HW4 solutions - Problem43p.172 DECISION VARIABLES: X1, X2...

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Problem 4-3 p. 172 X1 X2 Decision Variables 5 0 Objective Function 4 2 20 MAX Constraint 1 2 4 10 <= 20 Constraint 2 3 5 15 <= 15 Constraint 3 1 5 >= 0 Constraint 4 1 0 >= 0 DECISION VARIABLES: X1, X2 OBJECTIVE FUNCTION: MAX 4X1 +2X2 CONSTRAINTS: 2X1 + 4X2 <= 20 3X1 + 5X2 <= 15 X1, X2 >= 0 See tab Sensitivity Report 4-3 for answers to questions 4-3a,b,c,d,e,f
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Microsoft Excel 14.0 Sensitivity Report Worksheet: [HW4 solutions.xlsx]4-3 Report Created: 1/23/2011 9:21:55 AM Variable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $C$11 Decision Variables X1 5 0 4 1.00E+030 2.8 $D$11 Decision Variables X2 0 0 2 4.67 1.00E+030 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $E$13 Constraint 1 10 0 20 1.00E+030 10 $E$14 Constraint 2 15 1.33 15 15 15 $E$15 Constraint 3 5 0 0 5 1.00E+030 $E$16 Constraint 4 0 -4.67 0 3 0
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a) The objective function coefficient for X1 can decrease by 2.8 or incre amount without changing the optimal solution: 4-2.8 <= X1 <= infinity b) The optimal solution is unique. None of the allowable increase or de for the objective function coefficients are zero. c) The objective function coefficient for X2 has an allowable increase o d) The shadow price for X2 is -4.67; increasing X2 by one unit (from ze decrease the objective function value by 4.67, from 20 to 15.33. e) The allowable increase for the second constraint is 15, so increasing of the the second constraint by ten units (from 15 to 25) will improve t function by ten times the shadow price for that constraint, or 10 * 1.33 The new objective function value will be 20 + 13.33 = 33.33. f) The new reduced cost for X2 would be 2 - (4 × 0 + 1 × 1.333) = 0.6 would be profitable to increase the value of X2 and the current solutio longer be optimal.
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rease by any y ecrease values f 4.67. ro to one) will g the availability the objective 33 or 13.33. 67. Therefore, it on would no
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Decision Variables Miami Orlando Tallahassee Production Capacity Eustis 0 10 10 20 <= 20 Clermont 10 5 0 15 <= 20 Supply 10 15 10 >= >= >= Demand 10 15 10 Objective Function Miami Orlando Tallahassee Eustis  $260   $220   $290  Clermont  $230   $240   $310  Total Cost 8600 a) Decision Variables: How many tons to ship from each of the two processing plants distributors: X13 = Eustis to MiamiX14 = Eustis to Orlando X15 = Eustis to Tallahassee X23 = Clermont to Miami X24 = Clermont to Orlando X25 = Clermont to Tallahas Objective Function: MIN 260X13+220X14+290X15+230X23+240X24+310X25 Subject to the constraints: X13+X14+X15 <= 20 X23+X24+X25 <= 20 X13+X23 >= 10 X14+X24 >= 15 X15+X25 >= 10 ALL Xij >= 0 c) Optimal solution is Eus Orlando and 10 tons to Ta 10 tons to Miami and 5 to is $8600. f & g) see tab Sensitivity
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This note was uploaded on 04/17/2011 for the course BUSQOM 0050 taught by Professor Glowackia during the Spring '08 term at Pittsburgh.

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HW4 solutions - Problem43p.172 DECISION VARIABLES: X1, X2...

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