contoh soal pw dan aw pertemuan 11 dan 12
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Contoh Soal PW dan AW Pertemuan 11 dan 12. Matakuliah: D 0094 Ekonomi Teknik Tahun: 2007. Contoh-Contoh Soal PW dan variasinya. Contoh Soal. - PowerPoint PPT PresentationTRANSCRIPT
Contoh Soal PW dan AWPertemuan 11 dan 12
Matakuliah : D 0094 Ekonomi TeknikTahun : 2007
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Contoh-Contoh SoalPW dan variasinya
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Contoh SoalA 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.a) 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
b) 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
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Cash Flow Diagram
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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 ?
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Proyek Hydro Mr. Brcewell
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%
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$500,000 $500,000 $500,000 $500,000
$2,000,000
$50,000
0 15 30 45 60
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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
Alternate way to calculate P3
• Concept: Find the effective interest rate per payment period
• 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
15 30 45 600
$500,000 $500,000 $500,000 $500,000
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Membandingkan Proyek-Proyek Mutually Exclusive
� Mutually Exclusive Projects
� Alternative vs. Project
� Do-Nothing Alternative
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�Pendapatan ProyekProjects yang pendapatannya bergantung pada pilihan alternatif
�Pelayanan ProyekProjects yang pendapatannya tidak bergantung pada pilihan alternatif
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�Perioda AnalisaRentang waktu dimana pengaruh ekonomi dari investasi akan dievaluasi (study period or planning horizon).
�Perioda Pelayanan Yang DiperlukanRentang waktu dimana pelayanan suatu peralatan (or investment) akan dibutuhkan.
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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
Finite
Required service period
Infinite
Analysis = Requiredperiod service period
Projectrepeatabilitylikely
Project repeatabilityunlikely
Perioda Analysis Sama dengan Masa proyek
Perioda AnalysisTerlama diantara
Masa proyek dalam grup
Perioda AnalysisTerendah dari
common multipleof project lives
Perioda AnalysisSama dengan
satu dari masa proyek
Perioda Analysis lebih pendek dari
Masa proyek
Perioda Analysis lebih lama dari Masa proyek
Case 1
Case 2
Case 3
Case 4
Bagaimana memilih perioda analisa ?
Case 1: Analysis Period Equals Project Lives
Hitung PW untuk tiap proyek selama waktu proyek
$450$600
$500 $1,400
$2,075$2,110
0
$1,000 $4,000A B
PW (10%) = $283PW (10%) = $579
A
B
$1,000
$450$600
$500
Project A
$1,000
$600
$500$450
$3,000
3,993
$4,000
$1,400
$2,075
$2,110
Project BModifiedProject A
Membandingkan proyek dengan tingkat investasi berbeda – Asumsi bahwa dana yang tidak digunakan akan diinvestasikan pada MARR.
PW(10%)A = $283PW(10%)B = $579
This portionof investmentwill earn 10%return on investment.
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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
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Comparison of unequal-lived service projects when the required service period is shorter than the individual project
life
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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.
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Comparison for Service Projects with Unequal Lives when the required service period is longer than the individual project life
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Case 4: Analysis Period is Not Specified
• Project Repeatability UnlikelyProject Repeatability Unlikely
Use Use common service (revenue) period. period.
• Project Repeatability LikelyProject Repeatability Likely
Use the Use the lowest common multiple of of project lives.project lives.
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Proyek Berulang Yang Tak Serupa
Assume no revenues
PW(15%)drill = $2,208,470
PW(15%)lease = $2,180,210
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Proyek Berulang Yang Serupa
PW(15%)A=-$53,657
PW(15%)B=-$48,534
Model A: 3 YearsModel B: 4 yearsLCM (3,4) = 12 years
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Contoh-contoh Soal AW dan variasinya
Mutually Exclusive Alternativeswith Equal Project Lives
Standard Premium Motor Efficient Motor25 HP 25 HP$13,000 $15,60020 Years 20 Years$0 $089.5% 93%$0.07/kWh $0.07/kWh3,120 hrs/yr. 3,120 hrs/yr.
SizeCostLifeSalvageEfficiencyEnergy CostOperating Hours
(a) At i= 13%, determine the operating cost per kWh for each motor.(b) At what operating hours are they equivalent?
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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
Input power =output power
% efficiency
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• 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
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• 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
(b) break-evenOperating Hours = 6,742
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Model A: 0 1 2 3
$12,500
$5,000 $5,000$3,000
Model B: 0 1 2 3 4
$15,000
$4,000 $4,000 $4,000$2,500
Mutually Exclusive Alternatives with Unequal Project Lives
Required servicePeriod = Indefinite
Analysis period =LCM (3,4) = 12 years
Least common multiple)
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Model A:
$12,500$5,000 $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)]
= -$53,657AE(15%) = -$53,657(A/P, 15%, 12) = -$9,899
Model B:
$15,000$4,000 $4,000
$2,500
0 1 2 3 4
$4,000
• 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) = -$8,954
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Minimum Cost Analysis
• Concept: Total cost is given in terms of a specific design parameter• Goal: Find the optimal design parameter that will minimize the total cost• Typical Mathematical Equation:
where x is common design parameter• Analytical Solution:
AE i a bxc
x( )
xc
b*
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Typical Graphical Relationship
O & M Cost
Capital Cost
Total Cost
Design Parameter (x)
Optimal Value (x*)
Cost
($)
Optimal Cross-Sectional Area
Power Plant
Substation
1,000 ft.5,000 amps24 hours365 days
A copper conductor• Copper price: $8.25/lb• Resistance: 0.8145x10-5in2/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%
• Energy loss in kilowatt-hour (L)
Operating Cost (Energy Loss)
LI R
AT
2
1000
LA
A
5000 0 008145
100024 365
1 783 755
2 ( . )( )
, ,kwh
I = current flow in amperesR = resistance in ohmsT = number of operating hoursA = cross-sectional area
Energy loss cost kwh($0.05)
=$89,188
1 783 755, ,
A
A
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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)
1000 12 555
123 854
3
( )( ),
AA
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Capital Recovery Cost
31,797 A
2,890.6 A
0
25
CR A A A P
A
A
( = ( , - , . ) ( / , )
+ , . ( . )
= ,203
9%) 31 797 2 890 6 9%, 25
2 890 6 0 09
3
Given:Initial cost = $31,797ASalvage value = $2,890.6AProject life = 25 yearsInterest rate = 9%
Find: CR(9%)
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Total Equivalent Annual Cost
• Total equivalent annual cost AE = Capital cost + Operating cost = Material cost + Energy loss
• Find the minimum annual equivalent cost
AE AA
dAE
dA A
A
( ,203,
(,203
,
,
,203
.
*
9%) 389 188
9%)3
89 188
0
89 188
3
5 276
2
in2