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FE Review Engineering Economics. Things to Expect Multiple Choice Test Its Long and Brutal a morning and afternoon session of 4 hours each. –Don’t rely.

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Presentation on theme: "FE Review Engineering Economics. Things to Expect Multiple Choice Test Its Long and Brutal a morning and afternoon session of 4 hours each. –Don’t rely."— Presentation transcript:

1 FE Review Engineering Economics

2 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

3 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

4 Strategies Name of the Game is to get 70% 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

5 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

6 Approach I’ll try to review Then I’ll give you a problem that tests that Well time to see who can kill it in 90 seconds We’ll check the answer See if anyone needs to see how we got that.

7 The “Magic Numbers” Two most important questions –How much do I get –When do I get it I can add only money at the same point in time Concept of Equivalence

8 Looking For Easy Problems Lots of problems on first half of FE can have one shot solutions Take a number – multiply by a magic number Get the Answer Easy because have the form AXB =C

9 Basic Magic Numbers F/P – If you know how much you have now, how much will you have in the future P/F – If you know how much you have in the future what is that equal to today $known $to be found 0 n $known $to be found 0 n

10 Unit Cancelation Trick I know Future amount but I don’t know the present amount Similarly – I know what I have now how much will I have in the future

11 Two Ways to Get “Magic Number” Formula’s –These are found in your book

12 Using Tables Tables are designed for only one rate of interest given at the top of table

13 Question $500 is deposited into a bank savings account with 6% interest compounded annually. Most nearly how much will be in the account at the end of 3 years –(A) 550 –(B) $600 –(C) $650 –(D) $700 1.1910 Try this table look-up style.

14 The Answer How Many Picked B- $600

15 How Did I Do That? This is an AXB = C problem I have $500 now - that’s A I want to convert it to equivalent money in 3 years Present * F/P = Future (F/P for 3 years and 6% interest is B

16 Look Up F/P I’ve got my 6% interest Read over from N = 3 to F/P get 1.1910 Now $500 * 1.1910 = $595.50 very close to $600

17 Another Try If you need $800 in savings at the end of 4 years and your savings account yielded 5% interest paid annually, most nearly how much would you need to deposit today? –A- $570 –B- $600 –C- $660 –D- $770 P/F = (1+i)^-n This time we will do the formula method.

18 Checking Answer How Many Got C - $660?

19 How Did I Do That? I know I have $800 dollars in the future – I need the equivalent amount now. Future * P/F = Present P/F for 4 years and 5% interest –Plug in –(1+0.05)^-4 = 0.8227 –$800*0.8227 = $658.16 – close to $660

20 Another Easy One Is the Money Doubling type They ask you how long it takes for money to double at some interest rate Of course to make that happen you need F/P=2 Do a quick look up in the table for the first time F/P is about 2 –Look over to the value of n and report that as the answer

21 Try Another Most nearly, how long will it take a sum of money to double at a 5% annual percentage rate. –(A)- 6 years –(B)- 10 years –(C)- 11 years –(D)- 14 years 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 2 3 4 5 6 7 8 9 10 11 12 13 Oh yes they would Make you interpolate Between two tables.

22 Check Answer How Many Got (D) 14 years

23 How Did I Do That? First 2 at 18 years First 2 at 12 years

24 Picking My Answer Answer is about half way between 12 and 18 Choices are –A- 6 year –B- 10 years –C- 11 years –D- 14 years So which one is between 12 and 18?

25 The Annuity $5,700 per year for 5 years How much is that equal to right now if interest is 6% 0 1 5 I can’t just add up $5,700 * 5 because the money is at different Points in time

26 My Magic Number Friend P/A Present Value = P/A*amount of one annuity payment P/A is a function of interest rate and number of payments (6% interest, 5 payments) 4.212

27 Finish Up $5700 * 4.212 = $24008.4

28 Can Do That in Reverse Have $24,000 right now Want to know an annuity of 5 equal annual payments Present Amount * A/P = Annuity (This one will be your turn)

29 Try It You’ll Like It. Do $24,000 * A/P (for 6% interest and 5 payments)

30 The Answer Did you get A/P = 0.2374 And the Annuity is $5,698 per year

31 The Perpetual Annuity What happens if the annuity goes on forever –Can look at the value of A/P or P/A in the table for the biggest value of n Or can use formula for infinite annuity

32 Try A company will sell $100 worth of merchandise every year in perpetuity. How much are those sales worth today if the interest rate is 10%? –A- $1000 –B- $1100 –C-$1200 –D-$2,700

33 Check the Answer How many of you got A- $1000

34 How Did I Do That? $100/0.1 = $1,000

35 The Annuity Payments are the same repeating amount Payments start 1 compounding period in the future Payments occur at the end of each compounding period

36 Suppose the Annuity Starts at the Wrong Time? Whats it worth now? $1000 0 1 2 3 4,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,13

37 Try This $1000 0 1 2 3 4,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,13 P/A*$1,000 P/F

38 Interest Rates Interest is reported annually but often compounds more often –Result is that money sitting for a year will accumulate more interest than indicated by the annual rate To Find a Yield –Caution – Interest rates are in % orally and decimals in calculations –Step 1 – get a period interest rate Annual Rate # of Compounding Periods in a Year

39 To Get A Yield Use the F/P formula (1+i)^n –Where n is number of compounding periods Can substitute period interest rate into F/P formula Let r be the annual interest Rate m is the number of Compounding periods in a year Note yield is always higher than annual rate but usually not by a large amount Remember the 1 is there to preserve principle so its you’ll get 1+interest in Decimal form

40 Your Turn What is the effective annual rate of interest (yield) for money invested at 5% interest compounded quarterly –A 1.3% –B 5.0% –C 5.1% –D 20%

41 The Answer How Many had C - 5.1%

42 How Did I Do That Ie= ( 1 + (0.05/4))^4 -1 = 0.0509495 The 1 preserves my original investment 0.0509495 Convert back to % 5.09495 – very close to 5.1%

43 Another Interest Application Test writers like continuous compounding –Have all sorts of functions in books that no one uses You can use same old formula and just make m something big A Bank advertises 4.6% Interest with continuous Compounding – What is the Effective rate of interest (A) – 4.62% (B) – 4.71% (C) – 4.89% (D) - 4.94%

44 The Answer How Many Picked (B) – 4.71%

45 How Did I Do That? Make m some big number – say 1000 =(1+ (0.046/1000))^1000 = 1.047073 Get rid of the 1 0.047073 Convert back to % 4.7073% which is about 4.71%

46 The Total Life Cycle Cost Often want to know what is most cost effective –A cheaper short lived good –Or a longer lived but more expensive good

47 To Get a Total Life Cycle Cost Discount all money to the time the unit enters service (almost always 0 on FE exam) Now use A/P to spread cost over life of item

48 Move All the Money to Time Zero- Ie Get an NPV Spend $70,000 Spend $18,000 Recover $5,000 0 3 6 9 12 NPV = $70,000 + P/F(6%,3)*$18,000 +P/F(6%,6)*$18,000 +P/F(6%,9)*$18,000 +P/F(6%,12)*$13,000

49 Get the P/F Values from the Table You need for 3, 6, 9, and 12 years

50 Move All the Money to Time Zero- Ie Get an NPV Spend $70,000 Spend $18,000 Recover $5,000 0 3 6 9 12 NPV = $70,000 + P/F(6%,3)*$18,000 +P/F(6%,6)*$18,000 +P/F(6%,9)*$18,000 +P/F(6%,12)*$13,000 (-$114,917)

51 Now Convert to An Annuity Over the Service Life Get the Cost Per Year for 12 years NPV * A/P(6%,12)

52 Ok You Couldn’t Read it Get A/P(6%,12)

53 Now Get the Total Life Cycle Cost NPV * A/P(6%,12) = (-$13,707)

54 Try this Example Problem 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 $1000. The annual insurance cost is 2% of the purchase price. The electricity cost is $300/year and maintenance and replacement cost is $100/year. The effective annual interest rate is 6%. Neglect taxes

55 Continued What is most nearly the uniform annual cost of ownership –(A)- 1200 –(B)- 2100 –(C)- $2200 –(D)- $2300 Don’t do it just yet!

56 Pump Priming Notice some costs are already given to you as annual costs –$200 for insurance –$300 for electricity –$100 for repair parts –Total $600. –All I need to do is convert capital cost to a yearly cost

57 Get the NPV of Capital Cost -$10,000 0 8 $1000

58 Now Finish Take NPV convert to an Annuity 8 Also add the $600 annual cost

59 Pick Your Answer (A)- 1200 (B)- 2100 (C)- 2200 (D)- 2300

60 Check Your Answer How Many Got (B)- 2100?

61 Now Try This Warehouse A cost $100,000 now and has a salvage value of $10,000 after 10 years. Warehouse B costs $70,000 now, needs $18,000 of service every 3 years and is salvaged after 12 years for $5,000 Warehousing is needed forever and the interest rate is 6%. Which warehouse is the better deal –(A)A by $140 per year –(B) A by $190 per year –(C) B by $190 per year –(D) A by $880 per year

62 The Answer How Many Got (D) A by $880/per year

63 How Did We Do That $100,000 $10,000 0 1 2 ……………………………………………9 10 Move back with P/F $10,000 * 0.5584 = $5,584 NPV= -$100,000+$5,584 = -$94,416

64 Complete The Total Life Cycle Cost -$94,416 0 1 2 ……………………………………………………………..10 Spread to an annual cost with A/P -94,416*0.1359 = -$12,828

65 Finishing Same Method gets B at -$13,707 Difference -$13,707+$12,828 = $879 in favor of A

66 Depreciation In tax calculations where you pay taxes on earnings every year you need a way to spread out cost of something that lasts over a year. More than one depreciation scheme Only seen one on the FE – Straight-Line

67 Doing Straight-Line Take cost of the dilly-womper – say $100,000 Take amount dilly-womper is worth when its worn out – say 0 Take the life of the dilly-womper – say 10 years (Cost – Salvage)/Life = Annual Depreciation (100,000 – 0)/10 = 10,000

68 Book Values with Depreciation Book Value is Original Value – Depreciation taken to date My $700,000, 7 year life truck has a book value of $200,000 after 5 years of depreciation $700,000 – 5*100,000 = $200,000

69 Salvage Value Not Everything depreciates to zero value Suppose my truck cost $710,000. When its all worn out I can still get $10,000 out of scrap metal or salvage My Annual Depreciation is –(Cost – Salvage)/years of life –(710,000 – 10,000)/ 7 = 100,000

70 Book Value with Salvage Original Cost – Depreciation to Date = Book Value My Truck after 5 years $710,000 – 5*100,000 = $210,000

71 Now You Try It A $100,000 Asset It lasts 7 years It has a salvage value of $15,000 What is its book value after 3 years of depreciation –(A) 12,100 –(B) 36,400 –(C) 57,100 –(D) 63,100

72 What Answer Did You Get (D) 63,100

73 How Did I Do That? Get Annual Depreciation –(100,000 – 15,000)/ 7 = $12,173 Get Book Value –$100,000 – 3*$12,173 = $63,571 –Close to $63,600 which is answer D


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