Methods to solve Linear Programming Problem (LPP):
Graphical Method (this notes)
Simplex Method (notes coming soon)
Big-M Method (notes coming soon)
Two-phase Method (notes coming soon)
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It is the simplest method to solve LPP, with a limitation that it is used only for two variables. It is because we will be using Cartesian planes (XY-graph) to solve the problems.
Plot each constraints on graph: Plot constraints given in terms of either equalities or inequalities. To plot given constraints on graph:
Identify feasible solution: Based on inequalities/equalities on constraints follow the steps provided below to find feasible region.
Find out optimum solution: To find out the optimum solution from the feasible region so obtain, there are two methods:
Extreme corner point method
And that’s the solution of decision variables and eventually solution to given LPP by this method.
Iso-profit function line approach
Extreme corner point method | Iso-profit (or cost) method |
---|---|
Identify co-ordinates of each of the extreme points of the feasible region. | Determines the slope of the objective function and then join intercepts to reveal the profit. |
Compute the profit at each extreme point by substituting that points’ co-ordinates into the objective function. | In case of maximization, maintain the same slope through a series of parallel lines, and move towards the right until it touches the feasible region at only one point. |
In case of maximization, select the point where profit is maximize. In case of minimization, select the point where cost is minimize. | Compute the co-ordinates of the point touched by the line on the feasible region and calculate the profit or loss. |
Use graphical method to solve the following LPP
$\begin{aligned} \text{Maximize} \ \ &Z = 3x_1 + 9x_2 \\ \text{Subject to} \ \ &x_1 + 4x_2 \leq 8 \\ &x_1 + 2x_2 \leq 4 \\ \text{And} \ \ &x_1, x_2 \geq 0 \end{aligned}$
Solving the constraints by converting them into equation:
Equation | x₁ + 4x₂ ≤ 8 | x₁ + 2x₂ ≤ 4 | ||
---|---|---|---|---|
x₁ | 0 | 8 | 0 | 4 |
x₂ | 2 | 0 | 2 | 0 |
Plotting the values in graph and using and we will be using Extreme corner point method:
As inequalities of constraints in $\leq$, selecting the region on and towards the origin of constraint lines.
For optimal solution, computing the value of objective function using extreme points of feasible region obtain from graph:
Points | $Z = 3x_1 + 9x_2$ |
---|---|
(0,2) | $\bold{Z = 3(0) + 9(2) = 18}$ |
(0,0) | $Z = 3(0) + 9(0) = 0$ |
(4,0) | $Z = 3(4) + 9(0) = 12$ |
As objective is to maximize the profit, selecting maximum computed value of objective function, i.e. $Z = 18$
Solution: $x_1 = 0, \ \ x_2 = 2 \ \ \text{and} \ \ Z = 18$
Use graphical method to solve the following LPP
$\begin{aligned} \text{Maximize} \ \ &Z = 60x_1 + 80x_2 \\ \text{Subject to} \ \ &20x_1 + 30x_2 \geq 900 \\ &40x_1 + 30x_2 \geq 1200 \\ \text{And} \ \ &x_1, x_2 \geq 0 \end{aligned}$
Solving the constraints by converting them into equation:
Equation | 20x₁ + 30x₂ ≥ 900 | 40x₁ + 30x₂ ≥ 1200 | ||
---|---|---|---|---|
x₁ | 0 | 45 | 0 | 30 |
x₂ | 30 | 0 | 40 | 0 |
Plotting the values in graph and using and we will be using Extreme corner point method:
As inequalities of constraints in $\geq$, selecting the region on and away of the constraint lines.
For optimal solution, computing the value of objective function using extreme points of feasible region obtain from graph:
Points | $Z = 60x_1 + 80x_2$ |
---|---|
(0,40) | $Z = 60(0) + 80(40) = 3200$ |
(15,20) | $\bold{Z = 60(15) + 80(20) = 2500}$ |
(45,0) | $Z = 60(45) + 80(0) = 2700$ |
As objective is to minimize the cost, selecting minimum computed value of objective function, i.e. $Z = 2500$
Solution: $x_1 = 15, \ \ x_2 = 20 \ \ \text{and} \ \ Z = 2500$
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