contoh soal pw dan aw pertemuan 11 dan 12

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

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 ?

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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%

Bina Nusantara

$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

Bina Nusantara

�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.

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

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.

Bina Nusantara

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.

Bina Nusantara

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

Bina Nusantara

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?

Bina Nusantara

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

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

• 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

Bina Nusantara

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)

Bina Nusantara

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

Bina Nusantara

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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