2015 TIN205 EKONOMI TEKNIK

TIN205 - Ekonomi Teknik

Materi #10 Genap 2014/2015

6623 - Taufiqur Rachman

h t t p://ta ufiqurrach man.webl og.e sa unggu l.ac.id

Materi #10

TIN205 – EKONOMI TEKNIK

Pendahuluan … (1) http://taufiqurrachman.weblog.esaunggul.ac.id

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

 

Seandainya anda menginvestasikan senilai ($1.650) dalam rekening tabungan 6% per tahun. Kemudian anda hanya memperoleh $9.477 pada Januari tahun 2000. Apa arti dari 6% interest disini? Ini adalah opportunity cost jika menyimpan uang pada rekening tabungan, dan merupakan suatu tindakan yang terbaik yang dapat dilakukan saat itu.




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Pendahuluan … (2) http://taufiqurrachman.weblog.esaunggul.ac.id

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





Kemudian, pada tahun 1970, anda memperoleh tawaran untuk investasi lain dengan interest lebih dari 6% untuk investasi lain, anda akan mengambil investasi tersebut? Dalam hal ini, 6% dipandang sebagai minimum attractive rate of return/MARR (rate of return yang dibutuhkan). Maka, anda dapat menetapkan aturan keputusan berikut untuk memperoleh investasi yang diusulkan yang terbaik jika: ROR > MARR



Definisi #1 ROR http://taufiqurrachman.weblog.esaunggul.ac.id

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

Rate of return (ROR) merupakan interest

rate yang diperoleh pada unpaid balance dari angsuran suatu pinjaman.  Contoh: Sebuah bank meminjamkan $10.000 dan menerima pembayaran pertahun sebesar $4.021 selama 3 tahun.  Dalam hal ini, bank dikatakan memperoleh penghasilan kembali (return of) 10% dari pinjaman sebesar $10,000. TIN2005 - Ekonomi Teknik



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Loan Balance Calculation http://taufiqurrachman.weblog.esaunggul.ac.id

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A = $10.000 (A/P, 10%, 3) = $4.021 Year

Unpaid balance at beginning of year

Return on unpaid balance (10%)

Payment received

Unpaid balance at the end of year

0

–$10.000

–$10.000

1

–$10.000

–$1.000

+$4.021

–$6.979

2

–$6.979

–$698

+$4.021

–$3.656

3

–$3.656

–$366

+$4.021

0

A return of 10% on the amount still outstanding at the beginning of each year TIN2005 - Ekonomi Teknik


Contoh #1 (ROR) http://taufiqurrachman.weblog.esaunggul.ac.id

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John membeli barang seni seharga $80.000 dan menjual kembali seharga $53,9 juta pada 40 tahun kemudian.  Berapa rate of return dari investasi John ?




$53,9juta 0 40 $80.000 TIN2005 - Ekonomi Teknik



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Solusi Contoh #1 (ROR) http://taufiqurrachman.weblog.esaunggul.ac.id

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Diketahui: P = $80.000 ; F = $53,9juta ; n = 40 tahun  Ditanya: i  Jawab:




F  P1  i n $53,9juta  $80.0001  i  40

i  17,69% TIN2005 - Ekonomi Teknik


Contoh #2 (ROR) http://taufiqurrachman.weblog.esaunggul.ac.id

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



Pada awal 1985, suatu investasi dari 100 saham seharga $1.650 menjadikan sebuah perusahaan terbuka (go public). Investasi tersebut akan menjadi $13.312.000 pada 31 Januari 2015. Berapa rate of return pada investasi tersebut?

$13.312.000 0 30 $1.650 TIN2005 - Ekonomi Teknik



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Solusi Contoh #2 (ROR) http://taufiqurrachman.weblog.esaunggul.ac.id

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Diketahui: P = $1.650 ; F = $13.312.000 ; n = 30 tahun  Ditanya: i  Jawab:




F  P1  i n $13.312.000  $1.6501  i  30

i  34,97% TIN2005 - Ekonomi Teknik


Definisi #2 ROR http://taufiqurrachman.weblog.esaunggul.ac.id

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



Rate of return (ROR) is defined as breakeven interest rate (i*), which equates the present worth of a project’s cash outflows to the present worth of its cash inflows. Mathematical Relation:




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Definisi #3 ROR (Return on Invested Capital) http://taufiqurrachman.weblog.esaunggul.ac.id

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



Return on invested capital is defined as the interest rate earned on the unrecovered project balance of an investment project. It is commonly known as internal rate of return (IRR). Contoh: A company invests $10.000 in a computer and results in equivalent annual labor savings of $4.021 over 3 years. The company is said to earn a return of 10% on its investment of $10.000.



Project Balance Calculation http://taufiqurrachman.weblog.esaunggul.ac.id

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0

1

2

3

Beginning project balance

–$10.000

–$6.979

–$3.656

Return on invested capital

–$10.000

–$6.979

–$3.656

Payment received

–$10.000

+$4.021

+$4.021

+$4.021

Ending project balance

–$10.000

–$6.979

–$3.656

0

The firm earns a 10% rate of return on funds that remain internally invested in the project. Since the return is internal to the project, we call it internal rate of return. TIN2005 - Ekonomi Teknik



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Metode Perhitungan ROR http://taufiqurrachman.weblog.esaunggul.ac.id


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Metode Solusi Langsung (Log) n

Metode Solusi Metode Solusi Metode Trial & Langsung Grafik Error (Quadratic) Komputer

Proyek A

Proyek B

Proyek C

Proyek D

0

–$1,000

–$2,000

–$75,000

–$10,000

1

0

1,300

24,400

20,000

2

0

1,500

27,340

20,000

3

0

55,760

25,000

4

1,500



Metode Solusi Langsung (Log) Proyek A http://taufiqurrachman.weblog.esaunggul.ac.id

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$1,000 = $1,500(P/F, i, 4) 6623 - Taufiqur Rachman

$1,000 = $1,500(1 + i)–4

0.6667 = (1 + i)–4 ln 0.6667/–4 = ln (1 + i) 0.101365 = ln (1 + i)

ℯ–4 = (1 + i) i = ℯ0.101365 – 1 i = 10.67% TIN2005 - Ekonomi Teknik



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Metode Solusi Langsung (Quadratic) Proyek B http://taufiqurrachman.weblog.esaunggul.ac.id


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Metode Trial & Error Proyek C … (1) http://taufiqurrachman.weblog.esaunggul.ac.id


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

Langkah 1: Guess an interest rate, say, i = 15%



Langkah 2: Compute PW(i) at the guessed i value.



PW (15%) = $3,553



 

Langkah 3: If PW(i) > 0, then increase i. If PW(i) < 0, then decrease i.

PW(18%) = –$749 Langkah 4: If you bracket the solution, you use a linear interpolation to approximate the solution.




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Metode Trial & Error Proyek C … (2) http://taufiqurrachman.weblog.esaunggul.ac.id


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 3,553  i  15%  3%   3,553  749 

3,553

i  17.45%

0 -749

15%

i


18% Materi #10 Genap 2014/2015

Metode Solusi Grafik Komputer Proyek D http://taufiqurrachman.weblog.esaunggul.ac.id

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





Langkah (a): Create a NPW plot using Excel. Langkah (b): Identify the point at which the curve crosses the horizontal axis closely approximates the i*. Catatan: This method is particularly useful for projects with multiple rates of return, as most financial softwares would fail to find all the multiple i*s.




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Situasi Multiple Rates of Return http://taufiqurrachman.weblog.esaunggul.ac.id

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





Aturan keputusan dasar, jika ROR > MARR, maka terima proyek. Aturan ini tidak berlaku untuk situasi dimana investasi mempunyai “multiple rates of return”. Contoh: Cari rate(s) of return dari CF diagram berikut. $2,300 2

0

PW (i)  $1,000  0

1 $1,000

$2,300 $1,320  1 i (1  i)2

$1,320



Solusi Contoh Multiple ROR http://taufiqurrachman.weblog.esaunggul.ac.id

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1 . Then, 1 i $2,300 $1,320 PW (i )  $1,000   (1  i ) (1  i ) 2


Let x 

 $1,000  $2,300 x  $1,320 x 2 0 Solving for x yields, x  10 / 11 or x  10 / 12 Solving for i yields i  10% or 20% TIN2005 - Ekonomi Teknik



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Grafik Multiple ROR http://taufiqurrachman.weblog.esaunggul.ac.id


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NPW Plot for a Non-simple Investment with Multiple Rates of Return.



Kalkulasi Kesetimbangan Proyek http://taufiqurrachman.weblog.esaunggul.ac.id

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i = 20%

n=0

Begining Balance Interest

n=1

n=2

–$1,000

+$1,100

–$200

+$220

Payment

–$1,000

+$2,300

–$1,320

Ending Balance

–$1,000

+$1,100

$0

Cash borrowed (released) from the project is assumed to earn the same interest rate through external investment as money that remains internally invested. TIN2005 - Ekonomi Teknik



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Critical Issue Kesetimbangan Proyek http://taufiqurrachman.weblog.esaunggul.ac.id

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







Can the company be able to invest the money released from the project at 20% externally in period 1? If your MARR is exactly 20%, the answer is “yes”, because it represents the rate at which the firm can always invest the money in its investment pool. Then, the 20% is also true IRR for the project. Suppose your MARR is 15% instead of 20%. The assumption used in calculating i* is no longer valid. Therefore, neither 10% nor 20% is a true IRR.



How to Proceed Kesetimbangan Proyek http://taufiqurrachman.weblog.esaunggul.ac.id

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



If you encounter multiple rates of return, abandon the IRR analysis and use the NPW criterion.

If NPW criterion is used at MARR = 15%.  PW(15%)

= –$1,000 + $2,300 (P/F, 15%, 1) – $1,320 (P/F, 15%, 2 )

= $1.89 > 0  Investasi TIN2005 - Ekonomi Teknik


diterima. Materi #10 Genap 2014/2015

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Aturan Keputusan Non-Simple Investment http://taufiqurrachman.weblog.esaunggul.ac.id

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Kemungkinan multiple ROR  Jika PW(i) seperti Gambar 1, maka IRR = ROR.  Maka jika IRR > MARR, terima proyek. 

Jika PW(i) seperti Gambar 2, maka, IRR  ROR (i*).  Dapatkan IRR sebenarnya, atau  Gunakan method PW.


i*

i

PW (i)




i*

i* i


Aturan Keputusan Non-Simple Investment http://taufiqurrachman.weblog.esaunggul.ac.id


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Incremental Investment http://taufiqurrachman.weblog.esaunggul.ac.id

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n

Proyek A1

Proyek A2

Incremental Investment (A2 – A1)

0

–$1,000

–$5,000

–$4,000

1

$2,000

$7,000

$5,000

ROR

100%

40%

25%

PW(10%)

$818

$1,364

$546





Assuming MARR of 10%, you can always earn that rate from other investment source, i.e., $4,400 at the end of one year for $4,000 investment. By investing the additional $4,000 in A2, you would make additional $5,000, which is equivalent to earning at the rate of 25%. Therefore, the incremental investment in A2 is justified.



Incremental Analysis (Procedure) http://taufiqurrachman.weblog.esaunggul.ac.id


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Langkah 1: • Compute the cash flows for the difference between the projects (A,B) by subtracting the cash flows for the lower investment cost project (A) from those of the higher investment cost project (B). Langkah 2: • Compute the IRR on incremental investment (IRR).

this

Langkah 3: • Accept the investment B if and only if IRR B-A > MARR. TIN2005 - Ekonomi Teknik



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Contoh Incremental ROR http://taufiqurrachman.weblog.esaunggul.ac.id

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Diketahui MARR = 10%, proyek mana yang menjadi pilihan terbaik? n 0 1 2 3 IRR

B1

B2

–$3,000 –$12,000 1,350 4,200 1,800 6,225 1,500 6,330 25%

17.43%

B2 – B1 –$9,000 2,850 4,425 4,830 15%

Karena IRRB2-B1=15% > 10%, dan juga IRRB2 > 10%, pilih B2. TIN2005 - Ekonomi Teknik


IRR Increment Investment 3 Alternatif http://taufiqurrachman.weblog.esaunggul.ac.id

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n 6623 - Taufiqur Rachman

0

D1

D2

D3

–$2,000 –$1,000 –$3,000

1

1,500

800

1,500

2

1,000

500

2,000

3

800

500

1,000

IRR

34.37% 40.76% 24.81%

1. Langkah 1: Tetapkan IRR untuk tiap proyek untuk mngeliminasi tiap project yang gagal memenuhi MARR. 2. Langkah 2: Bandingkan D1 dan D2 berpasangan.

 IRRD1–D2=27.61% > 15%, pilih D1. 3. Langkah 3: Bandingkan D1 dan D3.  IRRD3–D1= 8.8% < 15%, pilih D1. 4. Kesimpulannya D1 adalah alternatif terbaik. TIN2005 - Ekonomi Teknik



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Incremental Borrowing Analysis http://taufiqurrachman.weblog.esaunggul.ac.id

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Principle: 



Decision Rule:

If the difference in flow (B-A) represents an increment of investment, then (A-B) is an increment of borrowing. When considering an increment of borrowing, the rate i*A-B is the rate we paid to borrow money from the increment.








If BRR B-A MARR, select B.

<

If BRR B-A = MARR, select either one. If BRR B-A MARR, select A.

>


Borrowing Rate of Return http://taufiqurrachman.weblog.esaunggul.ac.id


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n

B1

B2

B1–B2

0

–$3,000

–$12,000

+$9,000

1

1,350

4,200

–2,850

2

1,800

6,225

–4,425

3

1,500

6,330

–4,830




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Incremental Analysis for Cost-Only Projects http://taufiqurrachman.weblog.esaunggul.ac.id

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Items

CMS Option

FMS Option


Annual O&M costs:

Annual labor cost

$1,169,600

$707,200

832,320

598,400

3,150,000

1,950,000

Annual tooling cost

470,000

300,000

Annual inventory cost

141,000

31,500

1,650,000

1,917,000

Total annual costs

$7,412,920

$5,504,100

Investment

$4,500,000

$12,500,000

$500,000

$1,000,000

Annual material cost Annual overhead cost

Annual income taxes

Net salvage value TIN2005 - Ekonomi Teknik


Incremental Cash Flow (FMS – CMS) http://taufiqurrachman.weblog.esaunggul.ac.id

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n

CMS Option

FMS Option

Incremental (FMS–CMS)

0

–$4,500,000

–$12,500,000

–$8,000,000

1

–7,412,920

–5,504,100

1,908,820

2

–7,412,920

–5,504,100

1,908,820

3

–7,412,920

–5,504,100

1,908,820

4

–7,412,920

–5,504,100

1,908,820

5

–7,412,920

–5,504,100

1,908,820

6

–7,412,920

–5,504,100

Salvage

+ $500,000

+ $1,000,000



$2,408,820


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Solusi (FMS – CMS) http://taufiqurrachman.weblog.esaunggul.ac.id


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



PW(i)FMS–CMS = –$8,000,000 + $1,908,820(P/A, i, 5) + $2,408,820(P/A, i, 6) = 0

IRRFMS–CMS

= 12.43% < 15%, pilih CMS



Predicting Multiple RORs http://taufiqurrachman.weblog.esaunggul.ac.id

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–100% < i* < infinity

Net Cash Flow Rule of Signs: No. of real RORs (i*s) TIN2005 - Ekonomi Teknik


≤

No. of sign changes in the project cash flows Materi #10 Genap 2014/2015

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Contoh http://taufiqurrachman.weblog.esaunggul.ac.id

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Net Cash flow


n

Sign Change

0

–$100

1

–$20

2

$50

3

0

4

$60

5

–$30

1

6

$100

1

 

1


No. of real i*s = 3. This implies that the project could have (0, 1, 2, or 3) i*s but NOT more than 3.


Accumulated Cash Flow Sign Test http://taufiqurrachman.weblog.esaunggul.ac.id

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Find the accounting sum of net cash flows at the end of each period over the life of the project.




Period (n) Cash Flow (An)

Sum (Sn)

0 1 2

A0 A1 A2

S0 = A0 S1 = S0 + A1 S2 = S1 + A2

n

An

Sn = Sn–1 + An

If the series S starts negatively and changes sign ONLY ONCE, there exists a unique positive i*. TIN2005 - Ekonomi Teknik



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Contoh http://taufiqurrachman.weblog.esaunggul.ac.id

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n

An

Perubanhan tanda

Sn

0

–$100

–$100

1

–$20

–$120

2

$50

–$70

3

0

–$70

4

$60

–$10

5

–$30

–$40

6

$100

$60


Jumlah tanda berubah = 1, indicating a unique i*.

i* = 10.46%

1


Contoh Incremental B/C Ratios http://taufiqurrachman.weblog.esaunggul.ac.id

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A1

A2

A3

I

$5,000

$20,000

$14,000

B

12,000

35,000

21,000

C’

4,000

8,000

1,000

PW(i)

$3,000

$7,000

$6,000




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Solusi http://taufiqurrachman.weblog.esaunggul.ac.id

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A1

A2

A3

A1

I +C’

$9,000

A3

A2

$15,000 $28,000


BC(i) 1.33 1.25 1.40

Ranking Base



Cost-Effectiveness Studies … (1) http://taufiqurrachman.weblog.esaunggul.ac.id

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General Procedure 



Step 1: Establish the goals to be achieved by the analysis.

Step 2: Identify the imposed restrictions on achieving the goals, such as budget or weight.



Step 3: Identify all the feasible alternatives to achieve the goals.



Step 4: Identify the social interest rate to use in the analysis.



Step 5: Determine the equivalent life-cycle cost of each alternative, including research and development, testing, capital investment, annual operating and maintenance costs, and salvage value.




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Cost-Effectiveness Studies … (2) http://taufiqurrachman.weblog.esaunggul.ac.id

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





Step 6: Determine the basis for developing the costeffectiveness index. Two approaches may be used;  (1) the fixed-cost approach and  (2) the fixed-effectiveness approach.  If the fixed-cost approach is used, determine the amount of effectiveness obtained at a given cost.  If the fixed-effectiveness approach is used, determine the cost to obtain the predetermined level of effectiveness. Step 7: Compute the cost-effectiveness ratio for each alternative based on the selected criterion in Step 6. Step 8: Select the alternative with the maximum cost-effective index.



Cost-Effectiveness Decision Criterion http://taufiqurrachman.weblog.esaunggul.ac.id


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Fixed Cost Approach

Fixed Effectiveness Approach

Maximize Effectiveness

Minimize Cost

Subject to:

Subject to:

Budget Constraint

Must meet the minimum effectiveness




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Case Study – Selecting An Weapon System http://taufiqurrachman.weblog.esaunggul.ac.id


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Weapon System Alternatives http://taufiqurrachman.weblog.esaunggul.ac.id

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Alternative A j

Advantage

Disadvantage

Probability of Kill

A1: Inertial navigation system

Low cost, mature technology.

Accuracy, target recognition

0.33

A2: Inertial navigation system (Global positioning system)

Moderate cost, nature technology

Target recognition

0.70

A3: Imaging infrared (I2R)

Accurate, target recognition

High cost, bunkered target detection

0.90

A4: Synthetic aperture radar


High cost

0.99

A5: Laser detection / ranging


High cost, technical maturity

0.99

A6: Millimeter wave (MMW)

Moderate cost, accurate

Target recognition

0.80




23



Life-Cycle Costs for Weapon Development Alternative http://taufiqurrachman.weblog.esaunggul.ac.id

47

Expenditures in Million Dollars 6623 - Taufiqur Rachman

Phase Year FSD

IOC

A1*

A2

A3

A4

A5

A6

0

15

19

50

40

75

28

1

18

23

65

45

75

32

2

19

22

65

45

75

33

3

15

17

50

40

75

27

4

90

140

200

200

300

150

5

95

150

270

250

360

180

6

95

160

280

275

370

200

7

90

150

250

275

340

200

8

80

140

200

200

330

170

PW(10%)

315.92 492.22 884.27 829.64 1,227.23 612.70



Cost-Effectiveness Index http://taufiqurrachman.weblog.esaunggul.ac.id

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Type Cost/Unit Probability of Kill Cost/Kill

Kill/Cost

A1

$31,592

0.33

$95,733 0.0000104

A2

49,220

0.70

70,314 0.0000142

A3

88,427

0.90

98,252 0.0000102

A4

82,964

0.90

83,802 0.0000119

A5

122,723

0.99

123,963 0.0000081

A6

61,370

0.80

76,713 0.0000130




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Cost-Effectiveness Graph http://taufiqurrachman.weblog.esaunggul.ac.id

49

130.000

Unacceptable Region

A5

Maximize Effectiveness

110.000 Cost/kill


120.000

100.000

A3

A1

90.000 A4

80.000

A6 A2

70.000

Fixed Cost

60.000 300

400


500 600 700 800 900 1.000 1.100 1.200 1.300 Present value of life cycle cost ($ million) Materi #10 Genap 2014/2015


h t t p://ta ufiqurrach man.webl og.e sa unggu l.ac.id

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2015 TIN205 EKONOMI TEKNIK

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