IE343_HW9_Solutions

# IE343_HW9_Solutions - Homework#9Solutions P U R D U...

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Unformatted text preview: Homework#9Solutions P U R D U E U N I V E RSI T Y I E 343 : E ngi n e e r i ng E conom i cs F a l l 2010 I nst r u c to r : A . C a p pon i Missing name on the top of each page submitted Not stapled together 2 Not completed on the correct template 10 H om e w o r k A ssi gn m e n t #9 A ss i gn e d : 05 N ov 2010 D u e : 12 N ov 2010 2 P r ob l e m 1 (10 poi nts) C onsi d e r a 3-y e a r M A C R S p r op e r t y asse t w i t h a n i nst a l l e d cost b asi s of \$100. D e v e l op t h e M A C R S p e r c e n t a g e r a t e s , r k , f o r t h e a ss e t b a s e d o n t h e u n d e r l y i n g d e p r e c i a t i o n m e t h o d s . ( H i n t : y o u m a y w a n t t o f ol l ow t h e e x a m p l e s how n i n c l ass a n d use T a b l e 7.3 t o con f i r m you r a nsw e r .) Since it is a 3 year property, then we use the 200% DB method with switch to SL at the optimal point. Salvage value is zero, thus the entire cost basis of \$100 is depreciated. R = 2/3 Yr DDB SL M A C RS r t ( % ) C um D e p r e c i a t ion( % ) 1 33.33 ( D D B ) 33.33 2 44.44 ( D D B ) 77.78 3 14.81 (e i t h e r ) 92.59 4 7.4074 (S L ) 100 Homework#9Solutions P r ob l e m 2 (10 poi nts) H op p y H ops, I n c . p u r c h ase d hop h a r v est i ng m a c h i n e r y f o r \$150,000 f ou r y e a r s ago. D u e t o a c h a nge i n t h e m e t hod of h a r v est i ng, t h e m a c h i n e w as r e c e n t l y sol d f o r \$37,500. A ss u m e a 5 y e a r p r o p e r t y c l a ss . ( a ) D e t e r m i n e t h e M A C R S d e p r e c i a t i o n sc h e d u l e f o r t h e m a c h i n e r y f o r t h e f ou r y e a r s of ow n e r s h i p . Basis Cost = \$150,000. Using depreciation rates in Table 7.3, we have Year rt(%) dt=Basis Cost*rt Cumulative dt BVt 0 \$150,000 1 0.2 \$30,000 \$120,000 \$30,000 2 0.32 \$78,000 \$72,000 \$48,000 3 0.192 \$106,800 \$43,200 \$28,800 4 0.1152*0.5 = 0.0576 \$115,440 \$34,560 \$8,640 Since the machinery is sold in year 4 (before the full recovery period), then we apply the half-year convention in year 4. ( b) W h a t t h e ga i n o r l oss on t h e sa l e of t h e m a c h i n e r y? \$37,500 - \$34,560 = \$2,940 You ga i n \$2,940 on this sale. P r ob l e m 3 ( 10 po i n t s ) T e x t boo k p r ob l e m 7-12 : A const r u c t i on com p a n y is consi d e r i ng c h a ngi ng i ts d e p r e c i a t i on f r om t h e M A C R S m e t hod to t h e h ist o r i c a l S L m e t hod f o r a ge n e r a l p u r pose h a u l i ng t r u c k . T h e cost b asis of t h e t r u c k i s \$100,000, a n d t h e e x p e c t e d sa l v age v a l u e f o r d e p r e c i a t ion p u r poses is \$8,000. T h e c o m p a n y w i l l u se t h e t r u c k f o r e i g h t y e a r s a n d w i l l d e p r e c i a t e i t o v e r t h i s p e r i o d o f t i m e w i t h t h e S L m e t ho d . W h a t i s t h e d i f f e r e n c e i n t h e a m o u n t of d e p r e c i a t i o n t h a t w o u l d b e c l a i m e d i n y e a r f i v e ( i . e ., M A C R S v e r sus S L )? Using depreciation rates in Table 7.3, we know So if using MACRS, the depreciation in year 5 is If using SL method, the depreciation in year 5 is The difference in the amount of depreciation in year 5 is when recovery period is 5 year. Homework#9Solutions P r ob l e m 4 (10 poi nts) T e x t boo k p r ob l e m 7-15 : A m a n u f a c t u r e r of a e r os p a c e p r od u c t s p u r c h ase d t h r e e f l e x i b l e asse m b l y c e l l s f o r \$500,000 e a c h . D e l i v e r y a n d i nsu r a n c e c h a r ges w e r e \$35,000, a n d i nst a l l a t i on of t h e c e l l s cos t a not h e r \$50,000. a. D e t e r m i n e t h e cost b asi s of t h e t h r e e c e l l s. Three flexible assembly cells = 3(\$500,000) = \$1,500,000 Delivery and insurance charges = \$35,000 Installation of the cells = \$50,000  b. W h a t i s t h e c l ass l i f e of t h e c e l l s? From table 7-2 on page 310, you can find that the class life for Manufacture of aerospace products (asset class = 37.2) is 10 y e a r s. c. W h a t i s t h e M A C R S d e p r e c i a t i on i n y e a r f i v e? From table 7-2 on page 310, the GDS recovery period is seven years. GDS recovery rates (rk) can be found on page 312. From table 7-3, GDS recover rates, 0.0893, can be obtained.  d. I f t h e c e l ls a r e sol d to a not h e r com p a n y f o r \$120,000 e a c h a t t h e e n d of y e a r si x , how m u c h is t he r ec a p t u r e d d e p r e c i a t ion?  The depreciation recaptured is P r ob l e m 5 (15 poi nts) T e x t boo k p r ob l e m 7-20 : I f t h e i n c r e m e n t a l f e d e r a l i n com e t a x r a t e is 34 % a n d t h e i n c r e m e n t a l s t a t e i n c o m e t a x r a t e i s 6 % , w h a t i s t h e e f f e c t i v e c o m b i n e d i n c o m e t a x r a t e ( t )? I f s t a t e i n c o m e t a x es a r e 12 % of t a x a b l e i n com e , w h a t now i s t h e v a l u e of t? 37.96 % t = 0. 41.92 % Homework#9Solutions P r ob l e m 6 (10 poi nts) T e x t boo k p r ob l e m 7-22 : A \$125,000 t r a c t o r - t r a i l e r i s b e i ng d e p r e c i a t e d b y t h e S L m e t hod ov e r f i v e y e a r s t o a f i n a l B V o f z e r o . H a l f - y e a r c o n v e n t i o n d o e s n o t a p p l y t o t h i s a ss e t . A f t e r t h r e e y e a r s , t h e r i g i s sol d f o r ( a) \$70,000 o r ( b) \$20,000. I f t h e e f f e c t i v e i n com e t a x r a t e i s 40 % , w h a t i s t h e n e t c ash i n f l o w f r o m t h e s a l e f o r s i t u a t i o n ( a ) a n d s i t u a t i o n ( b )? Year 0 1 2 3 4 5 (a) (b) SL Depreciation Book Value \$125,000 \$25,000 \$100,000 \$25,000 \$75,000 \$25,000 \$50,000 \$25,000 \$25,000 \$25,000 \$0 (tax of \$8,000 owed on this gain) (tax credit of \$12,000 on this loss) ...
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## This note was uploaded on 01/20/2012 for the course INDUSTRIAL 343 taught by Professor Capponi during the Spring '11 term at Purdue.

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