IE343_HW12_Solutions

IE343_HW12_Solutions - Homework#12Solutions...

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Unformatted text preview: Homework#12Solutions PURDUEUNIVERSITY      HomeworkAssignment#12 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 A ssign e d : 3 D e c 2010 D u e : 10 D e c 2010 (10 points) P r ob l e m 1 T e x t b oo k p r ob l e m 8-33 (a) Two years from now, we will have that . This means that . Thus, in order to buy one dollar two years from now, the number of units of currency X to pay is 6.08, as opposed to 6.4 units needed today. (b) If the currency X is devaluing at a rate of 26% per year, then three years from now we will have that . Therefore, 6.91 units of X will be needed to buy one dollar three years from now, as opposed to 6.4 units needed today. (10 points) P r ob l e m 2 T e x t b oo k p r ob l e m 8-35  a)  Devaluation of currency A relative to dollar =  This is the IRR in Currency A b)  Devaluation of dollar relative to currency B =  This is the IRR in Currency B Homework#12Solutions (10 points) P r ob l e m 3 T e x t b oo k p r ob l e m 9-5 MARR = 10% Year EUAC 1  2 3 4 5 6  After 5 years, EUAC starts increasing. Therefore, choose N=5 years for the challenger. Homework#12Solutions (10 points) P r ob l e m 4 T e x t b oo k p r ob l e m 9-8 Following the example show in class, we have: Challenger EOY MV 0 1 2 3 4 $50,000 $40,000 $32,000 $24,000 $16,000 Loss in Market Value Cost of Capital Expenses Marginal Cost EUAC $10,000 $8,000 $8,000 $8,000 $5,000 $4,000 $3,200 $2,400 $13,000 $15,500 $18,000 $20,500 $28,000 $27,500 $29,200 $30,900 $28,000 $27,761* $28,196 $28,778 Loss in Market Value Cost of Capital Expenses Marginal Cost EUAC $10,000 $4,000 $4,000 $4,000 $3,500 $2,500 $2,100 $1,700 $18,500 $21,000 $23,500 $26,000 $32,000 $27,500 $29,600 $31,700 $32,000 $29,857 $29,779 $30,193 Defender EOY MV 0 1 2 3 4 $35,000 $25,000 $21,000 $17,000 $13,000 The economic life of challenger is 2 years, and the corresponding EUAC is $27,761. This is smaller than the marginal cost of keeping the Defender for one more year which is $32,000. Therefore, the defender should be replaced immediately. Homework#12Solutions (10 points) P r ob l e m 5 T e x t b oo k p r ob l e m 9-15 We want to determine the best time to abandon the centrifuge Year Present Worth  1  2 3   4 5  PW (10%) reaches a maximum at 3 years. Therefore, the centrifuge should be retained for three years before abandonment. Homework#12Solutions (20 points) P r ob l e m 6 T e x t b oo k p r ob l e m 9-16 EOY 0 MV L oss in MV C os t of C apit al =B2B3 =B3B4 =B4B5 =B5B6 =B6B7 =B7B8 =B8B9 =B9B10 =B2* 0.07 =B3* 0.07 =B4* 0.07 =B5* 0.07 =B6* 0.07 =B7* 0.07 =B8* 0.07 =B9* 0.07 A p p r ox. A f te r - T a x T ota l ( M a r gi na l) c os t M A C RS B V I n t e r est o n T a x C r edit A d j ust e d A f t e r -t a x T ota l( M a r g i n a l ) C os t 8000 A nn . E xp. 0 0 0.2 =H2*(1-G3) =0.07*H2*0.4 0.32 =H2*(1-G3-G4) 0.192 T o find P G i ve n F (P/ F ) P W (7 % ) =F3+I3 0.9346 =K3*J3 1.07 =M3*L3 =0.07*H3*0.4 =F4+I4 0.8734 =K4*J4 0.5531 =(L3+L4)*M4 =0.07*H4*0.4 =F5+I5 0.8163 =K5*J5 0.3811 =(L3+L4+L5)*M5 =0.07*H5*0.4 =F6+I6 0.7629 =K6*J6 0.2952 =0.07*H6*0.4 =F7+I7 0.713 =K7*J7 0.2439 0.0576 =H2*(1-G3-G4-G5) =H2*(1-G3-G4-G5G6) =H2*(1-G3-G4-G5G6-G7) =H2*(1-G3-G4-G5G6-G7-G8) =0.07*H7*0.4 =F8+I8 0.6663 =K8*J8 0.2098 0 0 =0.07*H8*0.4 =F9+I9 0.6227 0.1856 0 0 =0.07*H9*0.4 =F10+I10 0.582 =K9*J9 =K10*J1 0 0.1675 =(L3+L4+L5+L6)*M6 =(L3+L4+L5+L6+L7)* M7 =(L3+L4+L5+L6+L7+L 8)*M8 =(L3+L4+L5+L6+L7+L 8+L9)*M9 =(L3+L4+L5+L6+L7+L 8+L9+L10)*M10 G DS 8000 T o find A G i ve n P ( A /P) E U A C ( A f t e r T a x) 7750 =SUM(C3:E 3)*(1-0.4) =SUM(C4:E 4)*(1-0.4) =SUM(C5:E 5)*(1-0.4) =SUM(C6:E 6)*(1-0.4) =SUM(C7:E 7)*(1-0.4) =SUM(C8:E 8)*(1-0.4) =SUM(C9:E 9)*(1-0.4) =SUM(C10: E10)*(1-0.4) 8000 0 0 560 3000 4116 0 .2 6400 224 4340 0.9346 4056.16 1 .07 4340.09 1500 329 3000 2897.4 0 .32 3840 179.2 3076.6 0.8734 2687.1 0.5531 3729.71 2200 1000 224 3500 2834.4 0.192 2304 107.52 2941.92 0.8163 2401.49 0.3811 3485.07 4 1450 750 154 4000 2942.4 0.1152 1382.4 64.512 3006.91 0.7629 2293.97 0.2952 3376.71 5 950 500 101.5 4500 3060.9 0.1152 460.8 38.7072 3099.61 0.713 2210.02 0.2439 3328.93 6 600 350 66 . 5 5250 3399.9 0.0576 3.3307 E -13 12.9024 3412.8 0.6663 2273.95 0.2098 3340.58 7 300 300 42 6250 3955.2 0 0 9.3259 E -15 3955.2 0.6227 2462.9 0.1856 3412.37 21 775 0 4842.6 0 0 0 4842.6 0.582 2818.39 0.1675 3551.67 1 4700 2 3200 3 2200 4 1450 5 950 6 600 7 300 8 0 0 8000 1 4700 3300 2 3200 3 8 0 300 3000 3000 3500 4000 4500 5250 6250 0.1152 0.1152 T h e e con om i c l i f e of t h is e q u i p m e n t is 5 y e a r s, as E U A C (a f t e r t a x ) st a r ts i n c r e asi n g i n y e a r 6.  ...
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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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