Optimization using Calculus
Kuhn-Tucker
Conditions
1
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Introduction
2

Optimization with multiple decision variables and equality
constraint : Lagrange Multipliers.

Optimization with multiple decision variables and inequality
constraint : Kuhn-Tucker (KT) conditions

KT condition: Both necessary and sufficient if the objective function
is concave and each constraint is linear or each constraint function is
concave, i.e. the problems belongs to a class called the convex
programming problems.
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Kuhn Tucker Conditions: Optimization
Model
Consider the following optimization problem
Minimize f(X)
subject to
gj(X)  0
for j=1,2,…,p
where the decision variable vector
X=[x1,x2,…,xn]
3
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Kuhn Tucker Conditions
Kuhn-Tucker conditions for X* = [x1* x2* . . . xn*] to be a local minimum are
m
f
g
  j
0
xi j 1 xi
4
i  1, 2,..., n
j g j  0
j  1, 2,..., m
gj  0
j  1, 2,..., m
j  0
j  1, 2,..., m
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Kuhn Tucker Conditions …contd.

In case of minimization problems, if the constraints are
of the form gj(X)  0, then  j have to be non-positive

On the other hand, if the problem is one of maximization
with the constraints in the form gj(X) 0, then  j have to
be nonnegative.
5
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (1)
Minimize
subject to
f  x12  2 x22  3x32
g1  x1  x2  2 x3  12
g 2  x1  2 x2  3x3  8
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D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (1) …contd.
Kuhn – Tucker Conditions
g
g
f
 1 1  2 2  0
xi
xi
xi
j g j  0
gj  0
j  0
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2 x1  1  2  0
(2)
4 x2  1  22  0
(3)
6 x3  21  32  0
(4)
1 ( x1  x2  2 x3  12)  0
2 ( x1  2 x2  3x3  8)  0
(5)
(6)
x1  x2  2 x3  12  0
(7)
x1  2 x2  3x3  8  0
(8)
1  0
2  0
D Nagesh Kumar, IISc
(9)
(10)
Optimization Methods: M2L5
Example (1) …contd.
From (5) either 1 = 0 or x1  x2  2 x3  12 ,0
Case 1

From (2), (3) and (4) we have x1 = x2 =2 / 2 and x3 = 2 / 2

Using these in (6) we get 22  82  0,  2  0 or  8

From (10), 2  0 , therefore, 2 =0,

Therefore, X* = [ 0, 0, 0 ]
This solution set satisfies all of (6) to (9)
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D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (1) …contd.
Case 2:
x1  x2  2 x3  12  0
 Using (2), (3) and (4), we have 1  2  1  22  21  32  12  0
or 171  122  144
2
4
3
 But conditions (9) and (10) give us 1  0 and 2  0 simultaneously,
which cannot be possible with 171  122  144
Hence the solution set for this optimization problem is X* = [ 0 0 0 ]
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D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (2)
Minimize f  x12  x22  60 x1
subject to
g1  x1  80  0
g 2  x1  x2  120  0
10
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (2) …contd.
Kuhn – Tucker Conditions
2 x1  60  1  2  0
(11)
2 x2  2  0
(12)
1 ( x1  80)  0
2 ( x1  x2  120)  0
(13)
x1  80  0
(15)
gj  0
x1  x2  120  0
(16)
j  0
1  0
2  0
(17)
g
g
f
 1 1  2 2  0
xi
xi
xi
j g j  0
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D Nagesh Kumar, IISc
(14)
(18)
Optimization Methods: M2L5
Example (2) …contd.
From (13) either 1 = 0 or ( x1  80)  0 ,
Case 1


From (11) and (12) we have x1   2
2

 30 and x2   2
Using these in (14) we get 2  2  150  0
2
 2  0 or  150

Considering 2  0, X* = [ 30, 0]. But this solution set violates (15)
and (16)

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For 2  150 , X* = [ 45, 75]. But this solution set violates (15)
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (2) …contd.
Case 2: ( x1  80)  0
 Using x1  80 in (11) and (12), we have
2  2 x2
1  2 x2  220
(19)
 Substitute (19) in (14), we have 2 x2  x2  40   0
 For this to be true, either
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x2  0 or x2  40  0
D Nagesh Kumar, IISc
Optimization Methods: M2L5
Example (2) …contd.
 For x  0, 1  220
2
 This solution set violates (15) and (16)
 For x2  40  0,
1  140 and 2  80
 This solution set is satisfying all equations from (15) to (19) and hence
the desired
 Thus, the solution set for this optimization problem is X* = [ 80 40 ].
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D Nagesh Kumar, IISc
Optimization Methods: M2L5
Thank you
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D Nagesh Kumar, IISc
Optimization Methods: M2L5