Graphical Method | Methods to solve LPP | Linear Programming

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Methods to solve Linear Programming Problem (LPP):

  1. Graphical Method (this notes)

  2. Simplex Method (notes coming soon)

  3. Big-M Method (notes coming soon)

  4. Two-phase Method (notes coming soon)

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Graphical Method

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.

Steps to solve LPP by Graphical Method

  1. Represent variables on axis: As this method is applicable when we have two decision variables, represent x1x_1 and x2x_2 as XX and YY axis. Also as x1,x20x_1, x_2 \geq 0; x1x_1 and x2x_2 always lies in first quadrant.

  2. Plot each constraints on graph: Plot constraints given in terms of either equalities or inequalities. To plot given constraints on graph:

    • If inequalities are given then, convert them to equalities.
    • Use the method provided in video to find the values of x1x_1 & x2x_2 at different points.
    • Plot the determine co-ordinates by step-2 on the graph and connect them with straight lines.

  3. Identify feasible solution: Based on inequalities/equalities on constraints follow the steps provided below to find feasible region.

    • If constraints are \geq, then the area on or above the constraint line, i.e. away from the origin is feasible region.
    • If constraints are \leq, then the area on or below the constraint line, i.e. towards the origin is feasible region.
    • The area common to all constraints is called feasible region. Shade this area/region as feasible region. This region contains the optimal solution point.

  4. Find out optimum solution: To find out the optimum solution from the feasible region so obtain, there are two methods:

    1. Extreme corner point method
    2. Iso-profit function line approach

Let us learn how to use this two methods

  1. Extreme corner point method

    • Find the co-ordinates of each extreme point of feasible region.
    • Calculate and compare the value of the objective function at each extreme co-ordinates found above.
    • Identify the extreme co-ordinate at which optimal value of the objective function obtained.

And that’s the solution of decision variables and eventually solution to given LPP by this method.

  1. Iso-profit function line approach

    • Draw iso-profit line for an arbitrary but small value of objective function without violating any of the constraints.
    • Move iso-profit line parallel to the direction of increasing objective function. The farthest this line may intersect at one corner point providing a single optimal solution. Also, this line may intersect with one of the boundary lines of feasible area. Then at least two optimal solutions must lie on two adjoining corners and other will lie on boundary connecting them.
    • However, if iso-profit line goes on without the limit from the constraints, then an unbounded solution may exist. This usually indicates that an error has been made in formulating LP model.
    • If the line is iso-cost line (for minimization) drags the line near to the origin.

Difference between the two methods provided above

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.

Numerical #1

Use graphical method to solve the following LPP

Maximize  Z=3x1+9x2Subject to  x1+4x28x1+2x24And  x1,x20\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}

Solution #1

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:

extreme-point-corner-graph-plot-numerical-1

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=3x1+9x2Z = 3x_1 + 9x_2
(0,2) Z=3(0)+9(2)=18\bold{Z = 3(0) + 9(2) = 18}
(0,0) Z=3(0)+9(0)=0Z = 3(0) + 9(0) = 0
(4,0) Z=3(4)+9(0)=12Z = 3(4) + 9(0) = 12

As objective is to maximize the profit, selecting maximum computed value of objective function, i.e. Z=18Z = 18

Solution: x1=0,  x2=2  and  Z=18x_1 = 0, \ \ x_2 = 2 \ \ \text{and} \ \ Z = 18

Numerical #2

Use graphical method to solve the following LPP

Maximize  Z=60x1+80x2Subject to  20x1+30x290040x1+30x21200And  x1,x20\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}

Solution #2

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:

extreme-point-corner-graph-plot-numerical-2

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=60x1+80x2Z = 60x_1 + 80x_2
(0,40) Z=60(0)+80(40)=3200Z = 60(0) + 80(40) = 3200
(15,20) Z=60(15)+80(20)=2500\bold{Z = 60(15) + 80(20) = 2500}
(45,0) Z=60(45)+80(0)=2700Z = 60(45) + 80(0) = 2700

As objective is to minimize the cost, selecting minimum computed value of objective function, i.e. Z=2500Z = 2500

Solution: x1=15,  x2=20  and  Z=2500x_1 = 15, \ \ x_2 = 20 \ \ \text{and} \ \ Z = 2500

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