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## Special or Unique Cases in Graphical Method

Special or Unique Cases in Graphical MethodMultiple Optimal SolutionExample 1

Solve by means of graphical method

Max Z = 4x

_{1}+ 3x_{2}Subject to

4x

_{1}+ 3x_{2 }≤ 24x

_{1}≤ 4.5x

_{2}≤ 6x

_{1 }≥ 0 , x_{2 }≥ 0AnswerThe first constraint 4x

_{1}+ 3x_{2 }≤ 24, can be written in the form of equation4x

_{1}+ 3x_{2 }= 24Place x

_{1}=0, then x_{2}= 8Place x

_{2}=0, then x_{1}= 6Then coordinates are (0, 8) and (6, 0)

The second constraint x

_{1}≤ 4.5, can be written in the form of equationx

_{1}= 4.5The third constraint x

_{2}≤ 6, can be written in the form of equationx

_{2}= 6The corner positions of feasible region are A, B, C and D. Therefore the coordinates for the corner positions are

A (0, 6)

B (1.5, 6) (Solve the two equations 4x

_{1}+ 3x_{2 }= 24 and x_{2}= 6 to obtain the coordinates)C (4.5, 2) (Solve the two equations 4x

_{1}+ 3x_{2 }= 24 and x_{1}= 4.5 to obtain the coordinates)D (4.5, 0)

We are given that Max Z = 4x

_{1}+ 3x_{2}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 accomplished at both B and C corner positions. It can be attained not only at B and C but at every point between B and C. Thus the given problem has multiple optimal solutions.

No Optimal SolutionExample 1Work out or solve graphicallyMax Z = 3x

_{1}+ 2x_{2}Subject to

x

_{1}+ x_{2 }≤ 1x

_{1}+ x_{2 }≥ 3x

_{1 }≥ 0 , x_{2 }≥ 0AnswerThe first constraint x

_{1}+ x_{2 }≤ 1, can be written in the form of equationx

_{1}+ x_{2 }= 1Place x

_{1}=0, then x_{2}= 1Place x

_{2}=0, then x_{1}= 1Thus the coordinates are (0, 1) and (1, 0)

The first constraint x

_{1}+ x_{2 }≥ 3, can be written in the form of equationx

_{1}+ x_{2 }= 3Place x

_{1}=0, then x_{2}= 3Place x

_{2}=0, then x_{1}= 3Thus the coordinates are (0, 3) and (3, 0)

There is no common feasible region produced by two constraints combine, that is, we cannot find out even a single point which satisfies the constraints. Therefore there is no optimal solution.

Unbounded Solution

ExampleSolve through graphical method

Max Z = 3x

_{1}+ 5x_{2}Subject to

2x

_{1}+ x_{2 }≥ 7x

_{1}+ x_{2 }≥ 6x

_{1}+ 3x_{2 }≥ 9x

_{1 }≥ 0 , x_{2 }≥ 0AnswerThe first constraint 2x

_{1}+ x_{2 }≥ 7, can be written in the form of equation2x

_{1}+ x_{2 }= 7Place x

_{1}=0, then x_{2}= 7Place x

_{2}=0, then x_{1}= 3.5Hence, the coordinates are (0, 7) and (3.5, 0)

The second constraint x

_{1}+ x_{2 }≥ 6, can be written in the form of equationx

_{1}+ x_{2 }= 6Place x

_{1}=0, then x_{2}= 6Place x

_{2}=0, then x_{1}= 6Therefore, the coordinates are (0, 6) and (6, 0)

The third constraint x

_{1}+ 3x_{2 }≥ 9, can be written in the form of equationx

_{1}+ 3x_{2 }= 9Place x

_{1}=0, then x_{2}= 3Place x

_{2}=0, then x_{1}= 9Hence, the coordinates are (0, 3) and (9, 0)

The corner positions of feasible region are A, B, C and D. Therefore the coordinates for the corner positions are

A (0, 7)

B (1, 5) (Solve the two equations 2x

_{1}+ x_{2 }= 7 and x_{1}+ x_{2 }= 6 to obtain the coordinates)C (4.5, 1.5) (Solve the two equations x

_{1}+ x_{2 }= 6 and x_{1}+ 3x_{2 }= 9 to obtain the coordinates)D (9, 0)

We are given that Max Z = 3x

_{1}+ 5x_{2}At A (0, 7)

Z = 3(0) + 5(7) = 35

At B (1, 5)

Z = 3(1) + 5(5) = 28

At C (4.5, 1.5)

Z = 3(4.5) + 5(1.5) = 21

At D (9, 0)

Z = 3(9) + 5(0) = 27

The values of objective function at corner points are finding as 35, 28, 21 and 27. But there exists infinite or countless number of points in the feasible region which is unbounded. The value of objective function will be higher than the value of these four corner positions, that is, the maximum value of the objective function takes place at a point at ∞. Therefore the given problem has unbounded solution.

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