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Contoh Soal PW dan AW Pertemuan 11 dan 12

Contoh Soal PW dan AW Pertemuan 11 dan 12. Matakuliah : D 0094 Ekonomi Teknik Tahun : 2007. Contoh-Contoh Soal PW dan variasinya. Contoh Soal.

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Contoh Soal PW dan AW Pertemuan 11 dan 12

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  1. Contoh Soal PW dan AWPertemuan 11 dan 12 Matakuliah : D 0094 Ekonomi Teknik Tahun : 2007

  2. Contoh-Contoh SoalPW dan variasinya

  3. Contoh Soal A British food distribution conglomerate purchased a Canadian food store chain for $75 million (US) three years ago. There was a net loss of $10 million at the end of year 1 of ownership. Net cash flow is increasing with an arithmetic gradient of $+5 million per year starting the second year, and this pattern is expected to continue for the foreseeable future. This means that breakeven net cash flow was achieved this year. Because of the heavy debt financing used to purchase the Canadian chain, the international board of directors expects a MARR of 25% per year from the sale. • The British Conglomerate has just been offered $159.5 million (US) by a French company wishing to get a foothold in Canada. Use FW analysis to determine if the MARR will be realized at this selling price • If the British conglomerate continue to own the chain, what selling price must be obtained at the end of 5 years of ownership to make the MARR

  4. Cash Flow Diagram

  5. Contoh Permasalahan Investasi Mr. Bracewell • Membangun pabrik hydroelectric plant dengan menggunakan • simpanannya sendiri sebesar $800,000 • Kapasitas tenaga yang dihasilkan 6 juta kwhs • Tenaga listrik yang terjual stiap tahun setelah pajak • diperkirakansebesar - $120,000 • Perkiraan umur pelayanan 50 tahun • Apakah keputusan dari Bracewell menginvestasikan sebesar • $800,000 adalah tepat ? • Berapa lama modal dari Bracewell kembali dan kapan memberikan keuntungan ?

  6. Proyek Hydro Mr. Brcewell

  7. Equivalent Worth at Plant Operation • Equivalent lump sum investment • V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) + • . . . + $100K(F/P, 8%, 1) + $60K • = $1,101K • Equivalent lump sum benefits • V2 = $120(P/A, 8%, 50) • = $1,460K • Equivalent net worth • FW(8%) = V1 - V2 • = $367K > 0, Good Investment

  8. With an Infinite Project Life • Equivalent lump sum investment • V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) + • . . . + $100K(F/P, 8%, 1) + $60K • = $1,101K • Equivalent lump sum benefits assuming N = • V2 = $120(P/A, 8%,  ) • = $120/0.08 • = $1,500K • Equivalent net worth • FW(8%) = V1 - V2 • = $399K > 0 • Difference = $32,000

  9. Permasalahan Pembangunan Jembatan • Biaya Konstruksi = $2,000,000 • Biaya Perawatan Tahunan = $50,000 • Biaya Rrenovasi = $500,000 tiap 15 Tahun • Rencana untuk digunakan = perioda tak hingga • Interest rate = 5%

  10. 15 30 45 60 0 $50,000 $500,000 $500,000 $500,000 $500,000 $2,000,000

  11. Solution: • Construction Cost • P1 = $2,000,000 • Maintenance Costs • P2 = $50,000/0.05 = $1,000,000 • Renovation Costs • P3 = $500,000(P/F, 5%, 15) • + $500,000(P/F, 5%, 30) • + $500,000(P/F, 5%, 45) • + $500,000(P/F, 5%, 60) • . • = {$500,000(A/F, 5%, 15)}/0.05 • = $463,423 • Total Present Worth • P = P1 + P2 + P3 = $3,463,423

  12. 30 45 60 0 15 Alternate way to calculate P3 • Concept: Find the effective interest rate per payment period $500,000 $500,000 $500,000 $500,000 • Effective interest rate for a 15-year cycle • i = (1 + 0.05)15 - 1 = 107.893% • Capitalized equivalent worth • P3 = $500,000/1.07893 • = $463,423

  13. Membandingkan Proyek-Proyek Mutually Exclusive • Mutually Exclusive Projects • Alternative vs. Project • Do-Nothing Alternative

  14. Pendapatan Proyek • Projects yang pendapatannya bergantung pada pilihan alternatif • Pelayanan Proyek • Projects yang pendapatannya tidak bergantung pada pilihan alternatif

  15. Perioda Analisa • Rentang waktu dimana pengaruh ekonomi dari investasi akan dievaluasi (study period or planning horizon). • Perioda Pelayanan Yang Diperlukan • Rentang waktu dimana pelayanan suatu peralatan (or investment) akan dibutuhkan.

  16. Comparing Mutually Exclusive Projects • Prinsip: Proyek dibandingkan dalam jangka waktu yang sama • Aturan: Jika periode proyek diketahui, periode analisisnya harus sama dengan periode waktu analisisnya

  17. Bagaimana memilih perioda analisa ? Perioda Analysis Sama dengan Masa proyek Case 1 Perioda Analysis lebih pendek dari Masa proyek Case 2 Analysis = Required period service period Finite Perioda Analysis lebih lama dari Masa proyek Case 3 Case 4 Perioda Analysis Terlama diantara Masa proyek dalam grup Required service period Perioda Analysis Terendah dari common multiple of project lives Project repeatability likely Infinite Perioda Analysis Sama dengan satu dari masa proyek Project repeatability unlikely

  18. PW (10%) = $283 PW (10%) = $579 A B Case 1: Analysis Period Equals Project Lives Hitung PW untuk tiap proyek selama waktu proyek $2,110 $2,075 $600 $500 $1,400 $450 0 A B $1,000 $4,000

  19. Membandingkan proyek dengan tingkat investasi berbeda – Asumsi bahwa dana yang tidak digunakan akan diinvestasikan pada MARR. $600 $450 $500 $2,110 3,993 Project A $2,075 $1,000 $1,400 $600 $450 $500 Project B Modified Project A $4,000 $1,000 $3,000 This portion of investment will earn 10% return on investment. PW(10%)A = $283 PW(10%)B = $579

  20. Case 2: Analysis Period Shorter than Project Lives • Estimasikan salvage value pada akhir • perioda pelayanan yang ditentukan • Hitung PW untuk tiap proyek selama • perioda pelayanan yang ditentukan

  21. Comparison of unequal-lived service projects when the required service period is shorter than the individual project life

  22. Case 3: Analysis Period Longer than Project Lives • Mengajukan replacement projects yang • cocok atau melebihi perioda pelayanan • yang ditentukan • Hitung PW untuk tiap proyek selama • perioda pelayanan yang ditentukan.

  23. Comparison for Service Projects with Unequal Lives when the required service period is longer than the individual project life

  24. Case 4: Analysis Period is Not Specified • Project Repeatability Unlikely • Use common service (revenue) period. • Project Repeatability Likely • Use the lowest common multiple of project lives.

  25. Proyek Berulang Yang Tak Serupa PW(15%)drill = $2,208,470 Assume no revenues PW(15%)lease = $2,180,210

  26. Proyek Berulang Yang Serupa PW(15%)A=-$53,657 Model A: 3 Years Model B: 4 years LCM (3,4) = 12 years PW(15%)B=-$48,534

  27. Contoh-contoh Soal AW dan variasinya

  28. Standard Premium Motor Efficient Motor 25 HP 25 HP $13,000 $15,600 20 Years 20 Years $0 $0 89.5% 93% $0.07/kWh $0.07/kWh 3,120 hrs/yr. 3,120 hrs/yr. Size Cost Life Salvage Efficiency Energy Cost Operating Hours Mutually Exclusive Alternatives with Equal Project Lives (a) At i= 13%, determine the operating cost per kWh for each motor. (b) At what operating hours are they equivalent?

  29. Solution: (a): • Operating cost per kWh per unit Determine total input power Conventional motor: input power = 18.650 kW/ 0.895 = 20.838kW PE motor: input power = 18.650 kW/ 0.930 = 20.054kW

  30. Determine total kWh per year with 3120 hours of operation • Conventional motor: • 3120 hrs/yr (20.838 kW) = 65,018 kWh/yr • PE motor: • 3120 hrs/yr (20.054 kW) = 62,568 kWh/yr • Determine annual energy costs at $0.07/kwh: • Conventional motor: • $0.07/kwh  65,018 kwh/yr = $4,551/yr • PE motor: • $0.07/kwh  62,568 kwh/yr = $4,380/yr

  31. Capital cost: Conventional motor: $13,000(A/P, 13%, 12) = $1,851 PE motor: $15,600(A/P, 13%, 12) = $2,221 • Total annual equivalent cost: Conventional motor: AE(13%) = $4,551 + $1,851 = $6,402 Cost per kwh = $6,402/58,188 kwh = $0.1100/kwh PE motor: AE(13%) = $4,380 + $2,221 = $6,601 Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh

  32. (b) break-even Operating Hours = 6,742

  33. Model A: 0 1 2 3 $3,000 $5,000 $5,000 $12,500 Mutually Exclusive Alternatives with Unequal Project Lives Required service Period = Indefinite Analysis period = LCM (3,4) = 12 years Model B: Least common multiple) 0 1 2 3 4 $2,500 $4,000 $4,000 $4,000 $15,000

  34. 0 1 2 3 $3,000 $5,000 $5,000 $12,500 Model A: • First Cycle: • PW(15%) = -$12,500 - $5,000 (P/A, 15%, 2) - $3,000 (P/F, 15%, 3) • = -$22,601 • AE(15%) = -$22,601(A/P, 15%, 3) = -$9,899 • With 4 replacement cycles: • PW(15%) = -$22,601 [1 + (P/F, 15%, 3) • + (P/F, 15%, 6) + (P/F, 15%, 9)] • = -$53,657 • AE(15%) = -$53,657(A/P, 15%, 12) = -$9,899

  35. 0 1 2 3 4 $2,500 $4,000 $4,000 $4,000 $15,000 Model B: • First Cycle: • PW(15%) = - $15,000 - $4,000 (P/A, 15%, 3) - $2,500 (P/F, 15%, 4) • = -$25,562 • AE(15%) = -$25,562(A/P, 15%, 4) = -$8,954 • With 3 replacement cycles: • PW(15%) = -$25,562 [1 + (P/F, 15%, 4) + (P/F, 15%, 8)] • = -$48,534 • AE(15%) = -$48,534(A/P, 15%, 12) = -$8,954

  36. Minimum Cost Analysis • Concept: Total cost is given in terms of a specific design parameter • Goal: Find the optimal design parameter that will minimize the total cost • Typical Mathematical Equation: • where x is common design parameter • Analytical Solution:

  37. Typical Graphical Relationship Total Cost Capital Cost Cost ($) O & M Cost Design Parameter (x) Optimal Value (x*)

  38. Optimal Cross-Sectional Area Substation Power Plant A copper conductor • Copper price: $8.25/lb • Resistance: 0.8145x10-5in2/ft • Cost of energy: $0.05/kwh • density of copper: 555 lb/ft • useful life: 25 years • salvage value: $0.75/lb • interest rate: 9% 1,000 ft. 5,000 amps 24 hours 365 days

  39. Operating Cost (Energy Loss) • Energy loss in kilowatt-hour (L) I = current flow in amperes R = resistance in ohms T = number of operating hours A = cross-sectional area

  40. Material Costs • Material weight in pounds • Material cost (required investment) • Total material cost = 3,854A($8.25) • = 31,797A • Salvage value after 25 years: ($0.75)(31,797A)

  41. Capital Recovery Cost 2,890.6 A Given: Initial cost = $31,797A Salvage value = $2,890.6A Project life = 25 years Interest rate = 9% Find: CR(9%) 0 25 31,797 A

  42. Total Equivalent Annual Cost • Total equivalent annual cost AE = Capital cost + Operating cost = Material cost + Energy loss • Find the minimum annual equivalent cost

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