Learn & Review: Graphical Method | Linear Programming Problem | Learn OR with Asksia

Operations Research: Graphical Method for Linear Programming

This tutorial introduces the graphical method for solving linear programming problems, specifically focusing on cases with two decision variables. It outlines four potential outcomes: a unique optimal solution, infinite optimal solutions, an unbounded solution, and no solution. This lecture details the process for finding a unique optimal solution.

Problem Setup and Objective

The problem involves minimizing the objective function:

• Minimize Z = 20x₁ + 10x₂

Subject to the following constraints:

• x₁ + 2x₂ ≤ 40
• 3x₁ + x₂ ≤ 30
• 4x₁ + 3x₂ ≤ 60
• x₁ ≥ 0, x₂ ≥ 0

Steps for Graphical Solution

1. Convert Inequalities to Equations:

   • Replace all inequality signs (≤, ≥) with equal signs (=) to define the boundary lines of the feasible region.
   • Equation 1: x₁ + 2x₂ = 40
   • Equation 2: 3x₁ + x₂ = 30
   • Equation 3: 4x₁ + 3x₂ = 60

2. Determine Intercepts for Plotting Lines:

   • Method 1: Divide by the Right-Hand Side:
     • Divide each equation by its right-hand side constant to get the standard intercept form (x/a + y/b = 1).
     • Equation 1: (x₁/40) + (2x₂/40) = 1 => (x₁/40) + (x₂/20) = 1
       • x₁-intercept = 40, x₂-intercept = 20
     • Equation 2: (3x₁/30) + (x₂/30) = 1 => (x₁/10) + (x₂/30) = 1
       • x₁-intercept = 10, x₂-intercept = 30
     • Equation 3: (4x₁/60) + (3x₂/60) = 1 => (x₁/15) + (x₂/20) = 1
       • x₁-intercept = 15, x₂-intercept = 20
   • Method 2: Assume Zero for One Variable:
     • For each equation, set x₁ = 0 to find x₂, then set x₂ = 0 to find x₁.
     • Equation 1:
       • If x₁ = 0, 2x₂ = 40 => x₂ = 20. Point: (0, 20)
       • If x₂ = 0, x₁ = 40. Point: (40, 0)
     • (This method can be applied similarly for equations 2 and 3).

3. Plot the Lines on a Graph:

   • Draw the x₁ and x₂ axes. Since x₁ ≥ 0 and x₂ ≥ 0, focus on the first quadrant.
   • Choose an appropriate scale for the axes based on the intercept values (e.g., increments of 10).
   • Plot the intercept points for each equation and draw a line connecting them. Label each line corresponding to its equation number.

4. Identify the Feasible Region:

   • For each constraint, determine the direction of the inequality:
     • x₁ + 2x₂ ≤ 40: Arrows point towards the origin (below Line 1).
     • 3x₁ + x₂ ≥ 30: Arrows point away from the origin (above Line 2).
     • 4x₁ + 3x₂ ≤ 60: Arrows point towards the origin (below Line 3).
   • The feasible region is the area on the graph that satisfies all the constraints simultaneously, indicated by the overlapping directions of the arrows. This region is typically a polygon.

5. Determine the Corner Points (Extreme Points) of the Feasible Region:

   • The feasible region is bounded by the plotted lines and the axes. Identify the vertices of this polygon.
   • In this example, there are four corner points:
     • Point A: Intersection of the x₁-axis and Line 3 (or Line 2, depending on the exact shape).
     • Point B: Intersection of the x₁-axis and Line 1.
     • Point C: Intersection of Line 1 and Line 2.
     • Point D: Intersection of Line 2 and Line 3.

6. Calculate the Coordinates of the Corner Points:

   • Some points might be directly identifiable from the intercepts (e.g., Point B is (40, 0)).
   • For intersection points, solve the system of equations corresponding to the intersecting lines:
     • Point C (Intersection of Line 1 and Line 2):
       • x₁ + 2x₂ = 40
       • 3x₁ + x₂ = 30
       • Solving these yields: x₁ = 4, x₂ = 18. So, Point C is (4, 18).
     • Point D (Intersection of Line 2 and Line 3):
       • 3x₁ + x₂ = 30
       • 4x₁ + 3x₂ = 60
       • Solving these yields: x₁ = 6, x₂ = 12. So, Point D is (6, 12).
   • Note: The exact corner points depend on the specific graph and feasible region. In this example, the points are A=(15,0), B=(40,0), C=(4,18), and D=(6,12). Correction based on calculation: Point A is likely the origin (0,0) or another intersection. The provided calculation identifies A=(15,0) and B=(40,0) as intercepts on the x1 axis, and C=(4,18) and D=(6,12) as intersection points. The feasible region's corners are derived from these. The lecture identifies the corner points as A=(15,0), B=(40,0), C=(4,18), and D=(6,12).

7. Evaluate the Objective Function at Each Corner Point:

   • Substitute the (x₁, x₂) coordinates of each corner point into the objective function (Z = 20x₁ + 10x₂).
     • Point A (15, 0): Z = 20(15) + 10(0) = 300
     • Point B (40, 0): Z = 20(40) + 10(0) = 800
     • Point C (4, 18): Z = 20(4) + 10(18) = 80 + 180 = 260
     • Point D (6, 12): Z = 20(6) + 10(12) = 120 + 120 = 240

8. Determine the Optimal Solution:

   • Since the objective is to minimize Z, select the corner point that yields the smallest value of Z.
   • The minimum value of Z is 240, which occurs at Point D (6, 12).

Conclusion

• Optimal Solution: x₁ = 6, x₂ = 12
• Minimum Value of Z: 240

This represents a unique optimal solution. The next lecture will cover the other types of solutions (infinite, unbounded, no solution) using the graphical method.