Transcript Using Engineering Economics to Pass the FE Exam

Using Engineering Economics
to Pass the FE Exam
©2010 Dr. B. C. Paul
Problems discussed come from Practice FE exams from various sources.
Information on the FE Exam comes from information by reviewers and exam
preparers from various web sources.
Have 4 Simulated Engineering
Econ FE Exams
• The Exams can replace a homework
– If you pass an exam I’ll look for your lowest
homework and replace it with 100
– You can do up to 4 homework replacements this way
• Obviously 0’s for missed homework will be a primary target.
• An Exam can be traded for a 10% boost on the
Midterm score.
– If you got 78% then an FE will get you to 88%
– If you got 106% then an FE will get you to 116%
– You can trade in up to 2 FE exams for Midterm boosts
Things to Expect
• Multiple Choice Test
• Its Long and Brutal a morning and afternoon session of
4 hours each.
– Don’t rely on cramming the night before
– Eat well the day before – get a good nights sleep
– Decompress
• Your preping for a marathon as much as a test
– Bring a lunch because time between sessions is short
• Its Fast Paced
– Morning session is 120 questions in 4 hours
– Ie – 2 minutes each including time to look things up and
transcribe answers
– Afternoon session is 60 questions in 4 hours
• 4 minutes each
Things to Expect Continued
• Materials
– You will have a Formula book
– A test booklet for writing in
• May have an index telling you where the Engineering Econ
Question are but there is no indicating in the question set where one
subject begins or ends
– An answer sheet.
– Efficient movement is important
• Know the formula book but probably should not be your study guide
• May write answers down and then at end of the page (not the test)
transcribe them to the answer sheet
• Restricted to only certain calculators
– Is a financial or two available
– Can’t have preprogrammed formulas
– No “Class Assistant” on your laptop
Strategies
• Name of the Game is to get 70% equivilent of
the Questions right
– There is no penalty for wrong answers
– Thus – leave no question unanswered even if you fill
in blank circles by “the force” at the end
• Nail “low hanging fruit” quick
• Eliminate answers that are clearly wrong
– You can get a lot of questions down to 2 or 3 answers
by inspection
• Mark what you could come back to and improve
• Have another mark for things were you don’t
have anything but “the force” to guide you
Engineering Economics
• About 8% of the Morning Session
• Afternoon you can take specialized FE’s for
engineering disciplines, but if you go the general
route about 10% of afternoon
• Formula Section of Your Reference book is
mostly interest tables
– Most formulas are for the common “magic numbers” –
P/F, F/P, P/A, A/P, A/F, F/A
– Are a few tables or formulas for depreciation
– There are a few definitions which you should probably
know already and never have to look up
FE Exams for This Class
• Make-Up Work will be based on the FE Exam
(Engineering Economics Section Only)
• We will have a total of 4 mini FE exams
– You can take all or any lesser number
– They are optional
• FE exams will be timed sessions just like the
real FE
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You will be able to use a simple calculator
A Book of Formulas and Tables
No notes
No computer
No Class Assistant
Class FE Exams
• Exams will be timed
– Part 1 you get 18 minutes (2
minutes/question)
– Part 2 you get 24 minutes (4
minutes/question)
• Exams will be Pass/Fail
– If you get 75% or better you get your points
– If you get under 75% you get nothing
A Bonus Situation
• The real FE exam does not separate
scores by subject. Strength in one area
can cover weakness in another
– If you pass your FE morning in 13.5 minutes
instead of 18, and/or your FE afternoon in 18
minutes instead of 24 you get an extra 10
each (ie 110 or 120 instead of 100)
– If you break 90% on your FE instead of 75%
you get 110 instead of 100.
– If you do all of both you get 130.
Types of FE Exams Questions
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The Adjusting interest Questions
Simple Interest Problems
Simple Algebra
Magic Number One Shots
Simple Two Steps
Calculating Total Life Cycle Costs
Calculating NPVs
Depreciation
The Adjusting Interest Questions
• A credit card offers 1.4% effective monthly
interest. What is the effective yearly rate
with monthly compounding?
Principle to Solve
• This is almost identical to your homework #2
• Formula
(1 i )
n
(1+0.014)^12 = 1.18
subtract the 1
convert back to percent 18%
The Continuous Compounders
• Take the same formula and make n
something big.
Principle and Interest
• A bank loan of $25,000 bank loan is repaid
in yearly payments of $2745 over 15 years
at an effective annual interest rate of 7%.
What percentage of the first payment is
applied to principle?
Solution Principle
• $25,000 * 0.07 = Year 1 interest
– $1750 in interest
• Payment is $2745
• Subtract
– $995 goes to principle
• $995/ $2745 = 0.3625
– Or 36.25%
The Simple Interest Class
• A Bank Charges 10% simple interest on a
$300 loan. How much will have to be
repaid if the loan is paid back in a lump
sum at the end of 3 years?
The Solution Principle
• Interest Added to the Loan
– Principle*i*n = Interest
– 300*0.1*3 = $90
• Lump sum payback is Principle+Interest
– $390
Basic Algebra Class
• The cost of producing products from a
factory is
• C1= $25,000 + 0.03*P
– Where P is the number of items produced
• The items sell for $1.50 each
• How many items must be produced to
break-even
Solution
• Earnings = Cost (means break even)
• 1.5*P = 25000+0.03*P
• Solve for P
Magic Numbers One Shots
• A*B = C
– Where A and C are numbers from the problem
– B is one of the 6 magic numbers
Try One
• How long will it take a sum of money to
double at a 7% annual percentage rate?
What magic number is this playing
for?
• F/P
• Because P*F/P =2P
• Obviously we will be scanning the interest
table for an F/P equal to about 2
• Then we will find what value of n
corresponded to that F/P of 2
Another Try
• An investment of $40,000 results in a
savings of $8,000 per year. With 10%
interest how long will it take for the savings
to payback the initial investment?
What Magic Number is This Playing
For?
• P/A
• Because P/A*$8000 = $40,000
• Find out what value of n gives a P/A of 5
A limiting Case for an annuity
• If an annuity goes on forever its present value
is
P 
A
i
Simple Two Steps
• May have to use more than one principle
to get the answer
– As an example you may have both interest
compounding periods and a magic number
one shot
Try One
• Funds are deposited in an account at 12%
annual interest rate compounded
semiannually. How much must be
deposited in the account to have $10,000
in 10 years?
The Two Step
• Get a period interest rate
– 12%/2 = 6% per semi-annual compounding
period
• The magic number they need is P/F
– $10,000*P/F = Deposit Amount
• P/F is for 6% interest and 20 compounding periods
Try Two
•
A sum of $60,000 is needed for building
improvements in 5 years. To generate
this amount a sinking fund is established
into which 3 equal payments will be
made, one at the end of each of the first
3 years. After the 3rd year no further
payments will be made. If an effective
annual interest rate of 8% can be
expected, the amount that must be paid
into the fund each year is most nearly
The Cash Flow
$60,000
0
1
2
3
4
5
They are being tricky here because they want you to try to use F/A, but F/A
Requires payments up to and including the 5th year.
Instead
Use P/F to get the present value of the
$60,000
0
1
2
3
4
$60,000
5
Now you can use A/P to get the annual payments
It’s a two step (it also is effectively a total life cycle cost problem)
An Alternate Solution
Use P/F to discount back
To year 3
$60,000
0
1
2
3
4
5
Now I can use A/F to get the amount of the savings.
Calculating Total Life Cycle
Costs
• A company must purchase a machine that will
be used over the next 8 years. The purchase
price is $10,000 and the salvage value after 8
years is $1,000. The annual insurance is 2% of
the purchase price, the electricity cost is $300
per year, and maintenance and replacement
parts cost $100 each year. The effective annual
interest rate for economic analysis is 6%.
Neglect taxes. What is the effective annual
uniform cost of ownership?
The Cash Flow
$1,000
0
1
2
3
4
5
6
7
8
$600/year
Get the NPV
$10,000
-$10,000 - $600*P/A(8,6%) + $1000*P/F(8,6%)
Now convert that NPV to an annual cost
NPV*A/P(8,6%)
The NPV Class Problems
• A bank loan of $25,000 bank loan is repaid
in equal yearly payments of $2745 over 15
years. The $25,000 is invested in the
stock market at an effective annual return
of 10%. All money earned is kept in the
stock fund. What is the present day value
of the earnings over 15 years as a result of
the loan and investment venture?
The Cash Flow
Grows to ? In 15 years
At 10%.
$25,000
$2745/year for 15 years
Note that if I grow $25,000 to some future amount after 15 years and then
Discount it back to time 0 at 10% I’ll be right back to $25,0000
All I have to do is P/A(15, 10%)*2745 and subtract that from $25,000
Depreciation
•
A machine has an initial cost of $14,000,
a life of 15 years, and a straight line
depreciation value of $850. The salvage
value is most nearly
Solution Principle
• The initial purchase price is the cost basis
• The initial purchase price minus total
depreciation to date is the book value
• $14,000 – 15*850 = $1250
A Depreciation with a Discounted
Cash Flow Twist
•
A machine has an initial cost of $40,000
and operating costs of $3500 each year.
Its salvage value decreases by $4,000
each year. The machine is now 4 years
old. Assuming an effective annual
interest rate of 12%, the cost of owning
and operating the machine for one more
year is most nearly.
Solution Principle
• Need to recognize the book value of the
machine is $24,000
– $40,000 – 4*$4,000
• At 12% interest the value at one year is
– $24,000*1.12 = $26,880
•
•
•
•
At the end of the year book value is $20,000
By subtraction owning cost is $6,880
Given operation cost is $3,500
Total owning and operating is $10,380