equal payment economics

1. Interest Formulas for Time
Value of Money
Payment Series

2. The Time Value of Money
 Interest Rate
 Simple Interest
 Compound Interest
 Amortizing a Loan
 Compounding More Than Once per
Year

3.  “Compound interest is the eighth wonder of
the world. He who understands it, earns it ...
he who doesn't ... pays it.”
― Albert Einstein

4. Why TIME?
TIME allows you the opportunity to
postpone consumption and earn
INTEREST.
Why is TIME such an important
element in your decision?

5. Types of Interest
 Compound Interest
Interest paid (earned) on any previous interest
earned, as well as on the principal
borrowed.
 Simple Interest
Interest paid (earned) on only the original
amount, or principal, borrowed.

6. Simple Interest Formula
Formula SI = P0(i)(n)
SI: Simple Interest
P0: Deposit today (t=0)
i: Interest Rate per Period
n: Number of Time Periods

7. Simple Interest Example
 SI = P0(i)(n)
= $1,000(.07)(2)
= $140
 Assume that you deposit $1,000 in an
account earning 7% simple interest for 2
years. What is the accumulated interest at
the end of the 2nd year?

8. Simple Interest (FV)
FV = P0 + SI
= $1,000 + $140
= $1,140
 Future Value is the value at some future time
of a present amount of money, or a series of
payments, evaluated at a given interest rate.
 What is the Future Value (FV) of the
deposit?

9. Example : Generous Grandma
Your Grandma puts $1,000 in a bank for you, at 5% interest. Calculate the amount
after 20 years.
Simple interest:
A = 1000 (1 + 0.0520) = $2,000.00
Compounded annually:
A = 1000 (1 + .05)20
=$2,653.30
Compounded daily:
Compounded continuously:
A = 1000 e(.05*20)
= $2,718.28
10
.
718
,
2
$
365
05
.
1
1000 )
20
*
365
(









A

10. Annual percentage yield
 The simple interest rate that will produce the same amount in 1 year is
called the annual percentage yield (APY). To find the APY, proceed as
follows: This is also referred to as the effective rate.
int int
1 1
amount at amount at
simple erest compound erest
after year after year
   
   

   
   
   
(1 ) 1
1 1
1 1
m
m
m
r
P APY P
m
r
APY
m
r
APY
m
 
   
 
 
 
   
 
 
 
  
 
 

11. Calculate compound interest using this formula:
A—Total amount
p —principle
r —interest rate
n —number of compounding periods
t —time in years
nt
n
r
p
A 






 1

12. Example: $100 is invested at 10% interest
compounded yearly for 6 years
177.16

13. $250 invested at 6.5% for 8 years compounded
monthly.
419.92

14. Example……
 $500 invested at 12% for 10 years
compounded yearly.

15. Answer……
 Problem:
 $500 invested at 12% for 10 years compounded
yearly.
 Answer:
 
93
.
1552
12
.
1
500
1
12
.
1
500
1
10
10
1



















A
A
A
n
r
P
A
nt

16. Example……
 $1000 at 7.25% for 9 years compounded
monthly.

17. Answer……
 Problem:
 $1000 at 7.25% for 9 years compounded monthly.
 Answer:
57
.
1916
12
0725
.
1
1000
1
)
9
12
(


















A
A
n
r
P
A
nt

18. 18
Single Payment Future Amount
If you have $1 today and can earn an interest rate of 10% by the end of the
first year then you will have $1.10.
The$1.10 is calculated as the amount you start the period with plus the
product of what you start the period with times the rate of interest that
period the 1.10 = 1 + 1(.1).
If F is the amount at the end of the period, P is the amount at the beginning
of the period and i is the rate of interest during the period, then in general we
have
F = P + Pi = P(1 + i).

19. 19
Growth in general
If you start out with P and wait one period at rate i we just saw you have F =
P(1 + i). Now, if you again earn i, by the end of the second period you
would have
F = P(1 + i) + P(1 + i )i (start period with + start period with times i)
= P(1 + i)(1 + i)
= P(1 + i)2
.
In general, at the end of n periods, you have
F = P(1 + i)n
.

20. 20
The formula
F = P(1 + i)n
,
can be called the single payment future value (sometimes called the single
payment compound amount) formula.
Notice if we know or have specific numerical values for
-the present amount P,
-the constant interest rate each period i, and
-the number of periods n,
then we can solve for the future amount.
Example: Say your grandmother gives you $10 and wants you to pay her back
in 2 years and she will charge you 8% interest each year. How much do you
pay her back at the end of two years?
Let’s look at a time line on the next slide.

21. 21
0 1 2 3 … n
10 – you received $10
? – How much do you pay back?
F = 10(1 + .08)2
= 10(1.1664) = 11.66

22. 22
The Future Value Factor
We saw on the last page that the amount to be paid back was
F = 10(1 + .08)2
= 10(1.1664) = 11.66.
Remember in general we know
F = P(1 + i)n
.
(1 + i)n
is called the interest factor by which we multiple the present
amount to get the future amount.
The interest factor is part of the formula and includes the interest rate.

23. General Future Value Formula
F1 = A(1+i)1
F2 = A(1+i)2
General Future Value
Formula:
Fn = A (1+i)n
 How do we get n?
 …. Take logs

24. Investment with Variable interest
 The sum of $300,000 was placed in an
investment. The investors received $80,000
at the end of year 2, $110,000 at the end of
year 5, and the balance at the end of year 10.
The rate of investment was 8 percent for the
first 3 years and 9% thereafter. What sum did
the investors receive when the venture was
terminated?

25. Answer
 End of year 2: 300,000(1.08)2
- 80,000
=269,920
 End of year 3: 269,920(1.08) = 291,514
 End of year 5: 291,514(1.09) 2
-110,000
=236,348
 End of year 10: 236,348(1.09) 5
- X
=0
 X = $363,651

26. Annuity
 When the sum of principal and interest is
equal for each period.

27. Types of Annuities
 Ordinary Annuity: Payments or receipts
occur at the end of each period.
 Annuity Due: Payments or receipts occur at
the beginning of each period.
 An Annuity represents a series of equal
payments (or receipts) occurring over a
specified number of equidistant periods.

28. Examples of Annuities
 Student Loan Payments
 Car Loan Payments
 Insurance Premiums
 Mortgage Payments
 Retirement Savings

29. Parts of an Annuity
0 1 2 3
$100 $100 $100
(Ordinary Annuity)
End of
Period 1
End of
Period 2
Today
Equal Cash Flows
Each 1 Period Apart
End of
Period 3

30. Parts of an Annuity
0 1 2 3
$100 $100 $100
(Annuity Due)
Beginning of
Period 1
Beginning of
Period 2
Today Equal Cash Flows
Each 1 Period Apart
Beginning of
Period 3

31. Equal Payment Series
0 1 2 N
0 1 2 N
A A A
F
P
0 N

32. Equal Payment Series – Compound Amount
Factor
0 1 2 N
0 1 2 N
A A A
F
0 1 2
N
A A A
F

33. 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 N
F A i A i A
 
     


34. Uniform Series Compound Amount Factor
 This factor is used to calculate a future single
sum, F, that is equivalent to a uniform series
of equal end of period payments, A.

35. Equal Payment Series Compound Amount Factor (Future Value of
an annuity)
F A
i
i
A F A i N
N

 

( )
( / , , )
1 1
Example 2.9:
 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 3
N
F
A

36. Validation
F =?
0 1 2 3 4 5
$5,000 $5,000 $5,000 $5,000 $5,000
i = 6%
4
3
2
1
0
$5,000(1 0.06) $6,312.38
$5,000(1 0.06) $5,955.08
$5,000(1 0.06) $5,618.00
$5,000(1 0.06) $5,300.00
$5,000(1 0.06) $5,000.00
$28.185.46
 
 
 
 
 

37. 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 3
N
F
A = ?
A F
i
i
F A F i N
N

 

( )
( / , , )
1 1

38. Example 2.10 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

39. 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

40. Sinking Fund
 A company established a sinking fund for the
purpose to replace an existing asset at the
end of 6 years. The fund consisted of annual
end of year deposits of $5900 each, and
earned an annual interest of 8 %. However at
the end of the 6 year period, the asset was
still functioning satisfactory and the principal
was left intact. What was the principal in the
fund 3 years after the date of the last
deposit?

41. Solution
 F = 5900(7.336)
= 43,282
F (value of the interest) = 43,282 *(1.2597)= 54,522
F A
i
i
A F A i N
N

 

( )
( / , , )
1 1

42. Sinking Fund Deposit Factor
 To determine the amount of money, A, that
must be sunk into a fund at the end of each
period for n periods at i % interest rate per
period to accumulate F amount.
 The factor is used to calculate a uniform
series of equal end of period payments, A,
that are equivalent to a future sum, F.

43. Sinking Fund Factor
Example 2.11 – 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 3
N
F
A
)
1
)
1
(
( n



i
i
F
A

44. Excel Solution
 Given:
 F = $100,000
 i = 7%
 N = 8 years
0
1 2 3 4 5 6 7 8
$100,000
i = 8%
A = ?
Current age: 10 years old
• Find:
=PMT(i,N,pv,fv,type)
=PMT(7%,8,0,100000,0)
=$9,746.78

45. Capital Recovery Factor
 To relate a uniform
series of end of
period payments, A,
to a present sum, P,
we calculate as
given overleaf
 This factor is used to
calculate a uniform
series of end of
period payments, A,
that are equivalent
to a single sum of
money, P.

46. Capital Recovery Factor
Example 2.12: 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 3
N
P
A = ?
0
A P
i i
i
P A P i N
N
N


 

( )
( )
( / , , )
1
1 1

47. 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 2.14 Deferred Loan Repayment Plan

48. Two-Step Procedure
' $21,061.82( / ,6%,1)
$22,325.53
$22,325.53( / ,6%,5)
$5,300
P F P
A A P





49. Uniform Series Present Worth
 To determine the present single value sum of
money, P, that is equivalent to a uniform
series of equal payments, A, for n periods at i
% interest per period;
 This factor is used to calculate the present
sum, P, that is equivalent to a uniform series
of equal end of period payments, A.

50. Present Worth of Annuity Series
Example 2.14: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 3
N
P = ?
A
0
P A
i
i i
A P A i N
N
N

 


( )
( )
( / , , )
1 1
1

51. Problem
 Company A is required to make the following
end of year payments to company B: $12,000
for years 1 to 7, inclusive and $ 15,000 for
years 8 to 12, inclusive. To provide for these
payments, company A will deposit a sum of
money in a reserve fund at the beginning of
year 1. If the annual interest is 9%, What is
the present worth of the composite deposits.

52. Solution
 P = $12,000(5,03)
= $60,395
 P(7year) = 15,000 (3.8896)
= 58,344
 P (now) = 58,344/(1.09)^7
= 58,344/1.82
=31,916
P A
i
i i
A P A i N
N
N

 


( )
( )
( / , , )
1 1
1

53. Problems
 Given: A crankshaft company is trying to
decide whether to upgrade now for $150,000
or in 3years given that cost will increase at
12% per annum
 Find: How much will it cost them in 3 years?

54. Solution
 F = P(F/P, 12%, 3)
 =150,000*(1+.12)3
 = 150,000(1.4049)
 F = $210,735


55. Problem
 Given: The carmaker Renault will have to pay
$75M in 3 years at an interest of 15% per
annum
 Find: What is the $75M worth today.


56. Solution
 P = F(P/F, 15%, 3)
 =75m/ (1+.15)3
 = 75M(.6575)
 P = $49.31M


57. Problem
 Given: An amusement park intends to pays
$55,000 per year for 5 years at 15% per
annum interest.
 Find: Using the same amount of money, how
much is that today.

58. Solution
 P = A(P/A, 15%, 5)
 A = P i (1+i)n
((1+i)n
-1)
 = 55,000(3.3522)
 P = $184,371

59. Problem
 Given: A moving company wants to buy a
new truck for $250,000 in 4 years and
interest rates are 10% per year.
 Find: How much should they set aside each
year?


60. Solution
 A = F(A/F, 10%, 4)
 A = F i
((1+i)n
-1)
 = 250,000*(0.1/((1+0.1)4
– 1))
 = $250,000(.21547)
 A = $53,868