Lecture_4_Special Cases in Graphical Method

4.1 Multiple Optimal Solution
Example 1
Solve by using graphical method
Max Z = 4x1 + 3x2
Subject to
4x1+ 3x2 ? 24
x1 ? 4.5
x2 ? 6
x1 ? 0 , x2 ? 0
Solution
Lecture 4
Special Cases in Graphical Method
Linear Programming :
The first constraint 4x1+ 3x2 ? 24, written in a form of equation
4x1+ 3x2 = 24
Put x1 =0, then x2 = 8
Put x2 =0, then x1 = 6
The coordinates are (0, 8) and (6, 0)
The second constraint x1 ? 4.5, written in a form of equation
x1 = 4.5
The third constraint x2 ? 6, written in a form of equation
x2 = 6
1
The corner points of feasible region are A, B, C and D. So the coordinates for the corner points
are
A (0, 6)
B (1.5, 6) (Solve the two equations 4x1+ 3x2 = 24 and x2 = 6 to get the coordinates)
C (4.5, 2) (Solve the two equations 4x1+ 3x2 = 24 and x1 = 4.5 to get the coordinates)
D (4.5, 0)
We know that Max Z = 4x1 + 3x2
At A (0, 6)
Z = 4(0) + 3(6) = 18
At B (1.5, 6)
Z = 4(1.5) + 3(6) = 24
At C (4.5, 2)
Z = 4(4.5) + 3(2) = 24
At D (4.5, 0)
Z = 4(4.5) + 3(0) = 18
Max Z = 24, which is achieved at both B and C corner points. It can be achieved not only at B
and C but every point between B and C. Hence the given problem has multiple optimal solutions.