101879037-Class-11

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					           Simplex Method (Two Phase Technique)
Phase-I
Add the artificial variables to secure a starting solution.
Form a new objective function that seeks the minimization of the
sum of the artificial variables subject to the same constraints of
original problem.

If the minimum value of the new objective function is
zero(meaning that all artificial variables are zero), the problem
has a feasible solution space. Go to Phase-II. Otherwise, if the
minimum is positive, the problem has no feasible solution. STOP.

Phase-II

Use the optimum basic solution of Phase-I as starting solution for
the original problem
                      8/2/12                                    1
                                                                1
    Example

Objective Function

   Maximize z = 12x1 + 15x2 + 9x3

Constraints

    8x1 + 16x2 + 12x3 ≤ 250
    4x1 + 8x2 + 10x3 ≥ 80
    7x1 + 9x2 + 8x3 = 105

   x1 , x2 , x3 ≥ 0




                      8/2/12        2
                                    2
      Artificial variables
Artificial variables be added in constraints with (=) and (≥ ) signs.

        8x1 + 16x2 + 12x3 + s1 = 250
        4x1 + 8x2 + 10x3 – s2 + R1 =80
        7x1 + 9x2 + 8x3 + R2 = 105

R1 and R2 must be zero in above equations.
We introduce another objective function Minimum w = R1 + R2
during phase-I. If w becomes zero, the original problem will have
a feasible solution space.

Now,
   R1 = -4x1 - 8x2 – 10x3 + s2 + 80
   R2 = -7x1 - 9x2 - 8x3 +105

And     Minimum w = R1 + R2
                        8/2/12                                          3
                                                                        3
                         Phase-I (Min. w)




          n
          um
          col
          Key
Basi x1 x2 x3 s1 s2 R1 R2 b b/a
c                                            K
  w 11 17 18 0 -1 0 0 185                    e
                                             y
 R1 4 8 10 0 -1 1             0 80       8   r
                                             o
          nt
          me
          Ele
          Key
 R2 7 9   8     0   0     0
                       1 105 13.1            w
                              2
 s1 x1 x2 x3 s1 s2       250
Basi 8 16 12 1 0 0R10 R2 b20.8 b/a
c
  w 3.8 2.6 0 0 0.8 -1.8 0 41

                8/2/12                           4
 x3 0.4 0.8 1    0 -0.1 0.1 0        8           4
       n
       um
       col
       Key
      Basi x1 x2 x3 s1 s2 R1 R2 b b/a
      c
        w 3.8 2.6 0 0 0.8 -1.8 0 41              K
                                                 e
                                                 y
       x3 0.4 0.8 1   0 -0.1 0.1   0 8           r
           nt
           me
           Ele
           Key
                                         Optimumo
                                         Tableau w
     R2 3.8 2.6   0 0 0.8 0.8      1 41 for
Basi x1 x2 x3     s1 s2 R1         R2   b   b/a
                                         Phase-I
c    s1 3.2 6.4   0 1 1.2 -1.2     0 154
  w 0 0 0         0  0 -2.6        -1   0

 x3               0
       0 0.52 18/2/12 - 0.01 - 3.684            5
          6          0.18 6 0.10                5
    Basi x1 x2 x3 s1           s2      b    b/a
    c
      z -12 -15 -9 0            0      0


     x3    0 0.52 1       0        - 3.684
              6                 0.18
Before starting with the second phase we have to make
                                  4
the coefficients of basic variables in the z equation
      x1 1
equal to zero. 0.68 0 0 0.21 10.79

    Basi x1 x2 x3 s1 s2    b   b/a
     s1 0 4.21 0 1 1.87 119.4
    c
                          7
      z  0   - -9 0 2.52 129.4
           6.79       6    8
                8/2/12                                  6
                                                        6
Basi x1 x2 x3 s1 s2    b   b/a
c
  z  0   - -9 0 2.52 129.4
       6.79       6    8

 x3   0 0.52 1   0   - 3.684
         6         0.18
                    4
  x1 1 0.68 0 0 0.21 10.79
Basi x1 x2 x3 s1 s2       b   b/a
c
  s1 0 4.21 0 1 1.87 119.4
   z 0    -    0 0 0.87 162.6
                         7
        2.05              2
         6
  x3 0 0.52 1 0       - 3.684
         6 8/2/12  0.18             7
                     4              7
           n
           um
           col
           Key
Phase-II


   Basi x1     x2   x3 s1    s2    b       b/a
   c                                                 K
     z  0      -  0      0 0.87 162.6                e
             2.05                 2                  y
              6                                      r
               nt                                    o
               me
               Ele
               Key
     x3    0 0.52 1     - 3.684
                         0
              6       0.18                   Optimum w
                       4                     Tableau
    x1 x1 x2        0 0.21 10.79
    Basi 1 0.68 0 x3 s1 s2      b            b/a
    c
                    1 0 0.15 177
    s1 00 4.21 03.9 1.87 119.4
      z       0
                             7
     x2    0    1    1.9     0 -0.35   7
                8/2/12                               8
                                                     8
Solution:
x1= 6
X2=7
x3= 0


            Max. z = 12x1 + 15 x2 + 9 x3
               = 12*6 + 15*7 + 9*0
               = 72 + 105 +0
                 177
               =8/2/12                     9
                                           9
                 Class work
Mr. Akbar Lashari has two farms: Mango farm and
  Wheat farm. In this drought season he will need to
  purchase 35 acre-feet additional water for mango
  farm and 30 acre-feet for wheat farm. He can buy
  from a neighboring well of Mr. Ayaz Soomro. Mr.
  Soomro says, “I shall sell water on rate of Rs.4/0 per
  acre-feet.” Transportation of water will cost to Rs.
  1/0 for mango farm and Rs. 3/0 for wheat farm.
  However Mr. Soomro can sell 50 acre-feet of water
  in total.

Mr. Lashari can also buy from another well of Mr. Abdul
  Lateef Khatri. Mr. Khatri offers Rs. 5/0 per acre-feet,
  and transportation charges from there are: Rs. 1/0
  and Rs. 3/0 to mango farm and wheat farm
  respectively. Mr. Khatri can provide 60 acre-feet of
  water in total.

Help Mr. Lashari in deciding the amount of water from
                   8/2/12
  each well and which farm he should provide that           10
                                                            10

				
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