Presentasi sedang didownload. Silahkan tunggu

Presentasi sedang didownload. Silahkan tunggu

Operational Research Linear Programming With Simplex Method Minggu 2 Part 1.

Presentasi serupa


Presentasi berjudul: "Operational Research Linear Programming With Simplex Method Minggu 2 Part 1."— Transcript presentasi:

1 Operational Research Linear Programming With Simplex Method Minggu 2 Part 1

2 Introduction George Dantzig (1947) This is an iterative procedure that leads to the optimal solution in a finite number of steps. Begin with a basic feasible solution and then moves from one basic solution to the next until an optimal basic feasible solution is found.

3 Definisi Metode simplex adalah metode optimasi pemrograman linear dengan cara evaluasi sederetan titik-titik ekstrim sehingga nilai objektif dari suatu titik ekstrim lebih baik atau sama dengan nilai objektif suatu titik ekstrim yang dievaluasi sebelumnya

4 Contoh soal Suatu perusahaan skateboard akan memproduksi 2 jenis produk yaitu skateboard deluxe dan professional. Proses produksi terdiri dari 2 tahap yaitu proses perakitan dan proses penyesuaian. Waktu yang tersedia untuk proses perakitan adl 50 jam sedangkan utk proses penyesuaian adl 60 jam. Jumlah unit rakitan roda yang tersedia hanya 1200 unit. Setiap produk deluxe membutuhkan 1 unit rakitan roda, 2 menit proses perakitan, dan 1 menit proses penyesuaian. Setiap produk professional membutuhkan 1 unit rakitan roda, 3 menit proses perakitan, dan 4 menit proses penyesuaian. Jika dijual, keuntungan setiap produk deluxe dan professional berturut-turut adl $3 dan $4. Tentukan kombinasi produksi yang optimal!

5 Case x 1 = number of deluxe product x 2 = number of professional product Maximize Z = 3x 1 + 4x 2 (profit) subject to x 1 + x 2  x 1 + 3x 2  3000 x 1 + 4x 2  3600 with x 1, x 2  0

6 The Initial Simplex Tableau 1.Each constraint must be converted to an equation 2.In every equation there must be a variable that is basic in that equation. 3.The Right Hand Side (RHS) of every equation must be nonnegative constant.

7 Maximize Z = 3x 1 + 4x 2 + 0S 1 + 0S 2 + 0S 3 subject to x 1 + x 2 + S 1 = x 1 + 3x 2 + S 2 = 3000 x 1 + 4x 2 + S 3 = 3600 with x 1, x 2, S 1, S 2, S 3  0 Converting constraints into equations

8 ..contd c j = objective function coefficient for variable j b i = right-hand-side value for constraint i a ij = coefficient of variable j in constant i c row : a row of objective function coefficients b column : a column of RHS values of the constraint equations A matrix : a matrix with m rows and n columns of the coefficients of the variables in the constraint equations.

9 ..contd The input parameters for general linear programming model c1c1 c2c2...cncn a 11 a 21 : a m1 a 12 a 22 : a m2... : : :... a 1n a 2n : a mn b1b2:bmb1b2:bm

10 Initial Simplex Tableau cjcj cbcb BASIS x1x1 x2x2 S1S1 S2S2 S3S3 Solution S1S2S3S1S2S ZjZj c j - Z j 34000

11 In every equation there must be a variable that is basic in that equation If S 1, S 2, and S 3 are basic, x 1 and x 2 must be nonbasic. Therefore, the constraints are simply (1)(0) + (1) (0) + (1) S 1 + (0) S 2 + (0) S 3 = 1200 (2)(0) + (3) (0) + (0) S 1 + (1) S 2 + (0) S 3 = 3000 (1)(0) + (4) (0) + (0) S 1 + (0) S 2 + (1) S 3 = 3600 Or S 1 = 1200 S 2 = 3000 S 3 = 3600

12 Choosing the pivot column Rule : for maximization problem, the nonbasic variable with the largest c j -Z j value for all c j -Z j  0 is the pivot variable. The variable that is nonbasic and becomes a basic variable is often called the entering variable.

13 Choosing the pivot row If the j th the pivot column, compute all ratios b i /a ij, where a ij > 0. Select the variable basic in the row with the minimum ratio to leave the basis. The row with the minimum ratio is the pivot row

14 Initial Simplex Tableau cjcj cbcb BASIS x1x1 x2x2 S1S1 S2S2 S3S3 SolutionRatios S1S2S3S1S2S /1 = /3 = /4 = 900 ZjZj c j - Z j 34000

15 The pivot operation and the optimal solution Suppose the j th is the pivot column and the k th is the pivot row, then the element a kj is the pivot element. The pivot operation consists of m elementary row operations organized as follows: 1.Divide the pivot row (a k, b k ) by the pivot element a kj. Call the result (a k, b k ). 2.For every other row (a i, b i ), replace that row by (a, b) + (-a ij )(a k, b k ). In other words, multiply the revised pivot row by the negative of the a ij th element and add it to the row under consideration.

16 … pivot operation Step 1 : (1/4 4/4 0/4 0/4, 3600/4) Step 2* (modify row 2 by multiplying the revised pivot row by -3 adding it to row 2) ( , -2700) + ( , 3000) ( , 300)

17 Step 2* (modify row 1 by multiplying the revised pivot row by -1 adding it to row 1) ( , -900) + ( , 1200) ( , 300)

18 Second Simplex Tableau cjcj cbcb BASIS x1x1 x2x2 S1S1 S2S2 S3S3 Solution S1S2x2S1S2x ZjZj c j - Z j 2000

19 ..where Z 1 = (0)(0.75) + (0)(1.25) + (4)(0.25) = 1 Z 2 = (0)(0) + (0)(0) + (4)(1) = 4 Z 3 = (0)(1) + (0)(0) + (4)(0) = 0 Z 4 = (0)(0) + (0)(1) + (4)(0) = 0 Z 5 = (0)(-0.25)+ (0)(-0.75)+ (4)(0.25) = 1 and the objective value is (0)(300) + (0)(300) + 4(900) = 3600

20 Second Iteration Simplex Tableau cjcj cbcb BASIS x1x1 x2x2 S1S1 S2S2 S3S3 SolutionRatios S1S2x2S1S2x /0.75 = /1.25 = /0.25 = 3600 ZjZj c j - Z j 2000

21 Third Iteration Simplex Tableau cjcj cbcb BASIS x1x1 x2x2 S1S1 S2S2 S3S3 SolutionRatios S1x1x2S1x1x /0.2 = /0.4 = 2100 ZjZj c j - Z j

22 Fourth Simplex Tableau cjcj cbcb BASIS x1x1 x2x2 S1S1 S2S2 S3S3 Solution S1x1x2S1x1x ZjZj c j - Z j 00 0

23 Optimal Condition for Linear Programming The optimal solution to a linear programming problem with a maximization objective has been found when c j – Z j  0 for all variable columns in the simplex tableau.

24 Kesimpulan Solusi optimal didapatkan dengan nilai skateboard deluxe (X 1 )= 600; skateboard professional (X 2 )=600 dan keuntungan yang didapatkan adalah $4200

25 Review metode Simplex Mengubah bentuk batasan model pertidaksamaan menjadi persamaan. Membentuk tabel awal untuk solusi fisibel dasar pada titik origin dan menghitung nilai-nilai baris z j dan c j -z j Menentukan pivot column dengan cara memilih kolom yang memiliki nilai positif tertinggi pada baris c j -z j Menentukan pivot row dengan cara membagi nilai- nilai pada kolom solusi dengan nilai-nilai pada pivot column dan memilih baris dengan hasil bagi non negatif terkecil.

26 Review… Menghitung nilai pivot row yang baru menggunakan formula: nilai pivot row tabel lama dibagi dengan pivot elemen. Menghitung nilai pivot yang lain menggunakan formula: nilai baris tabel lama – (koef.pivot column yang berhubungan dikali dengan nilai pivot row yang berhubungan) Menghitung baris-baris z j dan c j -z j yang baru. Lakukan iterasi sampai nilai c j -z j adalah nol atau negatif. Diperolehlah solusi optimal.

27 Contoh soal: Selesaikan model program linear berikut ini menggunakan metode simplex! Maksimumkan Z= 4x 1 +5x 2 Constrains: x 1 +2x 2 ≤10 6x 1 +6x 2 ≤36 x 1 ≤4 X 1,x 2 ≥0

28


Download ppt "Operational Research Linear Programming With Simplex Method Minggu 2 Part 1."

Presentasi serupa


Iklan oleh Google