LPP: Formulation and graphical method of solution
Linear Programming and Graphical Method of Solution Linear programming is a powerful technique for optimizing the allocation of resources to achieve maxi...
Linear Programming and Graphical Method of Solution Linear programming is a powerful technique for optimizing the allocation of resources to achieve maxi...
Linear programming is a powerful technique for optimizing the allocation of resources to achieve maximum or minimum goals. We represent this allocation as a linear program, a set of linear inequalities and equalities.
Graphical method is a method for solving linear programs that uses graphical techniques to find the optimal solution. This method involves constructing the feasible region (the region that satisfies all inequalities) and finding the corner points of this region.
Formulas and notations:
Decision variables: x1, x2, ..., xn represent the amount of resource allocated to each component.
Constraints:
a11x1 + a12x2 + ... + a1nxn ≤ b1 (linear inequality)
a21x1 + a22x2 + ... + a2nxn ≤ b2 (linear inequality)
...
am1x1 + am2x2 + ... + amnxn ≤ bm (linear inequality)
Objective function: f(x) = c1x1 + c2x2 + ... + cnxn, where c1, c2, ..., cn are constants.
Feasible region: The region in the plane that satisfies all inequalities.
Corner points: The points where the boundary lines of the feasible region intersect.
Optimal solution: A point in the corner with the highest (minimum) value of the objective function.
Graphical method steps:
Draw the feasible region: This is the region where all constraints hold true.
Identify corner points: These are the points where the boundary lines of the feasible region intersect.
Choose the optimal solution: The point in the corner with the highest (or lowest) value of the objective function is the optimal solution.
Examples:
Maximize profit:
Maximize z = 2x1 + 3x2
Constraints:
x1 + x2 ≤ 10 (budget constraint)
2x1 + 3x2 ≤ 20 (production constraint)
x1 ≥ 0, x2 ≥ 0
Optimal solution: (5, 5)
Minimize cost:
Minimize z = c1x1 + c2x2
Constraints:
2x1 + 4x2 ≤ 10 (transportation constraint)
x1 + 3x2 ≤ 8 (production constraint)
x1, x2 ≥ 0
Optimal solution: (2, 4)