IE343_HW12_Solutions - Homework #12 Solutions PURDUE...

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Unformatted text preview: Homework #12 Solutions PURDUE UNIVERSITY Homework Assignment #12 IE 343: Engineering Economics Fall 2010 Instructor: A. Capponi Assigned: 3 Dec 2010 Due: 10 Dec 2010 Problem 1 Textbook problem 8-33 (10 points) ( ) (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. Problem 2 Textbook problem 8-35 a) (10 points) 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 #12 Solutions Problem 3 Textbook problem 9-5 (10 points) 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 #12 Solutions Problem 4 Textbook problem 9-8 (10 points) 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 #12 Solutions Problem 5 Textbook problem 9-15 (10 points) 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 #12 Solutions Problem 6 Textbook problem 9-16 EOY 0 MV Loss in MV Cost of Capit 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 (20 points) Approx. After-Tax Total (Marginal) cost MACRS BV Interest on Tax Credit Adjusted After-tax Total(Mar ginal) Cost 8000 Ann . Exp. 0 0 0.2 =H2*(1-G3) =0.07*H2*0.4 0.32 =H2*(1-G3-G4) 0.192 To find P Given F (P/F) PW(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 GDS 8000 To find A Given P (A/P) EUAC(After Tax) 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 3.3307E-13 12.9024 3412.8 0.6663 2273.95 0.2098 3340.58 7 300 300 42 6250 3955.2 0 0 9.3259E-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 The economic life of this equipment is 5 years, as EUAC (after tax) starts increasing in year 6. ...
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This note was uploaded on 09/19/2011 for the course IE 343 taught by Professor Vincent,g during the Spring '08 term at Purdue University-West Lafayette.

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