Session_ii_time Value of Money

04/17/23

Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don

Smith Texas A&M University 1

Time Value of Money

ENGINEERING ECONOMY

SESSION_II

Nilai Waktu Uang adalah menghitung nilai uang terhadap waktu

II. TIME VALUE OF MONEY

2.1. PENGERTIAN

Rp 15.000/ zakTh 1990

Rp 30.000/ zakTh 2000

Rp 50.00o/ zakTh 2010

Fenomena : • untuk mendapatkan suatu barang yang jenis dan jumlahnya sama diperlukan jumlah uang yang lebih banyak•Nilai uang senantiasa berubah (cenderung turun)• Daya beli uang senantiasa menurun (inflasi)

Fenomena : • untuk mendapatkan suatu barang yang jenis dan jumlahnya sama diperlukan jumlah uang yang lebih banyak•Nilai uang senantiasa berubah (cenderung turun)• Daya beli uang senantiasa menurun (inflasi)

Deposito AwalRp 10.000

Th 1990

Fenomena : Uang Rp 10.000 pada tahun 1990 dengan bunga 1 % /tahun menjadi 10.100 Uang 10.000 pd th 1990 secara finansial adalah sama dengan 10.100 pada tahun 1991 Kesamaan finansial inilah yang disebut dengan ekivalensi

Fenomena : Uang Rp 10.000 pada tahun 1990 dengan bunga 1 % /tahun menjadi 10.100 Uang 10.000 pd th 1990 secara finansial adalah sama dengan 10.100 pada tahun 1991 Kesamaan finansial inilah yang disebut dengan ekivalensi

TIME VALUE OF MONEY (Cont’)

Deposito AkhirRp 10.100

Th 1991


EkivalensiEkivalensi

Jumlah uang yang dipinjam atau yang diinvestasikan (P)

Jumlah uang yang dipinjam atau yang diinvestasikan (P)

Periode/ waktu peminjaman atau

investasi (n)

Periode/ waktu peminjaman atau

investasi (n)

Tingkat Bunga yang dikenakan (i)

Tingkat Bunga yang dikenakan (i)

Ekuivalensi

Time Value of Money

Ekuivalensi

Time Value of Money

Single PaymentSingle Payment Annual PaymentAnnual Payment

Future WorthFuture WorthPresent WorthPresent Worth Gradient SeriesGradient SeriesAnnual SeriesAnnual Series

n

F

PP

n i

A


B. ARUS KAS (CASH FLOW)

Diagram aliran kas adalah suatu ilustrasi grafis dari transaksi-transaksi finansial yang dilukiskan pada garis skala waktu.

Diagram arus kas menggunakan beberapa konvensi

Garis horisontal

Merupakan suatu skala waktu (time scale) dengan pergerakan waktu dari kiri ke kanan. Label periode (tahun, kuartal, bulan, dll),

1 19980 2 1999

Anak panah

Menyatakan arus kas dan ditempatkan pada akhir periode.Anak panah ke bawah menyatakan pengeluaran-pengeluaran (arus kas negatif/ arus kas keluar)Anak panah ke atas menyatakan penerimaan-penerimaan (arus kas positif/ arus kas masuk)

Aliran biaya (cash flow) : selisih antara total arus kas yang masuk (penerimaan) dan arus kas yang keluar (pengeluaran) untuk suatu periode waktu tertentu.

Panjangnya panah vertikal tidak selalu harus mencerminkan skala besarnya transaksi,

namun transaksi lebih besar harus digambarkan dengan panah lebih panjang

ARUS KAS (CASH FLOW)

Ditinjau dari pihak investor

Titik Tinjauan

Diagram arus kas tergantung pada titik tinjauan.Misal : Jika arus kas dilihat pada pihak peminjam maka uang yang dipinjam (arus kas masuk) pembayaran (arus kas keluar), tapi jika kita tinjau dari pihak bank uang pinjaman merupakan arus kas keluar dan pengembalian merupakan arus kas masuk

Arus kas keluar = uang diinvestasikan

Arus kas masuk = uang yang akan diterima/ hasil dari investasi

Ditinjau dari pihak peminjam

Arus kas keluar = uang diinvestasikan

Arus kas masuk = uang yang akan diterima/ hasil dari investasi

Peroide waktu : hari, bulan, semester, kuartal atau

tahun

0 3 421

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Blank & Tarquin: 5-th Edition Ch. 1 Authored by: Dr. Don Smith Texas

A&M University 8

1. Foundations: Overview1. F/P and P/F Factors

2. P/A and A/P Factors

3. F/A and A/F Factors

4. Interpolate Factor Values

5. P/G and A/G Factors

6. Geometric Gradient

7. Calculate i

8. Calculate “n”

9. Spreadsheets

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SESSION II Section 1

F/P and P/F Factors

ENGINEERING ECONOMY

F/P and P/F Factors

Pembayaran Tunggal(Single Payment)

Pembayaran Tunggal(Single Payment)

Future Worth Mencari Nilai Present Value

Jika Nilai Future Value Diketahui

Future Worth Mencari Nilai Present Value


Present WorthMencari Nilai Present Value


Present WorthMencari Nilai Present Value


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2.1 Basic Derivations: F/P factor

F/P Factor To find F given P

P0

Fn

N

………….

To Find F given P

Compound forward in time

Future Worth adalah proses matematis penambahan bunga pada induk sehingga terjadi penambahan jumlah induk secara nominal pada periode mendatang disebut pemajemukan

F = P (1 + i)

Dimana : P = nilai waktu sekarangF = nilai waktu yang akan datangn = jumlah periode pinjamani = suku bunga

n F = P (F/P, i,n)

FUTURE WORTH (FV) /NILAI UANG WAKTU YANG AKAN DATANG

Menggunakan Tabel bunga :Tabel i %, n tahun dan Kolom F/P

Mencari Nilai Future Worth (F) Jika Nilai Present Worth (P) Diketahui

0 2 3 n1

pemajemukan

..........

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2.1 Derivation by Recursion: F/P factor

F1 = P(1+i)

F2 = F1(1+i)…..but:

F2 = P(1+i)(1+i) = P(1+i)2

F3 =F2(1+i) =P(1+i)2 (1+i) = P(1+i)3

In general:

FN = P(1+i)n

FN = P(F/P,i%,n)

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2.2 Example- F/P Analysis

Example: P= $1,000;n=3;i=10%What is the future value, F?

0 1 2 3P=$1,000

F = ??

i=10%/year

F3 = $1,000[F/P,10%,3] = $1,000[1.10]3

= $1,000[1.3310] = $1,331.00

15

SOAL PERHITUNGAN NILAI FUTURE VALUE

SOAL 1 Bila Anda menyimpan uang sebesar Rp 100 juta rupiah pada akhir tahun 2006 dan 2007 dan menyimpan lagi pada akhir tahun 2009 sebesar 150 jt di salah satu bank. Berapakah besar nilai uang yang akan Anda dapatkan pada akhir tahun 2010 jika diketahui pada saat itu suku bunga sebesar 15 %/tahun Penyelesaian Diketahui : P(1) simpanan akhir th 2006 = Rp 100 jt P(2) simpanan akhir th 2007 = Rp 100 jt P(3) simpanan akhir th 2009 = Rp 150 jt

Suku bunga (i) = 15 % pertahun Periode (n1) = 4 tahun Ditanyakan : Penerimaan (F) di akhir tahun 2009

F= P1 + P2 + P3

F = P1 (F1/P1,i,n) + P2 (F2/P2,i,n) + P3 (F3/P3,i,n) F = P1(F/P;15%;4)+P2 (F/P;15%;3)+P3 (F/P;15%;1) F = 100(1,749) + 100(1,521) + 150(1,150)

F = 174,9 + 152,1 + 172,5 F = 499,5 juta

2006 2007 2008 2009 2010

P1 = 100 jt

Ft = ……..?

Lihat tabel bunga kolom F/P untuk i= 15 % n (1) = 4 tahun n(2) = 3 tahun n (3) = 1 tahun

P2 = 100 jt P3 = 150 jt

Present Worth : suatu proses untuk menghitung nilai uang waktu sekarang yang nilainya beberapa periode mendatang diketahui disebut dengan diskonting (discounting)Secara formula dapat ditulis sbb :

F PV = (1 + i)

Dimana : P = nilai waktu sekarangF = nilai waktu yang akan datangn = jumlah periode pinjamani = suku bunga

n

P = F (P/F, i,n)

Koefisien bunga :Tabel bunga Tabel i %, n tahun dan Kolom F/P

Mencari Nilai Present Worth (P) Jika Nilai Future Worth (F) Diketahui

0 2 3 n1

diskonting

..........

2.1 Present Worth Factor from F/P

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2.1 Present Worth Factor from F/P

Since FN = P(1+i)n

We solve for P in terms of FN

P = F{1/ (1+i)n} = F(1+i)-n

Thus:P = F(P/F,i%,n) where

(P/F,i%,n) = (1+i)-n

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2.1 P/F factor – discounting back in time

Discounting back from the future

P

Fn

N

………….

P/F factor brings a single future sum back to a specific point in time.

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2.2 Example – P/F Analysis

Assume F = $100,000, 9 years from now. What is the present worth of this amount now if i =15%?

0 1 2 3 8 9

…………

F9 =

$100,000

P= ??

i = 15%/yr

P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15)9)

= $100,000(0.1111) = $11,110 at time t = 0

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SESSION Section 2

P/A and A/P Factors

ENGINEERING ECONOMY

Pembayaran Tahunan(Annual Payment)

Pembayaran Tahunan(Annual Payment)

Gradient Series(G)

Gradient Series(G)

Mencari nilai Present Worth (P) jika di ketahui nilai Annual (A)

Mencari nilai Present Worth (P) jika di ketahui nilai Annual (A)

Annual Series(A)

Annual Series(A)

Mencari nilai Future Worth (F) jika di ketahui nilai Annual (A)

Mencari nilai Future Worth (F) jika di ketahui nilai Annual (A)

Mencari nilai Annual (A) jika di ketahui nilai Present Worth (P)

Mencari nilai Annual (A) jika di ketahui nilai Present Worth (P)

Mencari nilai Annual (A) jika di ketahui nilai Future Worth (F)Mencari nilai Annual (A) jika

di ketahui nilai Future Worth (F)

Bertambah setiap tahunBertambah setiap tahun

Berkurang setiap tahunBerkurang setiap tahun

Annual Payment

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2.2 Uniform Series Present Worth and Capital Recovery Factors

Annuity Cash Flow

$A per period

P = ??

0

………….. n 1 2 3 .. ..

n-1

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Desire an expression for the present worth – P of a stream of equal, end of period cash flows - A

0 1 2 3 n-1 n

A = given

P = ??

Adalah menghitung nilai waktu sekarang (PV) dari sejumlah uang tahunan (angsuran)

Dimana : F = nilai waktu sekarangA = pembayaran tahunan seragamn = jumlah periode pinjamani = suku bunga

P = A (P/A, i,n)

Menggunakan Tabel bunga

Mencari Nilai Present Worth (P) Jika Nilai Annual (A) Diketahui

(1 ) 1 0

(1 )

n

n

iP A for i

i i

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Write a Present worth expression

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

Term inside the brackets is a geometric progression.

Mult. This equation by 1/(1+i) to yield a second equation

[1]

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The second equation

[2]

To isolate an expression for P in terms of A, subtract Eq [1] from Eq. [2]. Note that numerous terms will drop out.

2 3 1

1 1 1 1..

1 (1 ) (1 ) (1 ) (1 )n n

PA

i i i i i

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Setting up the subtraction

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

[1]

2 3 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

[2]

-

1

1 1

1 (1 ) (1 )n

iP Ai i i

= [3]

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Simplifying Eq. [3] further

1

1 1

1 (1 ) (1 )n

iP Ai i i

1

11

(1 )nA

Pi i

(1 ) 1 0

(1 )

n

n

iP A for i

i i

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This expression will convert an annuity cash flow to an equivalent present worth amount one period to the left of the first annuity cash flow.

(1 ) 1 0

(1 )

n

n

iP A for i

i i

/ %, P A i n factor

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2.2 Capital Recovery Factor A/P, i%, n

Given the P/A factor

(1 ) 1 0

(1 )

n

n

iP A for i

i i

(1 )

(1 ) 1

n

n

i iA P

i

Solve for A in terms of P

Yielding….

A/P,i%,n factor

The present worth point of an annuity cash flow is always one period to the left of the first A amount

CONTOH SOAL MENCARI NILAI PRESENT VALUE JIKA NILAI ANNUAL DIKETAHUI 1. Hitunglah nilai sekarang (present value) sebuah pembayaran tahunan sebesar 5

jt setiap akhir tahun selama 10 tahun, jika bunga yang berlaku selama 10 tahun tersebut sebesar 8 %

Penyelesaian Diketahui : Pembayaran setiap akhir tahun selama 10 th = Rp 5 jt/ tahun Suku bunga (i) = 8 % pertahun Periode pembayaran = 10 tahun Ditanyakan : Nilai Present Value (P) Penyelesaian

P = F1 + F2…….+ F10 P = F1 (P/F,i,n1) + F2 (P/F,i,n2)…….+ F10(P/F,i,n10)

P = 5(0,9259) + 5(0,8573) + 5(0,7938) + 5(0,7350) + 5(0,6806) + 5(0,6302) + 5(0,5835) + 5(0,5403) 5(0,5002) + 5(0,4632) P = 33,5500 juta Menggunakan Tabel

Periode (n) Future Value (F)

(juta) Koefisien (P/F,8 %)

Present Value (P) (juta)

1 5 0,9259 4.6295 2 5 0,8573 4.2865 3 5 0,7938 3.9690 4 5 0,7350 3.6750 5 5 0,6806 3.4030 6 5 0,6302 3.1510 7 5 0,5835 2.9175 8 5 0,5403 2.7015 9 5 0,5002 2.5010 10 5 0,4632 2.3160

Total nilai (P) Present Value (juta) 33.5500

2 4 3 10 5 8 7 9 6 1

P = …….?

A = 5 jt setiap akhir tahun selama 10 tahun

Lihat tabel bunga i= 8 % kolom P/F u/ masing2 n

CARA I (PERTAMA) : Menggunakan Rumus Pembayaran Tunggal (Single Payment)

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Atau menggunakan cara yang lebih mudah dengan rumus pembayaran Annual Series P = A (P/A, i,n) P = 5 (P/A,8%,10) P = 5 (6,710) P = 33,5500 juta

CARA II (KEDUA) : Menggunakan Rumus Pembayaran Tahunan Seragam (Annual Payment)

SOAL PERHITUNGAN MENCARI NILAI ANNUAL JIKA DIKETAHUI PRESENT VALUE

1. Sebuah pinjaman sebesar Rp 33,55 juta dengan skenario pengembalian diangsur

tiap akhir tahun selama 10 tahun dengan nilai angsuran yang tetap. Jika suku bunga yang berlaku adalah 8 % hitunglah besar angsuran tahunan tersebut.

Penyelesaian Diketahui : Pinjaman awal (P) = Rp 33,55 juta Suku bunga (i) = 8 % pertahun Periode pembayaran = 10 tahun Ditanyakan : Nilai Pembayaran Tahunan (A) Penyelesaian

Menggunakan rumus A = P(A/P, i, n) A = 33,55 (A/P, 8 %, 10) A = 33,55 (0,14903)

A = 4.999957 juta , dibulatkan = 5 juta

2 4 3 10 5 8 7 9 6 1

P = 33,55 juta

A = ...........?

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F/A and A/F Factors

ENGINEERING ECONOMY Fifth Edition

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2.3 F/A and A/F Derivations

Annuity Cash Flow

0

………….. N

$A per period

$F

Find $A given the Future amt. - $F

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2.3 Sinking Fund and Series Compound amount factors (A/F and F/A)

Take advantage of what we already haveRecall:

Also:

1

(1 )nP F

i

(1 )

(1 ) 1

n

n

i iA P

i

Substitute “P” and simplify!

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2.3 A/F Factor

By substitution we see:

Simplifying we have:Which is the (A/F,i%,n) factor

1 (1 )

(1 ) (1 ) 1

n

n n

i iA F

i i

(1 ) 1nA

i

iF

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2.3 F/A factor from the A/F Factor

Given:

Solve for F in terms of A

(1 ) 1n

iA F

i

)=A

(1 1F

ni

i

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2.3 F/A and A/F Derivations

Annuity Cash Flow

0

………….. N

$A per period

$F

Find $F given thethe $A amounts

Adalah menghitung nilai waktu yang akan datang (FV) dari sejumlah uang tahunan (angsuran)

A(1+i) -1 F = i

Dimana : F = nilai waktu sekarangA = pembayaran tahunan n = jumlah periode pinjamani = suku bunga

n

F = A (F/A, i,n)n


Mencari Nilai Future Value (F) Jika Nilai Annual (A) Diketahui

F = ….?

n i

diketahui

Mencari Nilai Future Value (P) Jika Nilai Annual (A) Diketahui

F = ….?

n i

A A A

diketahuiA A A

Dimana : F = nilai waktu yang akan datangA = pembayaran tahunan seragam n = jumlah periode pinjamani = suku bunga

SOAL : Mencari F jika A diketahui

n i

A = …….?

Mencari Nilai Annual (A) Jika Nilai Future Value (F) Diketahui

F Diketahui

n i

A = ………?F Diketahui


SOAL : Mencari Annual jika F diketahui

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SOAL PERHITUNGAN NILAI FUTURE VALUE DARI ANNUAL 1. Hitunglah nilai yang akan datang (future value) pada akhir tahun 10 sebuah

pembayaran termin proyek yang dibayar secara tahunan sebesar 5 jt setiap akhir tahun selama 10 tahun, jika bunga yang berlaku selama 10 tahun tersebut sebesar 8 %

Penyelesaian Diketahui : Pembayaran setiap akhir tahun selama 10 th = Rp 5 jt/ tahun Suku bunga (i) = 8 % pertahun Periode pembayaran = 10 tahun Ditanyakan : Nilai Future Value (F) Penyelesaian Cara I (Pertama)

F = F1 + F2…….+ F10

F = P1 (F/P,i,n1) + P2 (F/P,i,n2)…….+ P10(F/P,i,n10) F = 5(1) + 5(1,080) + 5(1,166) + 5(1,260) + 5(1,360) + 5(1,469) + 5(1,587) + 5(1,714) + 5(1,851) 5(1,999) F = 72,430 juta

2 4 3 10 5 8 7 9 6 1

F = …….?

A = 5 jt setiap akhir tahun selama 10 tahun

Lihat tabel bunga i= 8 % kolom P/F u/ masing2 n

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Cara II (Kedua) Menggunakan Tabel

Periode (n) Nilai Present Value (F)

(juta) Koefisien (F/P,8

%) Nilai Future Value (P)

(juta) 0 5 1 5 1 5 1.080 5.400 2 5 1.166 5.830 3 5 1.260 6.300 4 5 1.360 6.800 5 5 1.469 7.345 6 5 1.587 7.935 7 5 1.714 8.570 8 5 1.851 9.255 9 5 1.999 9.995

Total nilai Present Value (P) 72,430

Cara III (Ketiga) Dapat pula kita menggunakan cara dengan mencari semua Present Valuenya dulu kemudian dicari nilai Future Value (lihat contoh soal Present Value jika Annual diketahui), diperoleh Nilai Present Value = 33,5500 juta F = P(F/P, i, n) F = 33.5500 (F/P, 8 %, 10) F = 33.5500 (2,159) F = 72,4344 juta Cara IV (Keempat) Atau menggunakan cara yang lebih mudah dengan rumus pembayaran Annual Series F = A (F/A, i,n) F = 5 (F/A,8%,10) F = 5 (14,487) F = 72,435 juta

SINKING FUND Mencari nilai tahunan (annual) bila diketahui nilai yang akan datang FV dengan tingkat bunga tertentu dan periode tertentu

F i A = (1 + i) - 1

Dimana : F = nilai waktu sekarangA = pembayaran tahunan n = jumlah periode pinjamani = suku bunga

nA = F (A/F, i,n)



n i

A = …….?


F Diketahui

n i

A = ………?F Diketahui


SOAL : Mencari Annual jika F diketahui

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SOAL PERHITUNGAN MENCARI NILAI ANNUAL JIKA DIKETAHUI FUTURE VALUE

1. Jika Anda berharap 10 tahun yang akan datang untuk mendapatkan uang sebesar Rp

72,430. Hitunglah besar angsuran tahunan yang harus Anda simpan dalam sebuah bank jika suku bunga tabungan yang berlaku selama 10 tahun adalah 8 %

Penyelesaian Diketahui : Pinjaman awal (P) = Rp 72,430 juta Suku bunga (i) = 8 % pertahun Periode pembayaran = 10 tahun Ditanyakan : Nilai Pembayaran Tahunan (A) Penyelesaian

Menggunakan rumus A = F(A/F, i, n) A = 72,430 (A/F, 8 %, 10) A = 72,430 (0,06903)

A = 4,999843 juta , dibulatkan = 5 juta

2 4 3 10 5 8 7 9 6 1

F = 72,430 juta

A = ...........?

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Interpolation in Interest Tables

ENGINEERING ECONOMY

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P/G and A/G Factors

ENGINEERING ECONOMY

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2.5 Arithmetic Gradient Factors

• In applications, the annuity cash flow pattern is not the only type of pattern encountered

•Two other types of end of period patterns are common

•The Linear or arithmetic gradient

•The geometric (% per period) gradient

•This section presents the Arithmetic Gradient

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• An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a contestant amount over n time periods.

•A linear gradient is always comprised of TWO components:

•The Gradient component

•The base annuity component

•The objective is to find a closed form expression for the Present Worth of an arithmetic gradient

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2.5 Linear Gradient Example

Assume the following:

0 1 2 3 n-1 N

A1+G

A1+2G

A1+n-2G

A1+n-1G

This represents a positive, increasing arithmetic gradient

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2.5 Example: Linear Gradient

• Typical Negative, Increasing Gradient: G=$50

The Base Annuity= $1500

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2.5 Example: Linear Gradient

• Desire to find the Present Worth of this cash flow

The Base Annuity= $1500

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• The “G” amount is the constant arithmetic change from one time period to the next.

•The “G” amount may be positive or negative!

•The present worth point is always one time period to the left of the first cash flow in the series or,

•Two periods to the left of the first gradient cash flow!

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2.5 Derivation: Gradient Component Only

Focus Only on the gradient Component

A1+G

A1+2G

A1+n-2G

A1+n-1G

0 1 2 3 n-1 N

“0” G

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2.5 Present Worth Point…

The Present worth point of a linear gradient is always: 2 periods to the left of the “1G” point or,

1 period to the left of the very first cash flow in the gradient series.

DO NOT FORGET THIS!

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0 1 2 3 4 5 6 7

$100$200

$300

$400

$500

$600

$700

X

The Present Worth Point of the Gradient

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2.5 Gradient Component

$0

$100$200

$300

$400

$500

$600

X


0 1 2 3 4 5 6 7

•The Gradient Component

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0 1 2 3 4 5 6 7

Base Annuity – A = $100

X


•PW of the Base Annuity is at t = 0

•PWBASE Annuity=$100(P/A,i%,7)

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2.5 Present Worth: Gradient Component

The PW of the Base Annuity is simply the Base Annuity –A{P/A, i%, n} factorWhat is needed is a present worth expression for the gradient component cash flow.We need to derive a closed form expression for the gradient component.

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2.5 The P/G factor for i and N

G (1 ) 1P=

i (1 ) (1 )

N

N N

i N

i i i

( / , %, )P G i N factor

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2.5 Further Simplification on P/G

2

(1 ) 1( / , %, )

(1 )

N

N

i iNP G i N

i i

Remember, the present worth point of any linear gradient is 2 periods to the left of the 1-G cash flow or, 1 period to the left of the “0-G” cash flow.

P=G(P/G,i,n)

BERTAMBAH SETIAP TAHUN Adalah mencari konversi nilai tahunan yang seragam/konstan (annual) dari pembayaran yang tidak konstan (gradient) yang bertambah setiap periode dengan pertambahan yang tetap

A = G (A/G,i,n)

Dimana : G = Gradien nilai angsuran A = pembayaran tahunan n = jumlah periode pinjamani = suku bunga


Pembayaran Tahunan GRADIENT SERIES PAYMENT

Mencari Nilai Annual (A) dari pembayaran Gradient (Tiap Tahun Bertambah)

F

Dimana : P = nilai waktu sekarangF = nilai waktu yang akan datangG = Gradient (nilai pertambahan yang tetapA = pembayaran tahunan seragam n = jumlah periode pinjamani = suku bunga

n iF

A

n iP P

SOAL_GRADIENT_BERTAMBAH

BERKURANG SETIAP TAHUN Adalah mencari nilai tahunan yang seragam/konstan (annual) dari pembayaran yang tidak konstan (gradient) yang berkurang setiap periode dengan pengurangan yang tetap

A = X - G (A/G,i,n)

Dimana : G = Gradien nilai angsuran X = Pembayaran yang paling besarA = pembayaran tahunan n = jumlah periode pinjamani = suku bunga


Pembayaran Tahunan GRADIENT SERIES PAYMENT

Mencari Nilai Annual (A) dari pembayaran Gradient (Tiap Tahun Berkurang)

F

Dimana : P = nilai waktu sekarangF = nilai waktu yang akan datangG = Gradient (nilai pertambahan yang tetap)X = nilai angsuran yang tertinggiA = pembayaran tahunan seragam n = jumlah periode pinjamani = suku bunga

n iF

A

n iP P

SOAL_GRADIENT_BERKURANG

X

RESUME EKUIVALENSI

FORMULASI DEFENISI KETERANGAN

P = F(P/F,i,n)

Mencari nilai sekarang (PV) jika nilai yang akan datang (FV) ditentukan

F = P(F/P,i,n)

Mencari nilai yang akan datang (FV) jika nilai sekarang (PV) ditentukan

P = ....?

FN=periodei= suku bunga

Pi= suku bunga

N=periode

F = ....?


A = P(A/P,i,n)

Mencari nilai angsuran (annual) jika nilai sekarang (PV) ditentukan

A = F(A/F,i,n)

Mencari nilai angsuran (annual) jika nilai yang akan datang (FV) ditentukan

P

A=...?

N i

iN

F

A=...?

RESUME EKUIVALENSI


P = A(P/A,i,n)

Mencari nilai sekarang (PV) dari nilai angsuran (annual) ditentukan

F = A(F/A,i,n)

Mencari nilai yang akan datang (FV) dari nilai angsuran (annual) ditentukan

P =….?

A

N i

iN

F=…..?

A

RESUME EKUIVALENSI


A = G(A/G,i,n)

Mencari nilai sekarang (PV) dari nilai angsuran (annual) ditentukan

A = X- G(A/G,i,n)

Mencari nilai yang akan datang (FV) dari nilai angsuran (annual) ditentukan

A=…?

N i

iN

N i

A=…?X

RESUME EKUIVALENSI

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2.5 Extension – The A/G factor

Some authors also include the derivation of the A/G factor.A/G converts a linear gradient to an equivalent annuity cash flow.Remember, at this point one is only working with gradient componentThere still remains the annuity component that you must also handle separately!

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2.5 The A/G Factor

Convert G to an equivalent A

( / , , )( / , , )A G P G i n A P i n

How to do it…………

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2.5 Gradient Example

• Consider the following cash flow

0 1 2 3 4 5

$100$200

$300$400

$500

Find the present worth if i = 10%/yr; n = 5 yrs

Present Worth Point is here!And the G amt. = $100/period

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2.5 Gradient Example- Base Annuity

• First, The Base Annuity of $100/period

0 1 2 3 4 5

A = +$100

•PW(10%) of the base annuity = $100(P/A,10%,5)

•PWBase = $100(3.7908)= $379.08

•Not Finished: We need the PW of the gradient component and then add that value to the $379.08 amount

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2.5 The Gradient Component

0 1 2 3 4 5

$0$100

$200$300

$400

We desire the PW of the Gradient Component at t = 0PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5)

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2.5 The Gradient Component

0 1 2 3 4 5

$0$100

$200$300

$400

PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5)

G (1 ) 1P=

i (1 ) (1 )

N

N N

i N

i i i

Could substitute n=5, i=10% and G = $100 into the P/G closed form to get the value of the factor.

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2.5 PW of the Gradient Component

PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5)

Calculating or looking up the P/G,10%,5 factor yields the following:

Pt=0 = $100(6.8618) = $686.18 for the

gradient PW

P/G,10%,5)

G (1 ) 1P=

i (1 ) (1 )

N

N N

i N

i i i

Sub. G=$100;i=0.10;n=5

6.8618

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2.5 Gradient Example: Final Result

• PW(10%)Base Annuity = $379.08

•PW(10%)Gradient Component= $686.18

•Total PW(10%) = $379.08 + $686.18

•Equals $1065.26

•Note: The two sums occur at t =0 and can be added together – concept of equivalence

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2.5 Shifted Gradient Example

• Consider the following Cash Flow

0 1 2 3 4 5 6 7

$600$550

$500$450

•The PW @ t = 0 requires getting the PW @ t =3;

•Then using the P/F factor move PW3 back to t=0

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Geometric Gradient

Mc

Hill

ENGINEERING ECONOMY

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2.4 Interpolation of Factors

• All texts on Engineering economy will provide tabulated values of the various interest factors usually at the end of the text in an appendix

• Refer to the back of your text for those tables.

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2.4 Interpolation of Factors

• Typical Format for Tabulated Interest Tables

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2.4 Interpolation (Estimation Process)

• At times, a set of interest tables may not have the exact interest factor needed for an analysis

• One may be forces to interpolate between two tabulated values

• Linear Interpolation is not exact because:

• The functional relationships of the interest factors are non-linear functions

• Hence from 2-5% error may be present with interpolation.

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2.4 An Example

• Assume you need the value of the A/P factor for i = 7.3% and n = 10 years.

• 7.3% is most likely not a tabulated value in most interest tables

• So, one must work with i = 7% and i = 8% for n fixed at 10

• Proceed as follows:

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2.4 Basic Setup for Interpolation

•Work with the following basic relationships

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2. 4 i = 7.3% using the A/P factor

• For 7% we would observe:

COMPOUND PRESENT SINKING COMPOUND CAPITAL

N AMT. FACTOR WORTH FUND AMOUNT RECOVERY

F/P P/F A/F F/A A/P

10 1.9672 0.5083 0.0724 13.8164 0.14238

A/P,7%,10) = 0.14238

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2. 4 i = 7.3% using the A/P factor

• For i = 8% we observe:

COMPOUND PRESENT SINKING COMPOUND CAPITAL

N AMT. FACTOR WORTH FUND AMOUNT RECOVERY

F/P P/F A/F F/A A/P

10 2.1589 0.4632 0.0690 14.4866 0.14903

(A/P,8%,10) = 0.14903

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2. 4 Estimating for i = 7.3%

• Form the following relationships

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2.4 Final Estimated Factor Value

• Observe for i increasing from 7% to 8% the A/P factors also increases.

• One then adds the estimated increment to the 7% known value to yield:

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2.4. The Exact Value for 7.3%

• Using a previously programmed spreadsheet model the exact value for 7.3% is:

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2.6 Geometric Gradients

• An arithmetic (linear) gradient changes by a fixed dollar amount each time period.

•A GEOMETRIC gradient changes by a fixed percentage each time period.

•We define a UNIFORM RATE OF CHANGE (%) for each time period

•Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next

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2.6 Geometric Gradients: Increasing

• Typical Geometric Gradient Profile

•Let A1 = the first cash flow in the series

0 1 2 3 4 …….. n-1 n

A1 A1(1+g)A1(1+g)2

A1(1+g)3

A1(1+g)n-1

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2.6 Geometric Gradients: Decreasing

• Typical Geometric Gradient Profile

•Let A1 = the first cash flow in the series

A1

A1(1-g)

A1(1-g)2

A1(1-g)3

A1(1-g)n-1

0 1 2 3 4 …….. n-1 n

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2.6 Geometric Gradients: Derivation

• First Major Point to Remember:

•A1 does NOT define a Base Annuity/

•There is not BASE ANNUITY for a Geometric Gradient!

•The objective is to determine the Present Worth one period to the left of the A1 cash flow point in time

•Remember: The PW point in time is one period to the left of the first cash flow – A1!

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2.6 Geometric Gradients: Derivation

• For a Geometric Gradient the following parameters are required:

•The interest rate per period – i

•The constant rate of change – g

•No. of time periods – n

•The starting cash flow – A1

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2.6 Geometric Gradients: Starting

• Pg = The Aj’s time the respective (P/F,i,j)

factor

•Write a general present worth relationship

to find Pg…. 2 11 1 1 1

1 2 3

(1 ) (1 ) (1 )...

(1 ) (1 ) (1 ) (1 )

n

g n

A A g A g A gP

i i i i

Now, factor out the A1 value and rewrite as..

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1 2 1

1 2 3

1 (1 ) (1 ) (1 )...

(1 ) (1 ) (1 ) (1 )

n

g n

g g gP A

i i i i

(1)

(1+g)Multuply both sides by to create another equation

(1+i)1 2 1

1 2 3

(1+g) (1+g) 1 (1 ) (1 ) (1 )...

(1+i) (1+i) (1 ) (1 ) (1 ) (1 )

n

g n

g g gP A

i i i i

(2)

Subtract (1) from (2) and the result is…..

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1 1

1+g (1 ) 11

1+i (1 ) 1

n

g n

gP A

i i

Solve for Pg and simplify to yield….

1

11

1 g i

n

g

gi

P Ai g

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2.6 Geometric Gradient P/A factor

• This is the (P/A,g,i,n) factor and is valid if g not equal to i.

1

11

1 g i

n

g

gi

P Ai g

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2.6 Geometric Gradient P/A factor

•Note: If g = i we have a division by “0” – undefined.

•For g = i we can derive the closed form PW factor for this special case.

•We substitute i for g into the Pg relationship to yield:

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2.6 Geometric Gradient: i = g Case

g 1

1 1 1 1P =A ...

(1+i) (1+i) (1+i) (1+i)

1

(1 )g

nAP

i

For the case i = g

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2.6 Geometric Gradients: Summary

•Pg = A1(P/A,g,i,n)

1

(1 )g

nAP

i

1

11

1 g i

n

g

g

iP A

i g

g not = to i

Case: g = i

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2.6 Geometric Gradient: Notes

•The geometric gradient requires knowledge of:

•A1, i, n, and g

•There exist an infinite number of combinations for i, n, and g: Hence one will not find tabulated tables for the (P/A, g,i,n) factor.

•You have to calculated either from the closed form for each problem or apply a pre-programmed spreadsheet model to find the needed factor value

•No spreadsheet built-in function for this factor!

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2.6 Geometric Gradient: Example

•Assume maintenance costs for a particular activity will be $1700 one year from now.

•Assume an annual increase of 11% per year over a 6-year time period.

•If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0.

•First, draw a cash flow diagram to represent the model.

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2.6 Geometric Gradient Example (+g)

•g = +11% per period; A1 = $1700; i = 8%/yr

0 1 2 3 4 5 6 7

$1700 $1700(1.11)1

$1700(1.11)2

$1700(1.11)3

$1700(1.11)5

PW(8%) = ??

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2.6 Solution

303: Use "g" 667: use f-bar

Geometric Gradients"E" or g or f-bar = 11%

i= 8%N= 7

P/A,g,i,n factor is…… 7.04732

First Amt= 1,700.00$ P. Value = 11,980.44$

• P = $1700(P/A,11%,8%,7)

•Need to calculate the P/A factor from the closed-form expression for a geometric gradient.

•From a spreadsheet we see:

1

11

1 g i

n

g

g

iP A

i g

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2.6 Geometric Gradient ( -g )

• Consider the following problem with a negative growth rate – g.

0 1 2 3 4

g = -10%/yr; i = 8%; n = 4

A1 = $1000 $900

$810$729

P0=??

We simply apply a “g” value = -0.10

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2.6 Geometric Gradient (-g value)

• Evaluate:

1

11

1 g i

n

g

g

iP A

i g

1

11

1 g i

n

g

g

iP A

i g

For a negative g value = -0.10

303: Use "g" 667: use f-bar

Geometric Gradients"E" or g or f-bar = -10%

i= 8%N= 4

P/A,g,i,n factor is…… 2.87637

First Amt= 1,000.00$ P. Value = 2,876.37$

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Determination of an Unknown Interest Rate

ENGINEERING ECONOMY

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2.7 When the i – rate is unknown

• A class of problems may deal with all of the parameters know except the interest rate.

•For many application-type problems, this can become a difficult task

•Termed, “rate of return analysis”

•In some cases:

•i can easily be determined

•In others, trial and error must be used

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2.7 Example: i unknown

• Assume on can invest $3000 now in a venture in anticipation of gaining $5,000 in five (5) years.

•If these amounts are accurate, what interest rate equates these two cash flows?

0 1 2 3 4 5

$3,000

$5,000

•F = P(1+i)n

•5,000 = 3,000(1+i)5

•(1+i)5 = 5,000/3000 = 1.6667

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2.7 Example: i unknown

• Assume on can invest $3000 now in a venture in anticipation of gaining $5,000 in five (5) years.

•If these amounts are accurate, what interest rate equates these two cash flows?

0 1 2 3 4 5

$3,000

$5,000

•(1+i)5 = 5,000/3000 = 1.6667

•(1+i) = 1.66670.20

•i = 1.1076 – 1 = 0.1076 = 10.76%

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2.7 For “i” unknown

• In general, solving for “i” in a time value formulation is not straight forward.

•More often, one will have to resort to some form of trial and error approach as will be shown in future sections.

•A sample spreadsheet model for this problem follows.

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2.7 Example of the IRR function

=IRR($D7:$D12)

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Determination of Unknown Number of Years

ENGINEERING ECONOMY

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2.8 Unknown Number of Years

• Some problems require knowing the number of time periods required given the other parameters

•Example:

•How long will it take for $1,000 to double in value if the discount rate is 5% per year?

•Draw the cash flow diagram as….

0 1 2 . . . . . . ……. n

P = $1,000

Fn = $2000

i = 5%/year; n is unknown!

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• Solving we have…..

0 1 2 . . . . . . ……. n

P = $1,000

Fn = $2000

•Fn=? = 1000(F/P,5%,x): 2000 = 1000(1.05)x

•Solve for “x” in closed form……

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• Solving we have…..

•(1.05)x = 2000/1000

•Xln(1.05) =ln(2.000)

•X = ln(1.05)/ln(2.000)

•X = 0.6931/0.0488 = 14.2057 yrs

•With discrete compounding it will take 15 years to amass $2,000 (have a little more that $2,000)

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2.8 No. of Years – NPER function

• From Excel one can formulate as:

=NPER(C23,C22,C20,C21)

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Spreadsheet Application – Basic Sensitivity Analysis

ENGINEERING ECONOMY

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2.9 Basic Sensitivity Analysis

• Sensitivity analysis is a procedure applied to a formulated problem whereby one can assess the impact of each input parameter relating to the output variable.

•Sensitivity analysis is best performed using a spreadsheet model.

•The procedure is to vary the input parameters within certain ranges and observe the change on the output variable.

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• By proper modeling, one can perform “what-if” analysis on one or more of the input parameters and observe any changes in a targeted output (response) variable

•Commercial add-in packages are available that can be linked to Excel to perform such an analysis

•Specifically: Palisade Corporation’s TopRank Excel add-in is most appropriate.

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• When you build your own models, devise an approach to permit varying at least one of the input parameters and store the results of each change in the output variable…then plot the results.

•If a small change in one of the input parameters represents a significant change in the output variable then…

•That input variable is “sensitive”

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• If an input parameter is deemed “sensitive” then some effort should go into the estimation of that parameter

•Because it does influence the response (output) variable.

•Less sensitive input parameters may not have as much effort required to estimate as those input parameters do not have that much impact on the targeted response variable.

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• When you build your own models, devise an approach to permit varying at least one of the input parameters and store the results of each change in the output variable…then plot the results.

•If a small change in one of the input parameters represents a significant change in the output variable then…

•That input variable is “sensitive”

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SESSION II Summary

Mc

GrawHill

ENGINEERING ECONOMY

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Chapter Summary

• This chapter presents the fundamental time value of money relationships common to most engineering economic analysis calculations

•Derivations have been presented for:

•Present and Future Worth- P/F and F/P

•Annuity Cash flows – P/A, A/P, F/A and A/F

•Gradients – P/G, A,G and P/A,g,i,n

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Chapter Summary

• One must master these basic time value of money relationships in order to proceed with more meaningful analysis that can impact decision making.

•These relationships are important to you professionally and in your personal lives.

•Master these concepts!!!

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SESSION II End of Slide Set

ENGINEERING ECONOMY

Session_ii_time Value of Money

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