Taylmod cff

Page 1

2. Relaxed solution: x1 = 5.71, Z = 17.13

b. Relaxed solution: x1 = 10.5, x2 = 23.7, Z = 1,473

x1 = 5, x2 = 1.67, Z = 25 Relaxed solution: x1 = 10.8, x2 = 23, s1 = 2.6

x1 = 6, s1 = 2, Z = 18

Relaxed solution: x1 = 10, x2 = 24.5, s2 = 4 1 17.13

x1 ≤ 5

2 25

x1 ≥ 6

1 1,473

x 1 ≤ 23

3 18

2 1,460

x 1 ≥ 24

3 1,460

3. a.

The rounded down solution would not be optimal. 4. a.

b.

Relaxed solution: x1 = 6, x2 = 3.2, s2 = 4.8,

Z = 2,720


1 2,720

x1 ≤ 3

x1 ≥ 4

2 2,700

3 2,400

Relaxed solution: x1 = 6, x2 = 3, s1 = 2, s2 = 6,

Z = 2,700

Relaxed solution: x1 = 5, x2 = 4, s2 = 2, s3 = 1,

Z = 2,400 5. a.

b. Node 1: x2 25 20 15 10

x1 = 6.25 x2 = 0

5

Relaxed solution: x1 = 6, x2 = 1, s1 = 8, Z = 310

Infeasible

0

5

10 15 20

25

30

x1


UB = 312.5 (x1 = 6.25, x2 = 0) LB = 300 (x1 = 6, x2 = 0) 1 312.5

x1 ≤ 6

x1 ≥ 7

Node 2: x2 25

25

20

20

15

Optimal integer solution: x1 = 6 x2 = 1

10 2 310

UB = 310 (x1 = 6, x2 = 1) LB = 310 (x1 = 6, x2 = 1)

3 ⬁

5

Infeasible Optimal solution at node 2: x1 = 6 x2 = 1 Z = 310

6. a.

Relaxed solution: x2 = 10.71, s2 = 1.28, s2 = 5,

Z = 5,785.7

Relaxed solution: x1 = 0.63, x2 = 10, s1 = 1.39, s2 = 4.37,

Z = 5,778

Node 3: x2

0

5

10 15 20

25

x1

15 Infeasible 10 5 0

5

10 15 20

25

x1


1 5,785.7

x 2 ≤ 10

2 5,778

x 2 ≥ 11

3 ⬁

Relaxed solution: x1 = 0, x2 = 10, x3 = 1, s1 = 1,

s2 = 5, Z = 5,775


Relaxed solution: x1 = 1, x2 = 9.57, s1 = 1.43, s2 = 4,

s4 = .43, Z = 5,767.8

1 5,785.7

x 2 ≥ 11

x 2 ≤ 10

2 5,778

x1 = 0

4 5,775

7. a.

3 ⬁

x1 ≥ 1

15 5,767.8

Infeasible


Relaxed solution: x 1 = 3.33, x 2 = 5, s 1 = 3.33,

Z = 366.5

Relaxed solution: x 1 = 2.5, x 2 = 6, s 2 = 0.5,

Z = 365

1 372.9

x2 ≤ 5

2 366.5

x2 ≥ 6

3 365.0


Relaxed solution: x 1 = 3, x 2 = 5, s 1 = 4, s 2 = 1,

Z = 350

Relaxed solution: x 1 = 4, x 2 = 4, s 1 = 7, s 3 = 1,

Z = 360

1 372.9

x2 ≥ 6

x2 ≤ 5

2 366.5

x1 = 3

4 350.0

3 365.0

x1 ≥ 4

15 360.0


Relaxed solution: x1 = 2, x2 = 6.25, s2 = 0.5, s3 = 0.25,

Z = 350

1 372.9

x2 ≥ 6

x2 ≤ 5

2 366.5

x1 = 3

4 350.0

3 365.0

x1 ≥ 4

5 360.0

x1 = 2

6 350.0

The rounded down solution would not be optimal.

x1 ≥ 3

7 ⬁

Infeasible


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