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Solving a Linear Programming Problem with Mixed Constraints Operation Research Minggu 3 Part 2.

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Presentasi berjudul: "Solving a Linear Programming Problem with Mixed Constraints Operation Research Minggu 3 Part 2."— Transcript presentasi:

1 Solving a Linear Programming Problem with Mixed Constraints Operation Research Minggu 3 Part 2

2 Maximize Z = 5x 1 + x 2 subject to x 1  10 x 1 – 2x 2  3 x 1 + x 2 = 12 With x 1, x 2  0

3  Ubah pertidaksamaan menjadi persamaan  Tambahkan variabel artifisial (A)  Tambahkan koefisien –M pada fungsi tujuan (karena fungsi maximasi)  Tambahkan slack variabel (S)

4 Aturan Koefisien BatasanPenyesuaianMaximasiMinimasi ≤Tambah variabel pengurang 00 =Tambah variabel artifisial -MM ≥Kurang variabel penambah 00 Dan tambah variabel artifisial -MM

5 Maximize Z = 5x 1 + x 2 + 0S 1 + 0S 2 – MA 1 – MA 2 subject to x 1 + S 1 = 10 x 1 – 2x 2 - S 2 + A 1 = 3 x 1 + x 2 + A 2 = 12 With x 1, x 2, S 1, S 2, A 1, A 2  0

6 Initial Simplex Tableau cjcj 5100-M cbcb BASISx1x1 x2x2 S1S1 S2S2 A1A1 A1A1 Solution 0 -M S1A1A2S1A1A ZjZj -2MM0M-M -15M c j - Z j 5+2M1-M0-M00

7 Second Simplex Tableau cjcj 5100-M cbcb BASISx1x1 x2x2 S1S1 S2S2 A1A1 A1A1 Solution 0 5 -M S1x1A2S1x1A ZjZj M0-5-M5+M-M15-9M c j - Z j 011+3M05+M-5-2M0

8 Third Simplex Tableau cjcj 5100-M cbcb BASISx1x1 x2x2 S1S1 S2S2 A1A1 A1A1 Solution S1x1x2S1x1x /3 -1/3 1/3 -1/3 1/3 -1/3 -2/3 2/3 1/ ZjZj 510-4/34/311/348 c j - Z j 0004/3-M-4/3-M-11/3

9 Modified Simplex Tableau for Artificial Variable Illustration cjcj 5100 cbcb BASISx1x1 x2x2 S1S1 S2S2 Solution S1x1x2S1x1x /3 -1/3 1/ ZjZj 510-4/348 c j - Z j 0004/3

10 Optimal Simplex Tableau for Artificial Variable Illustration cjcj 5100 cbcb BASISx1x1 x2x2 S1S1 S2S2 Solution S2x1x2S2x1x ZjZj c j - Z j 00-40

11 Solving the Minimization Problem

12  The pivot column : the nonbasic variable with the largest |c j – Z j | value, for c j – Z j < 0. (negatif terbesar)  Or choose the maximum value for zj-cj  The pivot row is as same as maximizing problem.

13 Minimize Z = 1200y y y 3 subject to y 1 + 2y 2 + y 3  3 y 1 + 3y 2 + 4y 3  4 with y 1, y 2, y 3  0

14 Minimize Z = 1200y y y 3 + MA 1 + MA 2 subject to y 1 + 2y 2 + y 3 – S 1 + A 1 = 3 y 1 + 3y 2 + 4y 3 - S 2 + A 2 = 4 with y 1, y 2, y 3, S 1, S 2, A 1, A 2  0

15 Initial Simplex Tableau for Minimization Problem cjcj MM cbcb BASISy1y1 y2y2 y3y3 S1S1 S2S2 A1A1 A2A2 Solution MMMM A1A2A1A ZjZj 2M5M -M MM7M c j - Z j M M M MM00

16 Second Simplex Tableau cjcj MM cbcb BASISy1y1 y2y2 y3y3 S1S1 S2S2 A1A1 A1A1 Solution M 3000 A1y2A1y2 1/ /3 4/3 0 2/3 -1/ /3 1/3 4/3 ZjZj (1/3M+ 1000) 3000 (-5/3M +4000) -M (2/3M ) M (-2/3M +1000) 1/3M c j - Z j (-1/3M +200) 0 (5/3M - 400) M (-2/3M +1000) 0 (5/3M )

17 Third Simplex Tableau cjcj MM cbcb BASISy1y1 y2y2 y3y3 S1S1 S2S2 A1A1 A1A1 Solution S1y2S1y2 ½½½½ /2 ½ -3/2 -1/ /2 ½ 0 ½ 3/2 ZjZj c j - Z j M-1500M

18 Optimal Simplex Tableau for Minimization Problem cjcj MM cbcb BASISy1y1 y2y2 y3y3 S1S1 S2S2 A1A1 A1A1 Solution y1y2y1y ZjZj c j - Z j (M-600)

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