#### 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 = 4x1 + 3x2

Subject to

4x1+ 3x2 ≤ 24

x1 ≤ 4.5

x2 ≤  6

x1 ≥ 0 , x2 ≥ 0

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

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

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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