Vector Worksheets With Answers - Math 125, Exam 3 Page 17
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19Now we need to use row operations to reinsert the standard basis vectors e
and e
back into
2
3
the columns were they were located in the last simplex table from Phase I – the x
-column
3
and x
-column respectively. This yields the simplex table
1
1
0
5/2 0
9/2
3/2
.
0
0
1/2 1
1/2
1/2
0
1
0
0
1
3
Now we being the minimum simplex algorithm. The only variable column starting with a
positive number is the column corresponding to the x
variable. There is only one positive
4
choice for pivoting in this column. So, we pivot on the 1. This yields the simplex table
1
9/2
5/2 0 0
15
.
0
1/2
1/2 1 0
2
0
1
0
0 1
3
Since there are no more x-variable columns starting with a negative number, we are done
with the minimum simplex algorithm. This results in the solution: the minimum z value is
15 attained at the basic feasible solution x
= 0, x
= 0, x
= 2, and x
= 3.
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