I cant get rid of the artificial variable in this problem using two phase method
In this problem I am trying to get rid of the artificial variable using the two phase method.
However all of the rows either have negatives or zeros and my final answer keeps coming
out to be a negative and none of the other answers plug into the constraints. The problem
is
Using Simplex method minimize C:
C=15x1+12x2
Subject to
3x1+x2>18
x1+2x2<12
x1+x2=10
The numbers behind the x's are the sub numbers(x sub 1, x sub2, etc...)
The greater than and less than signs are also equal to(the 1st constraint is greater than or
equal to)
My final answer for c = 144
Please help
Optimal Solution: C= 144; x 1= 8, x2= 2
________________________________________________________________________
The above LP can be rewritten as follows.
C=15x1+12x2
3x1+x2  18
x1+2x2  12
x1+x2  10
x1+x2  10
Subject to
We use four artificial variables s1, s2, s3 and s4 to rewrite the above system as follows.
C=15x1+12x2
Subject to
Tableau #1
x1 x2 s1
3x1+x2-s1
=18
x1+2x2 +s2
=12
x1+x2
-s3
=10
x1+x2
+s4 =10
s2
s3
s4
-C
3
1
1
1
15*
1
2
1
1
12
-1
0
0
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
1
0
0
0
0
0
1
18
12
10
10
0
s1,s2,s3,s4 are basic variables; x1 will be in and s1 will be out
Tableau #2
x1
x2
1
0.333333
0
1.66667
0
0.666667
0
0.666667
0
7*
s1
-0.333333
0.333333
0.333333
0.333333
5
s2
0
1
0
0
0
s3
0
0
-1
0
0
s4
0
0
0
1
0
-C
0
0
0
0
1
6
6
4
4
-90
x1,s2,s3 s4 are basic variables; x2 will be in and s2 out
Tableau #3
x1 x2 s1
1
0 -0.4
0
1 0.2
0
0
0.2
0
0
0.2
0
0 3.6*
s2
s3
-0.2 0
0.6
0
-0.4 -1
-0.4
0
-4.2
0
s4
0
0
0
1
0
-C
0 4.8
0
3.6
0
1.6
0
1.6
1
-115.2
x1, x2, s3 s4 are basic variables; s1 will be in and s3 out
Tableau #4
x1 x2 s1
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
s2
-1
1
-2
0
3
s3
-2
1
-5
1
18
s4
0
0
0
1
0
-C
0
8
0
2
0
8
0
0
1
-144
x1, x2, s1, s4 are basic variables;
Now all coefficients 3, 18 in the objective function become positive, see the last row
which means
-C= -144+3s2+18s3
So, the optimal solution is
x1=8, x2=2, s1=8, s4=0, s2=s3=0
The optimal value is 144.