ENGINEERING ECONOMICS
ECO - 130
Lecture # 3
Interest & Equivalence
Cash Flow Diagram
Simple & Compound Interest
 Time Value of Money
Present Value
Outline/Sequence
• Intrest rates (Simple & Compound)
• Compound intrest rates (Present value
of money)
• Compounding (Quaterly Compounding
& Semi annually Compounding)
• Economic Eqivalancy
INTEREST RATE/ TIME VALUE OF MONEY
Question: Would you prefer Rs. 5000 today or after 1 year?
• We know that receiving Rs. 5000 today is worth more than in the
future. This is due to OPPORTUNITY COSTS (Foregone
resource).
Interest Rate is always positive
• Opportunity cost of receiving Rs. 5000 in the future is the interest as it is the compensation for
we could have earned if we had received the Rs. 5000 sooner.
“using
money/
uncertainties
• There is a time value of money. Money is an asset, & people related to the future value of
would pay to have money available for use. The charge for its money.
use is called “Interest Rate”.
TYPES OF INTEREST
SIMPLE INTEREST
It is the interest that is computed on the original sum and entire duration of loan.
For example, if the loan amount is “P=Present Vale”, the duration of loan is “n” years and rate
of interest is “i %”, then after “n” years the lender will have an amount “F = Future Value”
F = P + n × (i P) = P (1 + i n)
given as:
Example: If an amount of Rs 5000/- is given on loan for 5 x years on simple interest of 8% per year.
Calculate the amount the lender have after 5 x years. The lender will have F= 5000(1+0.08 x 5) =7000/-
Analysis of Simple Interest: The interest and the
principle amount are paid at the end of the loan period
and the amount of interest that should actually be due
after each year is used by the borrower free of cost.
For Example: The interest on Rs 5000/ after 1st year of
borrowing should be = 0.08x5000= Rs 400/-. Being
simple interest the borrower do not give interest on this
amount to the lender & continue using it for next 4 x years
End of
Year
Beginning Balance
(Rs)
Interest
(Rs)
0
Ending
Balance
(Rs)
5,000
1
5000
400
5400
2
5400
400
5800
3
5800
400
6200
4
6200
400
6600
5
6600
400
7000
F = P (1 + i n), There are 4 variables, if 3 are given the 4th can easily be calculated
 P = $1,000
 i = 8%
 N = 3 years
what will be the ending balance that has to be returned at the end of 3
years?
End of
Year
Beginning Interest
Balance
earned
0
Ending
Balance
$1,000
1
$1,000
$80
$1,080
2
$1,080
$80
$1,160
3
$1,160
$80
$1,240
TYPES OF INTEREST
COMPOUND INTEREST
It is the interest charged on the original sum and un-paid interest. For example, if the loan
amount is “P”, the duration of loan is “n” years and rate of interest is “i %” per year, then
after “n” years the lender will have an amount “F” given as:
Single Payment Compound
Interest F = P (1+i)n
Example: If an amount of Rs 5000/- is given on loan for 5 x years on compound interest of 8% per year.
Calculate the amount the lender have after 5 x years. The lender will have F= 5000(1+0.08)5 =7346.64/n = 0:P
n= 1: F1= P (1+i)
End
of
Year
Beginning
Balance
(Rs)
Interest
(Rs)
Ending
Balance
(Rs)
n= 2: F2= F1(1+i) = P(1+i)2
0
n= 3: F3= F2(1+i) = P(1+i)3
1
5000
400
5400
2
5400
432
5832
3
5832
466.56
6298.56
4
6298.56
503.88
6802.44
5
6802.44
544.2
7346.64
n= 4: F4= F3(1+i) = P(1+i)4
n= 5: F5= F4(1+i) = P(1+i)5
n= n: Fn= Fn-1
(1+i) = P(1+i)n
5,000
Cash Flow Diagram
F = P (1+i)n , There are 4 variables, if 3 are given the 4th can easily be calculated
You put $500 in a bank for 3 years at 6% compound interest per year.What will
be the amount at the end each year?
F = P (1+i)n
 At the end of year 1 you have (1+0.06)  500 = $530
 At the end of year 2 you have (1+0.06)  530 = $561.80
 At the end of year 3 you have (1+0.06)  $561.80 = $595.51
Note:
$595.51 = (1.06)  561.80 = (1.06) (1.06) 530
= (1.06) (1.06) (1.06) 500 = 500 (1.06)3
COMPOUND INTEREST – PRESENT VALUE
Example: If you want to have $800 in savings at the end of four years, and compound
interest of 5% interest per year, how much do you need to put into the savings account
today?
F = P (1+i)n  P =F/ (1+i)n  P = 800/(1.05)4 = 800 x 0.8227=$658.16
P = F (P/F,i,n) This value is available in the compound interest table
P = 800(P/F, 5%,4) = 800 x 0.8227= $658.16
TABLE LOGIC
F=P (1+i)n  F/P = (1+i)n
P/F = 1/ (1+i)n
Compound Amount Factor: For generally used interest rates “i"
tables with various values of “n” have been developed to readily
provide the value of (1+i)n to the user to find F by multiplying it with P
Present Worth Factor: Similarly P/F or 1/ (1+i)n to readily calculate
the present value by multiplying it with “F”
COMPOUND INTEREST – PRESENT VALUE
Example: In 3 years, you need Rs 400 to pay a debt. In two more years, you need Rs
600 more to pay a second debt. How much should you put in the bank today to meet
these two needs if the bank pays an annual compound interest of 12% per year?
P = 400/ (1+0.12)3 +600/ (1+0.12)5
= 284.72 + 340.44 = Rs 625.16/-
P = 400 x 0.7118 + 600 x 0.5674
= 284.72 + 340.44 = Rs 625.16/-
COMPOUND INTEREST – INTEREST RATE & NUMBER OF YRS
Example: $800 is needed after F/P = 1.2155 i = (1.2155, i%, 4)
four years. What interest rate per P/F = 0.8227  i = (0.8227, i%, 4)
against n=4
year you should be looking for if Check any of these values in tables
Or use the formula i= [(1/(P/F))1/n -1] x 100
you have $658.16 for investment:F/P = 1.2155 n = (1.2155, i%, n)
Example: Investing $658.16 at the P/F = 0.8227  n = (0.8227, i%, n)
compound interest rate of 5% per Check any of these values in tables against i=5%
Or use the formula n= (ln F/P)/ln (1+i)
year. How many years will it take
to have $800 in your account :-
QUARTERLY COMPOUND INTEREST
Example: You put Rs 500 in a bank for 3 years at 6% compound interest per year.
Interest is compounded quarterly. Find the amount after 3 years.
F = P (1+i)n
What’s i = ???
Whats n =???
i = 6/4 = 1.5% / 3 months
n = 3 x 4 = 12
F = 500 ( 1+0.015)12 = Rs 597.81
COMPOUND INTEREST – NUMBER OF YRS
Example: How long will it take for $1,000 to double in value, at an interest
rate of 5% compounding semi annually?
F = P (1 + i)n
2000 = 1000 (1.025)n  1.025 n = 2
Taking ln on both sides
n ln(1.025) = ln (2)  n = 28.07
As it is compounding semi annually, it will take 14.5 years but
amount will be a bit higher than Rs 2000
COMPOUND INTEREST – NUMBER OF YRS
Example: How long will it take for $1,000 to double in value, at an interest
rate of 5% compounding annually?
F = P (1 + i)n
2000 = 1000 (1.05)n  1.05n = 2
Taking ln on both sides
n ln(1.05) = ln (2)  n = 14.2067
It will take 15 years but amount will be a bit higher than Rs 2000
ECONOMIC EQUIVALENCE
Which one would you prefer?
• $20,000 today
• $50,000 ten years from now
• $8,000 each year for the next ten years
We need to compare their economic worth!
Example: Ali borrowed Rs 5,000 from a bank at
8% interest rate and have to pay it back in 5 years.
There are many ways the debt can be repaid.
• Plan A: At end of each year pay Rs 1,000
principal plus interest due.
Economic equivalence exists between cash • Plan B: Pay interest due at end of each year and
flows if they have the same economic effect
principal at end of five years.
Which depends on
• Interest rate
• Amounts of money involved
• Time Period
BOTH SYSTEMS SHOULD HAVE SAME TIME
VALUE OF MONEY
• Plan C: Pay Rs 1250/- in five end-of-year
payments.
• Plan D: Pay principal and interest in one
payment at end of five years.
ECONOMIC EQUIVALENCE
Plan A: Pay Rs 1,000
principal + interest (Yrly)
Plan B: Pay Yrly interest
& principal after 5 yrs
Plan C: Pay in five endof-year payments
PLAN A
PLAN B
Total
Payment
Yr
Amt
Owed
Int
Owed
Total Principal
Owed Payment
1,000
1,400
1
5,000
400
5,400
852
1,252
4,320
1,000
1,320
2
4,148
332
4,480
920
1,252
240
3,240
1,000
1,240
3
3,227
258
3,485
994
1,252
2,000
160
2,160
1,000
1,160
4
2,233
179
2,412
1,074
1,252
1,000
80
1,080
1,000
1,080
5
1,160
93
1,252
1,160
1,252
5,000
6,200
SUM
5,000
6,260
Yr
Amt
Int
Total Principal
Owed owed Owed Payment
1
5,000
400
5,400
2
4,000
320
3
3,000
4
5
SUM
Plan D: Pay principal +
interest after five years
1,200
1,261
Total
Payment
ECONOMIC EQUIVALENCE
Plan A: Pay Rs 1,000
principal + interest (Yrly)
Plan B: Pay Yrly interest
& principal after 5 yrs
Plan C: Pay in five endof-year payments
PLAN C
PLAN D
Amt
Owed
Int
Owe
d
Total Principal
Total
Owed Payment Payment
1
5,000
400
5,400
0
400
2
5,000
400
5,400
0
400
3
5,000
400
5,400
0
4
5,000
400
5,400
5
5,000
400
5,400
Yr
SUM
2,000
Plan D: Pay principal +
interest after five years
Yr
End of Yr
payment
Amount
Owed
1
0
5,000
2
0
5,400
400
3
0
5,832
0
400
4
0
6,299
5,000
5,400
5
7,347
6,802
5,000
7,000
SUM
7,347
29,333
ECONOMIC EQUIVALENCE
Example: Ali borrowed Rs 5,000 from a bank at 8% interest rate and have to pay it back in 5 years. There
are many ways the debt can be repaid.
• Plan A: At end of each year pay Rs 1,000 principal plus interest due.
• Plan B: Pay interest due at end of each year and principal at end of five years.
• Plan C: Pay in five end-of-year payments.
• Plan D: Pay principal and interest in one payment at end of five years.
SUMMARY OF PAYMENT PLAN
a
b
c
d
Plan
Amount
Total Area Under Curve Ratio
Paid
Interest (Amount OWED)
b/c
A
6,200
1,200
15,000
0.08
B
7,000
2,000
25,000
0.08
C
6,260
1,260
15,767
0.08
D
7,347
2,347
29,333
0.08
What is the common property in all
Plans? It is Interest Rate=Total
Interest/Area under the curve
Or
Total Interest = Interest Rate x Area
under the curve
Ratio .08 is constant – all payment
methods are equivalent – equivalence
is established between all alternatives
Example: Assuming both the cash flows to be same calculate the value of ‘C’
assuming an interest rate of 12%.
$300
$100
0
1
$300
$300
$C
$100
2
3
4
Time Value of Money
100/1.12 + 100/ 1.12^2
+ 300/1.12^3 +
300/1.12^4 + 300/1.12^5 = $743
5
0
1
2
$C
$C
$C
3
4
5
Time Value of Money
743 = C/1.12 + C/1.12^2 + C/1.12^4 + C/1.12^5
743 = 0.8928C + 0.7971C + 0.6355C + 0.5674C
743 = 2.8928C  C= $257
?
Any Questions