# Special or Unique Cases in Graphical Method

Graphical Solution Procedure - Special or Unique Cases in Graphical Method**Special or Unique Cases in Graphical Method**

**Multiple Optimal Solution**

Example 1

Solve by means of graphical method

Max Z = 4x_{1} + 3x_{2}

Subject to

4x_{1}+ 3x_{2 }≤ 24

x_{1} ≤ 4.5

x_{2} ≤ 6

x_{1 }≥ 0 , x_{2 }≥ 0

**Answer**

The first constraint 4x_{1}+ 3x_{2 }≤ 24, can be written in the form of equation

4x_{1}+ 3x_{2 }= 24

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

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

Then coordinates are (0, 8) and (6, 0)

The second constraint x_{1} ≤ 4.5, can be written in the form of equation

x_{1} = 4.5

The third constraint x_{2} ≤ 6, can be written in the form of equation

x_{2} = 6

The 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 Solution**

** **

**Example 1**

**Work out or solve graphically**

Max Z = 3x_{1} + 2x_{2}

Subject to

x_{1}+ x_{2 }≤ 1

x_{1}+ x_{2 }≥ 3

x_{1 }≥ 0 , x_{2 }≥ 0

**Answer**

The first constraint x_{1}+ x_{2 }≤ 1, can be written in the form of equation

x_{1}+ x_{2 }= 1

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

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

Thus the coordinates are (0, 1) and (1, 0)

The first constraint x_{1}+ x_{2 }≥ 3, can be written in the form of equation

x_{1}+ x_{2 }= 3

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

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

Thus 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

**Example **

Solve through graphical method

Max Z = 3x_{1} + 5x_{2}

Subject to

2x_{1}+ x_{2 }≥ 7

x_{1}+ x_{2 }≥ 6

x_{1}+ 3x_{2 }≥ 9

x_{1 }≥ 0 , x_{2 }≥ 0

**Answer**

The first constraint 2x_{1}+ x_{2 }≥ 7, can be written in the form of equation

2x_{1}+ x_{2 }= 7

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

Place x_{2} =0, then x_{1} = 3.5

Hence, the coordinates are (0, 7) and (3.5, 0)

The second constraint x_{1}+ x_{2 }≥ 6, can be written in the form of equation

x_{1}+ x_{2 }= 6

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

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

Therefore, the coordinates are (0, 6) and (6, 0)

The third constraint x_{1}+ 3x_{2 }≥ 9, can be written in the form of equation

x_{1}+ 3x_{2 }= 9

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

Place x_{2} =0, then x_{1} = 9

Hence, 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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