Contoh Soal PW dan AW Pertemuan 11 dan 12

Presentasi berjudul: "Contoh Soal PW dan AW Pertemuan 11 dan 12"— Transcript presentasi:

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

2 Contoh-Contoh Soal PW dan variasinya
Bina Nusantara

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 Bina Nusantara

4 Cash Flow Diagram Bina Nusantara

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 ? Bina Nusantara

6 Proyek Hydro Mr. Brcewell
Bina Nusantara

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 $50,000 $500,000 $500,000 $500,000 $500,000 $2,000,000 Bina Nusantara

11 Solution: Construction Cost P1 = $2,000,000 Maintenance Costs
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 Bina Nusantara

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

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

14 Projects yang pendapatannya bergantung pada pilihan alternatif
Pendapatan Proyek Projects yang pendapatannya bergantung pada pilihan alternatif Pelayanan Proyek Projects yang pendapatannya tidak bergantung pada pilihan alternatif Bina Nusantara

15 Perioda Pelayanan Yang Diperlukan
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. Bina Nusantara

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 Bina Nusantara

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

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 Bina Nusantara

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

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. Bina Nusantara

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

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. Bina Nusantara

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

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 Bina Nusantara

27 Contoh-contoh Soal AW dan variasinya Bina Nusantara

28 Mutually Exclusive Alternatives with Equal Project Lives
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 (a) At i= 13%, determine the operating cost per kWh for each motor. (b) At what operating hours are they equivalent?

29 Solution: Operating cost per kWh per unit Determine total input power
Conventional motor: input power = kW/ = kW PE motor: input power = kW/ = kW Bina Nusantara

30 Determine total kWh per year with 3120 hours of operation
Conventional motor: 3120 hrs/yr ( kW) = 65,018 kWh/yr PE motor: 3120 hrs/yr ( 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 Bina Nusantara

31 Total annual equivalent cost: AE(13%) = $4,551 + $1,851 = $6,402
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: AE(13%) = $4,551 + $1,851 = $6,402 Cost per kwh = $6,402/58,188 kwh = $0.1100/kwh AE(13%) = $4,380 + $2,221 = $6,601 Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh Bina Nusantara

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

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

34 PW(15%) = -$12,500 - $5,000 (P/A, 15%, 2) - $3,000 (P/F, 15%, 3)
Model A: $12,500 $5,000 $3,000 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 Bina Nusantara

35 PW(15%) = - $15,000 - $4,000 (P/A, 15%, 3) - $2,500 (P/F, 15%, 4)
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 Bina Nusantara

37 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: Bina Nusantara

38 Typical Graphical Relationship
Total Cost Capital Cost Cost ($) O & M Cost Design Parameter (x) Optimal Value (x*) Bina Nusantara

39 Optimal Cross-Sectional Area
Substation Power Plant A copper conductor Copper price: $8.25/lb Resistance: x10-5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

40 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

41 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) Bina Nusantara

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

43 Total Equivalent Annual Cost
AE = Capital cost + Operating cost = Material cost + Energy loss Find the minimum annual equivalent cost Bina Nusantara