5. more interest formula (part ii)
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Transcript of 5. more interest formula (part ii)
5: More Interest Formulas
(continued…)
Dr. Mohsin Siddique
Assistant Professor
Ext: 29431
Date: 28/10/2014
Engineering Economics
University of SharjahDept. of Civil and Env. Engg.
2
Part I
Outcome of Today’s Lecture
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� After completing this lecture…
� The students should be able to:
� Understand geometric series compound interest formulas
More interest Formulas
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� Uniform Series
� Arithmetic Gradient
� Geometric Gradient
� Nominal and Effective Interest
� Continuous Compounding
Geometric Gradient Series
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� Instead of constant amount of increase, sometimes cash flows increase by a uniform rate of increase g (constant percentage amount) every subsequent period.
� For example: If the maintenance costs of car are $100 for the first year and they are increasing at a uniform rate, g, of 10% per year
100
110
121
133.1
146.41
1A
( )gAA += 112
( ) ( )2123 11 gAgAA +=+=
10 2 3 4 5
( ) ( )4145 11 gAgAA +=+=
( ) 11 1 −+= n
n gAA
...
Geometric Gradient Series
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� Let’s write a present worth value for each period individually, and add them up
� Recall:
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) nnnn igAigA
igAigAiAP−−+−−
−−−
++++++
++++++++=
1111
...111111
112
1
321
211
11
Eq. (1)
( )( ) n
n
iF
iP−+=
+=
1P
1F
Multiply Eq. (1) by (1+g)/(1+i) to obtain
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 1
11
1
431
321
211
11
1111
...11111111−−−−
−−−−
++++++
+++++++++=++nnnn igAigA
igAigAigAigP
Eq. (2)
Geometric Gradient Series
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� Subtract Equation (2) from Equation (1) to yield
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )[ ]
( ) ( )( )
−−+++−=
++−=−−+
++−=+−+
++−+=++−
−
−
−
−−−−
gi
igAP
igAgiP
igAAgPiP
igAiAigPP
nn
nn
nn
nn
11
111
11111
1111
11111
1
1
111
11
11
11
-
Eq. (3)
Eq. (4)Where gi ≠
Where is called geometric series present worth factor and has notation
( ) ( )( )
−−+++− −
gi
ig nn
11
111
( )nigAP ,,,/
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) nnnn igAigA
igAigAiAP−−+−−
−−−
++++++
++++++++=
1111
...111111
112
1
321
211
11
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 1
11
1
431
321
211
11
1111
...11111111−−−−
−−−−
++++++
+++++++++=++nnnn igAigA
igAigAigAigP
Geometric Gradient Series
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� 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%.
� Solution:
� 1. Repeated Present-Worth (Step-by-Step) Approach:
Geometric Gradient Series
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� 2. Using geometric series present worth factor formula
Multiple Compounding
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� The time standard for interest computations is One Year.
� Many banks compound interest multiple times during the year.
� e.g.: 12% per year, compounded monthly (1% interest is paid monthly)
� e.g.: 8% per year, compounded semi-annually (4% interest is paid twice a year or once every 6 months)
� Compounding is not less important than interest. You have to know all the info to make a good decision.
Multiple Compounding
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� You need to pay attention to the following terms:
� Time Period –The period over which the interest is expressed (always stated).
� e.g.: “6% per year”
� Compounding Period (sub-period) – The shortest time unit over which interest is charged or earned.
� e.g.: If interest is “6% per year compounded monthly”, compounding period is one month
� Compounding Frequency –The number of times (m) that compounding occurs within time period.
� Compounding semi-annually: m = 2; Compounding quarterly: m = 4
� Compounding monthly: m = 12; Compounding weekly: m = 52; compounding daily: m = 365
Nominal and Effective Interest Rate
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� Two types of interest are typically quoted:
� 1. Nominal interest rate, r, is an annual interest rate without considering the effect of (sub-period) compounding.
� 2. Effective interest rate, i or ia , is the actual rate that applies for a stated period of time which takes into account the effect of (sub-period) compounding.
� Sometimes one interest rate is quoted, sometimes another is quoted. If you confuse the two you can make a bad decision.
� Effective interest is the “real” interest rate over a period of time; Nominal rate is just given for simplicity (per year)
� All interest formulas use the effective interest rate
Nominal and Effective Interest Rate
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� Let r=nominal interest rate per interest period (usually one year)
� i=effective interest rate per interest period (sub-period compounding)
� m=Compounding frequency (No. of compounding sub period per time period)
� r/m=interest rate per compounding sub-period
1=P
( )11 /11 mrF +=
( ) ( )2112 /11/1 mrmrFF +=+=
1
0
2 3 4
( ) ( )4134 /11/1 mrmrFF +=+=
...
interest period (one year) sub period (quarter)
( )mmrF /11 +=
( ) ( )11 1111 aiiF +=+=
( ) 1/1 −+= ma mri
( ) ( )ma mri /11 +=+
Thus in general form we can write
Moreover, we also know that
Thus
Nominal and Effective Interest Rate
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� Example:
� Given an interest rate of 12% per year, compounded quarterly:
� Nominal rate=r = 12%
� Compounding frequency=m=4
� Effective (Actual) rate =r/m= 12%/4 = 3% per quarter
� Effective rate per year = [1+(0.12/4)]4-1= 0.1255=12.55%
� Investing $1 at 3% per quarter is equivalent to investing $1 at 12.55% annually
Nominal and Effective Interest Rate
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� Example: A bank pays 1.5% interest every three months. What are the nominal and effective interest rates per year?
� Solution:
Effective interest rate per three months=1.5%
Nominal interest rate per year = r = 4 x 1.5% = 6% a year
Effective interest rate per year= ia= (1 + r/m)m–1 = (1.015)4–1 =
=0.06136
=6.14% a year
Nominal and Effective Interest Rate
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� Example: $10K is borrowed for 2 years at an interest rate of 24% per year compounded quarterly. If the 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?
� Solution
Interest=F-P
� Interest charges for quarterly compounding:
� 10,000(1+24%/4)2x4-10,000 = $5938.48
� Interest charges for annually compounding:
� 10,000(1+24%)2-10,000 = $5376.00
� Savings: $5938.48 -$5376 = $562.48
Nominal and Effective Interest Rate
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� Example 4-15: A loan shark lends money on the following terms. “If I give you $50 on Monday, then you give back $60 the following Monday.”
� Solution
� 1.What is the nominal rate, r ?
� The loan shark charges i= 20% a week:
� 60 = 50 (1+i) [Note we have solved 60 = 50(F/P,i,1) for i]
� i= 0.2
� We know m = 52, so r = 52 x i= 10.4, or 1,040% a year
� 2. What is the effective rate, ia?
� ia= (1 + r/m)m–1 = (1+10.4/52)52–1 =13,104
� This means about 1,310,400 % a year !!!!
Nominal and Effective Interest Rate
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� Example 4-16: You deposit $5,000 in a bank paying 8% nominal interest, compounded quarterly. You want to withdraw the money in five equal yearly sums, beginning Dec. 31 of the first year. How much should you withdraw each year ?
08.0quarterly compounded %8 == yearlyr
4
0
8 12 16
(year)
w w w w w
20 20 months
1 2 3 4 5 5 years
( ) ( ) yearlymri ma %24.814/08.011/1 4 =−+=−+=
$5,000
Effective annual interest:
Nominal and Effective Interest Rate
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� This diagram may be solved directly to determine the annual withdrawal W with the capital recovery factor
� The depositor should withdraw $1260 per year
4
0
8 12 16
(year)
w w w w w
20 20 months
1 2 3 4 5 5 years$5,000
( ) ( )( ) 1260$
11
1%,,/ =
−++==
n
n
i
iiPniPAPW
Continuous Compounding
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� Continuous compounding is when m is infinite
� i.e. m�∞: r/m�0:
( )( )mn
ma
mrPF
mri
/1
1/1
+=
−+=
( )( ) rnmn
rma
PemrPF
emri
=+=
−=−+=
/1
11/1
Continuous Compounding
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� Continuous compounding can sometimes be used to simplify computations, and for theoretical purposes.
� The previous equation illustrates that (er – 1) is a good approximation of (1 + r/m)m -1 for large m (i.e., ∞).
� This means there are continuous compounding versions of the formulas we have seen earlier.
� F = P ern is analogous to F = P (F/P,r,n): (F/P,r,n)∞= ern
� P = F e-rn is analogous to P = F (P/F,r,n): (P/F,r,n)∞= e-rn
Continuous Compounding
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