Operational Research Linear Programming With Simplex Method Minggu

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Operational Research Linear Programming With Simplex Method Minggu 2 Part 1

Operational Research Linear Programming With Simplex Method Minggu 2 Part 1

Introduction n n George Dantzig (1947) This is an iterative procedure that leads to

Introduction n n 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.

Definisi n Metode simplex adalah metode optimasi pemrograman linear dengan cara evaluasi sederetan titik-titik

Definisi n 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

Contoh soal n Suatu perusahaan skateboard akan memproduksi 2 jenis produk yaitu skateboard deluxe

Contoh soal n 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!

Case x 1 = number of deluxe product x 2 = number of professional

Case x 1 = number of deluxe product x 2 = number of professional product Maximize Z = 3 x 1 + 4 x 2 (profit) subject to x 1 + x 2 1200 2 x 1 + 3 x 2 3000 x 1 + 4 x 2 3600 with x 1, x 2 0

The Initial Simplex Tableau 1. Each constraint must be converted to an equation 2.

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.

Converting constraints into equations Maximize Z = 3 x 1 + 4 x 2

Converting constraints into equations Maximize Z = 3 x 1 + 4 x 2 + 0 S 1+ 0 S 2 + 0 S 3 subject to x 1 + x 2 + S 1 = 1200 2 x 1 + 3 x 2 + S 2 = 3000 x 1 + 4 x 2 + S 3 = 3600 with x 1, x 2, S 1, S 2, S 3 0

. . contd cj = objective function coefficient for variable j bi = right-hand-side

. . contd cj = objective function coefficient for variable j bi = right-hand-side value for constraint i aij = 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.

. . contd The input parameters for general linear programming model c 1 c

. . contd The input parameters for general linear programming model c 1 c 2 a 11 a 21 a 12 a 22 : : am 1 am 2 . . : : : . . . cn a 1 n a 2 n : amn b 1 b 2 : bm

Initial Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x

Initial Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x 2 S 1 S 2 S 3 Solution 0 0 0 S 1 S 2 S 3 1 2 1 1 3 4 1 0 0 0 1 1200 3000 3600 Zj 0 0 0 c j - Zj 3 4 0 0 0

In every equation there must be a variable that is basic in that equation

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) 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   S 2 S 3 = 1200 = 3000 = 3600

Choosing the pivot column n n Rule : for maximization problem, the nonbasic variable

Choosing the pivot column n n Rule : for maximization problem, the nonbasic variable with the largest cj-Zj value for all cj-Zj 0 is the pivot variable. The variable that is nonbasic and becomes a basic variable is often called the entering variable.

Choosing the pivot row n If the jth the pivot column, compute all ratios

Choosing the pivot row n If the jth the pivot column, compute all ratios bi/aij, where aij > 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

Initial Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x

Initial Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x 2 S 1 S 2 S 3 0 0 0 S 1 S 2 S 3 1 2 1 1 3 4 1 0 0 0 1 Zj 0 0 0 c j - Zj 3 4 0 0 0 Solution Ratios 1200/1 = 1200 3000/3 = 1000 3600/4 = 900 0

The pivot operation and the optimal solution n Suppose the jth is the pivot

The pivot operation and the optimal solution n Suppose the jth is the pivot column and the kth is the pivot row, then the element akj is the pivot element. The pivot operation consists of m elementary row operations organized as follows: 1. Divide the pivot row (ak , bk) by the pivot element akj. Call the result (ak , bk). 2. For every other row (ai , bi), replace that row by (a , b) + (-aij)(ak , bk). In other words, multiply the revised pivot row by the negative of the aijth element and add it to the row under consideration.

…pivot operation Step 1 : (1/4 4/4 0/4, 3600/4) n Step 2* (modify row

…pivot operation Step 1 : (1/4 4/4 0/4, 3600/4) n Step 2* (modify row 2 by multiplying the revised pivot row by -3 adding it to row 2) (-0. 75 -3 0 0 -0. 75, -2700) +( 2 3 0 1 0 , 3000) ( 1. 25 0 0 1 -0. 75, 300) n

Step 2* (modify row 1 by multiplying the revised pivot row by -1 adding

Step 2* (modify row 1 by multiplying the revised pivot row by -1 adding it to row 1) (-0. 25 -1 0 0 -0. 25, -900) +( 1 1 1 0 0 , 1200) ( 0. 75 0 1 0 -0. 25, 300) n

Second Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x

Second Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x 2 S 1 S 2 S 3 Solution 0 0 4 S 1 S 2 x 2 0. 75 1. 25 0 0 1 1 0 0 0 1 0 -0. 25 -0. 75 0. 25 300 900 Zj 1 4 0 0 1 3600 c j - Zj 2 0 0 0 -1

. . where Z 1 = (0)(0. 75) + (0)(1. 25) + (4)(0. 25)

. . where Z 1 = (0)(0. 75) + (0)(1. 25) + (4)(0. 25) = 1 Z 2 = (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) + 4(900) = 3600

Second Iteration Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1

Second Iteration Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x 2 S 1 S 2 S 3 Solution Ratios 0 0 4 S 1 S 2 x 2 0. 75 1. 25 0 0 1 1 0 0 0 1 0 -0. 25 -0. 75 0. 25 300 900 300/0. 75 = 400 300/1. 25 = 240 900/0. 25 = 3600 Zj 1 4 0 0 1 3600 c j - Zj 2 0 0 0 -1

Third Iteration Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1

Third Iteration Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x 2 S 1 S 2 S 3 Solution 0 3 4 S 1 x 2 0 1 0 0 0 1 1 0 0 -0. 6 0. 8 -0. 2 -0. 6 0. 4 120 240 840 Zj 3 4 0 1. 6 -0. 2 4080 c j - Zj 0 0 0 -1. 6 0. 2 Ratios 120/0. 2 = 600 ----840/0. 4 = 2100

Fourth Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x

Fourth Simplex Tableau cj 3 4 0 0 0 cb BASIS x 1 x 2 S 1 S 2 S 3 Solution 0 3 4 S 1 x 2 0 1 0 0 0 1 5 3 -2 -3 -1 1 1 0 0 600 600 Zj 3 4 1 1 0 4200 c j - Zj 0 0 -1 -1 0

Optimal Condition for Linear Programming n The optimal solution to a linear programming problem

Optimal Condition for Linear Programming n The optimal solution to a linear programming problem with a maximization objective has been found when cj – Zj 0 for all variable columns in the simplex tableau.

Kesimpulan n Solusi optimal didapatkan dengan nilai skateboard deluxe (X 1)= 600; skateboard professional

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

Review metode Simplex n n Mengubah bentuk batasan model pertidaksamaan menjadi persamaan. Membentuk tabel

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

Review… n n Menghitung nilai pivot row yang baru menggunakan formula: nilai pivot row

Review… n n 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 zj dan cj-zj yang baru. Lakukan iterasi sampai nilai cj-zj adalah nol atau negatif. Diperolehlah solusi optimal.

Contoh soal: n n n Selesaikan model program linear berikut ini menggunakan metode simplex!

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

Thank You…

Thank You…