Vector Worksheets With Answers - Math 125, Exam 3 Page 9
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19The first step is now to place the standard basis vectors e
and e
in the columns corre-
2
3
sponding to y
and y
respectively. That is, pivot about the positive 1’s in the y
and y
1
2
1
2
columns. This yields the simplex table:
1
9
7
1 1
0 0
27
.
0
6
9
1 0
1 0
22
0
3
2
0
1
0 1
5
Now we begin the minimum simplex algorithm. Choosing the x-variable column with the
most positive entry, we choose the x
-column since it starts with a 9. We determine to pivot
1
5
22
on the “3” since the quotient
<
. This yields the next simplex table:
3
6
1
0 13
1
2
0
3
12
.
0
0 13
1
2
1
2
12
2
1
1
5
0
1
0
0
3
3
3
3
Continuing the minimum simplex algorithm, we choose the x
-column corresponding to the
2
leading entry 13. There is only one viable candidate for pivot in this columns – the 13 in
row 2. Pivoting here yields the final table in Phase I:
1
0 0
0
0
1
1
0
1
2
1
2
12
.
0
0 1
13
13
13
13
13
2
3
2
3
89
0
1 0
39
13
39
13
39
(c) (2 pts) Are we now able to move to Phase II of the generalized simplex method?
Clearly explain your answer.
Solution: Yes. The minimum simplex algorithm is done (since there are no positive entries
to begin any of the dependent variable columns) and the values of y
and y
corresponding
1
2
to the basic feasible solution are zero. (Remark: Saying that the standard basis vectors are
no longer in the columns corresponding to the y-columns is also correct.)
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