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Consider the following LP problem: Maximize Z = 2 x 1 - x 2 + x 3 s.t. 3 x 1 - 2

ID: 3027912 • Letter: C

Question

Consider the following LP problem:

            Maximize     Z = 2x1 - x2 + x3

            s.t.                       3x1   -   2x2 + 2x3   15

                                         -x1   +   x2   + x3     3

                                          x1   -    x2   + x3    4

                                   

                                                            xi 0, i = 1,2,3.

If we let x4 , x5 , and x6   to be the slack variables for the respective constraints, the simplex method yields the following final set of equations:

1. Conduct sensitivity analysis for the RHS, i.e., provide ranges for each bi , i = 1,2,3, in which the current solution remains optimal.

2. Find the solution for the problem if the RHS changes to:
      b = ( 10

4

2)

(one column matrix)

(0) Z +2x3 +x4 +xs =18 x2 + 5x3 +x4 + 3x5 =24 2x3 +x5 +x6= 7 (3) x1+4x3 +x4 + 2x5 =21. 18 5.5 ss 4.4 xx 3.3 0123

Explanation / Answer

1. solve by Simplex method

represent by tabular form:

Table 1
--------------------------------------------------------------
x1      x2    x3    s1      s2    s3    z           
-------------------------------------------------------------
3      -2     2      1      0      0      0      15   
-1     1      1      0      1      0      0      3    
1      -1     1      0      0      1      0      4    
-2     1      -1     0      0      0      1      0    

Table 2
--------------------------------------------------------------
x1      x2    x3    s1      s2    s3    z           
-------------------------------------------------------------
0      1      -1     1      0      -3     0      3    
0      0      2      0      1      1      0      7    
1      -1     1      0      0      1      0      4    
0      -1     1      0      0      2      1      8    

Table 3
--------------------------------------------------------------
x1      x2    x3    s1      s2    s3    z           
-------------------------------------------------------------
0      1      -1     1      0      -3     0      3    
0      0      2      0      1      1      0      7    
1      0      0      1      0      -2     0      7    
0      0      0      1      0      -1     1      11   

Table 4
--------------------------------------------------------------
x1      x2    x3    s1      s2    s3    z           
-------------------------------------------------------------
0      1      5      1      3      0      0      24   
0      0      2      0      1      1      0      7    
1      0      4      1      2      0      0      21   
0      0      2      1      1      0      1      18    

then the solution : z = 18; x1 = 21, x2 = 24, x3 = 0

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