Introduction to Engineering
Economic
Anastasia Lidya M.
1
Short Syllabus
• This course studies the basic concept of time value of money
and methods for alternative and investment evaluation. The
study covers definition and scope of engineering economics,
cash flow, concept of time value of money, present equivalent
value, annual value, internal rate of return, payback method,
profitability index method, sensitivity analysis, depreciation,
inflation and deflation, replacement analysis, tax analysis,
public investment cost-benefit analysis. The course provides
ability in conducting analysis and decision making for
alternative selection or investment evaluation using economic
criteria.
2
Grading
• Requirements:
– Min 80% Present in Class
• Score/Grade :
–
–
–
–
Project
Quiz
Mid test (UTS)
Final Test (UAS)
: 25%
: 25%
: 25%
: 25%
• No additional assignment to increase your score
• Project should be submitted on time
3
Outline
Meeting
Topics
1
Introduction to Engineering Economics and Time
Value of Money
2
Time Value of Money and More Interest Formula
3
Present Worth Analysis
4
Present Worth Analysis （Continued）
5
Annual Cash Flow Analysis
6
Rate of Return Analysis
7
Mid Term Exam
8
Incremental Analysis
9
Other Analysis Techniques
10
Other Analysis Techniques (Continued)
11
Depreciation and Income Taxes
12
Replacement Analysis
Selection of a Minimum Attractive Rate of Return
13
Economic Analysis in Government
4
The Role of Engineering Economic
Analysis
The problems most suitable for solution by EEA
have these qualities:
1. The problem is sufficiently important
2. The problem can’t be worked in one’s head ( a
careful analysis requires that we organize the
problem and all the various consequences)
3. The problem has economic aspects there are
sufficiently important to be a significant
component of the analysis leading to a decision
5
Examples of Engineering Economic
Analysis
• Engineering economic analysis focuses on
costs, revenues, and benefits that occur at
different times.
• For example, when a civil engineer designs a
road, a dam, or a building, the construction
costs occur in the near future;
• the benefits to users begin only when
construction is finished, but then the benefits
continue for a long time.
6
Engineering economic analysis is used to answer many
different questions
• Which engineering projects are worthwhile?
Has the mining or petroleum engineer shown
that the mineral or oil deposit is worth
developing?
• Which engineering projects should have a
higher priority? Has the industrial engineer
shown which factory improvement projects
should be funded with the available dollars?
7
Engineering economic analysis is used to answer many
different questions
• How should the engineering project be designed?
Has the mechanical or electrical engineer chosen
the most economical motor size? Has the civil or
mechanical engineer chosen the best thickness
for insulation? Has the aeronautical engineer
made the best trade-offs between :
1) lighter materials that are expensive to buy but
cheaper to fly and
2) heavier materials that are cheap to buy and
more expensive to fly?
8
Engineering economic analysis is used
to answer many different questions
• How to achieve long-term financial goals: How much
should you save each month to buy a house, retire,
or fund a trip around the world?
• How to compare different ways to finance purchases:
Is it better to finance your car purchase by using the
dealer's low interest rate loan or by taking the rebate
and borrowing money from your bank or credit
union?
9
Rational Decision-Making Process





Recognize the decision
problem
Collect all needed (relevant)
information
Identify the set of feasible
decision alternatives
Define the key objectives
and constraints
Select the best possible and
implementable decision
alternative
10
A Simple Illustrative Example: Car to Lease
– Saturn or Honda?





Recognize the decision
problem
Collect all needed (relevant)
information
Identify the set of feasible
decision alternatives
Define the key objectives
and constraints
Select the best possible and
implementable decision
alternative
•
Need to lease a car
•
Gather technical and
financial data
Select cars to consider
Wanted: small cash outlay,
safety, good performance,
aesthetics,…
Choice between Saturn and
Honda (or others)
Select a car (i.e., Honda,
Saturn or another brand)
•
•
•
•
11
Engineering Economic Decisions
Needed e.g. in the following (connected) areas:
Profit! Then continue
at the next stage…
Manufacturing
Design
Financial
planning
Investment
and loan
Marketing
12
What Makes Engineering Economic
Decisions Difficult? Predicting the Future
• Estimating the required
investments
• Estimating product
manufacturing costs
• Forecasting the demand for
a brand new product
• Estimating a “good” selling
price
• Estimating product life and
the profitability of
continuing production
13
Example 1:
Healthcare Service Improvement
1 Traditional Plan: Patients
visit the service providers
 2 New Strategy: Service
providers visit the patients
Which one of the two plans
is more economical? The
answer typically depends on
the type of patients and the
services offered. Examples?

service providers
patients
1
2
14
Example 2:
Equipment and Process Selection
• How do you choose between using alternative materials for
an auto body panel?
• The choice of material will dictate the manufacturing process
and the associated manufacturing costs
15
Example 3:
Equipment Replacement Problem
• Key question:
When is the right time to
replace an old machine or
equipment?
16
Example 4:
New Product and Product Expansion
• Shall we build or acquire a
new facility to meet the
increased (increasing
forecasted) demand?
• Is it worth spending money
to market a new product?
17
Example 5: MACH 3 Project
• R&D investment: $750 million(!)
• Product promotion through
advertising: $300 million(!)
• Priced to sell at 35% higher than
the preceding Sensor Excel model
(i.e., about $1.50 extra per razor)
• Question 1: Would consumers pay
$1.50 extra for a shave with greater
smoothness and less irritation?
• Question 2: What happens if the
blade consumption drops more
than 10% – due to the longer blade
life of the new razor?...
18
Example 6: Cost Reduction
• Should a company buy new
equipment to perform an
operation that is now done
manually?
• Should we spend money
now, in order to save more
money later?
• The answer obviously
depends on a number of
factors.
19
The Four Fundamental Principles of
Engineering Economics
1: An instant dollar is worth more than a distant dollar…
2: Only the relative (pair-wise) difference among the
considered alternatives counts…
3: Marginal revenue must exceed marginal cost, in order to
carry out a profitable increase of operations
4: Additional risk is not taken without an expected additional
return of suitable magnitude
20
Principle 1
An instant dollar is worth more than a
distant dollar…
Today
6 months later
21
Principle 2
Only the cost (resource) difference among
alternatives counts
Option
Monthly
Fuel Cost
Monthly
Maintenance
Cash paid
at signing
(cash
outlay )
Monthly
payment
Salvage
Value at end
of year 3
Buy
$960
$550
$6,500
$350
$9,000
Lease
$960
$550
$2,400
$550
0
The data shown in the green fields are irrelevant items for decision
making, since their financial impact is identical in both cases
22
Principle 3
Marginal (unit) revenue has to exceed
marginal cost, in order to increase
production
Marginal
cost
Manufacturing cost
1 unit
Marginal
revenue
Sales revenue
1 unit
23
Principle 4
Additional risk is not taken without a
suitable expected additional return
Investment Class
Potential
Risk
Expected
Return
Savings account
(cash)
Lowest
1.5%
Bond (debt)
Moderate
4.8%
Stock (equity)
Highest
11.5%
A simple illustrative example. Note that all investments imply
some risk: portfolio management is a key issue in finance
24
EXAMPLE
25
Solution
26
solution
• The shipping department would reduce its cost from
$793.50 to $688.50 by using the outside printer. In that
case, how much would the printing department's costs
decline? We will examine each of the cost components:
1. Direct Labor. If the printing department had been working
overtime, then the overtime could be reduced or eliminated.
But, assuming no overtime, how much would the saving be? It
seems unlikely that a printer could be fired or even put on less
than a 40-hour work week. Thus, although there might be a
$228 saving, it is much more likely that there will be no
reduction in direct labor.
27
Solution
2. Materials and Supplies. There would be a
$294 saving in materials and Supplies
3. Allocated Overhead Costs:
There will be no reduction in the printing
department’s monthly $5000 overhead, for
there will be no reduction in department floor
space.
28
ENGINEERING DECISION MAKING FOR CURRENT COSTS (1)
Example 1:
• A concrete aggregate mix is required to contain at
least 31% sand by volume for proper batching.
One source of material, which has 25% sand and
75% coarse aggregate, sells for $3 per cubic
meter (m3). Another source, which has 40% sand
and 60% coarse aggregate, sells for $4.40/m3.
• Determine the least cost per cubic meter of
blended aggregates.
29
ENGINEERING DECISION MAKING FOR CURRENT COSTS (2)
30
ENGINEERING DECISION MAKING FOR CURRENT COSTS (3)
Example 2:
• A machine part is manufactured at a unit cost of 40 cent for
material and 15 cent for direct labor. An investment of
$500,000 in tooling is required. The order calls for 3 million
pieces. Half-way through the order, a new method of
manufacture can be put into effect that will reduce the unit
costs to 34 cent for material and 10 cent for direct labor, but it
will require $100,000 for additional tooling. If all tooling costs
are to be amortized during the production of the order, and
other costs are 250% of direct labor cost, would it profitable
to make the change?
31
ENGINEERING DECISION MAKING FOR CURRENT COSTS (4)
32
exercise
33
Computing cash flows
The manager has decided to purchase a new
$30,000 mixing machine. The machine may be
paid for in one of two ways:
1. Pay the full price now minus a 3% discount.
2. Pay $5000 now; at the end of one year, pay
$8000; at the end of each of the next four
years, pay $6000.
• List the alternatives in the form of a table of
cash flows
34
Computing cash flows
35
Computing cash flows
• A man borrowed $1000 from a bank at 8%
interest. He agreed to repay the loan in two
end-of-year payments. At the end of the first
year, he will repay half of the $1000 principal
amount plus the interest that is due. At the
end of the second year, he will repay the
remaining half of the principal amount plus
the interest for the second year.
• Compute the borrower's cash flow !
36
Computing cash flows
37
Time Value of Money
38
TIME VALUE OF MONEY
• Which would you prefer, $100 cash today
or the assurance of receiving $100 a year
from now?
39
TIME VALUE OF MONEY
• You might decide you would prefer the $100
now because that is one way to be certain of
receiving it.
• If the current interest rate is 9% per year, and
you put $100 into the bank for one year, how
much will you receive back at the end of the
year?
40
TIME VALUE OF MONEY
• You will receive your original $100 together with $9
interest, for a total of $109. This example demonstrates
the time preference for money: we would rather have
$100 today than the assured promise of $100 one year
hence;
• but we might well consider leaving the $100 in a bank if
we knew it would be worth $109 one year hence. This is
because there is a time value of money in the form of
the willingness of banks, businesses, and people to pay
interest for the use of various sums.
41
What Do We Need to Know?
• To make such comparisons, 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.
42
Time Value of Money
• Money has a time value
because it can earn more
money over time (earning
power).
• 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
43
EQUIVALENCE
44
Equivalence
• Different sums of money at different times
may be equal in economic value
$106 one
year from now
0
Interest rate = 6% per
year
1
$100 now
$100 now is said to be equivalent to $106 one year from now, if the $100 is
invested at the interest rate of 6% per year.
45
Repayment Plans
End 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 30,772.48
P = $20,000, A = $5,141.85, F = $30,772.48
46
Cash Flow Diagram
47
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.
48
Simple Interest
• P = Principal amount
• i = Interest rate
• N = Number of interest
periods
• Example:
– P = $1,000
– i = 8%
– N = 3 years
End of
Year
Beginning
Balance
Interest
earned
0
Ending
Balance
$1,000
1
$1,000
$80
$1,080
2
$1,080
$80
$1,160
3
$1,160
$80
$1,240
49
Compound Interest
• P = Principal amount
• i = Interest rate
• N = Number of
interest periods
• Example:
– P = $1,000
– i = 8%
– N = 3 years
End of
Year
Beginning
Balance
Interest
earned
0
Ending
Balance
$1,000
1
$1,000
$80
$1,080
2
$1,080
$86.40
$1,166.40
3
$1,166.40
$93.31
$1,259.71
50
Comparing Simple to Compound Interest
51