Unconstrained and Constrained optimization
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Transcript Unconstrained and Constrained optimization
Fin500J: Mathematical Foundations in Finance
Topic 4: Unconstrained and Constrained
Optimization
Philip H. Dybvig
Reference: Mathematics for Economists, Carl Simon and Lawrence Blume,
Chapter 17, 18, 19
Slides designed by Yajun Wang
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Outline
Unconstrained Optimization
Functions of One Variable
o General Ideas of Optimization
o First and Second Order Conditions
o Local v.s. Global Extremum
Functions of Several Variables
o First and Second Order Conditions
o Local v.s. Global Extremum
Constrained Optimization
Kuhn-Tucker Conditions
Sensitivity Analysis
Second Order Conditions
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Unconstrained Optimization
An unconstrained optimization problem is one where
you only have to be concerned with the objective
function you are trying to optimize.
An objective function is a function that you are trying to
optimize.
None of the variables in the objective function are
constrained.
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General Ideas of Optimization
There are two ways of examining optimization.
Maximization (example: maximize profit)
In this case you are looking for the highest point on the
function.
Minimization (example: minimize cost)
In this case you are looking for the lowest point on the
function.
Maximization f(x) is equivalent to minimization –f(x)
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Graphical Representation of a Maximum
y
16
y = f(x) = -x2 + 8x
4
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x
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Questions Regarding the Maximum
What is the sign of f '(x) when x < x*?
Note: x* denotes the point where the function is at a
maximum.
What is the sign of f '(x) when x > x*?
What is f '(x) when x = x*?
Definition:
A point x* on a function is said to be a critical point if
f ' (x*) = 0.
This is the first order condition for x* to be a
maximum/minimum.
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Second Order Conditions
If x* is a critical point of function f(x), can we decide whether
it is a max, a min or neither?
Yes! Examine the second derivative of f(x) at x*, f ' '(x*);
x* is a maximum of f(x) if f ' '(x*) < 0;
x* is a minimum of f(x) if f ' '(x*) > 0;
x* can be a maximum, a minimum or neither if f ' '(x*) = 0;
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An Example of f''(x*)=0
Suppose y = f(x) = x3, then f '(x) = 3x2 and f ''(x) =6x,
This implies that x* = 0 and f ''(x*=0) = 0.
y=f(x)=x3
y
x
x*=0 is a saddle point where the point is neither a maximum nor a
minimum
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Example of Using First and Second
Order Conditions
Suppose you have the following function:
f(x) = x3 – 6x2 + 9x
Then the first order condition to find the critical
points is:
f’(x) = 3x2 - 12x + 9 = 0
This implies that the critical points are at x = 1 and x = 3.
4
2
0
-2
-4
-6
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-0.5
0
0.5
1
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3
3.5
4
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Example of Using First and Second
Order Conditions (Cont.)
The next step is to determine whether the critical points
are maximums or minimums.
These can be found by using the second order condition.
f
' '(x) = 6x – 12 = 6(x-2)
Testing x = 1 implies:
f ' '(1) = 6(1-2) = -6 < 0.
Hence at x =1, we have a maximum.
Testing x = 3 implies:
f ' '(3) = 6(3-2) = 6 > 0.
Hence at x =3, we have a minimum.
Are these the ultimate maximum and minimum of the
function f(x)?
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Local Vs. Global Maxima/Minima
A local maximum is a point that f(x*) ≥ f(x) for all x in some
open interval containing x* and a local minimum is a point
that f(x*) ≤ f(x) for all x in some open interval containing x*;
A global maximum is a point that f(x*) ≥ f(x) for all x in the
domain of f and a global minimum is a point that f(x*) ≤ f(x)
for all x in the domain of f.
For the previous example, f(x) as x and f(x) -
as x -. Neither critical point is a global max or min of f(x).
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Local Vs. Global Maxima/Minima (Cont.)
When f ''(x)≥0 for all x, i.e., f(x) is a convex function, then
the local minimum x* is the global minimum of f(x)
When f ''(x)≤0 for all x, i.e., f(x) is a concave function, then
the local maximum x* is the global maximum of f(x)
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Conditions for a Minimum or a Maximum
Value of a Function of Several Variables
Correspondingly, for a function f(x) of several
independent variables x
Calculate
f x and set it to zero.
Solve the equation set to get a solution vector x*.
Calculate 2 f x .
Evaluate it at x*.
Inspect the Hessian matrix at point x*.
Hx 2 f x
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Hessian Matrix of f(x)
f x is a C 2 functionof n variables,
2 f x
2
x1
Hx 2 f x
2 f x
xn x1
2 f x
x1xn
.
2 f x
2
xn
Since cross- partialsare equal for a C 2 function,H(x)
is a symmetricmatrix.
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Conditions for a Minimum or a Maximum Value of a
Function of Several Variables (cont.)
Let f(x) be a C2 function in Rn. Suppose that x* is a critical
point of f(x), i.e., f x * 0.
1.
If the Hessian Hx * is a positive definite matrix, then x*
is a local minimum of f(x);
If the Hessian Hx * is a negative definite matrix, then
x* is a local maximum of f(x).
If the Hessian Hx * is an indefinite matrix, then x* is
neither a local maximum nor a local minimum of f(x).
2.
3.
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Example
Find the local maxs and mins of f(x,y)
f ( x, y) x y 9 xy
3
3
Firstly, computing the first order partial derivatives (i.e.,
gradient of f(x,y)) and setting them to zero
f
2
3
x
9y
x
0
f ( x, y )
f 3 y 2 9 x
y
criticalpointsx*, y * is (0,0)and ( 3, -3 ).
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Example (Cont.)
We now compute the Hessian of f(x,y)
2 f 2 f
2
xy 6 x 9
x
2
f ( x, y ) 2
9 6 y .
2
f f
yx y 2
The first order leading principal minor is 6x and the second order
principal minor is -36xy-81.
At (0,0), these two minors are 0 and -81, respectively. Since the
second order leading principal minor is negative, (0,0) is a saddle
of f(x,y), i.e., neither a max nor a min.
At (3, -3), these two minors are 18 and 243. So, the Hessian is
positive definite and (3,-3) is a local min of f(x,y).
Is (3, -3) a global min?
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Global Maxima and Minima of a Function
of Several Variables
Let f(x) be a C2 function in Rn, then
When f(x) is a concave function, i.e., 2 f x is negative
semidefinite for all x and f x * 0 , then x* is a global
max of f(x);
When f(x) is a convex function, i.e., 2 f x is positive
semidefinite for all x and f x * 0 , then x* is a global
min of f(x);
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Example (Discriminating Monopolist)
A monopolist producing a single output has two types of customers. If it
produces q1 units for type 1, then these customers are willing to pay a price
of 50-5q1 per unit. If it produces q2 units for type 2, then these customers
are willing to pay a price of 100-10q2 per unit.
The monopolist’s cost of manufacturing q units of output is 90+20q.
In order to maximize profits, how much should the monopolist produce
for each market?
Profit is:
f (q1 , q2 ) q1 (50 5q1 ) q2 (100 10q2 ) (90 20(q1 q2 )).
T hecriticalpointsare
f
f
50 10q1 20 0 q1 3,
100 20q2 20 0 q2 4.
q1
q 2
2 f
2 f
2 f
2 f
10,
20,
0.
2
2
q1
q2
q1q2 q2 q1
2 f is negativedefinite (3,4)is theprofit- maximizingsupply plan.
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Constrained Optimization
Examples:
Individuals maximizing utility will be subject to a
budget constraint
Firms maximising output will be subject to a cost
constraint
The function we want to maximize/minimize is called
the objective function
The restriction is called the constraint
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Constrained Optimization (General Form)
A general mixed constrained multi-dimensional
maximization problem is
max f (x ) f (x 1 ,..., x n )
subject to
g1 (x 1 ,..., x n ) b1 , g 2 (x 1 ,..., x n ) b2 ,
, g k (x 1 ,..., x n ) bk ,
h1 (x 1 ,..., x n ) c 1 , h2 (x 1 ,..., x n ) c 2 ,
, hm (x 1 ,..., x n ) c m .
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Constrained Optimization (Lagrangian Form)
The Lagrangian approach is to associate a Lagrange
multiplier i with the i th inequality constraint and μi with
the i th equality constraint.
We then form the Lagrangian
L (x 1 ,..., x n , 1 ,..., k , 1 ,..., m )
k
f (x 1 ,..., x n ) i g i (x 1 ,..., x n ) bi
i 1
m
i hi (x 1 ,..., x n ) c i .
i 1
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Constrained Optimization (Kuhn-Tucker
Conditions)
If x * is a local maximum of f on the constraint set
defined by the k inequalities and m equalities, then,
there exists multipliers * , , * , * , , * satisfying
k
1
1
m
L( x* , * , * ) f ( x* ) k * gi ( x* ) m * hi ( x* )
i
i
0, j 1,
x j
x j
x j
x j
i 1
i 1
hi ( x* ) ci , i 1,
,n
,m
i* gi ( x* ) bi 0, i 1, , k
gi ( x* ) bi , i 1,
,k
i* 0, i 1, , k
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Constrained Optimization (Kuhn-Tucker
Conditions)
The first set of KT conditions generalizes the
unconstrained critical point condition
The second set of KT conditions says that x needs to
satisfy the equality constraints
The third set of KT conditions is
i* gi ( x* ) bi 0, i 1,
,k
That is to say
if i* 0 then gi (x * ) bi
if gi (x * ) bi then i* 0
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Constrained Optimization (Kuhn-Tucker
Conditions)
This can be interpreted as follows:
Additional units of the resource bi only have value if the
available units are used fully in the optimal solution, i.e., if
gi (x * ) bi the constraint is not binding thus it does not
make difference in the optimal solution and i*=0.
Finally, note that increasing bi enlarges the feasible region,
and therefore increases the objective value
Therefore, i0 for all i
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Example
max x y
2
subject to
x y 4
x 0, y 0.
2
2
Form the Lagrangian
L=x-y 2 (x 2 y 2 4) 1 x 2 y .
The first order conditions become:
L
1 2 x 1 0,
x
L
(2)
2y 2 y 2 0,
y
(1)
(3)x 2 y 2 4 0
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(4)1 x 0,
(5)2 y 0,
(6)1 0,
(7)2 0,
(8)x 0,
(8)y 0.
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Example (cont.)
By (1), 1 1 2 x , since 1 0, 1 1 0
0 and x 0, by (4), 1 0.
by (2), 2 2y (1 ), since 1 0
either both y and 2 are zero, or both
are positive, from (5) y 0, 2 0.
By (3) and (8) x=2, from (4), 1 0,
1
1
by (1), . So, (x , y , , 1 , 2 ) (2, 0, , 0, 0).
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Sensitivity Analysis
We notice that
L(x *, *, *) f (x *) * g (x *) b * '(h (x *) c ) f (x *)
What happens to the optimal solution value if the
right-hand side of constraint i is changed by a small
amount, say bi , i=1…k. or ci , i=1….m.
It changes by approximately * b or i* c i
i
i
i
*is
the shadow price of ith inequality constraint and i*
is the shadow price of ith equality constraint
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Sensitivity Analysis (Example)
• In the previous example, if we change the first
constraint to x2+y2=3.9, then we predict that the new
optimal value would be 2+1/4(-0.1)=1.975.
• If we compute that problem with this new constraint,
then x-y2= 3.9 1.9748.
• If, instead, we change the second constraint from x≥0
to x≥0.1, we do not change the solution or the
optimum value since 1* 0.
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Utility Maximization Example
T heutilityderivedfromexercise(X) and watchingmovies(M)
is described by thefunction
U X,M 100 e 2 X e M
Four hours per day are available to watch moviesand exercise.
Our Lagrangianfunctionis
L(X,M,λ) 100 e 2 X e M X M 4
First - Order Conditions:
LX 2e 2 X λ 0
LM e M λ 0
Lλ X M 4 0
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Utility Max Example Continued
First - Order Conditions:
LX 2e 2 X λ 0, (1) LM e M λ 0, (2) Lλ X M 4 0 (3)
From (2) we get thatλ e -M . Subst it uting into(1), we get
2e - 2 X e M 0 Solving (3) for M and substituting, we get
2e-2X e ( 4 X ) 0 ln(2)- 2X -4 X or 3X ln(2) 4
ln(2) 4
8 ln(2)
X*
and M *
3
3
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