Operational Research
Linear Programming
With Simplex Method
Minggu 2
Part 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!
Case
x1 = number of deluxe product
x2 = number of professional product
Maximize Z = 3x1 + 4x2 (profit)
subject to
x1 + x2  1200
2x1 + 3x2  3000
x1 + 4x2  3600
with
x1 , x2  0
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 = 3x1 + 4x2 + 0S1+ 0S2 + 0S3
subject to
x1 + x 2 + S1
= 1200
2x1 + 3x2 + S2
= 3000
x1 + 4x2
+ S3 = 3600
with
x1, x2, S1, S2, S3  0
..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
c1
a11
a21
c2
a12
a22
:
:
am1
am2
...
...
...
:::
...
cn
a1n
a2n
:
amn
b1
b2
:
bm
Initial Simplex Tableau
cb
0
0
0
cj
3
4
0
0
0
BASIS
x2
1
3
4
S1
1
0
0
S2
0
1
0
S3
0
0
1
Solution
S1
S2
S3
x1
1
2
1
Zj
0
0
0
0
0
0
cj - Zj
3
4
0
0
0
1200
3000
3600
In every equation there must be a
variable that is basic in that equation
If 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
S2
S3
= 1200
= 3000
= 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
0
0
0
S1
S2
S3
1
2
1
1
3
4
1
0
0
0
1
0
0
0
1
Zj
0
0
0
0
0
cj - Zj
3
4
0
0
0
Solution
Ratios
1200 1200/1 = 1200
3000 3000/3 = 1000
3600 3600/4 = 900
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 aijth 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
cb
0
0
4
cj
3
4
0
0
0
BASIS
x2
0
0
1
S1
1
0
0
S2
0
1
0
S3
-0.25
-0.75
0.25
Solution
S1
S2
x2
x1
0.75
1.25
0.25
Zj
1
4
0
0
1
3600
cj - Zj
2
0
0
0
-1
300
300
900
..where
Z1 = (0)(0.75) + (0)(1.25) + (4)(0.25) = 1
Z2 = (0)(0) + (0)(0)
+ (4)(1)
=4
Z3 = (0)(1) + (0)(0)
+ (4)(0)
=0
Z4 = (0)(0) + (0)(1)
+ (4)(0)
=0
Z5 = (0)(-0.25)+ (0)(-0.75)+ (4)(0.25) = 1
and
the 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
0
0
4
S1
S2
x2
0.75
1.25
0.25
0
0
1
1
0
0
0
1
0
-0.25
-0.75
0.25
300
300
900
300/0.75 = 400
300/1.25 = 240
900/0.25 = 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
0
3
4
S1
x1
x2
0
1
0
0
0
1
1
0
0
-0.6
0.8
-0.2
0.2
-0.6
0.4
120
240
840
Zj
3
4
0
1.6
-0.2
4080
cj - Zj
0
0
0
-1.6
0.2
Ratios
120/0.2 = 600
----840/0.4 = 2100
Fourth Simplex Tableau
cb
0
3
4
cj
3
4
0
0
0
BASIS
x2
0
0
1
S1
5
3
-2
S2
-3
-1
1
S3
1
0
0
Solution
S1
x1
x2
x1
0
1
0
Zj
3
4
1
1
0
4200
cj - Zj
0
0
-1
-1
0
600
600
600
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 nilainilai 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