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

Contoh-Contoh Soal PW dan variasinya Bina Nusantara

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

Cash Flow Diagram Bina Nusantara

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

Proyek Hydro Mr. Brcewell Bina Nusantara

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

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

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%

15 30 45 60 $50,000 $500,000 $500,000 $500,000 $500,000 $2,000,000 Bina Nusantara

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

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 = (1 + 0.05)15 - 1 = 107.893% Capitalized equivalent worth P3 = $500,000/1.07893 = $463,423

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Contoh-contoh Soal AW dan variasinya Bina Nusantara

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?

Solution: 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 Bina Nusantara

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 Bina Nusantara

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

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

Mutually Exclusive Alternatives with Unequal Project Lives Model A: 0 1 2 3 $12,500 $5,000 $3,000 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 Bina Nusantara

PW(15%) = -$12,500 - $5,000 (P/A, 15%, 2) - $3,000 (P/F, 15%, 3) Model A: $12,500 $5,000 $3,000 0 1 2 3 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

PW(15%) = - $15,000 - $4,000 (P/A, 15%, 3) - $2,500 (P/F, 15%, 4) 0 1 2 3 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

Bina Nusantara

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

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

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

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

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

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

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