CHAPTER 2.0
MONEY-TIME
RELATIONSHIP
1 CASH FLOW DIAGRAM
2 TIME VALUE OF MONEY
3 EQUIVALENCE
4 SINGLE PAYMENT COMPOUND I NTEREST
5 UNIFORM SERIES COMPOUND I NTEREST
6 ARITHMETIC AND GEOMETRIC GRADIENT
Cash Flow Diagrams
Cash flow diagrams (CFD) summarize
the costs and benefits of projects
Example:
Time Period
Size of Cash Flow
0 (today)
1
2
3
4
5
A CFD illustrates the size, sign,
and timing of individual cash flows
Periods may be months, quarters, years,
etc.
Receive $100 (positive CF)
Pay $100 (negative CF)
Positive CF of $100
Negative CF of $150
Negative CF of $150
Positive CF of $50
Tomorrow
COMMENTS:
• The end of one period is the
beginning of the next one
100
100
50
• Arrows point up for revenues or
benefits, down for costs
• One person’s payment (cash
outflow w. neg. sign) is another
person’s receipt (cash inflow w. pos.
sign)
It is essential to use only one
perspective in any CFD .
0
Today
1
100
2
3
4
150
150
5
Categories of Cash Flows
Category
Description
First Cost
Expense to build or to buy and install
Operations and
Maintenance
(O&M)
Annual expense, which often includes electricity,
labor, minor repairs, etc
Salvage Value
Receipt at project termination for sale or transfer of
the equipment (can be salvage lost)
Revenues
Annual receipts due to sale of products or services
Overhaul
Major capital expenditure that occurs during the
asset’s life
Prepaid
Expenses
Annual expenses, such as leases and insurance
payments, that must be pain in advance
Timing of Cash Flows
Time could start at:◦ At the beginning of analysis
◦ When the project is expected to be approved
◦ When design is started
◦ When the material are first ordered
◦ When construction begins
◦ When use begin
◦ The first cost
Cash Flow Diagram
Practice 1
Ernie’s Earthmoving is considering the purchase of
a piece of heavy equipment. What is the cash flow
diagram if the following cash flow are anticipated?
◦ First cost: RM129K
◦ O & M Cost:RM30K per year
◦ Overhaul Cost: RM 35K in year 3
◦ Salvage value: RM40K after 5 years
Cash Flow Diagram
Practice 2
Rather than purchasing the heavy equipment, as in
Example 2, Ernie’s Earthmoving is planning on
leasing it. The lease payments will be RM25K per
year. Ernie’s Earthmoving is responsible for the
overhaul and O & M costs. What is the cash flow
diagram?
Time Value of Money
Question: Would you prefer $100 today or $100 after 1 year?
There is a time value of money. Money is a valuable asset, and people
would pay to have money available for use. The charge for its use is called
interest rate.
Interest
The cost of money. More specifically, it is a cost to the borrower and an earning
to the lender, above and beyond the initial sum borrowed or loaned.
Interest Rate
A percentage periodically applied to a sum of money to determine the amount
of interest to be added to that sum.
Equivalence
Money is a valuable commodity. People will pay you to use your
money. You will have to pay someone to use his or her money.
For example, leaving RM100 in the bank for one year may generate
an interest of RM6 at the end of the year. That is, the bank has paid
you 6% interest to use your money. In the same vane, the bank will
turn around and lend the money to another individual and charge
10% interest. The borrower will have to pay back RM110 at the end
of one year. This means that your RM100 today is worth RM106 one
year from now because of the interest earned on the principal. With
the same token, RM100 lent by the bank is worth RM110 to the
bank one year from now. We can also say that, for you, RM100
today is equivalent to RM106 one year from now. This principle is
called as THEORY OF EQUIVALENCE.
Theory of Equivalence
In economics, because of the time value of
money, a certain amount at present is equal to
different future amounts given at an interest
rate.
Conversely, an amount of dollars invested now
will be worth more when the principal and its
accumulated interest are received in the future.
The total interest earned or charged is linearly
proportional to the initial amount of the loan (principal),
the interest rate and the number of interest periods for
which the principal is committed.
Simple Interest
When applied, total interest “i” may be found by
i = ( P ) ( N ) ( i ),
where
◦ P = principal amount lent or borrowed
◦ N = number of interest periods ( e.g., years )
◦ i = interest rate per interest period
Compounding
In some financial investments such as fixed
deposit (FD) which involves a deposit of a
certain amount of money, known as a principal,
at a given time, a specified interest rate
compounded periodically, and a specified
period of time;
F = P(1+i)n
Compound Interest
The practice of charging an interest rate to an initial sum and to any
previously accumulated interest that has not been withdrawn from
the initial sum. Compound interest is by far the most commonly
used system in the real world
Economic Equivalence
It exists between individual cash flows and/or pattern of cash flows
that have the save value. Even though the amounts and timing of the
cash flows may differ, the appropriate interest rate makes them
equal.
Example 1
An investor deposits RM1000 into FD account that
pays 10% a year, compounded annually, for 1 year.
F = P(1+i)n
F = 1000(1+0.10)1 = RM1100
If deposited for 2 years
F = 1000(1+0.10)2 = RM1210
If deposited for 5 years at 6%interest rate
F = 1000(1+0.06)5 = RM1338
Practice 3
If you invest RM4000 in a bank at an
interest rate of 6.25% per year, how much
money would you have at the end of three
years?
Annual Percentage Rate (APR)
Interest is most frequently quoted by financial institutions as
an APR. However, compounding often occurs more than once
annually, and the APR does not account for the effect of this
more frequent compounding. This situation leads to the
distinction between nominal and effective interest.
Nominal Interest.
A stated rate of interest for a given period (usually a year).
Effective Interest.
The actual rate of interest, which accounts for the interest
amount accumulated over a given period.
Example 2
If RM100 is invested in a bank at a rate of 12% per year to b compounded
monthly, what would be the effective interest rate?
Interest rate per period (month) = 12%/12month = 1%/month
1 year contains 12 compounding period
F = P(1+i)n = 100(1+0.01)12 = RM112.68
This is equivalent to applying an interest rate of 12.68% and compounding on a yearly
basis.
F = 100(1+0.1268)1 = RM112.68
In other words, the effective interest rate is 12.68% if the interest rate 12%/year and
is compounding monthly.
The 12% /year rate is called a nominal interest rate
Effective and Nominal
i = (1+j/m)m – 1
i = effective interest rate per period
j = nominal interest rate period
m = the number of compounding sub
periods
Practice 4
If RM100 invested in a bank at interest rate
of 12% per year to be compounded
quarterly, what would be the effective
interest rate?
Equivalence
When an organization is indifferent as to
whether it has a present sum of money now or
the assurance of some other money (or a series
of sums of money) in future, we say that the
present sum of money is equivalent to the
future sum or series of sums.
Equivalence
Given the choice of these two plans which would you
choose?
Year
1
2
3
4
5
Total
Plan 1
$1400
1320
1240
1160
1080
$6200
Plan 2
$400
400
400
400
5400
$7000
Theory of Equivalence
Determine a single equivalent value at a point in time
for plan 1
Determine a single equivalent value at a point in time
for plan 2
Both at the same interest rate
Judge the relative attractiveness of the two alternatives
from the comparable equivalent values
Interest Rate Formulas
P = present value of an amount of money
i = annual interest rate
n = number of annual periods
F = future value at the end of the period
A = annual amount, in series of equal payment or
receipts, made at the end of each year
Single Payment Compound Interest
General Notation
◦
F = P(1+i)n
Functional notation
◦ F = P(F/P, i, n)
◦ Read as follows – to find a future sum F, given a present sum,
P at an interest rate i per interest rate period and n interest
periods
◦ Called compound interest factor for single payment
Refer to table to get the factor value
Example 3
If RM1000 is invested for 5 years, interest rate
compounded annually, the future value is obtain by
multiplying RM1000 with the F/P for a single payment
at the given interest rates:
at i = 6%, F = RM1000(1.338) = RM1338
at i = 10%, F = 1000(1.611) = RM1611
at i = 14%, F = 1000(1.925) = RM1925
Single Payment Present Value Factor
General Notation
(P/F, i, n) is called single payment present value factor
 1 
F

P
 F 
n
n 
1  i 
 1  i  
Functional notation
◦ P = F(P/F, i, n)
◦ Read as follows – to find a present sum P, given a future sum, F at
an interest rate i per interest rate period and n interest periods
Refer to table to get the factor value
Example 4
What is a present value of RM1000 due 5 years from
now, discounted at an interest rate of 10%
compounded annually?
P = RM1000(P/F, 10%, 5) = 1000(0.6209)
P = RM620.90
Example 5
Question:
You borrowed RM1000 at 12% compounded annually. The loan is
paid back after 5 years. How much should be repaid?
Solution:
Using the compound interest tables for 12% and five periods, the
value of the (F/P, 12%, 5) factor is found to be 1.7623.
F = P(F/P, 12%, 5) = 1000(1+0.12)5
= 1762.30
Therefore the amount to be repaid equals to RM1762.30
Practice 5
If you invest RM1000 in a bank at an interest rate of
6.25% per year, how much money would you have at
the end of three years?
Practice 7 & 8
Practice 7
If $500 were deposited in a bank savings account, how much
would be in the account three years hence if the bank paid 6%
interest compounded annually?
Practice 8
Suppose the bank changed their interest policy in Example A to
6% interest, compounded quarterly. For this situation, how
much money would be in the account at the end of 3 years,
assuming a $500 deposit now?
Uniform Series Compound Interest Formulas
 Uniform Series Compound Interest
 F is required when A, i , n is known
 Functional Notation: F = A(F/A, i%, n)
 General Equation:
 1  i n  1
F  A

i


 Uniform Series Compound Amount/Sinking Fund Factors
 A is required when F, i%, n is known
 Functional Notation: A = F(A/F, i%, n)
 General Equation


i
A  F

n


1

i

1


Uniform Series Compound Interest
Formulas
Capital Recovery Factor
◦ A is required when P, i%, n is known
◦ Functional Notation: A = P (A/P, i%, n)
◦ General Equation
 i 1  i n 
A  P

n


1

i

1


Uniform Series Present Value Factor
◦ P is required when A, i%, n is known
◦ P = A(P/A, i, n)
◦ General Equation
 1  i n  1
P  A
n 
 i 1  i  
Uniform Series Compound Interest
It is often required to know the present value of a series of equal payments (or
receipt) over a number of years at a given interest rate.
Example: What is a present value of an investment which will generate RM1000
per year over the next 3 years if the interest rate is 10% compounded annually?
P = 1000[(P/F, 10, 1) + (P/F, 10, 2) + (P/F, 10, 3)]
P
A

A

A
1  i 1 1  i 2 1  i 3
P = 1000[(0.9091) + (0.8265) + (0.7513)]
= RM2486.90
OR
P = 1000*(P/A, 10, 3) = 1000*(2.4869) = RM2486.90
P/A is called uniform series present value factor
Practice 9
If you want to invest sufficient money in
the bank such that at an interest rate of
8%, you will receive RM20,000 per year
for the next 10 years, how much money
should you invest in the bank now?
Practice 10
If you take a home loan improvement loan of
RM10,000 to be paid over a five year period, what
would be the yearly payment if the interest rate is
12% per year?
Practice 11
You wish to deposit a single sum of money in a
savings account so that five equal annual
withdrawals of RM2000 can be made before
depleting the fund. If the first withdrawal is to
occur 1 year after the deposit and the fund pays
interest at a rate of 12% compounded annually,
how much should be deposited?
Practice 12
If you deposited RM10,000 at the end of each year,
how much would you accumulate at the end of 5 years
at an interest rate of 6%
Practice 13
If you need RM 100,000 ten years from now for your
son’s education, how much should you invest at the
end of each year at an interest rate of 8%
Practice 14
Jim Hayes read that out west, a parcel of land could
be purchased for $1000 cash. Jim decided to save a
uniform amount at the end of each month so that he
would have the required $1000 at the end of one
year. The local credit union pays 6% interest,
compounded monthly. How much would Jim have to
deposit each month?
Arithmetic and Geometric Gradient
In applications, the annuity cash flow pattern is not
the only type of pattern encountered
Two other types of end of period patterns are
common
◦ The Linear or arithmetic gradient
◦ The geometric (% per period) gradient
Arithmetic Gradient Factors
An arithmetic (linear) Gradient is a cash flow series
that either increases or decreases by a constant
amount over n time periods.
A linear gradient is always comprised of TWO
components:
Arithmetic Gradient Factors
•The Two Components are:
•The Gradient component
•The base annuity component
•The objective is to find a closed form
expression for the Present Worth of an
arithmetic gradient
Linear Gradient Example
A1+(n-1)G
Assume the following:
A1+(n-2)G
A1+2G
A1+G
0
1
2
3
n-1
N
This represents a positive, increasing arithmetic gradient
Example: Linear Gradient
• Typical Negative, Increasing Gradient: G=$50
The Base Annuity
= $1500
Example: Linear Gradient
• Desire to find the Present Worth of this cash flow
The Base Annuity
= $1500
Arithmetic Gradient Factors
• The “G” amount is the constant arithmetic
change from one time period to the next.
•The “G” amount may be positive or negative!
•The present worth point is always one time
period to the left of the first cash flow in the
series or,
•Two periods to the left of the first gradient cash
flow!
Derivation: Gradient Component Only
Focus Only on the gradient Component
(n-1)G
“0” G
(n-2)G
+2G
G
Removed Base annuity
0
1
2
3
n-1
N
Present Worth Point…
The Present worth point of a linear gradient is always:
◦2 periods to the left of the “1G” point or,
◦1 period to the left of the very first cash flow
in the gradient series.
DO NOT FORGET THIS!
Present Worth Point…
$700
$600
$500
$400
$300
$200
$100
X
0
1
2
3
4
5
The Present Worth Point of the
Gradient
6
7
Gradient Component
$600
•The Gradient Component
$500
$400
$300
$200
$100
$0
X0
1
2
3
4
The Present Worth Point of the
Gradient
5
6
7
Present Worth Point…
•PW of the Base Annuity is at t = 0
•PWBASE Annuity=$100(P/A,i%,7)
Base Annuity – A = $100
X
0
1
2
3
4
The Present Worth Point of the
Gradient
5
6
7
Gradient Series
1  1  ni   1  i  n 
P  G

2
i


or
P = G(P/G, i, n)
Where;
(P/G, I, n) is called gradient series, present worth factor
Uniform Series Equivalent to the Gradient
series
A = G(A/G, i, n)
Where;
(A/G, i, n) is called gradient to uniform series
conversion factor
Practice 15
A man has purchased a new automobile. He wishes to set aside enough money
in a bank account to pay the maintenance on the car for the first five years. It
has been estimated that the maintenance cost of an automobile is as follows:
Year
Maintenance Cost
1
$120
2
150
3
180
4
210
5
240
Assume the maintenance costs occur at the end of each year and that the bank
pays 5% interest. How much should the car owner deposit in the bank now?
Practice 16
On a certain piece of machinery, it is estimated that the
maintenance expenses will be as follows:
Year
Maintenance Cost
1
$100
2
200
3
300
4
400
What is the equivalent uniform annual maintenance cost for the
machinery if 6% interest is used?
Practice 17
A textile mill in India installed a number of new looms. It is expected that the
initial maintenances costs and expenses for repairs with high but then
decline for several years. The projected cost is:
Year
Maintenance and
Repair Costs
(rupees)
1
24000
2
18000
3
12000
4
6000
What is the projected equivalent annual maintenance and repair cost if
interest is 10%?
Practice 18
Maintenance costs for a particular production machine
increase by $1000/year over the 5 year life of the
equipment. The initial maintenance cost is $3000.
Using an interest rate of 8% compounded annually,
determine the present worth equivalent for the
maintenance costs.
Geometric Gradients
• An arithmetic (linear) gradient changes by a fixed dollar
amount each time period.
•A GEOMETRIC gradient changes by a fixed percentage
each time period.
•We define a UNIFORM RATE OF CHANGE (%) for each time
period
•Define “g” as the constant rate of change in decimal form
by which amounts increase or decrease from one period to
the next
Geometric Series
1  1  g n 1  i  n 
P  A

ig


nA1
P
1 i
ig
i g
Or
P = A1(P/A1, i, g, n)
Where, (P/A1, i,g,n) is called geometric series present value factor
Practice 19
A company is considering purchasing a new machine tool. In
addition to the initial purchase and installation costs,
company management is concerned about the machine’s
maintenance cost. The maintenance cost of the machine
tool are expected to be $1000 at the end of the first year of
the machine’s life and increase 8% per year thereafter. The
expected life of the machine tool is 15 years. Company
management would like to endow a maintenance fund for
expected costs. If the endowment account earns 10% per
year compounded annually, how much money must be
initially deposited in the account?
Geometric Series
 1  i n  1  g n 
F  A1 

ig


F  nA1 1  i 
n 1
ig
ig
Or
F = A1(F/A1, I, g, n)
Where, (F/A1, i,g,n) is called geometric series future value factor
Practice 20
You receive an annual bonus and deposits it in a
savings account that pays 8% compounded
annually. The size of the bonus increases by 10%
each year; the initial deposit was RM500.
Determine how much will be in the fund
immediately after tenth deposit.