10/15/99 Engineering Economics 1 1
ENGINEERING ECONOMICS I
TO INVEST,
OR NOT TO INVEST,
THAT IS THE QUESTION,
(With apologies to William Shakespeare)
t Given one or more potential projects,
– how do we decide in which to invest,if any?
t What criteria do we use to evaluate potential projects?
10/15/99 Engineering Economics 1 2
THE TWO BASIC INVESTMENT CONCERNS
uPROFITABILITY
(HOW MUCH MONEY WILL I MAKE?)
uRISK
(WHAT ARE MY CHANCES OF MAKING IT?)
10/15/99 Engineering Economics 1 3
PROFITABILITY EVALUATION
t KEY to PROFITABLITY ANALYSIS
e RETURN ON INVESTMENT
uKEY to RETURN ON INVESTMENT
e TIME VALUE OF MONEY
10/15/99 Engineering Economics 1 4
TIME VALUE OF MONEY I
SIMPLE INTEREST - SINGLE PAYMENT
t Example,$100 invested at 5% per annum
At beginning of Year 1,invest $100
At end of Year 1,receive (0.05)(100) = $5.00
At beginning of Year 2,leave the $100 on deposit
At end of Year 2,receive another $5.00
And so on,as long as we leave the $100 on deposit.
10/15/99 Engineering Economics 1 5
TIME VALUE OF MONEY II
COMPOUND INTEREST - SINGLE PAYMENT
t Example,$100 invested at 5% per annum
At beginning of Year 1,invest $100
At end of Year 1,add (0.05)(100) = $5.00
At beginning of Year 2,have $105.00 on deposit
At end of Year 2,add (0.05)(105.00) = $5.25
At beginning of Year 3,have $110.25 on deposit
At end of Year 3,add (0.05)(110.25) = $5.51
And so on.
10/15/99 Engineering Economics 1 6
TIME VALUE OF MONEY III
COMPOUND INTEREST - MULTIPLE PAYMENTS
t Example,$100 invested initially at 5% per annum
At end of Year 1,have $105.00 on deposit
Pay in another $200,Begin Year 2 with $305.00
At end of Year 2,add (0.05)(305.00) = $15.25
Pay in another $300,Begin Year 3 with $620.25.
At end of Year 3,add (0.05)(620.25) = $31.01.
Pay in $0,Begin Year 4 with $651.26.
And so on..
10/15/99 Engineering Economics 1 7
THE EQUATION OF MONEY
Let Q(n )= the amount of money available at the beginning
of the nth year
F(n) = the amount of money paid in at the end of the
nth year
i = the interest rate
Then,the Equation of Money is
Q(n+1) = (1+i) Q(n ) + F(n ),n = 0,…,N
Note,Q(0) = the initial payment
10/15/99 Engineering Economics 1 8
SOLVING THE EQUATION OF MONEY
1) Use a spreadsheet
2) For constant cash flows,there is an analytical solution,
i.e.,for
F(n+1) = F(n) = F
then
Q(N) = Q(0) [(1+i)N] + F {[(1+i)N] - 1}/i
10/15/99 Engineering Economics 1 9
ANNUITIES AND MORTGAGES
t MORTGAGE
Repay a Loan [Q(0)] in N Equal Payments,i.e.,
determine F so that Q(N) = 0
F = -{ i Q(0) [(1+i)N] } / {[(1+i)N] - 1}
t ANNUITY
Accumulate a Specified Amount [Q(N)] in N Equal
Payments F starting with Q(0) = 0.
F = - { i Q(N)} / {[(1+i)N] - 1}
10/15/99 Engineering Economics 1 10
PRESENT VALUE AND FUTURE VALUE
t PRESENT VALUE (PV)
Let FV be a amount of money to be received N years
from now? What is FV worth today?
PV = FV/(1 + i)N
t FUTURE VALUE (FV)
What will an investment PV made today be worth N Time
periods in the future?
FV = PV (1 + i)N
10/15/99 Engineering Economics 1 11
LOANS VERSUS INVESTMENTS
t LOAN
At the End of the Project Period You Get Your Money
Back,e.g.,a bank deposit or CD
t INVESTMENT
You Do Not Get Your Money Back per se.
The Income From the Investment Must Pay It Back
Along With a Profit.
10/15/99 Engineering Economics 1 12
OTHER COMPOUNDING INTERVALS
t Let M = Number of Compounding Intervals per Year and
t Let r = the Interest Rate for the Compounding Interval
t Then r = i/M and
t Nm = N x M,i.e.,Nm is the total number of compounding
intervals
Example
i = 6% compounded monthly for 2 years
Then,r = 0.06/12 = 0.005 and Nm = 2x12 = 24
10/15/99 Engineering Economics 1 13
CONTINUOUS VERSUS DISCRETE INTEREST
t DISCRETE INTEREST
Finite (and Small) Number of Compounding Intervals
t CONTINUOUS INTEREST
Interest Compounded Continuously - Limiting Case as the
Compounding Interval Becomes Small (say daily or less)
10/15/99 Engineering Economics 1 14
CONTINUOUS INTEREST
t EQUATION OF MONEY
dq/dt = i q + f(t)
t SOLUTION FOR CONSTANT f
q(T) = q(0) exp[iT] + f {exp[iT] - 1} / i