Matakuliah : D0762 – Ekonomi Teknik
Tahun
: 2009
Equivalence And Compound Interest
Course Outline 2
Outline
•
•
•
•
next
Time Value of Money
next
Mathematical Factor
next
How to Use Interest Table
next
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
Bina
University
- Nusantara
Engineering
Economy, William G. Sulivan, p.137-194, p.
135-140
2
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.
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.
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.
1
2
3
4
5
6
Year
Amnt.
Owed
Int. Owed
Total Owed
2+3
Princip.
Payment
Total
Payment
int* 2
1
5,000
400
5,400
1,000
1,400
2
4,000
320
4,320
1,000
1,320
3
3,000
240
3,240
1,000
1,240
4
2,000
160
2,160
1,000
1,160
5
1,000
80
1,080
1,000
1,080
Sum
15,000
1,200
16,200
5,000
6,200
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.
1
2
3
4
5
6
Year
Amnt.
Owed
Int. Owed
Total Owed
2+3
Princip.
Payment
Total
Payment
int*2
1
5,000
400
5,400
0
400
2
5,000
400
5,400
0
400
3
5,000
400
5,400
0
400
4
5,000
400
5,400
0
400
5
5,000
400
5,400
5,000
5,400
Sum
25,000
2,000
27,000
5,000
7,000
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.
1
2
3
4
Int. Owed
Total Owed
5
6
Total
Payment
Year
Amnt.
Owed
int*2
2+3
Princip.
Payment
1
5,000
400
5,400
852
1,252
2
4,148
332
4,480
920
1,252
3
3,227
258
3,485
994
1,252
4
2,233
179
2,412
1,074
1,252
5
1,160
93
1,252
1,160
1,252
Sum
15,768
1,261
17,029
5,000
6,261
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.
1
2
3
4
Int. Owed
Total Owed
5
6
Total
Payment
Year
Amnt.
Owed
int*2
2+3
Princip.
Payment
1
5,000
400
5,400
0
0
2
5,400
432
5,832
0
0
3
5,832
467
6,299
0
0
4
6,299
504
6,802
0
0
5
6,802
544
7,347
5,000
7,347
Sum
29,333
2,347
31,680
5,000
7,347
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).
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
Uniform Series
•
Uniform series formula derivation
A
n = 5 periods
0
1
A
A
2
3
A
4
A
5
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)]
F
(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
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.
Uniform Series
Uniform series compound amount factor :
Given F, Find A
(F/A,i,n) = [(1+i)n – 1]/i, i > 0
Uniform series sinking fund:
Given A, Find F
(A/F,i,n) = i/[(1+i)n – 1]
Uniform series capital recovery :
Given A, Find P
(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 P, Find A
How to Use Table
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)
(F/P,i,n) = 1/ (P/F,i,n)
Present Worth factor
Compound Amount factor
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)
Uniform Series Capital
Recovery Factor
F=A{[(1+i)n – 1]/i}=A(F/A,i,n)
Uniform Series
Compound
Amount Factor
(A/P,i,n) = 1/(P,A,i,n)
Uniform Series
Present Worth Factor
A=F{i/[(1+i)n–1]}=F(A/F,i,n)
(A/F,i,n) = 1/ (F/A,i,n)
Uniform Series
Sinking Fund Factor
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) Standard Form
Diagram for
= 519 + 247 = $766
Arithmetic Gradient:
240
210
180
n periods and n-1
nonzero flows in
increasing order
150
120
120
120
120
120
120 120
60
90
30
+
=
i= 5%
P?
A= 120
0
G= 30
Arithmetic Gradient
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
(n-1)G
F = G [(1+i)n-in-1]/i2
P = F (P/F, i, n) = G [(1+i)n-in-1]/[i2(1+i)n]
2G
A = F (A/F, i, n) =
G
…..
= G [(1+i)n-in-1]/i2 × i/[(1+i)n-1]
0
…..
A = G [(1+i)n-in-1]/[i(1+i)n-i]
0
1
2
3
…. n
F
17
Arithmetic Gradient
Arithmetic Gradient Uniform Series
(P/G,i,n) = { [(1+i)n – i n – 1] / [i2 (1+i)n] }
=1/(G/P,i,n)
Arithmetic Gradient Present Worth
(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)
(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.
18
Arithmetic Gradient
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%.
400 This is not in the standard form for using the gradient
300
200
100
equation, because the year-one cash flow is not zero.
We reformulate the problem as follows.
400
300
300
200
200
i= 6%
100
=
A A A A
A
A
A
100 100 100 100
A= 100
100
+
0
G= 100
A
A = A1+ G (A/G,6%,4) =100 + 100 (1.427) = $242.70
19
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:
Year n
1
2
3
4
5
Mnt. Cost
100 
110 
121 
133 
146.41 
(P/F,8%,n)
0.9259 =
0.8573
0.7938
0.7350
0.6806
PW of mnt.
$92.59
$94.30
$96.05
$97.83
$99.65
$480.42
20
Geometric Gradient
Geometric Gradient. At the end of year i, i = 1, ..., n, we incur a cost Aj = A(1+g)i-1.
A(1+q)n-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
A(1+q)2
A
A(1+q)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)
0
i≠q:
3 n(1+i)
….-n)/(i-q)n
P = A 2(1 - (1+q)
i=q:
P = A n(1+i)-1
21
Geometric Gradient
Example
Year n
1
2
3
4
5
n = 5, A1 = 100, g = 10%, i = 8%.
Mnt. Cost
100 
110 
121 
133 
146.41 
(P/F,8%,n)
0.9259 =
0.8573
0.7938
0.7350
0.6806
PW of mnt.
$92.59
$94.30
$96.05
$97.83
$99.65
$480.42
(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 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
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.
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
•
P  (1 + i) P  (1+r/2)2 P = (1+ 0.05/2)2 P = (1.050625) P
With i = 0.05/2, r = 0.05,
Nominal & Effective Interest
Terms the example illustrates:
r = 5%
is called the nominal interest rate per interest period (usually one year)
i = 2.5 %
is called the effective 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 subperiods per 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
Table Nominal & Effective Interest expressed in percent
Nominal rate Yearly
r
1
2
3
4
5
6
8
10
15
25
m=1
1.0000
2.0000
3.0000
4.0000
5.0000
6.0000
8.0000
10.0000
15.0000
25.0000
Effective rates, ia = (1 + r/m)m - 1
SemiMonthly
Daily
Continuously
ann.
m=2
m = 12
m = 365
1.0025
1.0046
1.0050
1.0050
2.0100
2.0184
2.0201
2.0201
3.0225
3.0416
3.0453
3.0455
4.0400
4.0742
4.0808
4.0811
5.0625
5.1162
5.1267
5.1271
6.0900
6.1678
6.1831
6.1837
8.1600
8.3000
8.3278
8.3287
10.2500
10.4713
10.5156
10.5171
15.5625
16.0755
16.1798
16.1834
26.5625
28.0732
28.3916
28.4025
Remark. The formula used for the continuous rate is er - 1, expressed
in percent. It is the limit of (1 + r/m)m as m goes to infinity.
27
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
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 Rate
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:
W
W
W
W
i = 2%, n = 20
$5000
30
Nominal and Effective Interest Rate
• 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.
i = 2%, n = 20
$5000
These withdrawals are equivalent to P = $5000
31
Nominal and Effective Interest Rate
Now consider the following:
A
W
W
W
W
Consider a one-year period:
i = 2%, n = 4
W
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.
32
Nominal and Effective Interest Rate
Solution 2 (Probably the easiest way)
W
$5000
ia = 8.24%, n =
5
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.
33
Continous Compounding
Table Nominal & Effective Interest expressed in percent
Nominal rate Yearly
r
1
2
3
4
5
6
8
10
15
25
m=1
1.0000
2.0000
3.0000
4.0000
5.0000
6.0000
8.0000
10.0000
15.0000
25.0000
Effective rates, ia = (1 + r/m)m - 1
SemiMonthly
Daily
Continuously
ann.
m=2
m = 12
m = 365
1.0025
1.0046
1.0050
1.0050
2.0100
2.0184
2.0201
2.0201
3.0225
3.0416
3.0453
3.0455
4.0400
4.0742
4.0808
4.0811
5.0625
5.1162
5.1267
5.1271
6.0900
6.1678
6.1831
6.1837
8.1600
8.3000
8.3278
8.3287
10.2500
10.4713
10.5156
10.5171
15.5625
16.0755
16.1798
16.1834
26.5625
28.0732
28.3916
28.4025
Remark. The formula used for the continuous rate is er - 1, expressed
in percent. It is the limit of (1 + r/m)m as m goes to infinity.
34
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.
r = nominal interest rate per year
m = number of compounding sub-periods per year
i = r/m = effective interest rate per compounding sub-period.


r
r

lim 1    1  lim 1  
m 
 m  m 
 m 
m
r
m
r
r

r

 1  lim 1  
m / r  
 m

ia = (1 + i)m – 1 =
= (1 + r/m)m – 1
m
r
r

 1 


r
1
1
r




 lim 1  x  x   1  lim 1  x  x   1   e  1
1/ x 


 x 0

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
i:
n:
P:
F:
effective interest rate per interest period (stated as a decimal)
number of interest periods
present sum of money
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
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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
Sinking Fund:
A
Capital Recovery:
A
Present Worth:
(P/A,i,n)
•
•
Arithmetic Gradient Formulas
Present Worth
P
(P/G,i,n)
Uniform Series
= G (A/G,i,n)
Geometric Gradient Formulas
If i  g,
P
(P/A,g,i,n)
= A{[(1+i)n –1]/i}
= A (F/A,i,n)
= F {i/[(1+i)n –1]}
= F (A/F,i,n)
= P {[i(1+i)n]/[(1+i)n – 1]
= P (A/P,i,n)
P
= A{[(1+i)n – 1]/[i(1+i)n]}
=A
= G {[(1+i)n – i n – 1]/[i2 (1+i)n]}
A
=G
= G {[(1+i)n – i n –1]/[i (1+i)n – i]}
= A {[1 – (1+g)n(1+i)-n]/(i-g)}
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=A
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
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