Learning Outcomes
• Mahasiswa dapat menyelesaikan PL dengan
menggunakan metode Karmarkar
Outline Materi:
•
•
•
•
Model Karmarkar (Interior point method)
Algoritma Karmarkar
Contoh penyelesaian &diskusi
Studi kasus
Interior Point Method
x2
40
•
•
•
20
•
Starts at feasible point
Moves through interior of feasible
region
Always improves objective
function
Theoretical interest
30
40
x1
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
1
1
X   ,...., 
n
n
T
(1)
(2)
(3)
Karmarkar’s Method
•
•
•
•
Suppose the LP is in the form (1) - (3)
T
1
1
Step 1: k = 0, start with the solution point X 0   n ,..., n 
Step 2: stop if CXk < ε, else go to Step 3
Step 3:
• Define
n-1
1
r
and  
• Compute
3n
n n  1
 ADk 

Dk  diag xk1 ,..., xkn ; P  
 1 
Where
T
cp
1
1
Ynew   ,...,   r
n
cp
n
X k 1 
DkYnew
1DkYnew

c p  [I  P PP
T

T 1
P] cDk 
T
and compute
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)
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
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.
Interior Point Method
Step 1: Choose any feasible interior point solution,
and set solution index t=0.
Step 2: If any component of x (t) 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
(0)
0
x
( t 1)
  X t Pt c
(t )
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.
Interior Point Method - contd
x
Step 4: If there is no limit on feasible moves in the 
direction
( t1)
(all components are nonnegative), stop ; the given model is
unbounded. Otherwise, construct the step size

1
x
( t 1)
X
1
t
.
Interior Point Method - contd
Step 4: compute the new solution
Then let t  t  1, and return to Step 2.
x
( t 1)
 x  x
(t )
( t 1)
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.
• Maximize
Subject to
The Formulation
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.
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