IE343_HW8_Solutions - Homework#8Solutions PU R D U...

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Unformatted text preview: Homework#8Solutions PU R D U E U N I V E RSI T Y H om e w o r k A ssign m e n t #8 I E 343 : E n gi n e e r i n g E con om 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 A ssign e d : 29 O c t 2010 D u e : 05 N ov 2010 2 P r ob l e m 1 (10 points) A se l e c t ion i s to b e m a d e b e t w e e n t w o st r u c t u r a l d esign s. A ssu m e a st u d y p e r iod of 25 y e a r s. T h e est i m a t e d m a r k e t v a l u e of A l t e r n a t i v e A a t e n d of y e a r 25 is $1,000. A ssu m e M A R R = 15 % p e r y e a r . C a p i t a l I n v es t m e n t A n n u a l E x p e n ses U s e f u l l i f e ( y e a r s) S a l v a ge V a l u e a t e n d o f u se f u l l i f e A lte r native A $12,000 $2,200 10 y e a r s $5,000 A lte r native B $30,000 $1,000 25 y e a r s $10,000 U si n g t h e cot e r m i n a t e d assu m p t ion w i t h s t u d y p e r iod 25 y e a r s, d e t e r m i n e w h i c h st r u c t u r e is b e t t e r b y u si n g e q u i v a l e n t w o r t h m e t hod s. P l e ase , st a t e c l e a r l y you r assu m p t ions. We first realize that the useful life of Alternative A is smaller than the study period of 25 years. We repeat the costs of alternative A from year 10 to year 25, and then use the estimated market value of $1,000 at the end of year 25. We then obtain the following cash flow diagram AlternativeA 5,000 1 10 1,000 1 2,200 2,200 12,000 25 12,000 We can now compare the two alternatives using the FW method. We have Homework#8Solutions Thus, A is more attractive than B. P r ob l e m 2 (10 points) A use f u l p i e c e of d e p r e c i a b l e p r op e r t y w as b ough t f o r $20,000. I f i t h as a u se f u l l i f e of 10 y e a r s a n d a sa l v age v a l u e of $5,000, h ow m u c h w i l l i t b e d e p r e c i a t e d i n t h e 9 t h y e a r , u si n g t h e 150 % d e c l i n i n g b a l a n c e sc h e d u l e? R=1.5/10 = 0.15 (150% DB method) B = $20,000  : Annual depreciation deduction in year k (1 k : Cumulative depreciation through year k : Book value at end of year k   We have As the salvage value is $5,000, the property in year 9 can only be depreciated by an amount . P r ob l e m 3 T e x t b oo k p r ob l e m 7-2 (10 points) For a property to be considered depreciable, it must satisfy the following conditions: a) It must be used in a business as a means of producing/generating income. b) c) It should lose value through wear and tear, and through obsolescence etc. Homework#8Solutions (15 points) P r ob l e m 4 T e x t b oo k p r ob l e m 7-6 The cost basis for the furniture = 100,000 + 20,000 = $120,000 a) Using straight line method (depreciation is same for all years)  b)  c) P r ob l e m 5 Textbook problem 7-7 a) The SL depreciation method is used. b) (15 points) Homework#8Solutions (10 points) P r ob l e m 6 Textbook problem 7-8 (only a and b) a) The SL method b) The 200% DB method with switchover to SL. B = $60,000  : Annual depreciation deduction in year k (1 : Cumulative depreciation through year k : Book value at end of year k  Year,k 0 1 2 3 4 5 dk(200%DB) 0 8,571.42 7,346.93 6,297.37 5,397.75 4,626.64 So no switchover to SL in 5 years. dk(SL) 0 3,428.571 3,032.967 2,673.469 2,344.023 2,038.651 BVk 60000 51,428.57 44,081.63 37,784.26 32,386.51 27,759.86 k ...
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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 University-West Lafayette.

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