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IE343_HW8_Solutions - Homework#8Solutions PU R D U E U N I...

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Homework # 8 Solutions ± Ǥ PURDUE UNIV ERSIT Y Hom e work A ss ignm e nt #8 IE 343: Engin ee ring E c onomi c s A ss ign e d: 29 O c t 2010 Fall 2010 Du e : 05 Nov 2010 In s tru c tor: A . Capponi Missing name on the top of each page submitted ± 2 Not stapled together ± 2 Not completed on the correct template ± 10 Probl e m 1 (10 poin ts ) A s e l ec tion i s to b e mad e b e tw ee n two s tru c tural d e s ign s . A ss um e a s tudy p e riod of 25 y e ar s . Th e e s timat e d mark e t valu e of Alt e rnativ e A at e nd of y e ar 25 i s $1 , 000 . A ss um e M ARR = 15% p e r y e ar . Alt e rnativ e A Alt e rnativ e B Capital Inv e s tm e nt $12 , 000 $30 , 000 Annual Exp e n s e s $2 , 200 $1 , 000 U s e ful lif e (y e ar s ) 10 y e ar s 25 y e ar s Salvag e Valu e at e nd of u s e ful lif e $5 , 000 $10 , 000 U s ing th e c ot e rminat e d a ss umption with s tudy p e riod 25 y e ar s , d e t e rmin e whi c h s tru c tur e i s b e tt e r by u s ing e quival e nt worth m e thod s . Pl e a s e , s tat e c l e arly your a ss umption s . 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 We can now compare the two alternatives using the FW method. We have
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