Special or Unique Cases in Graphical Method

Multiple Optimal Solution

Example 1

Solve by means of graphical method

Max Z = 4x₁ + 3x₂

Subject to

4x₁+ 3x₂ ≤ 24

            x₁ ≤ 4.5

x₂ ≤  6

 x₁ ≥ 0 , x₂ ≥ 0

Answer

The first constraint 4x₁+ 3x₂ ≤ 24, can be written in the form of equation

4x₁+ 3x₂ = 24

Place x₁ =0, then x₂ = 8

Place x₂ =0, then x₁ = 6

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

The second constraint x₁ ≤ 4.5, can be written in the form of equation

x₁ = 4.5

The third constraint x₂ ≤ 6, can be written in the form of equation

x₂ = 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 4x₁+ 3x₂ = 24 and x₂ = 6 to obtain the coordinates)

C (4.5, 2) (Solve the two equations 4x₁+ 3x₂ = 24 and x₁ = 4.5 to obtain the coordinates)

D (4.5, 0)

We are given 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 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 = 3x₁ + 2x₂

Subject to

x₁+ x₂ ≤ 1

x₁+ x₂ ≥ 3

x₁ ≥ 0 , x₂ ≥ 0

Answer

The first constraint x₁+ x₂ ≤ 1, can be written in the form of equation

x₁+ x₂ = 1

Place x₁ =0, then x₂ = 1

Place x₂ =0, then x₁ = 1

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

The first constraint x₁+ x₂ ≥ 3, can be written in the form of equation

x₁+ x₂ = 3

Place x₁ =0, then x₂ = 3

Place x₂ =0, then x₁ = 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 = 3x₁ + 5x₂

Subject to

2x₁+ x₂ ≥ 7

x₁+ x₂ ≥ 6

x₁+ 3x₂ ≥ 9

x₁ ≥ 0 , x₂ ≥ 0

Answer

The first constraint 2x₁+ x₂ ≥ 7, can be written in the form of equation

2x₁+ x₂ = 7

Place x₁ =0, then x₂ = 7

Place x₂ =0, then x₁ = 3.5

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

The second constraint x₁+ x₂ ≥ 6, can be written in the form of equation

x₁+ x₂ = 6

Place x₁ =0, then x₂ = 6

Place x₂ =0, then x₁ = 6

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

The third constraint x₁+ 3x₂ ≥ 9, can be written in the form of equation

x₁+ 3x₂ = 9

Place x₁ =0, then x₂ = 3

Place x₂ =0, then x₁ = 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 2x₁+ x₂ = 7 and x₁+ x₂ = 6 to obtain the coordinates)

C (4.5, 1.5) (Solve the two equations x₁+ x₂ = 6 and x₁+ 3x₂ = 9 to obtain the coordinates)

D (9, 0)

We are given that Max Z = 3x₁ + 5x₂

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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