Title: Intro to Engineering Economy
1Intro to Engineering Economy
- Objectives
- Understand the concept of time value of money
- Illustrate the basic equivalence calculations
- Understand Project Cash Flows
- Formulate mutually exclusive alternatives
- Perform Present Worth Analysis
2Engineering Economy Decisions
Planning
3Role of Engineering Economy
- Understand the Problem
- Collect all relevant data/information
- Define the feasible alternatives
- Evaluate each alternative
- Select the best alternative
- Implement and monitor
Tools Present Worth, Future Worth Annual Worth,
Rate of Return Benefit/Cost, Payback, Capitalized
Cost, Value Added
Major Role of Engineering Economy
4Interest The cost of money
- Interest rate a percentage that is periodically
applied and added to an amount of money over a
specified length of time - Interest rate for saving account
- Interest rate when borrowing money
- Time value of money
- A dollar today is worth more than a dollar one or
more years from now because of the interest
(profit) it can earn - Interest is the cost of money
- A cost to the borrower and an earning to the
lender
5Economic Equivalence
- Economic Equivalence exits between cash flows
that have the same economic effect and could be
traded for one another in the financial market
place - Cash flows can be converted to an equivalent
cash flow at any point in time - Two sums of money at two different points in
time can be made economically equivalent if we
consider - an interest rate and,
- No. of time periods between the two sums
Equality in terms of Economic Value
6Equivalence Illustrated
20,000 is received here
t 1 Yr
t0
21,800 paid back here
- 20,000 now is not equal in magnitude to 21,800
1 year from now - But, 20,000 now is economically equivalent to
21,800 one year from now if the interest rate is
9 per year. - If you were told that the interest rate is 9....
- Which is worth more?
- 20,000 now or
- 21,800 one year from now?
- The two sums are economically equivalent but not
numerically equal!
7Simple and Compound Interest
- A deposit of P dollars with interest rate of i
for N periods. - Simple Interest
- the practice of charging an interest rate only to
an initial sum (principal amount). - Total interest earned is
- I (iP)N
- Total future amount is
- F PI P(1iN)
- Compound Interest
- the practice of charging an interest rate to an
initial sum and to any previously accumulated
interest that has not been withdrawn. - Total future amount is
- F PI P(1i)N
8Ex Compound Interest
For compound interest, with 5 annual interest
rate for 3 years
Owe at t 3 years 1,000 50.00 52.50
55.13 1,157.63
9Terminology and Symbol
- P value or amount of money at a time designated
as the present or time 0. - Also, P is referred to as present worth (PW),
present value (PV), net present value (NPV),
discounted cash flow (DCF), and capitalized cost
(CC) dollars - F value or amount of money at some future time.
- Also, F is called future worth (FW) and future
value (FV) dollars - A series of consecutive, equal, end-of-period
amounts of money. - Also, A is called the annual worth (AW) and
equivalent uniform annual worth (EUAW) dollars
per year, dollars per month - n number of interest periods years, months,
days - i interest rate or rate of return per time
period percent per year, percent per month - t time, stated in periods years, months,
days, etc
10Cash Flow
- CASH INFLOWS
- Money flowing INTO the firm from outside
- Upward arrows (?)
- Revenues, Savings, Salvage Values, etc.
- CASH OUTFLOWS
- Disbursements
- Downward arrows (?)
- First costs of assets, labor, salaries, taxes
paid, utilities, rents, interest, etc. - NET CASH FLOW
- two or more receipts and disbursement at the
same time are summed and shown in a single arrow - Cash Inflows Cash Outflows
- END-OF-PERIOD CONVENTION
- placing all cash flow transactions at the end of
an interest period.
11Types of Cash Flows
- Single cash flow
- Equal (uniform) series
- Linear gradient series
- Geometric gradient series
- Irregular series
12Single-Payment Factors (F/P and P/F)
- Find F, given P , i, N
- Single-payment compound amount factor
- F P (1i)N P (F/P,i,N)
- Ex If you had 2,000 now an invested it at 10,
how much would it be worth in 8 years
- Find P, given F , i , N,
- Single-payment present worth factor
- P F (1 i )-N F (P/F,i,N)
- Ex Suppose that 1,000 is to be received in 5
years. At an annual interest rate of 12, what is
the present worth of this amount?
Also see Example 2.1 2.3
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15Example Adapted from Park (2004)
- A company has borrowed 250,000 to purchase an
equipment. A loan was offered with 8 interest
rate per year. The company plans to repay
installments in equal amounts over the next 6
years. - What is annual installment?
- What if the repayment is deferred for 1 year,
what would be the annual installment in this
case? - That is, the 1st loan payment starts at the end
of year 2 and continues to year 7.
16Shifted Uniform Series
- A shifted series is one whose present worth point
in time is NOT t 0. - Shifted either to the left of 0 or to the right
of t 0. - Dealing with a uniform series
- The PW point is always one period to the left of
the first series value - No matter where the series falls on the time
line.
i 10
A -500/year
P2
P0
17Series with other single cash flow
- It is common to find cash flows that are
combinations of series and other single cash
flows. - Solve for the series present worth values then
move to t 0. - Solve for the PW at t 0 for the single cash
flows. - Add the equivalent PWs at t 0.
i 10
18Ex Final Exam (2004)
- An engineering student who will soon receive his
B.S. degree is considering continuing his formal
education by working toward an M.S. degree. The
student estimates that his average earnings for
the next 6 years with a B.S. degree will be
40,000 per year. If he can get an M.S. degree
in one year, his earnings should average 44,000
per year for the subsequent 5 years. His
earnings while working on the M.S. degree will be
negligible and his additional expenses to be paid
out over this year will be 10,000. The student
estimates that his average per-year earnings in
the two decades following the initial 6-year
period will be 42,000 and 45,000, if he does
not stay for an M.S. degree. If he receives an
M.S. degree his earnings per year in the two
decades can be stated as 42,000 x and 45,000
x. The interest rate is assumed at 15. - Draw cash flow diagrams for two options working
and studying. The diagram must be clearly
labeled. - Find the value of x for which the extra
investment in formal education will pay for
itself. In other words, find the value of x in
which the two options (working and studying) are
break-even and equivalent. Note Use up to 4
decimal points throughout your calculation.
19Evaluating Alternatives
- Revenue/Cost the alternatives consist of cash
inflow and cash outflows - Select the alternative with the maximum economic
value - Service the alternatives consist mainly of cost
elements - Select the alternative with the minimum economic
value (min. cost alternative)
20Evaluating Alternatives
- Independent the decision on any one project has
no effect on the decision made on another project - Mutually exclusive Acceptance of one project
will automatically rejects all other projects
21Alternatives
Analysis
Selection
Problem
Execution
22Evaluating AlternativesPresent Worth Analysis
(PW)
A process of obtaining the equivalent worth of
future cash flows to some point in time
called the Present Worth (PW)
At an interest rate usually equal to or greater
than the Organizations established Minimum
Attractive Rate of Return (MARR).
Step 1 Calculate PW of each alternative at MARR
23Present Worth Analysis of Equal-Life
Alternatives
Step 1 Calculate PW of each alternative at MARR
- Ex Assume 2 investment alternatives with the
same useful life of 4 years. Due to limited
investment fund, the 2 alternatives are mutually
exclusive. Based on the following information,
which one should we select? Assume MARR 10
Also see Ex 5.1
24Ex PW Analysis (Equal-life)
Consider Machine A Machine B First
Cost 2,500 3,500 Annual Operating Cost
900 700 Salvage Value 200
350 Life 5 years 5 years i 10 per
year
Which alternative should we select?
25Present Worth Analysis of Different Life
Alternatives
- Comparison must be made over equal time periods
- Compare over the least common multiple, LCM, for
their lives - Assume repeatability. Alternative will repeat the
same manner over each life cycle - Cash flow estimates are the same in every life
cycle - Example 3,4, and 6 years. The lowest common
life is 12 years. - Evaluate all over 12 years for a PW analysis.
Also see Ex 5.2
26Ex PW Analysis (Different-life)
Assume MARR 10 per year.
Which alternative should we select?