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101879095-Class-13

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					           Post optimal or Sensitivity Analysis

1. Changes affecting feasibility
Ø  Feasibility of simplex solution is affected in one of the
two ways.

       üan increase or decrease in the resource (change in
    right hand side)

       A new constraint is added in the model
       ü




                      8/2/12                                   1
                                                               1
Change in resource

Let total glazing time by the company is reduced from 1499 to
400 hours. Find the maximum revenue.

  Basic x1
 Optimum Tableaux2     s1   s2    s3    s4     b

      Z     0     0    0  20/      0   32/3 1266
                           3                 0
     s1     0     0    1 -2/3      0   -5/3 84


     x1     1     0    0    1/3    0   -2/3 333
  s1    1 -2/3 0 -5/3                1500    2450/3
    s3   0    0    0 -1/3          1 -1/3 17
  x1    0 1/3 0 -2/3                  400 =  -100/3
  s3    0 negative and
 x1 becomes-1/3 1 0 -1/3 z         16000/3 1150/3
                                  600
                8/2/12                                          2
    x2   0    1        0          =0    1 250                   2
Dual Simplex
 Basic   x1 x2 s1       s2      s3    s4    b

    z    0     0    0  20/      0    32/3 16000
                        3                  /3
   s1    0     0    1 -2/3      0    -5/3 2450 /
                                            3

  x1     1  0  0 1/3          0 -2/3 -100 /
 Basic   x1 x2 s1 s2          s3 s4     b
                                        3
  s3     0  0       0   -1/3 1 -1/3 1150 /
    z    16 0       0     12 0    0   4800         Optim
                                        3
   s1 -5/2 0        1    -3/2 0   0    900         um
  x2     0  1       0     0   0  1    250
                                                   Tablea
 Ratio z/x1                      16 Min.
                                                   u
   s4    -3/2 0      0
                   8/2/12-1/2   0     1     50         3
                                                       3
If the available hours for polishing is 800, then




               8/2/12                               4
                                                    4
From above tableau,

x1 = 333- 1/3 s2 + 2/3 s4 &     x2 = 250- s4

Put the values of x1 and x2 in new constraint; i.e. x1 + 2x2 + s5 =
800

   we will get

   -1/3 s2 -4/3 s4 + s5 = -33
                      8/2/12                                    5
                                                                5
Using     dual
simplex




                 8/2/12   6
                          6
8/2/12   7
         7
        Post optimal or Sensitivity Analysis

2. Changes affecting Optimality
Ø  Optimality of simplex solution is affected in one of the
two ways.

    A change in objective function
    ü




    A decision variable is added
    ü




                    8/2/12                                    8
                                                              8
Change in objective function

Suppose plain tiles is sold for Rs. 40 instead of Rs. 20. The
objective function will become:

Z = 40x1 + 24x2

     y1 , y2 , y3 , y4 =


                1 -2/3 0               -5/3
0 , 40 , 0 , 24 0 1/3 0                 -2/3 = 0,40/3,0,-8/3
                0 -1/3 1               -1/3
 Basic x1 x2 s10 s20
                0                      s3 s4
                                       1        b

       Z      0     0       0    20/   0   32/3 1266
                                  3              0
Y1    s1      0     0       1 -2/3
                           8/2/12      0   -5/3 84              9
                                                                9
Find Optimum Tableau
 Basic   x1 x2 s1         s2   s3   s4     b

   z     0   0    0  40/       0    -8/3 12660
                      3
   s1    0   0    1 -2/3       0    -5/3   84


  x1     1  0  0 1/3           0 -2/3      333
 Basic   x1 x2 s1 s2           s3 s4        b

   s3    0   0   0 -1/3 1 -1/3 17
    z    0   8/3 0 40/ 0    0 39980/
                     3          3
   x2    0   1   0  0   0  1   250
   s1    0   5/3 1 -2/3 0   0 1502/3

                 8/2/12              Non optimal solution
                                                        10
                                                       10
8/2/12   11
         11
Adding a new activity




               8/2/12   12
                        12
                      1   -2/3   0   -5/3     1 1/3
x3 constraint column= 0    1/3   0    -2/3      1 =   1/3
                      0   -1/3   1   -1/3     2     5/3
                      0    0     0       1     00
 Basic x1 x2 x3 s1 s2 s3                s4     b

    Z     0    0    -   0 20/ 0 32/3 1266
                   10/3    3          0

   s1     0    0   1/3    1 -2/3 0 -5/3 84
  Basic x1 x2 x3 s1 s2 s3                s4     b
   x1     1    0 1/3   0 1/3 0 -2/3 333
    Z     0    0  0     0 6 2 46/9 1269
                                        4
   s3     0    0 5/3 0 -1/3 1 -1/3 17
   s1     0    0  0     1 -3/5 -1/5 - 403/
                   8/2/12          10/9 5                   13
                                                            13

				
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