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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: dean.jensen@sdsmt.edu•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