Interest Formulas – Equal Payment Series Engineering Economy.
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Transcript of Interest Formulas – Equal Payment Series Engineering Economy.
Interest Formulas – Equal Payment Series
Engineering Economy
Equal Payment Series
0 1 2 N
0 1 2 N
A A A
F
P
0 N
Using equal payment series you can find present value or future value
Compound Amount Factor
0 1 2 N 0 1 2 N
A A A
F
A(1+i)N-1
A(1+i)N-2
1 2(1 ) (1 )N NF A i A i A This is a geometric series with the first term as A and the constant r = (1+i)The formula for a geometric series = A (1- r^n)/1-r
Equal Payment Series Compound Amount Factor (Future Value of an annuity)
F Ai
iA F A i N
N
( )
( / , , )
1 1
Example:
Given: A = $5,000, N = 5 years, and i = 6%
Find: F
Solution: F = $5,000(F/A,6%,5) = $28,185.46
0 1 2 3N
F
A
Finding an Annuity Value
Example:• Given: F = $5,000, N = 5 years, and i = 7%• Find: A• Solution: A = $5,000(A/F,7%,5) = $869.50
0 1 2 3N
F
A = ?
A Fi
i
F A F i N
N
( )
( / , , )
1 1
Example: Handling Time Shifts in a Uniform Series
F = ?
0 1 2 3 4 5
$5,000 $5,000 $5,000 $5,000 $5,000
i = 6%
First deposit occurs at n = 0
5 $5,000( / ,6%,5)(1.06)
$29,876.59
F F A
Annuity Due
Excel Solution
=FV(6%,5,5000,0,1)Beginning period
Sinking Fund Factor
Example: College Savings Plan• Given: F = $100,000, N = 8 years, and i = 7%• Find: A• Solution:
A = $100,000(A/F,7%,8) = $9,746.78
0 1 2 3N
F
A
1)1( Ni
iFA
The term between the brackets is called the equal-payment-series sinking-fund factor.
Excel Solution
Given: F = $100,000 i = 7% N = 8 years
1)1( Ni
iFA
0
1 2 3 4 5 6 7 8
$100,000
i = 8%
A = ?
Current age: 10 years old
• Find: AUsing the equation:
Using built in Function:=PMT(i,N,pv,fv,type)=PMT(7%,8,0,100000,0)=$9,746.78
Capital Recovery Factor
1)1( Ni
iFA
NiPF )1(
1)1(
)1(N
N
i
iiPA
If we need to find A, given P,I, and N
Remember that:
Replacing F with its value
Capital Recovery Factor
Example: Paying Off Education Loan• Given: P = $21,061.82, N = 5 years, and i =
6%• Find: A• Solution: A = $21,061.82(A/P,6%,5) = $5,000
1 2 3N
P
A = ?
0
A Pi i
i
P A P i N
N
N
( )
( )
( / , , )
1
1 1
This factor is called capital recovery factor
P =$21,061.82
0 1 2 3 4 5 6
A A A A A
i = 6%
0 1 2 3 4 5 6
A’ A’ A’ A’ A’
i = 6%
P’ = $21,061.82(F/P, 6%, 1)
Grace period
Example: Deferred Loan Repayment Plan
Two-Step Procedure
' $21,061.82( / ,6%,1)
$22,325.53
$22,325.53( / ,6%,5)
$5,300
P F P
A A P
Adding the first year interest to the principal then calculating the annuity payment
Present Worth of Annuity Series
Example:Powerball Lottery• Given: A = $7.92M, N = 25 years, and i = 8%• Find: P• Solution: P = $7.92M(P/A,8%,25) = $84.54M
1 2 3N
P = ?
A
0
P Ai
i i
A P A i N
N
N
( )
( )
( / , , )
1 1
1
0 1 2 3 4 5 6 7 8 9 10 11 12
44
Option 2: Deferred Savings Plan
$2,000
Example: Early Savings Plan – 8% interest
0 1 2 3 4 5 6 7 8 9 10
44
Option 1: Early Savings Plan
$2,000
?
?
Option 1 – Early Savings Plan
10
44
$2,000( / ,8%,10)
$28,973
$28,973( / ,8%,34)
$396,645
F F A
F F P
0 1 2 3 4 5 6 7 8 9 10
44
Option 1: Early Savings Plan
$2,000
?
6531Age
Option 2: Deferred Savings Plan
44 $2,000( / ,8%,34)
$317,233
F F A
0 11 12
44
Option 2: Deferred Savings Plan
$2,000
?
At What Interest Rate These Two Options Would be Equivalent?
44
44
Option 1:
$2,000( / , ,10)( / , ,34)
Option 2:
$2,000( / . ,34)
Option 1 = Option 2
$2,000( / , ,10)( / , ,34) $2,000( / . ,34)
Solve for
F F A i F P i
F F Ai
F A i F P i F Ai
i
123456789101112131415161718192021224041424344454647
A B C D E F
Year Option 1 Option 201 (2,000)$ 2 (2,000)$ Interest rate 0.083 (2,000)$ 4 (2,000)$ FV of Option 1 396,645.95$ 5 (2,000)$ 6 (2,000)$ FV of Option 2 317,253.34$ 7 (2,000)$ 8 (2,000)$ Target cell 79,392.61$ 9 (2,000)$ 10 (2,000)$ 11 (2,000)$ 12 (2,000)$ 13 (2,000)$ 14 (2,000)$ 15 (2,000)$ 16 (2,000)$ 17 (2,000)$ 18 (2,000)$ 19 (2,000)$ 37 (2,000)$ 38 (2,000)$ 39 (2,000)$ 40 (2,000)$ 41 (2,000)$ 42 (2,000)$ 43 (2,000)$ 44 (2,000)$
Using Excel’s Goal Seek Function
Result
0 1 2 3 4 5 6 7 8 9 10
44
0 1 2 3 4 5 6 7 8 9 10 11 12
44
Option 1: Early Savings Plan
Option 2: Deferred Savings Plan
$2,000
$2,000
$396,644
$317,253
Interest Formulas(Gradient Series)
Linear Gradient Series
),,/(
1
112
NiGPGP
ii
iNiGP N
N
P
Gradient-series present –worth factor
Gradient Series as a Composite SeriesWe view the cash flows as composites of two series a uniform with a
payment amount of A1 and a gradient with a constant amount of G
$1,000$1,250 $1,500
$1,750$2,000
1 2 3 4 50
P =?
How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure?
Example:
Method 1:
$1,000$1,250 $1,500
$1,750$2,000
1 2 3 4 50
P =?
$1,000(P/F, 12%, 1) = $892.86$1,250(P/F, 12%, 2) = $996.49$1,500(P/F, 12%, 3) = $1,067.67$1,750(P/F, 12%, 4) = $1,112.16$2,000(P/F, 12%, 5) = $1,134.85
$5,204.03
Method 2:
P P A1 000 12%,5
604 80
$1, ( / , )
$3, .
P P G2 12%,5
599 20
$250( / , )
$1, .
P
$3, . $1, .
$5,204
604 08 599 20
0 1 2 3 4 5 6 7 25 26
$3.44 million
0 1 2 3 4 5 6 7 25 26
Cash Option
$175,000$189,000
$357,000
$196,000G = $7,000
Annual Payment Option
Example: Supper Lottery
Equivalent Present Value of Annual Payment Option at 4.5%
[$175,000 $189,000( / , 4.5%,25)
$7,000( / , 4.5%,25)]( / , 4.5%,1)
$3,818,363
P P A
P G P F
The gradient series is delayed by one period
To return the calculations to year zero
Geometric Gradient SeriesGeometric Gradient is a gradient series that is been determined by a fixed rate expressed as a percentage instead of a fixed dollar amount
For example the economic problems related to construction cost which involves cash flows that increase or decrease by a constant percentage
Present Worth Factor
giifi
NA
giifgi
igA
P
NN
,)1(
,)1()1(1
1
1
Geometric-gradient-series present-worth factor
Alternate Way of Calculating P
1
Let '1
( / , ', )(1 )
i gg
g
AP P A g N
g
Unconventional Equivalence CalculationsEGN3613
Ch2 Part IV
$50
$100 $100 $100
$150 $150 $150 $150
$200
Group 1 $50( / ,15%,1)
$43.48
P P F
Group 2 $100( / ,15%,3)( / ,15%,1)
$198.54
P P A P F
Group 3 $150( / ,15%,4)( / ,15%,4)
$244.85
P P A P F
Group 4 $200( / ,15%,9)
$56.85
P P F
$43.48 $198.54 $244.85 $56.85
$543.72
P
0
1 2 3 4 5 6 7 8 9
Composite Cash Flows
Unconventional Equivalence Calculations
Situation:
What value of A would make the two cash flow transactions equivalent if i = 10%?
Multiple Interest Rates
$300$500
$400
5% 6% 6% 4% 4%
Find the balance at the end of year 5.
0
12
34 5
F = ?
Solution1:
$300( / ,5%,1) $315
2 :
$315( / ,6%,1) $500 $833.90
3:
$833.90( / ,6%,1) $883.93
4 :
$883.93( / , 4%,1) $400 $1,319.29
5 :
$1,319.29( / , 4%,1) $1,372.06
n
F P
n
F P
n
F P
n
F P
n
F P
Cash Flows with Missing PaymentsP = ?
$100
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Missing paymenti = 10%
Solution
P = ?
$100
01 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Pretend that we have the 10th
paymenti = 10%
$100Add this cash flow tooffset the change
Approach
P = ?
$100
01 2 3 4 5 6 7 8 9 10 11 12 13 14 15
i = 10%
$100
Equivalent Cash Inflow = Equivalent Cash Outflow
$100( / ,10%,10) $100( / ,10%,15)
$38.55 $760.61
$722.05
P P F P A
P
P
Equivalence Relationship
Unconventional Regularity in Cash Flow Pattern
$10,000
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14
C C C C C C C
i = 10%
Payment is made every other year
Approach 1: Modify the Original Cash Flows
$10,000
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14
i = 10%
A A A A A A A A A A A A A A
$10,000( / ,10%,14)
$1,357.46
A A P
Relationship Between A and C
$10,000
01 2 3 4 5 6 7 8 9 10 11 12 13 14
i = 10%
A A A A A A A A A A A A A A
$10,000
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14
C C C C C C C
i = 10%
C
A =$1,357.46
A A
i = 10%
$10,000( / ,10%,14)
$1,357.46
( / ,10%,1)
1.1
2.1
2.1($1,357.46)
$2,850.67
A A P
C A F P A
A A
A
Solution
Approach 2: Modify the Interest Rate Idea: Since cash flows occur every other year, let's find out the
equivalent compound interest rate that covers the two-year period.
How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%.
(1+0.10)(1+0.10) = 1.21
Solution$10,000
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14
C C C C C C C
i = 21%
$10,000( / , 21%,7)
$2,850.67
C A P
1 2 3 4 5 6 7