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Chapter 4. Alternative Optimum Unbounded Solution Infeasible Solution . Chapter 4 unbound and infeasible solution. The Objectives this Chapter: Identify the problems which include more than one optimum solutions. Identify the problems which involve the unbounded solutions.
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Chapter 4 Alternative Optimum Unbounded Solution Infeasible Solution
Chapter 4 unbound and infeasible solution The Objectives this Chapter: • Identify the problems which include more than one optimum solutions. • Identify the problems which involve the unbounded solutions. • Identify the problems which are non feasible solution (no optimum solution and no feasible region solution)
Chapter 4 unbound and infeasible solution Alternative Optimum Alternative optimal solutions occur when the contour of the objective function is parallel to a constraint. In the case of a two decision variable model, when the contour of the objective function is parallel to a constraint, there are two optimal corner solutions, and in fact, all the points on the constraint between the two corners are also optimal solutions. If there is more than one solution to an LP (i.e., there exist alternative optimal solutions) then there are an infinite number of alternative optimal solutions.
To observe the connections among the Payoff slope, corner points, and any movement of the optimal solution from one corner point to another, right-click on the Payoff line in GLP and drag it. For example, observe the change in the objective function coordinates and OV as the line is dragged down.
Chapter 4 unbound and infeasible solution Alternative Optimum : Example /1 Consider the linear program model solve this problem using the graphic method. Max Z= 6x1 + 4x2 s.t. x1 + 4x2 ≤ 40 3x1 + 2x2 ≤ 30 3x1 + x2 ≤ 24 x1, x2 ≥ 0
Solution \ 1 Chapter 4 unbound and infeasible solution
Example 2 Chapter 4 unbound and infeasible solution Min Z= 3X + 3Y s.t. 1X + 2Y ≤ 16 1X + 1Y ≤ 10 5X + 3Y ≤ 45 X , Y ≥ 0
Solution/ 2 Chapter 4 unbound and infeasible solution
Chapter 4 unbound and infeasible solution Unbounded Models Unboundedmodels typically occur when one or more important constraints have been left out of the model. In this case, it results in an infinite number of allowable values for the decision variables that will improve the objective value. To illustrate this, remove the first four constraints from the Oak Products model and try to solve it using GLP.
Unbounded solution : Example / 1 Chapter 4 unbound and infeasible solution Max Z= 5X + 6Y s.t. 3X + Y ≥ 15 X + 2Y ≥ 12 3X + 2Y ≥ 24 X , Y ≥ 0
Solution / 1 Chapter 4 unbound and infeasible solution Points • (12,0) • (6,3) • (2,9) • (0,15)
Example / 2 Chapter 4 unbound and infeasible solution Max z = 2x1- x2 S.t: x1 - x2 <= 1 2x1 + x2 >= 6 x1; x2>= 0
Solution\2 Chapter 4 unbound and infeasible solution
Example 3 Chapter 4 unbound and infeasible solution Max Z = X1+1.5X2 S.t 2X1+3X2>= 20 X1+2X2>=12 X1,x2>=0
Solution /3 Chapter 4 unbound and infeasible solution
Infeasible ModelsInfeasibility (or inconsistency) refers to a model with an empty feasible region.This means that there is no combination of values for the decision variables that simultaneously satisfies all the constraints.To illustrate this, consider the following LP model: Chapter 4 unbound and infeasible solution
Infeasible solution : Example \1 Chapter 4 unbound and infeasible solution Min Z= X + Y s.t. 5X + 3Y ≤ 30 3X + 4Y ≥ 36 Y ≤ 7 X , Y ≥ 0
Solution\1 Chapter 4 unbound and infeasible solution
Example / 2 Chapter 4 unbound and infeasible solution Max z= 3x1 + 2x2 S.t: 1/40x1 +1/60x2 <= 1 1/50x1 +1/50x2 <= 1 x1 >=30 x2 >=20 X1,X2>=0
Solution /2 Chapter 4 unbound and infeasible solution
Example \3 Chapter 4 unbound and infeasible solution Max Z = 3X1-2X2 S.t X1+X2<=1 2X1+2X2>=4 X1, X2>=0
Solution \ 3 Chapter 4 unbound and infeasible solution
Important Notes Chapter 4 unbound and infeasible solution • Every linear program will fall into one of the following three mutually exclusive categories: • The model has an optimal solution. • There is no optimal solution, because the model is unbounded. • There is no optimal solution, because the model is infeasible.