Interest Formulas

1
Interest Formulas

• Uniform Series
• Factor Relationships
• Arithmetic Gradient
• Geometric Gradient

• Nominal and Effective Interest
• Continuous Compounding

2
INTEREST FORMULAS

• Example 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
Janes point of view
CUs point of view
F
3
UNIFORM SERIES FORMULA DERIVATION

• n 5 periods
•   F A (1i)4 A (1i)3 A
  (1i)2 A (1i) A
• F A (1i)4 (1i)3 (1i)2
  (1i) 1
• (1i) F A (1i)5 (1i)4 (1i)3
  (1i)2 (1i)
• (1i) F F A (1i)5 1
• F A (1i)5 1/i  
• A F i/(1i)5 1 F A (1i)5
  1/i

F
Uniform Series Factors (A/F,i,n)i/(1i)n
1. (F/A,i,n)(1i)n 1/i.
4
UNIFORM SERIES

• Uniform series compound amount factor
• (F/A,i,n) (1i)n
  1/i, i gt 0
• Uniform series sinking fund
• (A/F,i,n) i/(1i)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 (1i)n 1/i 500(1.05)5
  1/(0.05)
• 500 (5.5256) 2,762.82 ? 2,763.

5
UNIFORM SERIES

• Example A lot of land is available for 1000.
  Jim will put the same amount in the bank each
  month to have 1,000 at the end of the year. The
  bank pays 6 interest, compounded monthly. How
  much does Jim put in the bank each month?
• Solution
•  A 1000 (A/F, ½,12) 1000 (0.0811)
  81.10/month
• Example Suppose on January 1 Joe deposits 5000
  in a credit union paying 8 interest, compounded
  annually. He wants to withdraw all the money in
  five equal end-of year sums, beginning December
  31st of the first year.

A
A
A
A
A
1
2
3
4
5
P
6
UNIFORM SERIES

• We know
• A F i/(1i)n 1
• F P (1 i)n. 
• Hence
• A Pi (1 i)n/(1i)n 1 P(A/P,i,n)
• This is a formula to determine the value of a
  series of end-of-period payments, A, when the
  present sum P is known.
• (A/P,i,n) is called the uniform series capital
  recovery factor .

7
UNIFORM SERIES

•  Example Suppose on January 1 Joe deposits 5000
  in a credit union paying 8 interest, compounded
  annually. He wants to withdraw all the money in
  five equal end-of year sums, beginning December
  31st of the first year.
• P 5000 n 5 i 8 A unknown
• A P (A/P,8,5) Pi (1 i)n/(1i)n 1  
• 5000 (0.2504564545) 1252.28.
• The withdrawal amount is about 1252.
• Note This is the source of the 1252 in Plan C
  from Chapter 3.

A
A
A
A
A
1
2
3
4
5
P
8
UNIFORM SERIES

• Example
• An investor holds a time payment purchase
  contract on some machine tools. The contract
  calls for the payment of 140 at the end of each
  month for a five-year period. The first payment
  is due in one month. He offers to sell you the
  contract for 6800 cash today. If you otherwise
  can make 1 per month on your money, would you
  accept or reject the investors offer?
•  
• A 140, n60, i1
• A P (A/P,i,n)
• P A / (A/P,i,n)
• A (P/A,i,n) 140 (P/A,1,60) 140
  (44.955) 6293.70

Imagine 60 up arrows
9
UNIFORM SERIES

• The Reverse Point of View. We can solve
• A Pi (1 i)n/(1i)n 1
• for P to obtain
• P A (1i)n 1/i (1 i)n A (P/A,i,n)
• In this case
• (P/A,i,n) (1i)n 1/ i (1 i)n
• is called the uniform series present worth
  factor.

10
UNIFORM SERIES

• Uniform series compound amount factor
• (F/A,i,n) (1i)n 1/i, i gt 0
• Uniform series sinking fund
• (A/F,i,n) i/(1i)n 1
• Uniform series capital recovery
• (A/P,i,n) i (1 i)n/(1i)n 1
• Uniform series present worth
• (P/A,i,n) (1i)n 1/i (1 i)n

11
UNIFORM SERIES

• Example
•  Compute the value of the following cash flows at
  the first of year 5.
• Year Cash flow
• 1 100
• 2 100
• 3 100
• 4 0
• 5 - F
• The Sinking Fund Factor diagram is based on the
  assumption the withdrawal coincides with the last
  deposit. This does not happen in this example.

Note the zero
12
UNIFORM SERIES
What we have is
The standard approach is
A
A
A
A
A
A
1
2
3
1
2
3
4
5
-F
First Approach
A
A
A
Use the standard approach to compute F1. Then
compute the future value of F1 to get F. F1 100
(F/A,15,3) 100 (3.472) 347.20 F F1
(F/P,15,2) 347.20 (1.322) 459.00
F1
F
13
UNIFORM SERIES

• Second Approach Compute the future values of
  each deposit and add them.
• F F1 F2 F3
• 100(F/P,15,4) 100(F/P,15,3)
  100(F/P,15,2)
• 100 (1.749) 100 (1.521) 100 (1.322)
  459.20.

14
UNIFORM SERIES

• Example i 15
• This diagram is not in a standard form.
• Approach 1. Compute the present value of each
  flow and add them.
• P P1 P2 P3
• 20 (P/F,15,2) 30 (P/F,15,3) 20
  (P/F,15,4)
• 20 (0.7561) 30 (0.6575) 20 (0.5718)
  46.28

30
20
20
0
P
15
UNIFORM SERIES

• Approach 2. Compute the future value, F, of the
  flows at the end of year 4. Then compute the
  present value of the future value, F.
• Approach 3. Compute the present values of the
  flows at the end of year 1, P1. Then compute P,
  the present value of P1.
•  
• Other approaches will also work.

16
Relationships Between Compound Interest Factors

• F P(1i)n P(F/P,i,n) P F/(1i)n F
  (P/F,i,n)
•  
• PA(1i)n1/i(1 i)n A(P/A,i,n) A
  Pi(1 i)n/(1i)n1P(A/P,i,n)
•  
• FA(1i)n 1/iA(F/A,i,n)
  AFi/(1i)n1F(A/F,i,n)

(F/P,i,n) 1/ (P/F,i,n)
Present Worth factor
Compound Amount factor
Uniform Series Capital Recovery Factor
(A/P,i,n) 1/(P,A,i,n)
Uniform Series Present Worth Factor
Uniform Series Compound Amount Factor
(A/F,i,n) 1/ (F/A,i,n)
Uniform Series Sinking Fund Factor
17
Relationships Between Compound Interest Factors

• P A(P/A,i,n) A (1i)-1 (1i)-2
  (1i)-n A(P/F,i,1)(P/F,i,2)...(P/F,i
  ,n)
• ?
•  
• F A (F/A,i,n) A 1 (1i) (1i)2
  ... (1i)n-1
• A 1 (F/P,i,1) (F/P,i,2) ...
  (F/P,i,n-1)

(P/A,i,n) (P/F,i,1) (P/F,i,2) ...
(P/F,i,n)
A
A
A
A
A
A
A
A
..
..
0
1
2
3
n
0
1
2
3
n
F
P
(F/A,i,n) 1 (F/P,i,1) (F/P,i,2) ...
(F/P,i,n-1)
18
Relationships Between Compound Interest Factors

• (A/P,i,n) i(1 i)n/(1i)n1
• (A/F,i,n) i/(1i)n1
• We start with an identity
• i (1i)n i i (1i)n i i i (1i)n 1
• Now divide by (1i)n 1 to get 
• i (1i)n/(1i)n 1 i/ (1i)n 1 i.
• This gives

(A/P,i,n) (A/F,i,n) i
19
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.

240
210
180
150
120
20
Arithmetic Gradient

• We break up the cash flows into two components

120
90
A 120
60
30
0
and
1
2
3
4
5
G 30
Standard Form Diagram for Arithmetic Gradient n
periods and n-1 nonzero flows in increasing order
P1
P2
P1 A (P/A,5,5) 120 (P/A,5,5) 120 (4.329)
519 P2 G (P/G,5,5) 30 (P/G,5,5) 30
(8.237) 247 P P1 P2 766.
Note 5 and not 4. Using 4 is a common mistake.
21
Arithmetic Gradient

• F G(1i)n-2 2G(1i)n-3 (n-2)G(1i)1
  (n-1)G(1i)0
• F G (1i)n-2 2(1i)n-3
  (n-2)(1i)1 n-1
• (1i) F G (1i)n-1 2(1i)n-2 3(1i)n-3
  (n-1)(1i)1
• iF G (1i)n-1 (1i)n-2 (1i)n-3
  (1i)1 n 1
• G (1i)n-1 (1i)n-2 (1i)n-3
  (1i)1 1 nG
• G (F/A, i, n) - nG G (1i)n-1/i nG
• F G (1i)n-in-1/i2
• P F (P/F, i, n) G (1i)n-in-1/i2(1i)n
  
• A F (A/F, i, n)
• G (1i)n-in-1/i2 i/(1i)n-1
• A G (1i)n-in-1/i(1i)n-i

(n-1)G
2G
G
.
..
0
F
22
Arithmetic Gradient

• Arithmetic Gradient Uniform Series
• Arithmetic Gradient Present Worth
• (P/G,5,5)
• (1i)n i n 1/i2 (1i)n
• (1.05)5 0.25
  1/0.052 (1.05)5
• 0.026281562/0.003190703
  8.23691676.

(P/G,i,n) (1i)n i n 1 / i2 (1i)n
1/(G/P,i,n)
(A/G,i,n) (1/i ) n/ (1i)n 1
1/(G/A,i,n)
(F/G,i,n) G (1i)n-in-1/i2
1/(G/F,i,n)
23
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
300
200
100
This is not in the standard form for using the
gradient equation, because the year-one cash
flow is not zero. We reformulate the problem as
follows.
A
A
A
A
24
Arithmetic Gradient
400
300
G 100
300
200
200
100

A1100

100
0
0 1 2 3 4
0 1 2 3 4
0 1 2 3 4
The second diagram is in the form of a 100
uniform series. The last diagram is now in the
standard form for the gradient equation with n
4, G 100. A A1 G (A/G,6,4) 100
100 (1.427) 242.70
25
Arithmetic Gradient

• Example  
• With i 10, n 4, find an equivalent uniform
  payment A for
• This is a problem with decreasing costs instead
  of increasing costs.
• The cash flow can be rewritten as the
  DIFFERENCE of the following two diagrams, the
  second of which is in the standard form we need,
  the first of which is a series of uniform
  payments.

24000
18000
12000
6000
1
2
3
4
0
26
Arithmetic Gradient
A124000
G6000
A1 A1 A1 A1
24000
3G
18000
2G

-
12000
G
6000
0 1 2 3 4
0 1 2 3 4
0 1 2 3 4
A A1 G(A/G,10,4)
24000 6000 (A/G,10,4) 24000
6000(1.381) 15,714.
27
Arithmetic Gradient

• Example Find P for the following diagram with i
  10.

150
100
50
1
2
3
4
5
6
P
J
This is not in the standard form for the
arithmetic gradient. However, if we insert a
present value J at the end of year 2, the
diagram from that point on IS in standard form.
Thus J 50 (P/G,10,4) 50 (4.378) 218.90 P
J (P/F,10,2) 218.90 (0.8264) 180.9
28
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
•  

29
Geometric Gradient

• Geometric Gradient. At the end of year i, i 1,
  ..., n, we incur a cost Aj A(1g)i-1.
• P A(1i)-1A(1q)1(1i)-2A(1q)2(1i)-3
• A(1q)n-2(1i)-n1A(1q)n-1(1i)-n
•  
• P(1q)1(1i)-1 A(1q)1(1i)-2A(1q)2(1i)-3
  
• A(1q)n-1(1i)-nA(1q)n(1i)-n-1
•  
• P - P(1q)1(1i)-1 A(1i)-1 -
  A(1q)n(1i)-n-1
• P (1i-1-q) A (1 - (1q)n(1i)-n)
• i?q P A (1 - (1q)n(1i)-n)/(i-q)
• iq P A n(1i)-1

A(1q)n-1
A(1q)2
A(1q)1
A
0
1
2
3
.
n
30
Geometric Gradient

• Example n 5, A1 100, g 10, i 8.

(P/A,g,i,n) (P/A,10,8,5) 4.8042 P
A(P/A,g,i,n) 100 (4.8042) 480.42.
31
NOMINAL AND EFFECTIVE INTEREST RATE
32
Nominal and Effective Interest Rate
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(124/4)24-10,000 5938.48
• interest charges for annually compounding
• 10,000(124)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
33
Nominal and Effective Interest Rate
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 With i 0.05/2, r 0.05, P ? (1
i) P ? (1r/2)2 P (1 0.05/2)2 P (1.050625) P
34
Nominal and Effective Interest Rate

• 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 (10.05/2)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.

ia (1 r/m)m 1 (1 i)m 1

• The term i we have used up to now
• is more precisely defined as the
• effective interest rate
• per interest period.
• If the interest period is one year
• (m 1) then i r.

35
Nominal and Effective Interest Rate

• Example A bank pays 1.5 interest every three
  months. What are the nominal and effective
  interest rates per year?
• Solution 
• Nominal interest rate per year is
• r 4 ? 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.

36
Nominal and Effective Interest Rate
37
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 (1i)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
  (110.4/52)52 1 ? 13,104. This means about
  1,310,400 a year.

38
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(1i)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.

39
Nominal and Effective Interest Rate

• Example 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
W
i 2, n 20
5000
40
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 dont, but suppose we do.). We
  compute 
• A P (A/P,i,n) 5000 (A/P,2,20) 5000
  (0.0612) 306.
• These withdrawals are equivalent to P 5000

i 2, n 20
5000
41
Nominal and Effective Interest Rate

• Now consider the following
• Consider a one-year period
• 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.

A
W
W
W
W
i 2, n 4
W
42
Nominal and Effective Interest Rate

• Solution 2 (Probably the easiest way)
• 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 (1i)n/(1i)n 1
  5000(0.252)
• 1260 per year.

W
5000
ia 8.24, n 5
43
Continuous Compounding
44
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.
• 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

ia (1 i)m 1 (1 r/m)m 1
45
Summary Notation

• i effective interest rate per interest period
  (stated as a decimal) 
• n number of interest periods 
• P present sum of money  
• F 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

46
Summary Formulas

• Single Payment formulas
• Compound amount F P (1i)n P (F/P,i,n)
• Present worth P F (1i)-n F (P/F,i,n)
• Uniform Series Formulas
• Compound Amount F A(1i)n 1/i A
  (F/A,i,n)
• Sinking Fund A F i/(1i)n 1 F
  (A/F,i,n)
• Capital Recovery A P i(1i)n/(1i)n
  1 P (A/P,i,n)
• Present Worth P A(1i)n 1/i(1i)n
  A (P/A,i,n)
• Arithmetic Gradient Formulas
• Present Worth P G (1i)n i n 1/i2
  (1i)n G (P/G,i,n)
• Uniform Series A G (1i)n i n 1/i
  (1i)n i G (A/G,i,n)
• Geometric Gradient Formulas 
• If i ? g, P A 1 (1g)n(1i)-n/(i-g)
  A (P/A,g,i,n)
• If i g, P A n (1i)-1 A (P/A,g,i,n)

47
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 (1i)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