Question: FIGURE 3 . 1 9 THE SOLUTION FOR THE TUCKER, INC., PROBLEM table [ [ 4 0 9 0 0 . 0 0 0

FIGURE 3.19 THE SOLUTION FOR THE TUCKER, INC., PROBLEM
\table[[40900.00000],[Variable,Value,Reduced Cost],[--------------,---------------,-----------------],[S,100.00000,0.00000],[SC,150.00000,0.00000],[D1,40.00000,0.00000],[D2,0.00000,-10.00000],[Constraint,Slack/Surplus,Dual Value],[--------------,------,------,----------------],[1,0.00000,15.00000],[2,20.00000,0.00000],[3,0.00000,34.50000],[4,60.00000,0.00000],[5,0.00000,-35.00000],[6,75.00000,0.00000],[,Objective,Allowable,Allowable],[Variable,Coefficient,Increase,Decrease],[-----------,-------------,----------,----------],[S,190.00000,35.00000,Infinite],[SC,150.00000,Infinite,23.33333],[D1,-15.00000,15.00000,172.50000],[D2,-10.00000,10.00000,Infinite],[,RHS,Allowable,Allowable],[Constraint,Value,Increase,Decrease],[----------,------------,----------,----------],[1,200.00000,40.00000,60.00000],[2,180.00000,Infinite,20.00000],[3,1200.00000,133.33333,200.00000],[4,100.00000,Infinite,60.00000],[5,100.00000,50.00000,100.00000],[6,75.00000,75.00000,Infinite]]
94
$200 per unit for model DRB and $280 per unit for model DRW. The linear programming model for this problem is as follows:
Max 200DRB+280DRW,
s.t.
,20DRB+25DRW40,000 Steel available
,40DRB+100DRW120,000 Manufacturing minutes
,60DRB+40DRW96,000 Assembly minutes
,DRB,DRW0,
 FIGURE 3.19 THE SOLUTION FOR THE TUCKER, INC., PROBLEM \table[[40900.00000],[Variable,Value,Reduced Cost],[--------------,---------------,-----------------],[S,100.00000,0.00000],[SC,150.00000,0.00000],[D1,40.00000,0.00000],[D2,0.00000,-10.00000],[Constraint,Slack/Surplus,Dual

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