The following linear programming problem described the manufacturing two products (X, & X2) by using two raw materials (R1 & R2); Max. Z= 3X, + 4X2 Subject to: (Constraint of raw material R,) 2X1 + 3X2 s 1200 2X1 + X2s 1000 (Constraint of raw material R2) X2s 200 (Demand for product X2) X1 X2 2 0 1- By using the Algebraic Analysis, find whether a change in the right-hand side for constraint of raw material R2, (1000), and the current basis remains optimal. 2- Change this linear programming model to Dual Problem Model, and find the values of the variables and W by using the optimal simplex tableau which shown below. B.V. X1 X2 S2 S3 K X1 1 - 1/4 3/4 450 X2 0 1 1/2 - 1/2 100 S3 -1/2 1/2 1 200

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The following linear programming
problem described the manufacturing
two products (X, & X2) by using two raw
materials (R1 & R2);
Мах. Z3D 3X1 +
4X2
Subject to:
(Constraint of raw material R1)
2X1 + 3X2< 1200
2X1 + X2 s 1000
(Constraint of raw material R2)
X2s 200
(Demand for product X2)
X1 X2 z 0
1- By using the Algebraic Analysis, find
whether a change in the right-hand side
for constraint of raw material R2, (1000),
and the current basis remains optimal.
2- Change this linear programming
model to Dual Problem Model, and find
the values of the variables and W by
using the optimal simplex tableau which
shown below.
B.V. X1 X2 S1
S2
S3
K
X1
1
- 1/4 3/4
450
X2
1
1/2 - 1/2
100
S3
-1/2 1/2
1
200
Transcribed Image Text:The following linear programming problem described the manufacturing two products (X, & X2) by using two raw materials (R1 & R2); Мах. Z3D 3X1 + 4X2 Subject to: (Constraint of raw material R1) 2X1 + 3X2< 1200 2X1 + X2 s 1000 (Constraint of raw material R2) X2s 200 (Demand for product X2) X1 X2 z 0 1- By using the Algebraic Analysis, find whether a change in the right-hand side for constraint of raw material R2, (1000), and the current basis remains optimal. 2- Change this linear programming model to Dual Problem Model, and find the values of the variables and W by using the optimal simplex tableau which shown below. B.V. X1 X2 S1 S2 S3 K X1 1 - 1/4 3/4 450 X2 1 1/2 - 1/2 100 S3 -1/2 1/2 1 200
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