Quiz 3 with Answer key - Engineering Economy | ISYE 3025, Quizzes of Systems Engineering

Download Quiz 3 with Answer key - Engineering Economy | ISYE 3025 and more Quizzes Systems Engineering in PDF only on Docsity! t cert; that | have abide, d by the Teference ch rules Governing Student Conduct on exams ang by the SPecific rules on Materials for this exam, and that) Ve Neither given nor received assistance during this examination, t Certify that | have "ead this Statoment and understand it, teas recetenetng, Signatures long tttteeetentstseettnis. General instructions: St seen, Up the Solution method, ining 10-20%, for calculations, ISYE 3025 A (Sharp) Quiz 3, 29 March 2002 Page 2of4, Name or initiats ari la, [20] Compute the depreciation expense that would be taken using classical or pre-198] 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: 77,000 Estimated salvage value: 18,000 Depreciation life: 6 Number of years kept: 8 SL, year 1 SL, year 2 SL, year 7 SYD, year 1 SYD, year 2 SYD, year 7 12953 | 12633 Y 4600 3200 1000 Assume d= 9 OB Z.(1_.¥) 236 5 1or a“ 277 _ 772 © 1b. [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. LE m=1,6 and nok Z the ekhed ie dopre lation ts greaber a, 2. [25] The following questions relate to the net cash proceeds from the sale of an asset that had been depreciated using one of the MACRS methods (either % or alternative SL method). The sale occurred Vy before the end of the depreciation life. The firm that owned and operated the asset is a large, profitable firm in the top tax bracket (range). The tax rates are shown in the table. Write answers in the boxes. [10] Compute the net cash proceeds from sale of the asset, using values in the table for "part a.” [10] Compute the net cash proceeds from sale of the asset, using values in the table for "part b." c. (5] What type of adjustment in the depreciation expense did the firm have to make in the year of sale? Explain in the box below. of Tax rate for ordinary income 39% Tax rate for capital gains 20% Original cost of asset, $ 70,000 Book vaiue at end of year of sale 28,000 Net cash proceeds Selling price, part a 32,000 ares 20, 470 Selling price, part b 25,000 1g P pi 2 6 , | 70 Ll Answer to part o. Only tral the depreciation expense is computed Sor the year of sale, of & premlure sell. “ SHALL PRILE 2 PRC EyOTAK <—-CRRgcemscier 32,600 — ia 1, 280 Corporate income taxes for U.S. corporations. Range lower | Range upper | Marginal Range limit, $ limit, $ tax rate, % Tax computation, X is taxable income 1 0 50.000 15 O+ 0.15(X%) 2 50.001 75 000 25 7500+ 0.25(X - 50 000) 3 75001 100 600 34 13 750+ 0.34{X - 75 000) 4 100 001 335 000 3445 22 250+ 0.39(X — 100 000) 5 335 001 | 10 000 000 34 113900+ 0.34(X — 335 000) OR_0.34(X) 6 10 000 00% | 15 000.000 35 3400 000+ 0.35(X - 10 000 000) 15 000 001 | 18333 333 35+3 5150 000+ 0.38(X - 15 000 000) 8 18 333 334 no limit 35 6416 666+ 0.35(X -18 333 333) OR_0.35(X) Annual depreciation percentages under MACRS specified percentages method (with half-year convention) Recovery |Recovery period or property class Year 3-year 5-year 7-year 10-year 15-year 20-year 1 33.33 20.00 14.29 10.00 5.00 3.750 2 44.45 32.00 24.49 18.00 9.50 7.219 3 14.81 19.20 17.49 14.40 8.55 6.677 4 741 11.52 12.49 11.52 7.70 6.177 5 11.52 8.93 9.22 6.93 5.713 6 5.76 8.92 7.37 6.23 §.285 7 8.93 6.55 5.90 4.888 8 4.46 *6.55 5.90 4.522 9 6.56 5.91 4.462 10 8.55 5.90 4.461 1 3.28 5.91 4.462 12 5.90 4.461 13 5.91 4.462 14 5.90 4461 15 5.91 4AG2 16 2.95 4.461 17 4462 18 4.461 19 4.462 20 4.461 21 2.231 ISYE 3025 Engineering Economy Interest Factor Formulas Form. Short Numerical Cash flow diagram, equivalent flow nr name Paramaters, example results interest factor formula shown in dashed line taf re] eFiepliin 1.403 Fy Yo 140 F=P(1+i% 1 Convert a single amount at fime Oto a single amount at time N. i O14 2 nN to|eprl Ff ieptiin 0.7130 140 0.07] 5 100 PoP +iyN P Convert a single amount at time Nto a single amount at ime Q. 2a | FIA | FEAL EEN 1258 | FS Afi + iN- 4] F4 100 | 0.05| 10 1,258 i ' Convert a uniform series from time 1 to ime Nio a single amount at time N. The last A element coincides 1 with F oo 1 2 °3.. WN 2] AFT Fi AT iin 0.07950 | {a= Fi { 1258 0.05} 10 100 (+ph-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 | PIA Alitn 9.7791 | |p= Al(1 +8 - 4] —Y 1 104] 40 10 ia +8 AP ‘Convert a uniform series from time 1 fo time N to a single amount af time 0. P occurs 1 period before the : first A element : Or solar} pilaliitn 0.18744 | |, = PIG +9} 1 O14 8 (1+i)N~4 Convert a single amount at time Oto a uniform series from time 1 to time N. P occurs 1 Period before the F 4 first A element. 4a | FIG G i |N 159.6 Fe GI(1+ i) iN - 1) Yj 0 | 0.09] 15 7,978 ? Convert a linear gradient from time 2 to ime Nio a single value at time N- The first G element occurs at time 2. * ape] epie!littn 4381 | [P= Gi +i iN- WY, 7 7 YY | 009) 15 2,190 Fa +r Convert a linear gradient from time 2 to time N to a single value at time 0. The first G element occurs at time 2. 4c) NG] A Gain 5.435 A= Gi(t+i%-in- 1] 50 | 0.091 15 272 (+i -i Convert a tinear gradient from time 2 to time Noa uniform series from time 1 to time (N-1}. ‘The first G element occurs at time 2. The {N-1) elements convert to N elements. safroil F Pat itn] g | ava | [ra AI +9" +g) 100 {0.12} 20 0.03} 3,741 i-g Convert a geometric gradient from time 1 to ima: N to a single value af time N_ Use (NA,)(1+)"" when i = g. t Sb] Pai] P iALd i EN{ g { 9034 P= _Ali-(+g1+i-y 100 | 0.12] 20} 0.03; 903 i-g Convert a geometric gradient from time 7 to lime N fo a single value al time 0. Use (NA V(1+) when? = g. Copyright 2001, G. Sharp, All rights reserved