5 more interest formulas
TRANSCRIPT
5. More Interest Formulas
Dr. Mohsin Siddique
Assistant Professor
Ext: 29431
Date: 21/10/2014
Engineering Economics
University of SharjahDept. of Civil and Env. Engg.
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Part I
Outcome of Today’s Lecture
3
� After completing this lecture…
� The students should be able to:
� Understand uniform series compound interest formulas
More interest Formulas
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� Uniform Series
� Arithmetic Gradient
� Geometric Gradient
� Nominal and Effective Interest
� Continuous Compounding
Uniform Series
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� In chapter 3 (i.e., interest and equivalence), we dealt with single payments compound interest formula:
Functional NotationAlgebraic Equivalent
� Examples:
� ______________________________________________________________________________________________________________________________________________________________________________
Uniform Series
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� Quite often we have to deal with uniform (equidistant and equal-valued) cash flows during a period of time:
� Remember: A= Series of consecutives, equal, end of period amounts of money (Receipts/disbursement)
� Examples: _______________________________________________
� ______________________________________________________
Deriving Uniform Series Formula
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� Let’s compute Future Worth, F, of a stream of equal, end-of-period cash flows, A, at interest rate, i, over interest period, n
0 1 2 3 4
F
0 1 2 3 4
A
F1
0 1 2 3 4
A
F2
0 1 2 3 4
A
F4
0 1 2 3 4
A
F3
0 1 2 n-1 n
A
F
= +
++F=F1+F2+F3+F4
Recall
( )31F1 iA += ( )21F2 iA +=
( )11F3 iA += ( )01F4 iA +=
Let n=4
Deriving Uniform Series Formula
0 1 2 3 4
A FF=F1+F2+F3+F4
( )31F1 iA += ( )21F2 iA +=
( )11F3 iA += ( )01F4 iA +=F =
+ +
+
( ) ( ) ( ) AiAiAiA ++++++= 123 111F
For general case, we can write that
Multiplying both sides with (1+i)
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )[ ]iiiiAi
iAiAiAiAinnn
nnn
++++++++=+
++++++++=+−−
−−
1...1111F
1...1111F21
21
( ) ( ) ( )( ) ( ) ( )[ ]1...111
...111321
321
+++++++=
+++++++=−−−
−−−
nnn
nnn
iiiAF
AiAiAiAFEq. (1)
Eq. (2)
Deriving Uniform Series Formula
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� Eq. (2)-Eq. (1)
( ) ( ) ( ) ( ) ( )[ ]iiiiAi nnn ++++++++=+ −− 1...1111F 21
( ) ( ) ( )[ ]1...111 321 +++++++= −−− nnn iiiAF
Eq. (2)
Eq. (1)-
( )[ ]11iF −+= niA
( ) [ ]niAFAi
iA
n
%,,/11
F =
−+=
Eq. (3)
Eq. (4)
( )
−+i
i n 11Where is called uniform series compound
amount factor and has notation [ ]niAF %,,/
Deriving Uniform Series Formula
10
� Eq. (4) can also be written as
( ) [ ]niFAFi
iF
n%,,/
11A =
−+=
( )
−+ 11 ni
iWhere is called uniform series sinking fund
factor and has notation [ ]niFA %,,/
Eq. (5)
ni
given
Find%,,
Example 4-1
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Example 4-2
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( ) =
−+=
11A
ni
iF
Example 4-2
13
Deriving Uniform Series Formula
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� If we use the sinking fund formula (Eq. 5) and substitute the single payment compound amount formula, we obtain
( ) ( ) ( )
−++=
−+=
111
11A
n
n
n i
iiP
i
iF ( )niPF += 1Q
( )( ) ( )niPAP
i
iiP
n
n
%,,/11
1A =
−++=
� It means we can determine the values of A when the present sum P is known
Where is called uniform series capital
recovery factor and has notation
( )( )
−++
11
1n
n
i
ii
( )niPAP %,,/
Eq. (6)
Deriving Uniform Series Formula
15
� Eq. (6) can be rewritten as
( )( ) ( )niAPA
ii
iA n
n
%,,/1
11P =
+−+=
� It means we can determine present sum P when the value of A is known
Where is called uniform series present
worth factor and has notation
( )( )
+−+n
n
ii
i
1
11
( )niAPA %,,/
Eq. (7)
Example 4-3
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Example 4-6
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Example 4-6
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Example 4-6
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Example 4-7
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0 1 2 3 4
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Problem
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� Determine n based on 3.5% interest rate?
Solution:P = A (P/A, 3.5%, n)$1,000 = $50 (P/A, 3.5%, n)
(P/A, 3.5%, n) = 20
From the 3.5% interest table, n = 35.
Problem
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� A sum of money is invested at 2% per 6 month period (semi-annually) will double in amount in approximately how much years?
P = $1 n = unknown number of
semiannual periods
i = 2% F = 2
F = P (1 + i)n
2 = 1 (1.02)n
2 = 1.02n
n = log (2) / log (1.02)= 35
Therefore, the money will double in 17.5 years.
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Part II
(from next class bring interest tables and formula sheet)
Outcome of Today’s Lecture
25
� After completing this lecture…
� The students should be able to:
� Understand arithmetic gradient interest formulas
More interest Formulas
26
� Uniform Series
� Arithmetic Gradient
� Geometric Gradient
� Nominal and Effective Interest
� Continuous Compounding
Arithmetic Gradient Series
27
� It’s frequently happen that the cash flow series is not constant amount.
� It probably is because of operating costs, construction costs, and revenues to increase of decrease from period to period by a constant percentage
Arithmetic Gradient Series
28
� Let the cash flows increase/decrease by a uniform fixed amount Gevery subsequent period
Eq. (1)
Recall
( ) ( ) ( )( ) ( ) ( )[ ])1(1)2(...121F
)1()1(1)2(...121F132
0132
−++−+++++=
+−++−+++++=−−
−−
niniiG
iGniGniGiGnn
nn
Arithmetic Gradient Series
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� Multiplying Eq. (1) with (1+i), we get
( ) ( ) ( ) ( )[ ]1221 )1)(1(1)2(...121Fi1 ininiiG nn +−++−+++++=+ −−
( ) ( ) ( ) ( )[ ]1221 )1)(1(1)2(...121Fi1 ininiiG nn +−++−+++++=+ −−
( ) ( ) ( )[ ])1(1)2(...121F 132 −++−+++++= −− niniiG nn
( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( )[ ] nGiiiiG
niiiiGnn
nn
−+++++++++=
+−++++++++=−−
−−
111...11iF
111...11iF1221
1221
� Eq. (2)-Eq. (1)
Eq. (2)
-
( )
( ) ( )
−−+=
−−+=
−
−+=
2
1111
11iF
i
niiGn
i
i
i
GF
nGi
iG
nn
n
( ) [ ]niGFGi
iniGF
n
%,,/11
2=
−−+=
Eq. (3)
Eq. (4)
Arithmetic gradient future worth factor
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( ) ( ) ( ) ( ) ( ) ( )[ ] ( )( ) ( ) ( ) ( )[ ]
( )[ ]( )
nGi
iGiF
nGiiGiiF
nGiiiiG
inGiiiiiGi
n
n
nn
nn
−
−+=
−−+=
−+++++++++=
+−++++++++++=+−−
−
11
11
111...11iF
1111...111iF1221
231
-
Arithmetic Gradient Series
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� Substituting F from single payment compound formula, we can write Eq.(4) as
( )( ) [ ]niGPG
ii
iniGP
n
n
%,,/1
112
=
+−−+=
Eq. (5)
� (P/G ,i%, n) is known as Arithmetic gradient present worth factor
� Now substituting value of F from uniform series compound amount factor, we can write Eq. (4) as
Recall
( ) ( )
( )( )( )( )
( )niGAGA
ii
iniiGA
i
iA
i
iniGF
n
n
nn
%,,/
11
11
1111
2
2
=
−+−−+=
−+=
−−+=
( )
−+=i
iA
n 11FQ
� (A/G, i%, n) is known as Arithmetic gradient uniform series factor
Arithmetic Gradient Series
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Example 4-8
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� 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. i = 5%.
Thus, estimated costs by year are $120, $150, $180, $210, $240.
=
Example 4-8
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Example 4-9
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� 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%.
=
A=?
Example 4-9
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Example 4-10
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Example 4-10
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Example
39
(Hint: look for differences in time axis.)
