.Contemporary Engineering Economics.4) = −$10.5) − $280( P / F . 649 • For A3: AE (10%)3 = [−$3. 4)]( A / P. • For A2: PW (10%) 2 = −$8.. by Chan S. or likewise. Page | 20 . Upper Saddle River. ISBN: 0-13-611848-8 © 2011 Pearson Education.53 Note: It is assumed that a new oil filter be required at the end of year 4 to expect a salvage value of $600.10%. Inc. we could imagine replacing the machines every year or two could be the best thing. 000 − $3.4) − $200(P / F. NJ. For information regarding permission(s).300 ∴ A2 is a better choice.000 + $100)( P / F.976.10%. photocopying. All rights reserved..5) = −$2. 000( P / A. So really there are more possibilities than five alternatives we just looked.) (b) With lease.895.976.500 + $600(P / F.5) = −$2.10%.10%.10%.33( A / P.10%. NJ 07458.10%. 042 AE (10%) 2 = −$10.33 AE(10%)1 = −$10.10%. electronic. or transmission in any form or by any means. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. It is also assumed that the lease payment would be made at the beginning of the year.5) −$520( P / A. Pearson Education.10%.5) = −$3. 042( A / P. Inc.500 + $1. write to: Rights and Permissions Department.10%. Both A4 and A5 are feasible but we do not consider these alternatives because we need to know the salvage values of the machines after oneyear use. recording. Park. storage in a retrieval system.2) − ($3. mechanical. the O&M costs will be paid by the leasing company: • For A1: PW (10%)1 = −$6. (Note: If the salvage value of the machines after one or two years were high enough. Fifth Edition.10%.10%. 000( P / F . 10%. Upper Saddle River.700.52 Given: Required service period = indefinite.000.000 / 180. NJ 07458.576 Supporting equipment: AE(10%)2 = $75.2650 per lb ∴ Option 1 is a better choice. ∞)] 0. i.878 cost/lb = $46.20) +($0.10%. NJ.000 = $0.000. ∞) = 100 .10%. or transmission in any form or by any means.15) +$62. ∞) = $40. 000( P / G.000 cost/lb = $47.000. by Chan S. Upper Saddle River. Fifth Edition. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. For information regarding permission(s). Inc.5)( P / F .51 • Option 1: AE(18%)1 = $200.Contemporary Engineering Economics. 000( P / A.000)(180)( A / F.10%. recording.30) = $456 • Operating cost: AE (10%)3 = [$31.08)($200. mechanical.305. or likewise.215)(180. storage in a retrieval system.10%.10%.000.000) = $47.10%. write to: Rights and Permissions Department. 056 Note that ( P / G.215)(180.005 + 0..10%. analysis period = indefinite Plan A: Incremental investment strategy: • Capital investment : AE (10%)1 = [$400. ∞) = 1/ i 2 or ( P / G. ISBN: 0-13-611848-8 © 2011 Pearson Education.878 / 180.000 = $0. All rights reserved.305. 000( P / F .18%. Park. 6. Page | 21 . 20)]( A / P.000) = $46. electronic.15)] $63. Inc.000( A / F..05 + $0.15)]( A / P.10 ×( P / F .10%. Pearson Education. photocopying. 000 + $400. 000( P / A. 000 +[ + $1.20) −(0. 6.18%.10%.2573 per lb • Option 2: AE(18%)2 = ($0. ∞) • = $49.700.000(180)( A / P. Inc. 788 Total equivalent annual worth: AE(10%) B = $55..15)] • ×( A / P. ISBN: 0-13-611848-8 © 2011 Pearson Education. All rights reserved.088 Plan B: One time investment strategy: • Capital investment: AE (10%)1 = $550. ∞) • = $55. Fifth Edition.10%. photocopying. 000( A / P.516 $4.53 (a) • Energy loss in kilowatt-hour: 6. Minimum Cost Analysis 6.Contemporary Engineering Economics. or likewise.0825) = A A • Material weight in pounds: A = (770. 709. electronic.30) = $912 • Operating cost: AE (10%)3 = [$35.15) +$55.000 + $912 + $39. 000( P / A.576 + $456 + $40.10%. recording. by Chan S. or transmission in any form or by any means.. Upper Saddle River. • Total equivalent annual worth: AE(10%) A = $49.10%. NJ 07458. storage in a retrieval system.056 = $90. ∞)( P / F . 000( P / A.10%. Upper Saddle River. NJ. ∞) = $39.83) A($6) = $4. 000( A / F .11 (24 × 365)($0. Pearson Education.788 = $95. mechanical. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.10%. write to: Rights and Permissions Department. 000 Supporting equipment: AE (10%) 2 = $150. Park.83) A 123 • Total material costs: (200)(12)(555) (770.700 ∴ Plan A is a better choice.10%. 625 A Page | 22 . For information regarding permission(s). Inc. All rights reserved.196.11%. or likewise. 25) + $1× 770. Inc.11 dAE(11%) = 542. write to: Rights and Permissions Department. ISBN: 0-13-611848-8 © 2011 Pearson Education.51 2. Inc. electronic. storage in a retrieval system.11 A • Optimal cross-sectional area: 4. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. 6.83 A]( A / P.44 A + 4.9464 (c) Note: increasing energy cost increases the energy –loss term coefficient and moves the curve up and to the right.11 = $3.83 A × 0. 709. by Chan S.020.44 A • Total equivalent annual cost: AE (11%) = 542. indicating a larger-diameter conductor being optimal.. • Capital recovery cost: CR(11%) = [$4. or transmission in any form or by any means.9464) + 4. NJ.Contemporary Engineering Economics.54 We assume the friction factor is 0.. Page | 23 .709. mechanical. 709.9464 inches 2 (b) Minimum annual equivalent total cost: AE (11%) = 542. Upper Saddle River.44 − =0 dA A2 A = 2.44(2. For information regarding permission(s).11 = 542. Upper Saddle River. NJ 07458. 625 A − $1× 770. photocopying. Fifth Edition. Park. recording. Pearson Education. .70 (10)5 = $158.70 1. electronic. we need to convert the capital expenditure into the required capital recovery cost per unit. or transmission in any form or by any means.10) ⎛ 10 ⎞ = $52. storage in a retrieval system. 000 = $7. ISBN: 0-13-611848-8 © 2011 Pearson Education. NJ.1⎟⎠ where i = g = $479. 031 per year $22.038. NJ 07458.05/kWh) 1 (0.454.Contemporary Engineering Economics. Fifth Edition. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.02) × 1705 0. (a) (600)3 (10. write to: Rights and Permissions Department.000) (8.000) (8. Park.55 ST 6. All rights reserved.39 (b) 1 (600)3 (10. Since all operating costs are already given in dollars per part. For information regarding permission(s). photocopying.740 ⎜ ⎝ 1 + 0. 031 unit capital recovery cost = 3.47 PW (10%)2 = $52.10) = $324. Inc. We will first calculate the unit cost under each production method. Pearson Education. by Chan S. recording.760)($0. mechanical.10%.02) × Cost ($) = 0. or likewise.05/kWh) Cost ($) = (0.10%. 480( A / P. Upper Saddle River.740( P / A1 . Upper Saddle River.10) = $22. PW (10%)1 = $52.1 We assume i = 10% . • Conventional method: CR (16%) = $106.760)($0. 064.821..16%.75 Short Case Studies ST 6. 740( P / A.2 This case problem appears to be a trivial decision problem as one alternative (laser blanking method) dominates the other (conventional method).10%. A problem of this nature (from an engineer’s point of view) involves more strategic planning issues than comparing the accounting data.34 per part Page | 24 . Inc.538.705 (6)5 = $2. If Ford relies on an outside supplier. 000 BTUs. it will take 6 months to get up to the required production volume. mechanical. Upper Saddle River..3 Discuss the topic “Replacement analysis.” ST 6. All rights reserved. 000 BTUs (0.204. If Ford decides to make the window frame in house..519.6 lb/ton = 6.19 $0. Inc. 000. Park. What option should Ford exercise to satisfy the production need during this start-up period? ST 6.4 Given: annual energy requirement = 145. Upper Saddle River. Pearson Education.34 $23.Contemporary Engineering Economics.73 It appears that the window frame production by the laser blanking technique would save about $8. the subcontracting work should be in this cost range.67 $0.300) = 13. In this case. Inc. 000.40 $5. or likewise. photocopying.173 per year $17. 000 (a) Annual fuel costs for each alternative: • Alternative 1: Weight of dry coal = 145.814 lb = 2.75)(14. • Laser blanking method: CR(16%) = $83. NJ. 000. electronic. 000 = $5. Fifth Edition. For information regarding permission(s). depending upon the blanking method adopted.49 Laser $8. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.98 $0. the part cost would range between $14. storage in a retrieval system. or transmission in any form or by any means. net proceeds from demolishing the old boiler unit = $1. 1 ton=2.132.6 lbs. write to: Rights and Permissions Department. 000. by Chan S. 204. recording.10) = $17.16%.173 Unit capital recovery cost = 3.49. NJ 07458. If Ford decides to go with the laser blanking.519.50 $7.814 lb 13. the die investment would have now been made. If Ford were producing the window frames by the conventional method.73 and $23.72 $14.55 ton Page | 25 . ISBN: 0-13-611848-8 © 2011 Pearson Education. one of the important issues is to address if it is worth switching to the laser blanking now or later.42 $0.72 per part Description Steel cost/part Transportation cost/part Blanking cost/part Capital cost/part Total unit cost Blanking Method Conventional $14. 000( A / P.76 for each part produced. For information regarding permission(s). 4%. Pearson Education. NJ.000.5 (0. NJ 07458. 20) = $4.868. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.000. recording.000.000)( A / P.362 Unit cost = $591.000.764.000.017 Annual fuel cost = $1.000 = $0.48 If the pipe had a 50-year service life with a zero salvage value when it was placed in service 20 years ago. or transmission in any form or by any means.10%.017 = $1. 208(1. photocopying.04) −20 = $1. $4. by Chan S.000.53 Alternative 2: 145.94) Gas cost = $9.868.400) = $104.660.000(0.081 (b) Unit cost per steam pound: • Alternative 1: Assuming a zero salvage value of the investment AE (10%) = ($1. mechanical. electronic.20) +$1.0041 per steam lb • Alternative 2: AE (10%) = ($889. the annual capital recovery cost to the owner would be as follows.55×$55.81)(139.770. Fifth Edition.300 + $100. 208( P / F .064 145. Inc.064 + $104..795 = $591.000.06) Oil cost = $1.Contemporary Engineering Economics. storage in a retrieval system.920. Park.000) = $1. (It is assumed that the owner’s interest rate would be 5% Page | 26 .20) +$371. or likewise.764. Inc..000 = $0.000(0.000 −$1.5 = $340. • Annual fuel cost = 6.409 / 145.000)( A / P. ISBN: 0-13-611848-8 © 2011 Pearson Education.0129 per steam lb (c) Select alternative 1.356.78)(1. ST 6. Upper Saddle River. All rights reserved.35 (0.409 Unit cost = $1.10%.362 / 145.081 = $1.660. I could estimate the cost of the pipe 20 years ago as follows.200 − $1. Upper Saddle River. write to: Rights and Permissions Department.132.5 Assuming that the cost of your drainage pipe has experienced a 4% annual inflation rate. recording. Upper Saddle River.30) = $1. or likewise.545. In other words. Pearson Education. per year. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.20 per year You can view this number as the annual amount he expects to recover from his investment considering the cost of money. if he did not purchase the pipe.15.98 $1. NJ. Upper Saddle River.920.89 $1.617.15 Page | 27 . by Chan S. Fifth Edition. Assumed interest rate 0% 3% 4% 5% Claim cost $1..5%. so there is some room for negotiation. he still has 30 more years to go.20( P / A. So. On paper. For information regarding permission(s). With only a 20-year’s usage.) CR (5%) = $1.5%. 617.152.. electronic. 48( A / P. but the city’s interest rate could be different from the owner’s.462. Inc.Contemporary Engineering Economics. mechanical. storage in a retrieval system. write to: Rights and Permissions Department. All rights reserved. ISBN: 0-13-611848-8 © 2011 Pearson Education.24 $1. NJ 07458.48 at 5% annual interest. or transmission in any form or by any means. the owner could invest his $1. Inc. photocopying. the owner could claim this number. Park. the unrecovered investment at the current point is $105.920.50) = $105.


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