Graphical Method

 Graphical Method:

Feasible Region:

The points lying in the common region that satisfies all the constraints simultaneously.

The common closed region thus obtained is called the Feasible Region.

Examples for Graphical Method:

Q1. Solve the following LPP by Graphical Method:

Maximum z = 3x1+4x2;

Subjected to 5x1+4x2 <= 200

                    3x1+5x2 <= 150

                    5x1+4x2 >= 100

                    8x1+4x2 >= 80

and x1,x2 >=0

Soln:

STEP I:

Let  5x1+4x2 =200 --- (1)

       3x1+5x2 =150 ---(2)

       5x1+4x2 = 100 ---(3)

       8x1+4x2 = 80  ---(4)

(1) 5x1+4x2 = 200                          (2) 3x1+5x2=150                 (3)5x1+4x2=100

Put x1=0; 4x2=200                           Put x1=0; 5x2=150                 Put x1=0; 4x2=100

        [x2=50]   (0,50)                              [x2=30]  (0,30)                       [x2=25]  (0,25)

Put x2=0; 5x1=200                           Put x2=0; 3x1=150                 Put x2=0; 5x1=100

        [x1=40] (40,0)                                [x1=50] (50,0)                        [x1=20]  (20,0)

(1) --(40,50)                                      (2) --(50,30)                             (3) --(20,25)

(4)8x1+4x2=80

Put x1=0; 4x2=80

      [x2=40]  (0,20)

Put x2=0; 8x1=80

      [x1=10]  (10,0)

(4) --(10,20)

STEP II:

STEP III:

 C is the intersection of  3x1+5x2=150 --(i)

                                     5x1+4x2=200 --(ii)

(i) x 4 => 12x1+20x2=600

(ii) x 5 =>25x1+20x2=1000

               (-)     (-)       (-)

               -----------------------

               -13x1         = -400

                       [x1= 30.8]

 3x1+5x2=150

3(30.8) + 5x2 =150

5x2=57.6

  [x2=11.52]

C(30.8,11.52)

     CORNER POINTS                   VALUE OF Z

     A(20,0)                                            60

     B(40,0)                                           120

     C(30.8,11.52)                                 (138.48)

     D(0.30)                                           120

     E(0,25)                                            100

The unique optimal solution  is given by max z= 138.48 for x1= 30.8 and x2=11.52.

Explanation:

As per the given problem , first assume all the constraints as equality form. Next "Put x1=0 and x2=0" and find out their respective points in the form of (x1,0) (0,x2) and (x1,x2).

 Develop a graphical representation by substituting (x1,x2) for all the subjected constraints.In this graph locate the points where the lines intersect with x1 and x2 .

The corner points of the feasible region is mentioned as A(x1,x2) B(x1,x2) C(x1,x2) D(x1,x2) and E(x1,x).

Here all the corner points will have their respective values except the intersecting point "C".

To find their values , identify the lines which intersect to form the point "C" and find the values of C.

Finally provide all the corner points of the feasible region and mention their corresponding optimal solution by substituting the values to the maximize equation.