Contemporary Engineering Economics, 4 th edition ©2007 Time Value of Money Lecture No.4 Chapter 3 Contemporary Engineering Economics Copyright © 2006
Dec 21, 2015
Contemporary Engineering
Economics, 4th edition ©2007
Time Value of Money
Lecture No.4Chapter 3Contemporary Engineering EconomicsCopyright © 2006
Contemporary Engineering
Economics, 4th edition © 2007
Chapter Opening Story —Take a Lump Sum or Annual Installments
Mrs. Louise Outing won a lottery worth $5.6 million.
Before playing the lottery, she was offered to choose between a single lump sum $2.912 million, or $5.6 million paid out over 20 years (or $280,000 per year).
She ended up taking the annual installment option, as she forgot to mark the “Cash Value box”, by default.
What basis do we compare these two options?
Contemporary Engineering
Economics, 4th edition © 2007
Year Option A
(Lump Sum)
Option B
(Installment Plan)
0
1
2
3
19
$2.912M $283,770
$280,000
$280,000
$280,000
$280,000
Contemporary Engineering
Economics, 4th edition © 2007
What Do We Need to Know?
To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different point in time.
To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis.
Contemporary Engineering
Economics, 4th edition © 2007
Time Value of Money Money has a time value
because it can earn more money over time (earning power).
Money has a time value because its purchasing power changes over time (inflation).
Time value of money is measured in terms of interest rate.
Interest is the cost of money—a cost to the borrower and an earning to the lender This a two-edged sword whereby earning
grows, but purchasing power decreases (due to inflation), as time goes by.
Contemporary Engineering
Economics, 4th edition © 2007
Cash Flow Transactions for Two Types of Loan RepaymentEnd of Year Receipts Payments
Plan 1 Plan 2
Year 0 $20,000.00 $200.00 $200.00
Year 1 5,141.85 0
Year 2 5,141.85 0
Year 3 5,141.85 0
Year 4 5,141.85 0
Year 5 5,141.85 30,772.48The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)
Contemporary Engineering
Economics, 4th edition © 2007
Methods of Calculating Interest Simple interest: the practice of charging an
interest rate only to an initial sum (principal amount).
Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.
Contemporary Engineering
Economics, 4th edition © 2007
Simple Interest
P = Principal amount i = Interest rate N = Number of
interest periods Example:
P = $1,000 i = 10% N = 3 years
End of Year
Beginning
Balance
Interest earned
Ending Balance
0 $1,000
1 $1,000 $100 $1,100
2 $1,100 $100 $1,200
3 $1,200 $100 $1,300
Contemporary Engineering
Economics, 4th edition © 2007
Simple Interest Formula
( )
where
= Principal amount
= simple interest rate
= number of interest periods
= total amount accumulated at the end of period
F P iP N
P
i
N
F N
$1,000 (0.10)($1,000)(3)
$1,300
F
Contemporary Engineering
Economics, 4th edition © 2007
Compound Interest
P = Principal amount i = Interest rate N = Number of
interest periods Example:
P = $1,000 i = 10% N = 3 years
End of Year
Beginning Balance
Interest earned
Ending Balance
0 $1,000
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331
Contemporary Engineering
Economics, 4th edition © 2007
Compounding Process
$1,000
$1,100
$1,100
$1,210
$1,210
$1,3310
1
2
3
Contemporary Engineering
Economics, 4th edition © 2007
0
$1,000
$1,331
1 2
3
3$1,000(1 0.10)
$1,331
F
Cash Flow Diagram
Contemporary Engineering
Economics, 4th edition © 2007
Relationship Between Simple Interest and Compound Interest
Contemporary Engineering
Economics, 4th edition © 2007
Compound Interest Formula
1
22 1
0 :
1: (1 )
2 : (1 ) (1 )
: (1 )N
n P
n F P i
n F F i P i
n N F P i
Contemporary Engineering
Economics, 4th edition © 2007
Some Fundamental Laws
2
F m a
V i R
E m c
The Fundamental Law of Engineering Economy
(1 )NF P i
Contemporary Engineering
Economics, 4th edition ©2007
Compound Interest
“The greatest mathematical discovery of all time,”
Albert Einstein
Contemporary Engineering
Economics, 4th edition © 2007
Practice Problem: Warren Buffett’s Berkshire Hathaway Went public in 1965: $18
per share Worth today (June 22,
2006): $91,980 Annual compound growth:
23.15% Current market value:
$115.802 Billion If his company continues to
grow at the current pace, what will be his company’s total market value when reaches 100? ( lives till 100 (76 years as of 2006)
Contemporary Engineering
Economics, 4th edition © 2007
Market Value
Assume that the company’s stock will continue to appreciate at an annual rate of 23.15% for the next 24 years.
24$115.802 (1 0.2315)
$17.145 trillions
F M
Contemporary Engineering
Economics, 4th edition © 2007
EXCEL Template
In 1626 the Indians sold Manhattan Island to Peter Minuit of the Dutch West Company for $24.
• If they saved just $1 from the proceeds in a bank account that paid 8% interest, how much would their descendents have now?
• As of Year 2006, the total US population would be close to 300 millions. If the total sum would be distributed equally among the population, how much would each person receive?
Contemporary Engineering
Economics, 4th edition © 2007
Excel Solution
380
$1
8%
380 years
$1(1 0.08) $5,023,739,194,020
P
i
N
F
=FV(8%,380,0,1)= $5,023,739,194,020
$5,023,739,194,020Amount per person
300,000,000
$16,746
Contemporary Engineering
Economics, 4th edition © 2007
Practice Problem
Problem Statement
If you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10?
Contemporary Engineering
Economics, 4th edition © 2007
Solution
0 1 2 3 4 5 6 7 8 9 10
$100$200
F
10
8
$100(1 0.10) $100(2.59) $259
$200(1 0.10) $200(2.14) $429
$259 $429 $688F
Contemporary Engineering
Economics, 4th edition © 2007
Practice problem
Problem Statement
Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn a 10% interest, what would be the balance at the end of 4 years?
$1,000 $1,500
$1,210
0 1
2 3
4?
$1,000
Contemporary Engineering
Economics, 4th edition © 2007
$1,000$1,500
$1,210
0 1
2
3
4
?
$1,000
$1,100
$2,100 $2,310
-$1,210
$1,100
$1,210
+ $1,500
$2,710
$2,981
$1,000
Contemporary Engineering
Economics, 4th edition © 2007
SolutionEnd of Period
Beginning
balance
Deposit made
Withdraw Ending
balance
n = 0 0 $1,000 0 $1,000
n = 1 $1,000(1 + 0.10) =$1,100
$1,000 0 $2,100
n = 2 $2,100(1 + 0.10) =$2,310
0 $1,210 $1,100
n = 3 $1,100(1 + 0.10) =$1,210
$1,500 0 $2,710
n = 4 $2,710(1 + 0.10) =$2,981
0 0 $2,981