Matakuliah
Tahun
: K0414 / Riset Operasi Bisnis dan Industri
: 2008 / 2009
Fuzzy Linear Programming
Pertemuan 7 (GSLC)
Learning Outcomes
• Mahasiswa dapat menghitung solusi model linear
programming dengan menggunakan metode Fuzzy
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Outline Materi:
•
•
•
•
Interior point Method
Karmarkar’s Method
Transform any LP to the Karmarkar’s Form
Example.
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Interior Point Method
x2
40
• Starts at feasible point
• Moves through interior of feasible
region
• Always improves objective function
• Theoretical interest
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40
x1
5
Karmarkar’s Method
• The LP form of Karmarkar’s method
minimize z = CX
subject to
AX = 0
1X = 1
X≥0
This LP must also satisfy

satisfies AX = 0
 Optimal z-value = 0
• Where X= (x1, x2, …., xn)T , A is an m x n matrix
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1
1
X   ,...., 
n
n
(1)
(2)
(3)
T
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Karmarkar’s Method
•
•
•
•
Suppose the LP is in the form (1) - (3)
T
1
1
,..., 
Step 1: k = 0, start with the solution Xpoint
0 
n
n
Step 2: stop if CXk < ε, else go to Step 3
Step 3:
• Define
• Compute
and compute

1
n-1
r
and  
3n
n n  1
 ADk 

Dk  diag xk1 ,..., xkn ; P  
 1 
T
Where
cp
1
1
Ynew   ,...,   r
n
cp
n
X k 1 
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DkYnew
1DkYnew

c p  [I  P PP
T

T 1
P] cDk 
T
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How to Transform any LP to the Karmarkar’s Form
• Step 1: set up the dual form of the LP
• Step 2: apply the dual optimal condition to form the combined
feasible region “=“ form
• Step 3: Convert the combined feasible region to the homogeneous
form: AX = 0
 Add the “ sum of all variables ≤ M” constraint
 Convert this constraint to “=“ form
 Introduce new dummy variable d2 = 1 to the system to convert the
system to AX = 0 and 1X = M + 1
• Step 4: convert the system to the form (1)-(3)
 Introduce the set of new variables xj = (M +1)xj’ to convert the
system to the form AX’ = 0 and 1X’ = 1
 Introduce new dummy variable d3’ to ensure (2) and (3)
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Examples
• Example 1: convert the following LP to the Karmarkar’s
LP
Maximize z = 3x1 + x2
Subject to 2x1 – x2 ≤ 2
x1 + 2x2 ≤ 5
x1, x2 ≥ 0
• Example 2: Perform one iteration of Karmarkar’s method
for the following LP
Minimize z = 2x1 + 2x2 – 3x3
s.t.
- x1 – 2x2 + 3x3 = 0
x1 + x2 + x3 = 1
x1, x2, x3 ≥ 0
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Computational Method
• Interior Point Methods
Barrier or interior-point methods, by contrast, visit points
within the interior of the feasible region. These methods
derive from techniques for nonlinear programming that
were developed and popularized in the 1960s by Fiacco
and McCormick, but their application to linear
programming dates back only to Karmarkar's innovative
analysis in 1984.
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Interior Point Method
Step 1: Choose any feasible interior point solution,
and set solution index t=0.
(t)
Step 2: If any component of x is 0, or if recent steps
have made no significant change in the solution value,
stop. Current point is either optimal or very nearly so.
Step 3: Construct the next move direction
x
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(0)
0
x
( t 1)
  X t Pt c
(t )
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Interior Point Method -contd
Where
 x1(t )

0

Xt 


 0
0
x2(t )
,
0
0 


0 

(t ) 
xn 
Pt  I  ATt (At ATt ) 1 A t , ct  X t c.
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Interior Point Method - contd
Step 4: If there is no limit on feasible moves in the direction
( t1) (all components are nonnegative), stop ; the
x
given model is unbounded. Otherwise, construct the step
size

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x
( t 1)
X
1
t
.
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Interior Point Method - contd
Step 4: compute the new solution
Then let t  t  1, and return to Step 2.
x
( t 1)
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 x  x
(t )
( t 1)
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A Simple Example
The Marriott Tub Company manufactures two lines of
bathtubs, called Model A and model B. Every tub requires
a certain amount of steel and zinc; the company has
available a total of 25,000 pounds of steel and 6,000
pounds of zinc. Each model A bathtub requires a total of
125 pounds of steel and 20 pounds of zinc, and each yields
a profit of $90. Each model B bathtub can be sold for a
profit of $70; it in turn requires 100 pounds of steel and 30
pounds of zinc. Find the best production mix of the
bathtubs.
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The Formulation
• Maximize
Subject to
P  90 x  70 y
125 x  100 y  25000
20 x  30 y  6000
x, y  0
Where x and y are the numbers of model A and model B
bathtubs that the company will make, respectively.
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Solving by Interior Point Method
x1
x2
x3
x4
Obj
Initial
100
100
2500
1000
16000
1st
129.4
86.4
178.8
818.8
17698
2nd
152.8
58.7
3.3
1184
17861
3 rd
157
53.6
5.7
1250
17882
4 th
189
13.60
8.3
1810
17962
5 th
199.6
0.4
6.5
1995
17992
6 th
199.7
0.3
0.1
1994
17994
7 th
199.9
0.1
0.1
1999
17998
8 th
200
0
0
2000
18000
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