Transcript Contoh Soal PW dan AW Pertemuan 11 dan 12 Tahun

Matakuliah : D 0094 Ekonomi Teknik
Tahun : 2007
Contoh Soal PW dan AW
Pertemuan 11 dan 12
Contoh-Contoh Soal
PW dan variasinya
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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.
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
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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
$500,000
$500,000
$500,000
$500,000
0
$50,000
$2,000,000
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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
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Alternate way to calculate P3
• Concept: Find the effective interest rate per payment period
0
15
$500,000
30
$500,000
45
$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
60
Membandingkan Proyek-Proyek
Mutually Exclusive
•
Mutually Exclusive Projects
•
Alternative vs. Project
•
Do-Nothing Alternative
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Pendapatan Proyek
•
Projects yang pendapatannya
bergantung pada pilihan alternatif
Pelayanan Proyek
•
Projects yang pendapatannya tidak
bergantung pada pilihan alternatif
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Perioda Analisa
•
Rentang waktu dimana pengaruh
ekonomi dari investasi akan
dievaluasi (study period or
planning horizon).
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Perioda Pelayanan Yang Diperlukan
•
Rentang waktu dimana pelayanan
suatu peralatan (or investment)
akan dibutuhkan.
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
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Bagaimana memilih
perioda analisa ?
Finite
Required service
period
Analysis = Required
period
service
period
Case 1
Perioda Analysis
Sama dengan
Masa proyek
Case 2
Perioda Analysis
lebih pendek dari
Masa proyek
Case 3
Perioda Analysis
lebih lama dari
Masa proyek
Case 4
Perioda Analysis
Terlama diantara
Masa proyek dalam grup
Perioda Analysis
Project
repeatability
likely
common multiple
of project lives
Project
repeatability
unlikely
Perioda Analysis
Sama dengan
satu dari masa proyek
Terendah dari
Infinite
Case 1: Analysis Period Equals Project
Lives
Hitung PW untuk tiap proyek selama waktu proyek
$600
$450
$2,075
$500
$1,400
0
$1,000
A
PW (10%)A= $283
PW (10%)B= $579
$4,000
B
$2,110
Membandingkan proyek dengan tingkat
investasi berbeda – Asumsi bahwa dana
yang tidak digunakan akan diinvestasikan
pada MARR.
$600
$450
$500
$2,110
Project A
$2,075
$1,000
3,993
$1,400
$600
$450
$500
Project B
$1,000
This portion
of investment
will earn 10%
return on
investment.
Modified
Project A
PW(10%)A = $283
$3,000
$4,000
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
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 Unlikely
Use common service (revenue) period.
• Project Repeatability Likely
Use the lowest common multiple of
project lives.
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Proyek Berulang Yang Tak Serupa
PW(15%)drill = $2,208,470
PW(15%)lease = $2,180,210
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Assume no
revenues
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
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Contoh-contoh Soal
AW dan variasinya
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Mutually Exclusive Alternatives
with Equal Project Lives
Standard
Motor
Size
25 HP
Cost
$13,000
Life
20 Years
Salvage
$0
Efficiency
89.5%
Energy Cost
$0.07/kWh
Operating Hours 3,120 hrs/yr.
Premium
Efficient Motor
25 HP
$15,600
20 Years
$0
93%
$0.07/kWh
3,120 hrs/yr.
(a) At i= 13%, determine the operating cost per kWh for each motor.
(b) At what operating hours are they equivalent?
Solution:
(a):
•
Operating cost per kWh per unit
Input power =
output power
% efficiency
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
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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
Bina Nusantara
• 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
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(b) break-even
Operating
Hours = 6,742
Mutually Exclusive Alternatives with Unequal Project Lives
Model A:
0
1
2
Required service
Period = Indefinite
3
$3,000
$5,000
$5,000
Analysis period =
LCM (3,4) = 12 years
$12,500
Model B:
Least common
multiple)
0
1
2
3
4
$2,500
$4,000
$15,000
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$4,000
$4,000
0
1
2
3
Model A:
$3,000
$5,000
$5,000
$12,500
• 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
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0
1
2
3
4
Model B:
$2,500
$4,000
$4,000
$4,000
$15,000
• 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) = -$8,954
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Minimum Cost Analysis
• Concept: Total cost is given in terms of a specific design
parameter
• Goal: Find the optimal design parameter that will minimize
the total cost
• Typical Mathematical Equation:
c
AE (i )  a  bx 
x
where x is common design parameter
• Analytical Solution:
x* 
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c
b
Typical Graphical Relationship
Total Cost
Capital Cost
Cost ($)
O & M Cost
Design Parameter (x)
Optimal Value (x*)
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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)
I2R
L
T
1000 A
I = current flow in amperes
R = resistance in ohms
T = number of operating hours
A = cross-sectional area
5000 2 ( 0.008145)
L
(24  365)
1000 A
1,783,755

kwh
A
1,783,755
Energy loss cost 
kwh($0.05)
A
$89,188
=
A
Material Costs
• Material weight in pounds
1000(12)(555) A
 3,854 A
3
12
• Material cost (required investment)
Total material cost = 3,854A($8.25)
= 31,797A
• Salvage value after 25 years: ($0.75)(31,797A)
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Capital Recovery Cost
Given:
Initial cost = $31,797A
Salvage value = $2,890.6A
Project life = 25 years
Interest rate = 9%
2,890.6 A
0
25
31,797 A
Find: CR(9%)
CR (9%) = (31,797 A - 2,890.6 A) ( A / P, 9%, 25)
+ 2,890.6 A (0.09)
= 3,203 A
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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
89,188
A
dAE (9%)
89,188
 3,203 
dA
A2
 0
AE ( 9%)  3,203 A 
A* 
89,188
3,203
 5.276 in 2
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