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IENG 301 – FALL 2011IENG 301 – FALL 2011
• Meetings:•Tu, Th: 8:00 – 8:50 AM
• Instructor: Dr. Dean Jensen
•Phone: 394-1278•E-mail: [email protected]•Office Hrs:
•M*, Tu, W*, F*: 11:00 – 11:50AM* These offices hours are held in CM 320
• Class website:http://webpages.sdsmt.edu/~djensen/IENG301
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Other Course Other Course ObjectivesObjectives
Other Course Other Course ObjectivesObjectives
1. Solve problems in a manner expected on the Fundamentals of Engineering exam.
2. Evaluate personal finance choices.
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Materials
Engineering Notebook – 9-3/4" x 7-1/2", 5x5 quad-ruled, 80-100 pp. (approx.).
FE Supplied-Reference Tables for Eng. Econ.
Engineering Problems Paper – 8-1/2" x 11", three hole drilled, ruled five squares/division, 50 pp.
OPTIONAL: any engineering econ text such as Blank, L. & Tarquin, A. (2005). Engineering Economy (6th ed.). New York NY: McGraw – Hill. 759pp. ISBN 0-07-320382-3.
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FE Supplied-Reference Tables
Go to www.ncees.org Study Materials Fundamentals of Engineering FE Supplied-Reference
Free Preview Read & Accept Terms FE Supplied-Reference Handbook as multiple PDF
files Engineering Economics
Save the file to your computer Print these out, cut & paste into your Eng. Notebook
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Course StructureCourse Structure
• Grading: Percentage
• Weighting:•Assignments 25%
•Exam I 25%•Exam II 25%
•Exam III 25%
•Bonus Points 5%
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PoliciesPolicies• Bonuses: – 105%, but no make-ups
• Assignments:•Due at class (or earlier), all equal wt. (%)•No late work – drop lowest scoring HW
• Exams:•Open engineering notebook•Closed text, etc.
•Put FE reference tables in notebook•Make-up Exams
•Sponsored activities schedule ahead of time•Otherwise, add extra weight to next midterm
•No make-up Final
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Assignment StructureAssignment Structure
• Format for most problems:•Find (objective)•Given (organize relevant data, only)•Cash Flow Diagram (rarely dropped)•Soln. (steps to solve):
•Write equation in Table Factor Form•Convert to values (or equation forms)•Double underline answer to question
• Turn in on EP Paper•Not graded if illegible!
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Data CollectionData Collection
Name Course IDPreferred name Term / YearYour SDSM&T E-mail address
Your major and anticipated graduation dateYour hometown
Anything else the instructor should know about you
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Transport Data Collection
Transport Data Collection
Course: IENG 301
This course meets at Kansas City St. on: Tuesdays 8:00 to 8:50 Thursdays 8:00 to 8:50
When and where is your next class held at?
Would you likely use a shuttle for $20/mo.?
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Engineering Econ Engineering Econ ProcessProcess
Engineering Econ Engineering Econ ProcessProcess
• Identify alternative uses for limited resources
• Obtain needed data
• Analyze data to determine preferred alternative:
•Screening decisions(meets minimum acceptable?)
•Preference decisions (Select from competing
alternatives)
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Typical DecisionsTypical DecisionsTypical DecisionsTypical Decisions
• Cost reduction (e.g., equipment, tooling, facility layout)
• Capacity expansion (e.g., to increase production, sales)
• Equipment / Project selection
• Lease or buy decisions
• Make or buy decisions
• Equipment replacement
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Fundamental Concept:
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Lets Get Started…Lets Get Started…Lets Get Started…Lets Get Started…
• Would you rather have $10 000 today or $10 000 five years from now?
• If you don’t need it right now, what could you do with it?
• Would it be worth the same in five years?
• Money changes value with time!
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Rate of ReturnRate of ReturnRate of ReturnRate of Return
• (ROR) is the rate of change in value earned over a specific period of time – expressed as a percentage of the original amount
Period Ending Amount – Period Starting AmountPeriod Starting Amount
• The Rate of Return is a measure of how much risk there is in an investment
Higher Risk Higher ROR
x 100%ROR =
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Rate of Return and Rate of Return and InterestInterest
Rate of Return and Rate of Return and InterestInterest
• The Interest Rate (i) is the percentage change in value earned over a specific period of time.
• For simple interest, a return is earned only on the original amount (principal, p) each period.
• If the principal is invested for n periods:Total Money Returned = p + (p)(n)(i)
Total Interest Earned = (p)(n)(i)
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Compound Compound vsvs Simple Simple InterestInterest
Compound Compound vsvs Simple Simple InterestInterest
• For simple interest, a return is earned only on the original principal each period.
• For compound interest, a return is earned on the entire amount (principal + total interest already earned) invested at the beginning of the current period.
• Effectively, you are also earning interest on your interest (and on your investment principal)!
• Unless explicitly stated otherwise, this course uses compound interest.
(And so does the rest of the world!)
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Using Compound Interest to Using Compound Interest to Make Economic Decisions …Make Economic Decisions …Using Compound Interest to Using Compound Interest to Make Economic Decisions …Make Economic Decisions …
• Paid $100,000 for it - 3 years ago
• Don’t need it now• Option 1 – Sell it for $50,000• Option 2 – Lease it for
$15,000 for 3 years. Sell
it for $10,000 at the
end of the lease.
Note:Leases typically pay at the beginning of a time period.Loans typically pay at the end of a time period.
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Questions?Questions?Questions?Questions?
• What about the $100,000?•The $100 K is irrelevant - it is a sunk cost, and makes no difference in the decision at this point in time.
• How do we select between the options?
•We need to know under which conditions we would be economically indifferent (equivalent) - have the same amount of money at the same time - and then if the conditions are better for one option, we will select that option.
• Any other factors?•Since we need to account for the time value of money - we need to know the interest rate and the compounding period.
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Cash Flow DiagramsCash Flow DiagramsCash Flow DiagramsCash Flow Diagrams
$15 k
0n =
$15 k
1
$15 k
2
$10 k
3YRS
OPTION 2:
$50 k
0n = YRS
OPTION 1:
31 2
F3?
F3?
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QuestionQuestionQuestionQuestion
• Under what conditions would I be indifferent between Options 1 & 2?• Indifferent means Economically
Equivalent:– Have the same amount of money
at same point in time, after accounting for all of the cash flows.
– In this case, 3 years from now.
• Interest Rates…– Percentage– Compounding annually
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Future Value in 3 Future Value in 3 years…years…
Future Value in 3 Future Value in 3 years…years…
I% Option 1 Option 22.5% $53,844 $57,2885.0% $57,881 $59,6527.5% $62,115 $62,09410% $66,550 $64,615
At what interest rate, am I indifferent between the two options?• They are economically equivalent at
an interest rate just a little less than 7.5%
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Option 1Option 1Option 1Option 1
50,000 nowi = 10% compounded annually
F1 = 50,000 + 50,000 (.10) = 55,000
F2 = 55,000 + 55,000 (.10) = 50,000 (1 + .10)2 = 60,500
F3 = 60,500 + 60,500 (.10) = 50,000 (1 + .10)3 = 66,550
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Generalizing …Generalizing …Generalizing …Generalizing …
P = Present value at the beginning of first
period.Fn = Future value at end
of n periods in the future.
Fn = P (1 + i)n = P (F/P,i,n)
(F/P,i,n) = (1+i)n
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Standard Factors Used Standard Factors Used to Solve ECON to Solve ECON
ProblemsProblems
Standard Factors Used Standard Factors Used to Solve ECON to Solve ECON
ProblemsProblems( F / P, i, n) Find F Given P( P / F, i, n) Find P Given F( F / A, i, n) Find F Given A( A / F, i, n) Find A Given F( P / A, i, n) Find P Given A( A / P, i, n) Find A Given P ( P / G, i, n) Find P Given G( A / G, i, n) Find A Given G( F / G, i, n) Find F Given G
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Tables…Tables…Tables…Tables…
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Tables…Tables…Tables…Tables…
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Tables…Tables…Tables…Tables…
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Tables…Tables…Tables…Tables…
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Future Given Present
P is the present value at Time 0 F is the future value at Time n
(n compounding periods in the future)
i is the effective interest rate
0 n
P
F ?
1 2 3
F = P(F/P,i,n)
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Tables…Tables…Tables…Tables…
= i
F3 = 50 000(F/P,10%,3)F3 = 50 000(F/P,10%,3) = 50 000(1.3310)F3 = 50 000(F/P,10%,3) = 50 000(1.3310) = $66 550
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Formulas…Formulas…Formulas…Formulas…
F3 = 50 000(F/P, 10%,3)F3 = 50 000(F/P, 10%,3) = 50 000(1+.10)3F3 = 50 000(F/P, 10%,3) = 50 000(1+.10)3 = 50 000(1.3310)F3 = 50 000(F/P, 10%,3) = 50 000(1+.10)3 = 50 000(1.3310) = $66 550
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