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Matakuliah
Tahun
: D0762 – Ekonomi Teknik
: 2009
Introduction to Engineering Economy & Basic
Concepts
Course Outline 1
References
• Engineering Economy – Leland T. Blank, Anthoy J.
Tarquin, Fourth Edition, McGraw-Hill International
Edition, 1998
• Engineering Economic Analysis, Donald G. Newman,
Third Edition, Engineering Press. Inc, 1988
• Engineering Economy, William G. Sulivan, Elin M.
Wicks, C. Patrick Koelling, 14th edition, Pearson Prentice
Hall, 2009
Bina Nusantara University
3
Outline
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Role of Engineering Economy in decision making
Interest Calculation
next
Equivalence
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Simple and Compound Interest
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Symbols and Their Meaning
next
Cash Flow Diagram
next
References :
- Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p.2-37
- Engineering Economic Analysis, Donald G. Newman, p. 1-51
- Engineering Economy, William G. Sulivan, p.21-35, p. 135-140
next
Role of Engineering Economic
• Engineering Economy is making decision
• Engineering Economy about determining the economic factors and the economic
criteria utilized when one or more alternatives are considered for selection.
• We can say as mathematical techniques which simplify economic comparisons.
• Technique and models of engineering economy assist people in making decision
• Help to decide which one is the best among available alternatives based on economic
factors
• It is based on the systematic evaluation of the costs and benefits of proposed
technical projects
Engineering Economic Analysis
Key Questions:
• Which engineering projects are worthwhile?
• Which projects should have higher priority?
• How the project should be designed?
• How to achieve long-term financial goals?
• How to compare different ways to finance purchases?
• How to make short and long-term financial decisions?
Engineering Economic Analysis
Examples of Non-Monetary Factors:
• Meeting customer expectations consistently
• Maximization of employee satisfaction
• Maintaining flexibility to meet changing demand
• Maintenance of a desired public image
• Leveling cyclic fluctuations in production
• Improvement of safety in operations
• Reduction of pollutants
Decision Making Process
•
•
•
•
Procedure for decision making Process
1. Understand the problem and goal
2. Collect relevant information
3. Define the alternative solution
4. Evaluate each alternative
5. Select the best alternative using some criteria
6. Implement the solution and monitor the result
Engineering economy has a major role in steps 2, 3 and 5. And it is the primary technique in step 4
to perform the economic-based analysis of each alternatives
Steps 2 and 3 set up the alternatives, and engineering economy helps structure the estimates of
each one
Step 4 utilizes one or more engineering economy models this terms to complete the economic
analysis upon which a decision is made
Interest Calculation
• The manifestation of the time value of money is termed interest, which is the
increase between an original sum of money borrowed and the final amount owed
Interest = total amount now- original principal
Interest rate = Interest accrued per time unit
x 100%
Original amount
Interest Calculation
• Example 1.3
The Oracle Investment Group Invested $100.000 on May 1 and withdrew a total
of $106.000 exactly 1 year later. Compute (a) the interest gained and (b) the
interest rate on the investment
Solution
Interest : $106.000 – 100.000 = $6000
Interest rate = $6000 per year
x 100% = 6 % per year
Interest Calculation
•
Example 1.4
Stereophonic, Inc. plans to borrow $20,000 from a bank for 1 year at 9% interest for new recording equipment. Compute (a) the interest and
(b) the total amount due after 1 year. (c) construct a graph which shows the numbers that would be used to compute the loan interest rate of
9% per year
Solution
(a) Total interest = $20,000 (0.09)
= $1800
(b) Total due
= $20,000 + $1800
= $21,800
Note :
Total amount due may also be computed
as:
Total = principal (1+interest rate)
= $20,00(1,09)
= $21,800
Equivalence
• Equivalence is an essential factor in engineering economic analysis.
• Different sum of money at different times are equally in economic value on certain
interest rate.
• Equivalence concept help us to do comparison of alternatives option
• Changing the interest rate destroy the equivalence between two series of
payment/cashflow
Equivalence
• Example
– If the interest rate (i) 6% per year; $100 to day (present time) would be
equivalent to $106 one year from to day
Amount accrued = 100 + 100(0,6) = $106
- If we have $100 now, it is equivalent to $100/1.06 = 94.34 one year ago at i =
6%
Simple and Compound Interest
• Simple interest is calculated using the principal only, ignoring any interest accrued
in preceding interest periods. The total simple interest over several periods is
computed as
Interest = (principal) (number of periods) (interest rate)
• For compound interest the interest accrued for each interest period is calculated on
the principal plus the total amount of interest accumulated in all previous periods.
interest = (principal + all accrued interest) (interest rate)
Simple and Compound Interest
Symbol and Their Meaning
• The relations in engineering economy commonly include the following symbols and their
(sample) units :
– P = value or amount of money at a time denoted as the present,
called the present worth or present value;
– F = value or amount of money at some future time, called future
worth or future value
– A = series of consecutive, equal, end-of-period amount of money,
called the equivalent value per period or annual worth;
– n = number of interest periods ; years, month, days
– i = interest rate per interest period; percent per year, percent
per month
– t = time stated in periods; years, months, days
Example 1.10
• A College student about to graduate plans to borow $200 now and repay the
entire loan principal plus accrued interet at 10% per year in 5 years. List the
engineering economy symbols involved and their values if the student wants to
know the total amount owed after 5 years
Solution
In this case P and F, but not A, are involved, since all transactions are single
payments. Time t is expressed in years
P = $2000; i = 10% per year; n = 5 years; F = ?
The future amount F is the unknown amount
Example 1.11 & 1.12
•
•
Assume you borrow $2000 now at 12% per year for 5 years and must repay the loan in equal yearly
payments. Determine the symbols involved and their values
Solution
Time t in in years
P = $2000 ; A = ? Per year for 5 years
i = 12% per year ; n = 5 years
There is no future value F involved
On May 1, 1988 you deposited$500 into an account which paid interest at 10% per year and withdrew
an equal annual amount for the following 10 years. List the symbols and their values
Solution
Time t is in years; the P(deposit) and A amounts (ten withdrawals) are
P = $500; A= ? Per year
i = 10% per year; n = 10 years.
Cash flow Diagram
Samples of Cash flow inflows








Revenue
Operating cost reduction
Asset salvage value
Receipt of loan principal
Income tax savings
Receipts from stock and bond sales
Construction and facility cost savings
Saving or return of corporate capital funds
Samples of Cash Outflows








First cost of assets
Engineering design costs
Operating costs (annual and incremental)
Periodic Maintenance and rebuild costs
Loan Interest and principal payments
Major expected / unexpected upgrade costs
Income taxes
Expenditure of corporate capital funds.
Cash Flow Diagram
Example of positive and negative cashflow diagram
Cash flow
+
1
2
3
4
5
6
time
Example 1.15
Reconsider Example 1.10, where P = $2000 is borrowed and F is
sought after 5 years. Construct the cash-flow diagram for this case
assuming an interest rate of 10% per year
$2000
Cash flow
+
-
1
2
i= 10%
3
4
5
6
F?
year
Cash flow Diagram
• Example 1.16
Assume that Mr. Ramos stars now and make five equal deposits of A = = $1000 per
year into a 17% per year investment and withdraws the accumulated total
immediately after the last deposit. Construct the cash flow diagram
Cash flow
+
0
F?
i= 17%
1
2
3
4
A = $1000
5
6
year
Cash flow Diagram
• Example 1.16
Assume that you want to deposit an unknown amount
into an investment opportunity 2 years from now that is
large enough to withdraw $400 per year for 5 years
starting 3 years from now. If the rate of return is 15% per
year, construct the cash flow diagram
Cash flow
+
0
F?
i= 17%
1
2
3
4
5
6
year
A = $1000
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