problems in the simplex method
DESCRIPTION
Problems in The Simplex Method. Minggu 3 Part 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). - PowerPoint PPT PresentationTRANSCRIPT
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Problems in The Simplex Method
Minggu 3Part 3
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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)
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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
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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
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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
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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)
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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
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Infeasible LP Ketidaklayakan solusi Kesalahan memformulasi program
linear
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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
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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
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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
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Suatu masalah yang tidak terbatas diidentifikasikan dalam simplex pada saat pemilihan pivot row tidak mungkin dilakukan saat nilai pivot row negatif atau tak terhingga
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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
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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
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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
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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
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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
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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
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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
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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
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Degenerasi muncul ketika terdapat pivot row yang seri
Pilih salah satu pivot row secara acak dan lakukan iterasi lanjut secara normal