Lecture Notes 6 Lecturer: Andrea Galeotti Non-Linear Programming ...

Lecture Notes 6
Lecturer: Andrea Galeotti
Non-Linear Programming
So fare we have studied constrained optimization problems
with equality constrained:
max
−→x
© F ¡ −→x ¢ ,s.t. g ¡ −→x ¢ = b ª
Here you solve the FOCs to find candidates
Then you check Hessian matrix at each candidate to
establish whether the candidate is indeed an optimum

Constrained Optimization with Inequality Constraints
s.t. g ¡ −→ x ¢ ≤ b
Two possible solutions:
max
−→x F ¡ −→ x ¢
xi ≥ 0, i =1, ..., n
Solution at which the constraint binds
Solution at which the constraint does not bind

BC
IC
The LHS graph describes a situation were at the optimum
the constraint binds.
The RHS graph describes a situation were at the optimum
the constraint does not bind.
Note that in the RHS there is a point were the BC
and IC are tangent. However, that is not an optimum
becauseyoucouldmovetowardstheoriginanddoing
so the constraint will still hold (will be unbind) and you
will move to a highest indifference curve.
IC
BC

Example:
y 16.25
15
13.75
12.5
11.25
10
0
1.25
2.5
LHS
3.75
x
5
max F
−→
x ≥0 ¡ −→ ¢
x
y
10
8
6
4
2
0
0.5
1
1.5
RHS
LHS: F 0 (x) =0forsomex>0, say x ∗ . This is an
Interior Solution
RHS: F 0 (x) < 0foranyx ≥ 0. The optimum here is
x ∗ =0. This is a corner solution.
2
2.5
x
3

Hence, in the first order conditions for a maximum we
must include both possibilities.
Necessary conditions: −→ x ∗ such that for any i =1,...,n
1) x ∗ i ∂F
∂xi
2) x ∗ i
3) ∂F
∂xi
≥ 0
¡ −→x ∗ ¢ =0
¡ −→x ∗ ¢ ≤ 0
Observe that:
if x ∗ i
is interior: ∂F
∂xi
If x∗ i is a corner Solution, i.e. x∗i satisfied.
How to solve:
¡ −→x ∗ ¢ =0. Thus 1,2,3 are satisfied
= 0. Thus, 1,2,3 are
We guess a solution and then we impose the FOCs
which derive from that solution. This gives us a system.
We solve it and we check that the solution is consistent
with the solution of the first order conditions. In this
case we have a candidate. Otherwise we know that our
guess cannot be an optimum.

Kuhn-Tucker Conditions for Optimality
max F (x)
x
s.t. g (x) ≤ b
x ≥ 0
Define a slack variable s = b − g (x) ≥ 0, then problem
can be rewritten as:
s.t. g (x)+s =
max F (x)
x,s
b
x ≥ 0, s ≥ 0
We define the Lagrangian:
$ (x, s) =F (x)+λ [b − g (x) − s]

Therefore:
Necessary conditions:
(i) conditions for x
x ∂$
∂x
(ii) conditionsfors
s ∂$
∂s
max $ (x, s)
x,s≥0
=0, ∂$
∂x
=0, ∂$
∂s
≤ 0, x ≥ 0
≤ 0, s ≥ 0

Note that (i) isthesameas:
x ∂$
∂x
∂$
∂x
x
=
=
≥
· ¸
∂F
x − λ∂g
∂x ∂x
∂F
− λ∂g ≤ 0
∂x ∂x
0
Note that (ii) isthesameas:
=0
s ∂$
∂s
λ
=
≥
−sλ = −λ [b − g (x)] = 0
0
s = b − g (x) ≥ 0

Thus we have two possibilities:
I) either the constraint does not bind g (x)

We really do not need to bother about the slack variable
s.
We can simply rewriting the original problem as:
And therefore:
FOCs
$ (x, s) =F (x)+λ [b − g (x)]
max
x≥0
x ∂$
∂x
λ ∂$
∂λ
= 0 and λ∂$
∂λ
slackness conditions.
x ∂$
∂x
F (x)+λ [b − g (x)]
∂$
= 0,
∂x
∂$
= 0,
∂λ
≤ 0, x ≥ 0
≥ 0, λ ≥ 0
= 0 are called complementarity

Note: in case of inequality constraints the way in which
the constraint is written matters for the necessary conditions.
To avoid any confusion you should transform each inequality
constraint in the following form:
Thus if the constraint is
function ≤ constant
h (x) ≥ b ⇐⇒ −h (x) ≤−b
And you always write the Lagrangian by inserting the
constraint in the following form:
constant-function

Again, if you must solve:
Then you write as:
And then you write:
s.t. h (x) ≥
max
x≥0
b
s.t. − h (x) ≤
max
x≥
−b
F (x)
F (x)
$ (x) =F (x)+λ [−b + h (x)]

Example:
s.t. px + qy ≤
max U (x, y)
M
x, y ≥ 0
Assume that ∂U/∂x,∂U/∂y > 0. That is an increase
in consumption leads to an increase in utility.
Write the Lagrangian:
$ (x, y, λ; M, p, q) =U (x, y)+λ [M − px − qy]
Necessary conditions
x ∂$
∂x
λ ∂$
∂λ
∂$
= 0,
∂x
∂$
= 0,
∂λ
≤ 0, x ≥ 0
≥ 0, λ ≥ 0
It is easy to see that the constraint should bind at the
candidate, i.e. λ>0.

Indeed note that
∂$
∂x
∂$
∂y
∂U
=
∂x
∂U
=
∂y
− λp =0
− λq =0
Suppose the constraint does not bind. This implies that
λ = 0 and therefore it must be the case that at the
candidate point (x ∗ ,y ∗ )
∂$
∂x
∂$
∂y
= ∂U
∂x (x∗ ,y ∗ ) ≤ 0
= ∂U
∂y (x∗ ,y ∗ ) ≤ 0
which contradicts the fact that U is monotonic in x and
y.

Just to repeat again:
Lagrangian:
Two Constraints
s.t. g1 (x, y) ≤
max F (x, y)
x,y≥0
b1, g2 (x, y) ≤ b2
$ (x, y, λ, µ) = F (x, y)+λ [b1 − g1 (x, y)] +
+µ [b2 − g1 (x, y)]
K.T. Necessary conditions:
(i) conditions for x and y :
x ∂$
∂x
y ∂$
∂y
∂$
= 0; x ≥ 0;
∂x
∂$
= 0; y ≥ 0;
∂y
(ii) conditions for the constraints:
≤ 0
≤ 0
λ [b1 − g1 (x, y)] = 0; λ ≥ 0; g1 (x, y) ≤ b1
µ [b2 − g2 (x, y)] = 0; µ ≥ 0; g2 (x, y) ≤ b2

Suppose the second constraint is a binding constraint
(with equality), i.e. g2 (x, y) =b2.
s.t. g1 (x, y) ≤ b1
g2 (x, y) = b2
max F (x, y)
x,y≥0
In this case µ can take any value because we do not
require it to be strictly or equal to zero. Thus, we
can write this constraint in the Lagrangian function no
matter the way.

Another example: A Linear Programming Problem:
max U (S, D)
S,D≥0
= 4S + D
s.t. S + D ≤ 10
S +2D ≤ 12
The first inequality constraint represents a time constraint:
An individual has at maximum 10 hours a week to
spend in leisure. Leisure activity can be diversified in
S =sailing and D =diving.

The second inequality constraint is a cash constraint:
The individual has a maximum of 12 pounds a week to
spend in leisure.
S
S+2D=12
S+D=10
What is the combination of sailing and diving which is
feasible and which maximize the utility?
D

First: write the Lagrangian in the right way:
$ (S, D, λ, µ) = 4S + D + λ [10 − S − D]+
+µ [12 − S − 2D]
Second: write down the KT necessary conditions:
S ∂$
∂S
∂$
∂S
D
=
=
S[4 − λ − µ] =0; S ≥ 0;
4−λ− µ ≤ 0
∂$
∂D
∂$
∂D
λ [10 − S − D]
=
=
=
D[1 − λ − 2µ] =0; D ≥ 0;
1−λ− 2µ ≤ 0
0; λ ≥ 0; S + D ≤ 10
µ [12 − S − 2D] = 0; µ ≥ 0; S +2D ≤ 12

Third: Solve by guessing the type of solution.
We know that either there is an interior or a corner
solution
a) Guess an interior solution, i.e. S, D > 0
Thenitmustbethecasethat:
∂$
∂S
∂$
∂D
=
=
4−λ− µ =0
1−λ− 2µ =0
Solving yields to µ = −3, which contradicts the condition
µ ≥ 0.
Thus, the solution should be a corner solution.
Yet, there are many candidates for the corner solution.
We know that one of the two constraints should not
bind.

) Guess that the cash constraint does not bind, i.e.
µ =0
b1) Guess a corner solution S =0andD>0 µ =0,
λ>0
S ∂$
∂S
∂$
∂S
D
=
=
S[4 − λ − µ] =0; S ≥ 0;
4−λ− µ ≤ 0
∂$
∂D
∂$
∂D
λ [10 − S − D]
=
=
=
D[1 − λ − 2µ] =0; D ≥ 0;
1−λ− 2µ ≤ 0
0; λ ≥ 0; S + D ≤ 10
µ [12 − S − 2D] = 0; µ ≥ 0; S +2D ≤ 12

Use your guess:
Violated....
∂$
= 4−λ≤ 0
∂S
D ∂$
= 0 ⇐⇒ λ =1;
∂D
∂$
= 1−λ≤ 0
∂D
D = 10, S +2D ≤ 12

2) Then guess a corner solution: S>0, D=0,µ=0
and λ>0. Then
∂$
∂S
∂$
∂D
λ [10 − S − D]
=
=
=
4−λ =0 ⇐⇒ λ =4
1−λ≤ 0
0 ⇐⇒
10 − S − D = 0 ⇐⇒ S =10
S ≤ 12
Nowyoucanverifythatifyoutake
(S =10,D =0,λ=4,µ=0)
all the K.T. conditions hold.
S
S+2D=12
IC
S+D=10
D

Kuhn-Tucker Sufficiency Theorem
Consider the following programming problem:
s.t. g ¡ −→ x ¢ ≤ b
max F
−→
x ≥0 ¡ −→ ¢
x
After having solve the first order conditions we obtain
a candidate. How do we know that the candidate is a
global maximum?
Theorem: If
(i) F is a concave function
(ii) g isaconvexfunction
(iii) The qualification constraint is satisfied
Then the solution to the K.T. conditions describes a
global maximum.

Note that if the constrained is a linear function, condition
(iii) holds.
In general the qualification constraint is that the Jacobian
matrix of the binding constraints has full rank.
(evaluated at the candidate point.)
Example:
s.t. g (x, y) ≤
max U (x, y)
x,y≥0
b
Suppose U is concave and the constraint is linear. Then
(i) and(ii) and(iii) holds.

Theorem: Arrow-Enthoven Sufficiency Theorem
If
s.t. g ¡ −→ x ¢ ≤ b
(a) F is a quasiconcave function
(b) g is a quasiconvex function
max F
−→
x ≥0 ¡ −→ ¢
x
(c) thequalification constraint is satisfied
(d) ∂F
∂xi 6=0foratleastonei
Then the solution to the K.T. conditions describes a
global maximum.

Take the problem we have analysed before:
Note that:
max U (S, D)
S,D≥0
= 4S + D
s.t. S + D ≤ 10
S +2D ≤ 12
U is concave, so it is quasiconcave
The two constraints are convex, so they are quasiconvex
The qualification constraint is satisfied (the constraints
are both linear)
∂U
∂S =46= 0
So the solution we have found before is a global maximum.

Equivalent to
Now you solve it.
Minimization problem
s.t. F (x) ≥
min C (x)
Q
x ≥ 0
s.t. − F (x) ≤
max −C (x)
−Q
x ≥ 0