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Lecture 4

# Lecture 4 - 540:311 DETERMINISTIC MODELS IN OPERATIONS...

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540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 4: Chapter 3.5 – 3.12 Class Meeting: Thu Feb 2 nd 10:20-11:40am Prof. W. Art Chaovalitwongse

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Exam ple 1 Each soldier built: Sell for \$27 and uses \$19 worth of raw materials. Increase ABC ` s variable labor/overhead costs by \$14. Requires 2 hours of finishing labor. Requires 1 hour of carpentry labor. Each train built: Sell for \$21 and used \$9 worth of raw materials. Increases ABC ` s variable labor/overhead costs by \$10. Requires 1 hour of finishing labor. Requires 1 hour of carpentry labor. ABC, Inc., manufactures wooden soldiers and trains. Conditions: • All needed raw material. • Only 100 finishing hours. • Only 80 carpentry hours. • Demand for the trains is unlimited. • At most 40 soldiers are bought each week. ABC wants to maximize weekly profit (revenues – expenses).
Decision Variables x 1 = number of soldiers produced each week x 2 = number of trains produced each week Objective Function: Aim to maximize weekly profit in some function of the decision variables. The weekly profit can be expressed in terms of the decision variables x1 and x2: Weekly profit = weekly revenue – weekly raw material costs – the weekly variable costs Weekly revenue = 27x 1 + 21x 2 ; Weekly raw material costs = 10x 1 + 9x 2 ; Weekly variable costs = 14x 1 + 10x 2 Weekly profit = (27x 1 + 21x 2 ) – (10x 1 + 9x 2 ) – (14x 1 + 10x 2 ) = 3x 1 + 2x 2 OBJECTIVE FUNCTION: Maximize z = 3x 1 + 2x 2 CONSTRAINTS: Constraint 1 Each week, no more than 100 hours of finishing time may be used. Constraint 2 Each week, no more than 80 hours of carpentry time may be used. Constraint 3 Because of limited demand, at most 40 soldiers should be produced. Constraint 1: 2 x 1 + x 2 100 Constraint 2: x 1 + x 2 80 Constraint 3: x 1 40

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Problem Formulation Max z = 3x 1 + 2x 2 (objective function) Subject to (s.t.) 2 x 1 + x 2 100 (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 40 (constraint on demand for soldiers) x 1 0 (sign restriction) x 2 0 (sign restriction)
2-D Graphical Representation X1 X2 10 20 40 50 60 80 20 40 60 80 100 finishing constraint carpentry constraint demand constraint z = 60 z = 100 z = 180 Feasible Region G A B C D E F H From figure, we see that the set of points satisfying the ABC LP is bounded by the five sided polygon DGFEH. Any point on or in the interior of this polygon (the shade area) is in the feasible region.

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Case 1: Unique LP Solution The last isoprofit intersecting (touching) the feasible region indicates the optimal solution for the LP. For the ABC problem, this occurs at point G (x 1 = 20, x 2 = 60, z = 180). X1 X2 10 20 40 50 60 80 20 40 60 80 100 finishing constraint carpentry constraint demand constraint z = 60 z = 100 z = 180 Feasible Region G A B C D E F H
Case 2: Infinite Number of Solutions max z = 3x1 + 2x2 s.t. 1 40 x 1 1 60 x 2 + 1 1 50 x 1 1 x 2 + 1 x 1 x 2 0 , Any point (solution) falling on line segment AE will yield an optimal solution of z =120.

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Lecture 4 - 540:311 DETERMINISTIC MODELS IN OPERATIONS...

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