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Matakuliah: D0762 – Ekonomi Teknik Tahun: 2009. Equivalence And Compound Interest Course Outline 2. Outline. Time Value of Money Mathematical Factor How to Use Interest Table Nominal and Effective Interest References : Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p.44-99

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Matakuliah d0762 ekonomi teknik tahun 2009

Matakuliah: D0762 – Ekonomi Teknik

Tahun: 2009

Equivalence And Compound InterestCourse Outline 2


Matakuliah d0762 ekonomi teknik tahun 2009

Outline

  • Time Value of Money

  • Mathematical Factor

  • How to Use Interest Table

  • Nominal and Effective Interest

    References :

  • Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p.44-99

  • Engineering Economic Analysis, Donald G. Newman, p. 41-86

  • Engineering Economy, William G. Sulivan, p.137-194, p. 135-140

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Matakuliah d0762 ekonomi teknik tahun 2009

Time Value of Money

Question: Would you prefer $100 today or $100 after 1 year?

There is a time value of money. Money is a valuable asset, and people would pay to have money available for use. The charge for its use is called interest rate.

Question: Why is the interest rate positive?

  • Argument 1:Money is a valuable resource, which can be “rented,” similar to an apartment. Interest is a compensation for using money.

  • Argument 2:Interest is compensation for uncertainties related to the future value of the money.


Matakuliah d0762 ekonomi teknik tahun 2009

Time Value of Money (example)

  • Example: You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years. There are many ways the debt can be repaid

    • Plan A: At end of each year pay $1,000 principal plus interest due.

    • Plan B: Pay interest due at end of each year and principal at end of five years.

    • Plan C: Pay in five end-of-year payments.

    • Plan D: Pay principal and interest in one payment at end of five years.


Matakuliah d0762 ekonomi teknik tahun 2009

Time Value of Money (example)

  • You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years.

    Plan A: At end of each year pay $1,000 principal plus interest due.


Matakuliah d0762 ekonomi teknik tahun 2009

Time Value of Money (example)

  • You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years.

    Plan B: Pay interest due at end of each year and principal at end of five years.


Matakuliah d0762 ekonomi teknik tahun 2009

Time Value of Money (example)

  • You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years.

    Plan C: Pay in five end-of-year payments.


Matakuliah d0762 ekonomi teknik tahun 2009

Time Value of Money (example)

  • You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years.

    Plan D: Pay principal and interest in one payment at end of five years.


Matakuliah d0762 ekonomi teknik tahun 2009

Mathematical Factor

  • Single Payment Compound Formula

    If you put P in the bank now at an interest rate of i% for n years,

    the future amount you will have after n years is given by

    F = P (1+i)n

    The term (1+i)n is called the single payment compound factor.

    The factor is used to compute F, given P, and given i and n.

    Handy Notation.

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

    F = P (1+i)n = P (F/P,i,n).


Matakuliah d0762 ekonomi teknik tahun 2009

Mathematical Factor

  • Present Value

    Example

    If you want to have $800 in savings at the end of four years, and 5% interest is paid annually, how much do you need to put into the savings account today?

    We solve P (1+i)n = F for P with i = 0.05, n = 4,

    F = $800.

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

    = 800/(1.05)4 = 800 (1.05)-4 = 800 (0.8227) = $658.16.

    Single Payment Present Worth Formula P = F/(1+i)n = F(1+i)-n


Matakuliah d0762 ekonomi teknik tahun 2009

A

A

A

A

A

0

1

2

3

4

5

Uniform Series

  • Uniform series formula derivation

n = 5 periods

F = A (1+i)4 + A (1+i)3 + A (1+i)2 + A (1+i) + A

F = A [(1+i)4 + (1+i)3 + (1+i)2 + (1+i) + 1 ]

(1+i) F = A [(1+i)5 + (1+i)4 + (1+i)3 + (1+i)2 + (1+i)]

(1+i) F – F = A [(1+i)5 – 1]

F = A [(1+i)5 – 1]/i

A = F i/[(1+i)5 – 1]F = A [(1+i)5 – 1]/i

F


Matakuliah d0762 ekonomi teknik tahun 2009

Uniform Series

Uniform series compound amount factor:

(F/A,i,n) = [(1+i)n – 1]/i, i > 0

Uniform series sinking fund

(A/F,i,n) = i/[(1+i)n – 1]

Example 4-1. Jane deposits $500 in a credit union at the end of each year for five years. The CU pays 5% interest, compounded annually. At the end of five years, immediately following her fifth deposit, how much will Jane have in her account?

F = A (F/A,i,n) = A [(1+i)n – 1]/i = $500[(1.05)5 – 1]/(0.05)

= $500 (5.5256) = $2,762.82  $2,763.


Matakuliah d0762 ekonomi teknik tahun 2009

Uniform Series

Uniform series compound amount factor:

(F/A,i,n) = [(1+i)n – 1]/i, i > 0

Uniform series sinking fund:

(A/F,i,n) = i/[(1+i)n – 1]

Uniform series capital recovery:

(A/P,i,n) = [i (1 + i)n]/[(1+i)n – 1]

Uniform series present worth:

(P/A,i,n) = [(1+i)n – 1]/[i (1 + i)n]

Given F, Find A

Given A, Find F

Given A, Find P

Given P, Find A


Matakuliah d0762 ekonomi teknik tahun 2009

How to Use Table


Matakuliah d0762 ekonomi teknik tahun 2009

Relationships Between Compound Interest Factors

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

P=A[(1+i)n–1]/[i(1 + i)n] =A(P/A,i,n) A =P[i(1 + i)n]/[(1+i)n–1]=P(A/P,i,n)

F=A{[(1+i)n – 1]/i}=A(F/A,i,n) A=F{i/[(1+i)n–1]}=F(A/F,i,n)

(F/P,i,n) = 1/ (P/F,i,n)

Present Worth factor

Compound Amount factor

Uniform Series Capital Recovery Factor

(A/P,i,n) = 1/(P,A,i,n)

Uniform Series

Present Worth Factor

Uniform Series Compound

Amount Factor

(A/F,i,n) = 1/ (F/A,i,n)

Uniform Series Sinking Fund Factor


Matakuliah d0762 ekonomi teknik tahun 2009

240

210

180

150

120

120

120

120

120

120

120

90

60

30

+

=

0

i= 5%

A= 120

G= 30

P?

Arithmetic Gradient

Suppose you buy a car. You wish to set up enough money in a bank account to pay for standard maintenance on the car for the first five years. You estimate the maintenance cost increases by G = $30 each year. The maintenance cost for year 1 is estimated as $120.

Thus, estimated costs by year are $120, $150, $180, $210, $240.

P = A(P/A,5%,5) + G(P/G,5%,5)

= 120(P/A,5%,5) + 30 (P/G,5%,5)

= 120 (4.329) + 30 (8.237)

= 519 + 247 = $766

Standard Form Diagram for Arithmetic Gradient: n periods and n-1 nonzero flows in increasing order


Arithmetic gradient

Arithmetic Gradient

0

0

1

1

2

2

3

3

….

….

n

n

F = G(1+i)n-2 + 2G(1+i)n-3 + … + (n-2)G(1+i)1 + (n-1)G(1+i)0

F= G [ (1+i)n-2 + 2(1+i)n-3 + … + (n-2)(1+i)1 + n-1]

(1+i) F = G [(1+i)n-1 + 2(1+i)n-2 + 3(1+i)n-3 + … + (n-1)(1+i)1]

iF= G [(1+i)n-1 + (1+i)n-2 + (1+i)n-3 + … + (1+i)1 – n + 1] =

= G [(1+i)n-1 + (1+i)n-2 + (1+i)n-3 + … + (1+i)1 + 1] – nG =

= G (F/A, i, n) - nG = G [(1+i)n-1]/i – nG

F= G [(1+i)n-in-1]/i2

P= F (P/F, i, n) = G [(1+i)n-in-1]/[i2(1+i)n]

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

= G [(1+i)n-in-1]/i2 × i/[(1+i)n-1]

A= G [(1+i)n-in-1]/[i(1+i)n-i]

(n-1)G

2G

G

…..

0

…..

F

17


Arithmetic gradient1

Arithmetic Gradient

Arithmetic Gradient Uniform Series

Arithmetic Gradient Present Worth

(P/G,5%,5) =

= {[(1+i)n – i n – 1]/[i2 (1+i)n]}

= {[(1.05)5 – 0.25 – 1]/[0.052 (1.05)5]}

= 0.026281562/0.003190703 = 8.23691676.

(P/G,i,n) = { [(1+i)n – i n – 1] / [i2 (1+i)n] }

=1/(G/P,i,n)

(A/G,i,n) = { (1/i )– n/ [(1+i)n –1] }

=1/(G/A,i,n)

(F/G,i,n) = G [(1+i)n-in-1]/i2

=1/(G/F,i,n)

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Arithmetic gradient2

Arithmetic Gradient

400

300

300

200

200

i= 6%

100

100

100

100

100

100

+

=

0

A= 100

G= 100

A

A

A

A

This is not in the standard form for using the gradient equation, because the year-one cash flow is not zero.

We reformulate the problem as follows.

400

300

200

100

A

A

A

A

A = A1+ G (A/G,6%,4) =100 + 100 (1.427) = $242.70

Example 4-6. Maintenance costs of a machine start at $100 and go up by $100 each year for 4 years. What is the equivalent uniform annual maintenance cost for the machinery if i = 6%.

19


Geometric gradient

Geometric Gradient

Example Suppose you have a vehicle. The first year maintenance cost is estimated to be $100. The rate of increase in each subsequent year is 10%. You want to know the present worth of the cost of the first five years of maintenance, given i = 8%.

Repeated Present-Worth (Step-by-Step) Approach:

20


Geometric gradient1

Geometric Gradient

Geometric Gradient. At the end of year i, i = 1, ..., n, we incur a cost Aj = A(1+g)i-1.

P= A(1+i)-1+A(1+q)1(1+i)-2+A(1+q)2(1+i)-3+…

…+A(1+q)n-2(1+i)-n+1+A(1+q)n-1(1+i)-n

P(1+q)1(1+i)-1= A(1+q)1(1+i)-2+A(1+q)2(1+i)-3+…

…+A(1+q)n-1(1+i)-n+A(1+q)n(1+i)-n-1

P - P(1+q)1(1+i)-1= A(1+i)-1- A(1+q)n(1+i)-n-1

P (1+i-1-q) = A (1 - (1+q)n(1+i)-n)

i≠q:P = A (1 - (1+q)n(1+i)-n)/(i-q)

i=q:P = A n(1+i)-1

A(1+q)n-1

A(1+q)2

A(1+q)1

A

0

2

3

….

n

21


Geometric gradient2

Geometric Gradient

Example n = 5, A1 = 100, g = 10%, i = 8%.

(P/A,g,i,n) = (P/A,10%,8%,5) = 4.8042

P = A(P/A,g,i,n) = 100 (4.8042) = $480.42.

Go Back

22


Nominal effective interest

Nominal & Effective Interest

Nominal Interest Rate

Ten thousand dollars is borrowed for two years at an interest rate of 24% per year compounded quarterly.

If this same sum of money could be borrowed for the same period at the same interest rate of 24% per year compounded annually, how much could be saved in interest charges?

interest charges for quarterly compounding:

$10,000(1+24%/4)2*4-$10,000 = $5938.48

interest charges for annually compounding:

$10,000(1+24%)2 - $10,000 = $5376.00

Savings:

$5938.48 - $5376 = $562.48

Compounding is not less important than interest

You have to know all the information to make a good decision

23


Matakuliah d0762 ekonomi teknik tahun 2009

Effective Interest

  • The effective interest rates you pay are a function of how much money you have available and how much money you give up for the use of these funds.

  • In the simplest form of borrowing, a one-year loan of $ 10,000 at 12% interest will costs $ 1,200.

    The effective interest rate is $ 1,200 / $ 10,000 or 12%.

    As we change the costs and/or amount of funds available, the effective interest rate will change.


Matakuliah d0762 ekonomi teknik tahun 2009

Nominal & Effective Interest

  • This topic is very important. Sometimes one interest rate is quoted, sometimes another is quoted. If you confuse the two you can make a bad decision. 

  • A bank pays 5% compounded semi-annually. If you deposit $1000, how much will it grow to by the end of the year?

  • The bank pays 2.5% each six months.

  • You get 2.5% interest per period for two periods.

  • 1000  1000(1.025) = 1,025  1025(1.025) = $1,050.60

  • With i = 0.05/2, r = 0.05,

  • P  (1 + i) P  (1+r/2)2 P = (1+ 0.05/2)2 P = (1.050625) P


Matakuliah d0762 ekonomi teknik tahun 2009

Nominal & Effective Interest

Terms the example illustrates:

r = 5% is called the nominal interest rateper interest period(usually one year)

i = 2.5 % is called theeffective interest rate per interest period

ia = 5.0625% is called the effective interest rate per year

In the example:m = 2 is the number of compounding subperiodsper time period.

ia = (1.050625) – 1 = (1.025)2 – 1 = (1+0.05/2)2 – 1

  • r = nominal interest rate per year 

  • m = number of compounding sub-periods per year

  • i = r/m = effective interest rate per compounding sub-period.


Nominal and effective interest rate

Nominal and Effective Interest Rate

27


Nominal and effective interest rate1

Nominal and Effective Interest Rate

Example Joe Loan Shark lends money on the following terms. “ Duh, If I gib you 50 dolla on Monday, den youse guys owes me 60 dolla da following Monday.”

1.What is the nominal rate, r?

We first note Joe charges i = 20% a week,

since 60 = (1+i)50  i = 0.2.

Note we have solved F = 50(F/P,i,1) for i.  

We know m = 52, so r = 52  i = 10.4, or 1,040% a year.  

2.What is the effective rate, ia ?

From ia = (1 + r/m)m – 1

we have ia = (1+10.4/52)52 – 1  13,104.

This means about 1,310,400 % a year.

28


Matakuliah d0762 ekonomi teknik tahun 2009

Nominal and Effective Interest Rate

Suppose Joe can keep the $50, as well as all the money he receives in payments, out in loans at all times? How much would Joe L. Shark have at the end of the year?

We use F = P(1+i)n to get F = 50(1.2)52 $655,232

(not bad, but probably illegal) 

Words of Warning. When the various compounding periods in a problem all match, it makes calculations much simpler.

When they do not match, life is more complicated.

Recall Example. We put $5000 in an account paying 8% interest, compounded annually. We want to find the five equal EOY withdrawals.

We used A = P(A/P,8%,5) = 5000  (0.2505)  $1252.

Suppose the various periods are not the same in this problem


Nominal and effective interest rate2

Nominal and Effective Interest Rate

W

W

W

W

i = 2%, n = 20

$5000

Sally deposits $5000 in a CU paying 8%nominal interest, compounded quarterly. She wants to withdraw the money in five equal yearly sums, beginning Dec. 31 of the first year. How much should she withdraw each year? 

Note : effective interest is i = 2% = r/4 = 8%/4 quarterly, and there are 20 periods.

Solution:

30


Nominal and effective interest rate3

the withdrawal periods and the compounding periods are not the same.

If we want to use the formula

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

then we must find a way to put the problem into an equivalent form

where all the periods are the same.

Solution 1. Suppose we withdraw an amount A quarterly. (We don’t, but suppose we do.). We compute 

A = P (A/P,i,n) = 5000 (A/P,2%,20) = 5000 (0.0612) = $306.

These withdrawals are equivalent to P = $5000

Nominal and Effective Interest Rate

i = 2%, n = 20

$5000

31


Nominal and effective interest rate4

Nominal and Effective Interest Rate

Now consider the following:

Consider a one-year period:

This is now in a standard form that repeats every year.

W = A(F/A,i,n) = 306 (F/A,2%,4) = 306 (4.122) = $1260.

 Sally should withdraw $1260 at the end of each year.

A

W

W

W

W

i = 2%, n = 4

W

32


Nominal and effective interest rate5

Nominal and Effective Interest Rate

Solution 2 (Probably the easiest way)

ia = (1 + r/m)m – 1 = (1 + i)m – 1 = (1.02)4 – 1 = 0.0824  8.24%

Now use:

W = P(A/P,8.24%,5) = P {[i (1+i)n]/[(1+i)n – 1]} = 5000(0.252)

= $1260 per year.

W

$5000

ia = 8.24%, n = 5

33


Continous compounding

Continous Compounding

34


Continuous compounding

Continuous Compounding

We will pay little attention to continuous compounding in this course. You are supposed to read the material on continuous compounding in the book, but it will not be included in the homework or tests.

ia = (1 + i)m – 1 =

= (1 + r/m)m – 1

r = nominal interest rate per year 

m = number of compounding sub-periods per year

i = r/m = effective interest rate per compounding sub-period.

Continuous compounding can sometimes be used to simplify computations, and for theoretical purposes. The table above illustrates that er - 1 is a good approximation of (1 + r/m)m for large m. This means there are continuous compounding versions of the formulas we have seen earlier.

For example,

F = P ern is analogous to F = P (F/P,r,n): (F/P,r,n)inf= ern

P = F e-rn is analogous to P = F (P/F,r,n): (P/F,r,n)inf= e-rn

35


Summary notation

Summary Notation

i: effective interest rate per interest period (stated as a decimal) 

n: number of interest periods 

P: present sum of money  

F: future sum of money: an amount, n interest periods from the present, that is equivalent to P with interest rate i 

A: end-of-period cash receipt or disbursement amount in a uniform series, continuing for n periods, the entire series equivalent to P or F at interest rate i. 

G: arithmetic gradient: uniform period-by-period increase or decrease in cash receipts or disbursements 

g: geometric gradient: uniform rate of cash flow increase or decrease from period to period

r: nominal interest rate per interest period (usually one year)

ia: effective interest rate per year (annum) 

m: number of compounding sub-periods per period

36


Summary formulas

Summary: Formulas

Single Payment formulas:

Compound amount:F = P (1+i)n = P (F/P,i,n)

Present worth:P = F (1+i)-n = F (P/F,i,n)

Uniform Series Formulas

Compound Amount:F = A{[(1+i)n –1]/i} = A (F/A,i,n)

Sinking Fund:A = F {i/[(1+i)n –1]}= F (A/F,i,n)

Capital Recovery:A = P {[i(1+i)n]/[(1+i)n – 1]= P (A/P,i,n)

Present Worth:P = A{[(1+i)n – 1]/[i(1+i)n]} = A (P/A,i,n)

Arithmetic Gradient Formulas

Present WorthP = G {[(1+i)n – i n – 1]/[i2 (1+i)n]} = G (P/G,i,n)

Uniform SeriesA = G {[(1+i)n – i n –1]/[i (1+i)n – i]} = G (A/G,i,n)

Geometric Gradient Formulas

If i  g, P = A {[1 – (1+g)n(1+i)-n]/(i-g)}= A (P/A,g,i,n)

If i = g,P = A [n (1+i)-1] = A (P/A,g,i,n)

37


Summary formulas1

Summary: Formulas

Nominal interest rate per year, r: the annual interest rate without considering the effect of any compounding  

Effective interest rate per year, ia:

ia = (1 + r/m)m – 1 = (1+i)m – 1 with i = r/m

Continuous compounding, :

r – one-period interest rate, n – number of periods

(P/F,r,n)inf= e-rn

(F/P,r,n)inf= ern

38


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