Operational ResearchLinear Programming With Simplex Method

Minggu 2Part 1

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.

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

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!

Casex1 = number of deluxe productx2 = number of professional product

Maximize Z = 3x1 + 4x2 (profit)subject to x1 + x2 12002x1 + 3x2 3000 x1 + 4x2 3600

with x1, x2 0

The Initial Simplex Tableau1. Each constraint must be

converted to an equation2. 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.

Maximize Z = 3x1 + 4x2 + 0S1+ 0S2 + 0S3 subject to x1 + x2 + S1 = 12002x1 + 3x2 + S2 = 3000 x1 + 4x2 + S3 = 3600with x1, x2, S1, S2, S3 0

Converting constraints into equations

..contdcj = objective function coefficient for variable jbi = right-hand-side value for constraint iaij = coefficient of variable j in constant i

c row : a row of objective function coefficientsb column : a column of RHS values of the constraint 　

　　 equationsA 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

c1 c2 . . . cn

a11

a21

:

am1

a12

a22

:

am2

. . .

. . .: : :. . .

a1n

a2n

:

amn

b1

b2

:bm

Initial Simplex Tableau

cj 3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3 Solution

000

S1

S2

S3

121

134

100

010

001

120030003600

Zj 0 0 0 0 0 0

cj - Zj 3 4 0 0 0

In every equation there must be a variable that is basic in that equationIf S1, S2, and S3 are basic, x1 and x2 must be nonbasic. Therefore, the constraints are simply (1)(0) + (1) (0) + (1) S1 + (0) S2 + (0) S3 = 1200

(2)(0) + (3) (0) + (0) S1 + (1) S2 + (0) S3 = 3000 (1)(0) + (4) (0) + (0) S1 + (0) S2 + (1) S3 = 3600

Or

S1 = 1200 　 S2 = 3000

S3 = 3600

Choosing the pivot column 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 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 x1 x2 S1 S2 S3 Solution Ratios

000

S1

S2

S3

121

134

100

010

001

120030003600

1200/1 = 12003000/3 = 10003600/4 = 900

Zj 0 0 0 0 0 0

cj - Zj 3 4 0 0 0

The pivot operation and the optimal solution 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 aij

th element and add it to the row under consideration.

…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) (-0.75 -3 0 0 -0.75, -2700)

+ ( 2 3 0 1 0 , 3000) ( 1.25 0 0 1 -0.75, 300)

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)

Second Simplex Tableau

cj 3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3 Solution

004

S1

S2

x2

0.751.250.25

001

100

010

-0.25-0.750.25

300300900

Zj 1 4 0 0 1 3600

cj - Zj 2 0 0 0 -1

..whereZ1 = (0)(0.75) + (0)(1.25) + (4)(0.25) = 1Z2 = (0)(0) + (0)(0) + (4)(1) = 4Z3 = (0)(1) + (0)(0) + (4)(0) = 0Z4 = (0)(0) + (0)(1) + (4)(0) = 0Z5 = (0)(-0.25)+ (0)(-0.75)+ (4)(0.25) = 1andthe objective value is (0)(300) + (0)(300) + 4(900) = 3600

Second Iteration Simplex Tableau

cj 3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3 Solution Ratios

004

S1

S2

x2

0.751.250.25

001

100

010

-0.25-0.750.25

300300900

300/0.75 = 400300/1.25 = 240900/0.25 = 3600

Zj 1 4 0 0 1 3600

cj - Zj 2 0 0 0 -1

Third Iteration Simplex Tableau

cj 3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3 Solution Ratios

034

S1

x1

x2

010

001

100

-0.60.8-0.2

0.2-0.60.4

120240840

120/0.2 = 600-----

840/0.4 = 2100

Zj 3 4 0 1.6 -0.2 4080

cj - Zj 0 0 0 -1.6 0.2

Fourth Simplex Tableau

cj 3 4 0 0 0

cb BASIS x1 x2 S1 S2 S3 Solution

034

S1

x1

x2

010

001

53-2

-3-11

100

600600600

Zj 3 4 1 1 0 4200

cj - Zj 0 0 -1 -1 0

Optimal Condition for Linear Programming 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 Solusi optimal didapatkan dengan

nilai skateboard deluxe (X1)= 600; skateboard professional (X2)=600 dan keuntungan yang didapatkan adalah $4200

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 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-nilai pada kolom solusi dengan nilai-nilai pada pivot column dan memilih baris dengan hasil bagi non negatif terkecil.

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 zj dan cj-zj yang baru. Lakukan iterasi sampai nilai cj-zj adalah nol

atau negatif. Diperolehlah solusi optimal.

Contoh soal: Selesaikan model program linear berikut

ini menggunakan metode simplex! Maksimumkan Z= 4x1+5x2 Constrains:

x1+2x2 ≤10 6x1+6x2 ≤36 x1 ≤4 X1,x2 ≥0

Thank You…