moduleBsolutions

moduleBsolutions - MODULE B END-OF-MODULE PROBLEMS B.1 Let...

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Unformatted text preview: MODULE B END-OF-MODULE PROBLEMS B.1 Let x = number of standard model to produce y = number of deluxe model to produce Maximize 40 x + 60 y Subject to 30 30 450 10 15 180 6 x y x y x x y + + , B.2 2 4 6 8 10 12 14 16 18 20 x 2 4 6 8 10 12 14 16 18 20 Optimal x y = = ( 29 10 , y Feasible corner points ( x , y ): (0, 3), (0, 10), (2.4, 8.8), (6.75, 3). Maximum profit is 100 at (0, 10). Quantitative Module B: Linear Programming 1 B.3 2 4 6 8 10 12 14 16 18 20 x 2 4 6 8 10 12 14 16 18 20 Optimal x y = = ( 29 4 8 , y Feasible corner points ( x , y ): (0, 2), (0, 10), (4, 8), (10, 2). Maximum profit is 52 at (4, 8). B.4 (a) Corner points (0, 50), (50, 50), (0, 200), (75, 75), (50, 150). (b) Optimal solutions: (75, 75) and (50, 150). Both yield profit of $3,000. 50 100 150 200 250 300 350 400 450 x 50 100 150 200 250 300 350 400 Feasible Region x 2 1 2 Instructors Solutions Manual t/a Operations Management B.5 (a) Adding a new constraint will reduce the size of the feasible region unless it is a redundant constraint. It can never make the feasible region any larger. (b) A new constraint can only reduce the size of the feasible region; therefore the value of the objective function will either decrease or remain the same. If the original solution is still feasible, it will remain the optimal solution. B.6 (a) Let x 1 = number of liver flavored biscuits in a package x 2 = number of chicken flavored biscuits in a package Minimize x x 1 2 2 + Subject to x x x x x x x 1 2 1 2 1 1 2 40 2 4 60 15 + + , (b) Corner points are (0, 40) and (15, 25). Optimal solution is (15, 25) with cost of 65. (c) Minimum cost = 65 cents. B.7 20 40 60 80 100 120 20 40 60 80 100 120 140 160 Feasible Region Drilling Constraint Wiring Constraint c b d Number of Air Conditioners: x 1 Let x 1 = number of air conditioners to be produced x 2 = number of fans to be produced Maximize 25 15 1 2 x x + Subject to wiring drilling nonnegativity 3 2 240 2 1 140 1 2 1 2 1 2 x x x x x x + ( 29 + ( 29 ( 29 , Profit: @ a : ( x 1 = , x 2 = ) Obj = $0 @ b : ( x 1 = , x 2 120 = ) Obj = + = 25 15 120 800 $1, @ c : ( x 1 40 = , x 2 60 = ) Obj = + = 25 40 15 60 900 $1, * @ d : ( x 1 70 = , x 2 = ) Obj = + = 25 70 15 750 $1, The optimal solution is to produce 40 air conditioners and 60 fans each period. Profit will be $1,900. Quantitative Module B: Linear Programming 3 B.8 50 100 150 200 250 300 50 100 150 200 250 300 Feasible Region c b a x 1 x 2 Let x 1 = number of Model A tubs produced x 2 = number of Model B tubs produced Maximize 90 70 1 2 x x + Subject to steel zinc nonnegativity 125 100 25 000 20 30 6 000 1 2 1 2 1 2 x x x x x x + + ( 29 ( 29 ( 29 , , , Profit: @ a : ( x 1 = , x 2 200 = ) Obj = + = 90 70 200 000 00 $14, ....
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This note was uploaded on 02/15/2012 for the course BA 252 taught by Professor Jamescampbell during the Winter '03 term at UMSL.

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moduleBsolutions - MODULE B END-OF-MODULE PROBLEMS B.1 Let...

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