Problems in The Simplex Method

Minggu 3Part 3

Solusi yang tidak fisibel (infeasible solution)

Masalah yang tidak terbatas (The unbounded linear programming)

Solusi optimal majemuk (The alternate optimal solution)

Pivot row yang seri (degeneracy)

The infeasible linear programming problem No solution that satisfies the constraints

and non-negativity conditions for the problem.

Maximize Z = 2x1 + 4x2

subject to2.5x1 + 3x2 300 5x1 + 2x2 400 2x2 150 x1 60

x2 60 with x1 , x2 0

Initial Simplex Tableau for Infeasibility Problemcj 2 4 0 0 0 0 -M 0 -M

cb BASIS x1 x2 S1 S2 S3 S4 A1 S5 A2 Solution

000

-M-M

S1

S2

S3

A1

A2

2.55010

32201

10000

01000

00100

000-10

00010

0000-1

00001

3004001506060

Zj -M -M 0 0 0 M -M M -M -120M

cj - Zj 2+M 4+M 0 0 0 -M 0 -M 0

Second Simplex Tableau for Infeasibility Problemcj 2 4 0 0 0 0 -M 0 -M

cb BASIS x1 x2 S1 S2 S3 S4 A1 S5 A2 Solution

000

-M4

S1

S2

S3

A1

x2

2.55010

00001

10000

01000

00100

000-10

00010

3220-1

-3-2-201

120280306060

Zj -M -M 0 0 0 M -M -4 4 240-6M

cj - Zj 2+M M 0 0 0 -M 0 4 -M-4

Third Simplex Tableau for Infeasibility Problemcj 2 4 0 0 0 0 -M 0 -M

cb BASIS x1 x2 S1 S2 S3 S4 A1 S5 A2Solutio

n

200

-M4

x1

S2

S3

A1

x2

10000

00001

0.4-20

-0.40

01000

00100

000-10

00010

1.2-42

-1.2-1

-1.24-21.21

4840301260

Zj 2 4(0.8 +

0.4M)

0 0 M -M (-1.6 + 1.2M)

(1.6 - 1.2M

336 -12M

cj - Zj 0 0(-0.8 -

0.4M)

0 0 -M 0 (1.6 - 1.2M

(-1.6 + 0.2M)

Final Simplex Tableau for Infeasibility Problemcj 2 4 0 0 0 0 -M 0 -M

cb BASIS x1 x2 S1 S2 S3 S4 A1 S5 A2 Solution

2-M0

-M4

x1

A2

S3

A1

x2

10000

00001

-0.2-0.5

10.20.5

0.30.250.5-0.3

-0.25

00100

000-10

00010

1.2-42

-1.2-1

-1.24-21.21

120280306060

Zj 2 4(1.6 +

0.3M)

(-0.4 + 0.05M) 0 M -M M -M 320 –

10M

cj - Zj 0 0(-1.6 -

0.3M)

(0.4 - 0.05M) 0 -M 0 -M 0

Infeasible LP Ketidaklayakan solusi Kesalahan memformulasi program

linear

The unbounded linear programming If the objective function can be

made infinitely large without violating any of the constraints.

Maximize Z = 2x1 + 3x2

subject to x1 - x2 2

-3x1 + x2 4

with x1 , x2 0

Initial Simplex Tableau for Unbounded Illustrationcj 2 3 0 0

cb BASIS x1 x2 S1 S2 Solution

00

S1

S2

1-3

-11

10

01

24

Zj 0 0 0 0 0

cj - Zj 2 3 0 0

Second Simplex Tableau for Unbounded Illustration

cj 3 4 0 0

cb BASIS x1 x2 S1 S2 Solution

03

S1

x2

-2-3

01

10

01

64

Zj -9 0 0 0 12

cj - Zj 11 0 0 -3

Suatu masalah yang tidak terbatas diidentifikasikan dalam simplex pada saat pemilihan pivot row tidak mungkin dilakukan saat nilai pivot row negatif atau tak terhingga

The alternate optimal solution If two or more solutions

yield the optimal objective value.

Maximize Z = 10/3x1 + 4x2

subject to 2.5x1 + 3x2 300

5x1 + 2x2 400

2x2 150

with x1 , x2 0

Initial Simplex Tableau for Alternate Optimal Solution cj 10/3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3Solutio

n

000

S1

S2

S3

2.550

322

100

010

001

300400150

Zj 0 0 0 0 0 0cj - Zj 10/3 4 0 0 0

Second Simplex Tableau for Alternate Optimal Solution cj 10/3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3Solutio

n

004

S1

S2

x2

2.550

001

100

010

-1.5-10.5

7525075

Zj 0 4 0 0 2 0cj - Zj 10/3 0 0 0 -2

Fourth Simplex Tableau for Alternate Optimal Solution cj 10/3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3Solutio

n

10/304

x1

S3

x2

100

001

-0.2-10.5

0.30.5

-0.25

010

605050

Zj 10/3 4 4/3 0 0 400cj - Zj 0 0 -4/3 0 0

Third Simplex Tableau for Alternate Optimal Solution cj 10/3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3Solutio

n

10/304

x1

S2

x2

100

001

0.4-20

010

-0.62

0.5

3010075

Zj 10/3 4 4/3 0 0 400cj - Zj 0 0 -4/3 0 0

Jika melakukan iterasi lanjut pada tabel simplex yang sudah memenuhi syarat optimal

Diindikasikan oleh nilai 0 pada baris cj-zj untuk variabel bukan dasar

Memberi keleluasaan pada perusahaan untuk memilih kombinasi produk

The concept of degeneracy in linear programming If one or more of the basic variables has

a value of zero. Occurs whenever two rows satisfy the criterion of selection as pivot row.

Maximize Z = 4x1 + 3x2

subject to x1 - x2 2

2x1 + x3 4 x1 + x2 + x3 3

with x1 , x2 , x3 0

Initial Simplex Tableau for Degeneracy Illustration cj 4 0 3 0 0 0

cb BASIS x1 x2 x3 S1 S2 S3 Solution Ratio

000

S1

S2

S3

121

-101

011

100

010

001

243

2/1 = 24/2 = 23/1 = 3

Zj 0 0 0 0 0 0 0cj - Zj 4 0 3 0 0 0

Second Simplex Tableau for Degeneracy Illustration cj 4 0 3 0 0 0

cb BASIS x1 x2 x3 S1 S2 S3 Solution Ratio

400

x1

S2

S3

100

-122

011

1-2-1

010

001

201

-0/21/2

Zj 4 -4 0 4 0 0 8cj - Zj 0 4 3 -4 0 0

Third Simplex Tableau for Degeneracy Illustration cj 4 0 3 0 0 0

cb BASIS x1 x2 x3 S1 S2 S3 Solution Ratio

400

x1

x2

S3

100

010

0.50.50

0-11

0.50.5-1

001

201

2/0.50/0.5

-

Zj 4 0 2 0 2 0 8cj - Zj 0 0 1 0 -2 0

Fourth Simplex Tableau for Degeneracy Illustration cj 4 0 3 0 0 0

cb BASIS x1 x2 x3 S1 S2 S3 Solution Ratio

430

x1

x3

S3

100

-120

010

1-21

01-1

001

201

2/1-

1/1

Zj 4 2 3 -2 3 0 8cj - Zj 0 -2 0 2 -3 0

Fifth Simplex Tableau for Degeneracy Illustration cj 4 0 3 0 0 0

cb BASIS x1 x2 x3 S1 S2 S3Solution

Ratio

430

x1

x3

S1

100

-120

010

001

11-1

-121

121

Zj 4 2 3 0 1 2 10cj - Zj 0 -2 0 0 -1 -2

Degenerasi muncul ketika terdapat pivot row yang seri

Pilih salah satu pivot row secara acak dan lakukan iterasi lanjut secara normal