Optimization Techniques
Amrita Vishwa Vidyapeetham
October 16, 2020
Outline
Multivariable Optimization
Optimization Problem
Outline
Multivariable Optimization
Optimization Problem
STATEMENT OF AN OPTIMIZATION PROBLEM
An optimization or a mathematical programming problem can be stated as
follows.


x1
 x2 


Find X =  .  which minimizes f (X)
 .. 
xn
subject to the constraints
gj (X) ≤ 0, j = 1, 2, . . . , m
lj (X) = 0, j = 1, 2. . . . , p,
(1)
where X is an n-dimensional vector called the design vector, f (X) is termed
the objective function, and gj (X) and lj (X) are known as inequality and
equality constraints, respectively. The number of variables n and the number
of constraints m and/or p need not be related in any way. The problem stated
in Eq. (1) is called a constrained optimization problem.
Constrained Optimization-Kuhn-Tucker Condition
Some optimization problems do not involve any constraints and can be
stated as


x1
 x2 


Find X =  .  which minimizes f (X).
.
 . 
xn
Constrained Optimization-Kuhn-Tucker Condition
Some optimization problems do not involve any constraints and can be
stated as


x1
 x2 


Find X =  .  which minimizes f (X).
.
 . 
xn
Such problems are called unconstrained optimization problems.
Constrained Optimization
In this lecture the Kuhn-Tucker conditions will be discussed with examples
for a point to be a local optimum in case of a function subject to inequality
constraints.
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Kuhn-Tucker Condition
Applications
Example
A manufacturing firm producing small refrigerators has entered into a
contract to supply 50 refrigerators at the end of the first month, 50 at the end
of the second month, and 50 at the end of the third. The cost of producing x
refrigerators in any month is given by $(x2 + 1000). The firm can produce
more refrigerators in any month and carry them to a subsequent month.
However, it costs $20 per unit for any refrigerator carried over from one
month to the next. Assuming that there is no initial inventory, determine the
number of refrigerators to be produced in each month to minimize the total
cost.
Solution:
Applications
Example
A manufacturing firm producing small refrigerators has entered into a
contract to supply 50 refrigerators at the end of the first month, 50 at the end
of the second month, and 50 at the end of the third. The cost of producing x
refrigerators in any month is given by $(x2 + 1000). The firm can produce
more refrigerators in any month and carry them to a subsequent month.
However, it costs $20 per unit for any refrigerator carried over from one
month to the next. Assuming that there is no initial inventory, determine the
number of refrigerators to be produced in each month to minimize the total
cost.
Solution:
Let x1 , x2 , and x3 represent the number of refrigerators produced in the first,
second, and third month, respectively. The total cost to be minimized is
given by
Applications
Example
A manufacturing firm producing small refrigerators has entered into a
contract to supply 50 refrigerators at the end of the first month, 50 at the end
of the second month, and 50 at the end of the third. The cost of producing x
refrigerators in any month is given by $(x2 + 1000). The firm can produce
more refrigerators in any month and carry them to a subsequent month.
However, it costs $20 per unit for any refrigerator carried over from one
month to the next. Assuming that there is no initial inventory, determine the
number of refrigerators to be produced in each month to minimize the total
cost.
Solution:
Let x1 , x2 , and x3 represent the number of refrigerators produced in the first,
second, and third month, respectively. The total cost to be minimized is
given by
total cost = production cost + holding cost
Applications
f (x1 , x2 , x3 ) =(x12 + 1000) + (x22 + 1000) + (x32 + 1000) + 20(x1 − 50)
+ 20(x1 + x2 − 100)
=x12
+
x22
+
x32
+ 40x1 + 20x2
The constraints can be stated as
(2)
Applications
f (x1 , x2 , x3 ) =(x12 + 1000) + (x22 + 1000) + (x32 + 1000) + 20(x1 − 50)
+ 20(x1 + x2 − 100)
=x12
+
x22
+
x32
+ 40x1 + 20x2
The constraints can be stated as
g1 (x1 , x2 , x3 ) = x1 − 50 ≥ 0,
(2)
Applications
f (x1 , x2 , x3 ) =(x12 + 1000) + (x22 + 1000) + (x32 + 1000) + 20(x1 − 50)
+ 20(x1 + x2 − 100)
=x12
+
x22
+
x32
+ 40x1 + 20x2
The constraints can be stated as
g1 (x1 , x2 , x3 ) = x1 − 50 ≥ 0,
g2 (x1 , x2 , x3 ) = x1 + x2 − 100 ≥ 0,
(2)
Applications
f (x1 , x2 , x3 ) =(x12 + 1000) + (x22 + 1000) + (x32 + 1000) + 20(x1 − 50)
+ 20(x1 + x2 − 100)
=x12
+
x22
+
x32
(2)
+ 40x1 + 20x2
The constraints can be stated as
g1 (x1 , x2 , x3 ) = x1 − 50 ≥ 0,
g2 (x1 , x2 , x3 ) = x1 + x2 − 100 ≥ 0,
g3 (x1 , x2 , x3 ) = x1 + x2 + x3 − 150 ≥ 0.
(3)
Applications
f (x1 , x2 , x3 ) =(x12 + 1000) + (x22 + 1000) + (x32 + 1000) + 20(x1 − 50)
+ 20(x1 + x2 − 100)
=x12
+
x22
+
x32
(2)
+ 40x1 + 20x2
The constraints can be stated as
g1 (x1 , x2 , x3 ) = x1 − 50 ≥ 0,
g2 (x1 , x2 , x3 ) = x1 + x2 − 100 ≥ 0,
g3 (x1 , x2 , x3 ) = x1 + x2 + x3 − 150 ≥ 0.
Note
For this problem, we obtain the minimum X ∗ = [50, 50, 50] using the
Kuhn-Tucker conditions.
(3)
Exercise
I
Minimize
f (x1 , x2 ) = (x1 − 1)2 + (x2 − 5)2
subject to the constraints
g1 = −x12 + x2 ≤ 4,
g2 = −(x1 − 2)2 + x2 ≤ 3
using Kuhn-Tucker conditions.
(4)