Quiz 3 with Solution Key - Engineering Economy | ISYE 3025, Quizzes of Systems Engineering

Download Quiz 3 with Solution Key - Engineering Economy | ISYE 3025 and more Quizzes Systems Engineering in PDF only on Docsity! tSyE 3025 A (Sharp) Quiz 3, Nov 1, 2002 Page 1 of 4, Name or intiats Se i Don't use. For use of graders only. : Roll sequence ; i ! quence 26 Student name (print) : Original! C 2 P.O. Box number or ! grade ; Gt number : DO NOT WRITE YOUR STUDENT ID (or SOC. SECURITY) NUMBER ! i De nol use,” For use of graders ony.” ‘ Problem #, points #t: 20 #2: 20 #3:_20 #4: 20 #5: 25 Total: 105 Points deducted -? 2 | -¢ -—B | -20 | -us Points awarded [ (3 1% 4 iz s 62 Certification: | certify that | have abided by the Georgia Tech rutes governing student conduct on exams and by the specific rules on reference materials for this exam, and that | have neither given nor recsived assistance during this examination. | certify that | have read this statement and understand it, Signature: iis... General instructions: 1. One 8.5x11 inch sheet of notes allowed with written or typed information on both sides, but no photo reductions, no books. Financial or scientific calculators allowed, 2. Write your name on first page, your name or initials on each subsequent page DO NOT write your Student ID number or Social Security number on any page. 3. Generally, no questions will be answered except for illegible or incomplete documents. If you think a problem is incompletely stated, make a typical assumption and continue working. 4. Show intermediate work and answers Gf applicable) so that partial credit may be given. Set up each problem. Put answers in places shown, If there is no place shown, circle or underline final answers. 5. You may round to the nearest dollar for intermediate and final answers. For percentages, rates, and factors use an accuracy of four significant digits or more, unless stated otherwise. 6. Attached are some formulas, interest tables, a federal corporate income tax table, and the MACRS specified percentages table, You may detach them and throw them away when finished, 7. There are 105 points, of which 5 points are for extra credit. 8. Make sure your paper is placed in the large box. 9. Grades will be posted on the WebCT page. Papers will be brought to the next Friday help session, Any papers not picked up then will be placed in a box in the corridor to ISyE Room 118 and left there until the next quiz. If not picked up by the next quiz, they will be discarded. 10. If your calculator malfunctions, dont despair. Most of the problem value is for setting up the solution method, You should be able to perform most of the calculations by hand if your calculator fails: ae CAL SL‘ £22 ($1000 - Noe) "6,666.07 ! -\ sSyE 3025 A (Sharp) Quiz3,Nov1,2002 Page 2 of 4, Name or initials aii, Ja. [20] Compute the depreciation expense that would be taken using classical or pre-1981 methods, without the half- year convention for each of the years shown, and enter the values in the table. The asset is kept for the number of years shown. You may round to the nearest dollar. [Do not enter book values in the table.] Purchase cost: 31,000 Estimated salvage value: 11,000__ Depreciation life: 4 Number of years kept: 6 SL, year 1 SL. year 2 SL, year 5 SYD, year 5 (6,666.67}/ 6606 .7)/ 6 o66."7| 2,428 .™\ Pe a - — ee Jorn fn reece 4H4Z33.3 37,666. fF ain, @§ 5 2 = 20,979.99 54-200 Ys 2433332 Saget 52 17666,65 "Ghee z = 37bbe BYP» GtT+ 4. S41 241 = 21 6 fein» |= B (10002 14,534.4% $ 4, (51000) = 12,2 66 Hai) THAT , tor = 485914 Fhe = F295, Ars 2428.59 lb. [5] Why is it that with the classical or pre-1981 declining balance method it sometimes happens that the depreciation expense in the last year of the depreciation lifetime is zero? Explain in box. DOB = MO°RMS/N ... Because He purchase fs 40 Sagar worth enytimns , $0 i+ can't cepreciet... TH has abeeky depreciated ty, 2. [20] A small company, operating in Country X, has the revenues and expenses shown in the table (there are no other Tevenues or expenses). The tax structure in Country X is shown in the right part of the table. Compute taxable income, income tax, profit after tax, marginal tax rate, effective tax tate, and cash flow after tax. Express tax rates to 3 significant digits. Revenues 60,000 tange |range lower|range upper{marginal tax limit limit rate, % Expenses 1 (J 10,000 0% 1 Labor 5,000 2 10,001 20,000 15% Materials 9,000 3 20,001 30,000 20% Depreciation expense 7,000 4 30,001 40,000 25% Rent 1,000 5 40,001 no limit 30% insurance 1,000 Exp * [6,009 3 23,000 Answers: dep + F002 taxable income $ 57000 60000 income tax 56,950 xX 7 2Beee profit after tax $ 37.00% “ z) 20,280 ik Tax! 10,000(0S + €0,008./5) * B,c0ot: > Ax: ’ marginal tax rate 25° ty ; z000(. 25) 2 6,750 effective tax rate 18,24 3 ty cash flow after tax $23,250 uw tn 5 15%, ~*~ Senge 7 ~ 6,758 | efbectue ™™* > Zaae0 - 1827 CPtatr) + (bee) Cash iow A/T> 30,2807 7000 = 37,280  ISye ou25 Fall 2002 ISyE3025Fall2002uuiz3Tables.xls.Sheet1 10/31/62 09:4 Corporate income taxes for U.S. corporations Interest factors for T= 11% Range lawer | Range upper | Tax Single cash flows Unifarm series cash flows ‘Arithmetic gradient Range Fimit, 3 diesit, 5 rate, % Tax comput. X is taxable income N FIP PIF FA AIF PIA AIP AIG PIG 1 q 50000) 15 0+ 0.15(X) 1 | 1.1100 90090 ; 1.0000 1.0000 | as009 = 1.1400 oO Q 2 50 004 75000] 25 7500+ _ 0.25(x - 50 000) 2 | 1.2321 81162} 2.1100 47393 | 1.7126 58393 | .47393 at tez 3 75 001 100 000} 34 13750+ 0.34 - 75 000) 3 | 1.3676 73119 | 3.3421 29921 | 2.4437 409211 93055 2.2740 4 100 004 335.000] 34+5 22 250+ —0.39(X - 100 000) 4 | 1.518 65873 | 4.7097 21233 | 3.1024 32293] 1.3700 4.2502 5 335 0017 10 000 Of “34 113 800+ 0.34(K - 335 000) 5_| 1.6851__59345 | 6.2278 16057 | 3.6959 27057 | 1.7923 6.6240 OR 0.34%) 6 | 1.8704 53464 | 7.9129 12638 | 4.2305 23638 | 2.1976 9.2072 6 10.000 001} 15000000] 35 3400 000+ —0.35(X - 10 000 000) 7 | 20762 48166 | 9.7833 10222 | 4.7122 .21222| 25863 12.187 7 15.000 G01] 18333 333] 35 +3 5150000 + —0,38(X - 15 600 000) 8 | 23045 43393 | 11.859 .084321| 5.1461 19432] 29585 15,225 8 18 333 334) no limi 35 6416 666+ 0.35(X -18 333 333) 9 | 2.5580 39092 | 14.164 070602] 5.5370 .18060| 3.3144 18,352 OR 0.35(X) 10 _|_2.8394 25218 | 16.722 .osaeot| 5.8802 16980 | 3.6544 21.522 Annual depreciation percentages under MACRS specified Percentages method interest factors for $= 12% (with half-year convention) Single cash flows Uniform series cash flows Arithmetic gradient Recov. [Recovery period or property class N | FP Pr FIA A PIA AP AG PIG. Year | 3-year Syear 7-year 10-year 18-year 1 | 1.1200 89286 | 1.0000 4.0000 | 0.8928 1.1200 0 0 1 33.33 20.00 14.29 10.00 5.00 2 | 1.2544 79719 | 2.1200 47170 | 16901 59170] .47170 79718 2 44.45 32,00 24.49 18.00 9.50 3 | 1.4049 71178 | 3.3744 29635 | 2.4018 41635 | 92461 2.2208 3 14.81 19.20 17.49 14.40 8.55 4 | 1.8735 63552 | 4.7793 20923 | 3.0373 .32923 | 1.3589 4.1273 4 74t 11.52 12.49 11.52 7.70 5_| 1.7623 .56743 | 6.3628 .15741 | 3.6048 27741} 1.7746 6.3970 5 11.52 8.93 9.22 6.93 6 | 1.9738 50663 | 8.1152 42323 | 4.1114 24323 | 2.1720 8.9302 6 5.76 8.92 7.37 6.23 7 | 22107 45235 | 10.0890 .o9912 | 4.5638 21912 | 25515 11.644 7 8.93 6.55 5.90 8 | 24760 40388 | 12.300 081303] 4.9676 20130] 29131 14.471 8 4.46 6.55 5.90 9 | 2.7731 .38061 | 14.776 .067679| 5.3282 18768 | 3.2574 17.356 9 6.56 5.91 io | 3.1956 32197 | 17.549 _.oseee4| 5.6502 17698 | 3.5847 20,254 10 6.55 5.90 "1 3.28 5.91 Interest factors for = 13% 12 5.90 Single cash flows Uniform series cash flows Arithmetic gradient 13 5.91 N | FP PE FIA AE PIA AP AG PIG 14 5.90 1 | 1.1300 88496 | 1.0000 1.0000 | 0.8850 1.1300 0 a 15 5.91 2 | 1.2769 78315 | 2.1300 .4eo4a | 1.668t 59948] 4eoas 78315, 16 2.95 3 | 1.4428 69305 | 3.4069 29352 | 23612 42352 | 91872 2.1602 4 | 1.6305 61332 | 4.8498 20619 | 2.9745 33619} 1.3479 4.0092 S_| 4.6424 54276 | 6.4803 15431 | 3.5172 28431 | 1.7571 6.1802 6 | eoe2g- ssos2 | 8.3227 12015 | 3.9975 C5015} 2.1468 © 8.5818 7 | 2.3526 42508 | 10.4047 09611 | 4.4226 22611] 25171 14.132 8 | 2.6584 37616 | 12.757 078387] 4.7988 20839 | 28685 13.765 9 | 3.0040 33288 | 15.416 osasag | 5.1317 19487] 3.2014 46.428 to | 3.3946 29459 | 18.420 054290} 5.4262 18429 | 3.5162 19.080  ISyE 3025 Engineering Economy Interest Factor Formulas Form, Short Numerical Cash flow diagram, equivalent nr. _name Paramaters, example results Interest factor formula flow shown in dashed line fa] FP] FiPiain 1.403 Fa 100 | 0.07] 5 140 F=P(i+i)% ' Convert a single amount at time 0 to a single amount at time N. oO 84 2 NI fb | PIF F P i N 0.7130 140 10.071 5 100 P=F(1+iy% P i. Convert.a single amount at time Nip a single amount at time Od ccoceeeceeceeeeeercen ee 2a| FA; FLiaAii dn 12.58 Fe Al +i%~ 4) FA 100 } 0.051 10 1258 i t Convert a uniform series from time 1 fo time Nto a single amount at time N- ' __The last A element coincides with F. —_ _ ve gO 4 203... Nt aflar]leialaéin 0.07950 | | as Fi { | | | } 1258 [5 4 0.05| 10 100 (+ih-4 A Convert a single Value at time N to a uniform series from time 1 to time N. — ~ ~ ” The last_A element coincides with F, 3a | PA | P N 9.7791 | |p= Al(1+i)X- 1] ee 1} 40 244 ie ays Ap Convert a uniform series from time 1 t6 time N fo a singlé amount at time 0. P ! i occurs 1 period before the first A element Oo, 1 °2~°3~ N a i[N 0.18744 | | a= Pra+i t on 4 a f Lt i 4000 011 8 187 a+iX-4 A | Convert a single amount at time 0 to 4 uniform series from time J to time N. Py . - F + occurs 1 period before the first A element. ad 1 1 4a; FIG] FiGciiain 159.6 Fe Gli +i)%-an- 4] 012 3. 4 50 | 0.09} 15 7978 2? el | Convert a linear gradient from ti The first G element occurs at time 2, ime 2 to time N to a single value at time N. N 4381 | |pe Gli +i NaNn-4 ajpcl pie]: [Ey 50 [008 15 2190 Pa sie Convert a linear gradient from ti The first G element occurs at time 2. ime 2 to time N to a single value at time 0. 4c | AG A!lGii N ere Git +i)N-iN- 1] 50 10.09 6 272 i(1 +i); convert linear gradient to: 0 N Convert a linear gradient from ti The first G element occurs at time 2. ime 2 to time N to a uniform serles from time 1 to time N. The (N-1) elements convert to N elements. er cal Fit] FOP AT i Nig 146,08 Fe Ald +n"- (1+ 9) 1 $0.09 26 | 0.05 146 i-g Use (NA1)(142)*"! whens = g. Convert a geometric gradient from time 1 to time N to a single value at time N. P Fa i 4 convert to P or F ' v ' L N oO Sb} Pigi | Pot Ay tint g 9.031 Ps Aft-(+oMt+i) 100 } 0.12 20] 0.03 903 feg Use (NA,)/(14/) when? = g. Convert a geometric gradient from time 1 to time N to a single value at time 0. 1 Copyright 2001, G. Sharp, All rights reserved  a fling ae ite any oron) .05 } en 7 Tax. nd 8 = Ea ee Geers * Otewisy a 9G; 600 jae qricaiane + os waht sv Break eueng 986 (454 285, 4g B,o00 ,'y , see b) t bit is “Bred © 3300, echoes Faden ed th ity aes ahs fe npuel Morty Sy 28) Feat BA, a! S96 (25-976 +61 a 2 ahad is, riecnade®, Gro reek c, y 16) mee Meledgad ! "nga MR. 184 ea "Sha +GL96 K 7 ci Abul LF ty by nf Shen © mi} a Si te ze Gs Cy i: cs ia eps ‘f 1200 1400 "Bioon + cas. 354 $3 deewat beg Alpy Crack So xaos cont Se vcan? eat te! Bregected 4080 fey: 320L 2/26ve% Ss tae : fee's Sen “pate vi, 06, £83 ood Yin ‘ is & we? One Riod What NPY costs neg i an fo would Yow. deed 4, h 0%, Noy: Psoneleta forts ena iat ee \* 4) Ass, é mnt ‘of % get a ? ne Ae fon fate fo breakeven e 3 . “Wey » but Gnora resale ip Mhgeddere | app, (4/ iia) 508 Ca fn 3,2) a FR tele ante ogee bese this amy neal oggt 000 Ap pg 2) henge