System optimization – Assignment 1
Do Dinh Truong – 2110418
a. Feasible region
b. Standard form
Minimize z,
z + 2x1 + x2 = 0
Subject to x1 + 2x2 - x3
=2
x1
+ x4
=3
x2
+ x5
=2
x1 + x 2
+ x6 = 4
x1, x2, x3, x4, x5, x6 ≥ 0
c. Simplex method
Two-Phase method:
Minimize w,
w + x1 + 2x2 – x3= 2
Subject to
z + 2x1 + x2
x1 + 2x2 - x3
=0
+ x7 = 2
x1
+ x4
x2
+ x5
x1 + x 2
+ x6
x1, x2, x3, x4, x5, x6 ≥ 0
w
z
x7
x4
x5
x6
2
0
2
3
2
4
x1
1
2
1
1
0
1
w
z
x2
x4
x5
x6
0
-1
1
3
1
3
x1
0
3/2
[1/2]
1
-1/2
1/2
z
x1
x4
x5
x6
z
x1
x3
x5
x6
z
x1
=3
=2
=4
x2
2
1
[2]
0
1
1
x3
-1
0
-1
0
0
0
x4
0
0
0
1
0
0
x5
0
0
0
0
1
0
x6
0
0
0
0
0
1
x7
0
0
1
0
0
0
x2
0
0
1
0
0
0
x3
0
1/2
-1/2
0
1/2
1/2
x4
0
0
0
1
0
0
x5
0
0
0
0
1
0
x6
0
0
0
0
0
1
x7
-1
-1/2
½
0
-1/2
-1/2
-4
2
1
2
2
x1
0
1
0
0
0
x2
-3
2
-2
1
-1
x3
2
-1
[1]
0
1
x4
0
0
1
0
0
x5
0
0
0
1
0
x6
0
0
0
0
1
-6
3
1
2
1
x1
0
1
0
0
0
x2
1
0
-2
1
[1]
x3
0
0
1
0
0
x4
-2
1
1
0
-1
x5
0
0
0
1
0
x6
0
0
0
0
1
-7
3
x1
0
1
x2
0
0
x3
0
0
x4
-1
1
x5
0
0
x6
-1
0
x3
x5
x2
3
1
1
0
0
0
0
0
1
1
0
0
-1
1
-1
0
1
0
Optimal solution: x1 = 3, x2 = 1
d. Maximize (2 + β)x1 + x2
Since we minimize (-2 – β) x1 - x2 in the standard form.
x2 x3 x5 x1
ΔcBT = [0
1
𝑎̅x4 = [−1]
1
−1
0 0 -β]
0
𝑎̅x6 = [ 2 ]
−1
1
Δpx4 = [0
1
0 0 -β] [−1] – 0 = β
1
−1
Δpx5 = [0
0
0 0 -β] [ 2 ] – 0 = -β
−1
1
px4 + Δpx4 = -1 + β ≤ 0
px5 + Δpx5 = -1 - β ≤ 0
Therefore,
-1 ≤ β ≤ 1
2
-1
1