For this LP max 8.t.. = 4x1+xz <6 ≤10 3x1 + 2x2 6x1 + 3x2 1, 20 Betsy says that Rowo of the optimal Primal tableau is: 31 y $1 32 rhs 20 0 1 3 1 0 2 Read y = [1 ya] = cayB-¹ knowing that y should be dual optimal and that = Z, Betsy is wrong because then the true optimal value of should be y b = cgy B-¹b = Remember y = [3/1 Y2 Ym] = CByB-¹ is the dual optimal solution and that the values are readable from rowo of the optimal tableau as follows: y, is the value under 8; or the negative under e; or, if needed, the value under a after deleting Big-M. = from the optimal tableau and

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For this LP
max
8.t.
= 4x₁+xz
3x1 + 2x1
6x1 +3x2
1, 20
Betsy says that Rowo of the optimal Primal tableau is:
Z 1 y $1 s rhs
20
1
0
2
0 1
3
≤6
<10
Read y = [y1 92] = CÂYB™
knowing that y should be dual optimal and that = 2,
Betsy is wrong because then the true optimal value of should be
y b = cgyB¹b =
Remember y = [y1 Y2
· [Y₁ Y2 ・・・ Ym] = CyB¹ is the dual optimal solution and that
the values are readable from rowo of the optimal tableau as follows: y, is the
value under ; or the negative under e; or, if needed, the value under after
deleting Big-M.
=
from the optimal tableau and
Transcribed Image Text:For this LP max 8.t. = 4x₁+xz 3x1 + 2x1 6x1 +3x2 1, 20 Betsy says that Rowo of the optimal Primal tableau is: Z 1 y $1 s rhs 20 1 0 2 0 1 3 ≤6 <10 Read y = [y1 92] = CÂYB™ knowing that y should be dual optimal and that = 2, Betsy is wrong because then the true optimal value of should be y b = cgyB¹b = Remember y = [y1 Y2 · [Y₁ Y2 ・・・ Ym] = CyB¹ is the dual optimal solution and that the values are readable from rowo of the optimal tableau as follows: y, is the value under ; or the negative under e; or, if needed, the value under after deleting Big-M. = from the optimal tableau and
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