Section 2.1
Solutions Chapter 2
1.
SECTION 2.1
2.1.9
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From Prop. 2.1.2(a), if x is a local minimum, then
f (x ) (x x ) 0,
or
n
f (x )
i=1
xi
x X,
(xi xi ) 0.
If xi = i , then xi xi , xi . Letting xj = xj , for j = i, we have
f (x )
0.
xi
Similarly, i

Section 3.2
Solutions Chapter 3
SECTION 3.2
3.2.6
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Assume that the matrix
J=
2xx L(x , ) h(x )
h(x )
0
is invertible, but the suciency conditions do not hold for x and . Since x and satisfy
the rst and the second order necessary conditions of Prop. 3.2

Solutions Chapter 5
SECTION 5.1
5.1.4
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Throughout this exercise we will use the fact that strong duality holds for convex quadratic
problems with linear constraints (cf. Section 3.4).
The problem of finding the minimum distance from the origin to a lin

Lecture 2
Zhi-Quan Luo
Lecture 2: Gradient Methods
Gradient Methods - Motivation
Principal Gradient Methods
Gradient Methods - Choices of Direction
Asymptotic Convergence
Local Convergence Rate
1
Lecture 2
Zhi-Quan Luo
Motivation
Consider the minimiz

Lecture 5: Second Order Methods
Newtons Method
Convergence Rate of the Pure Form
Global Convergence; Variants of Newtons Method
Trust Region Methods
Lecture 5
Zhi-Quan Luo
Newtons Method
x
r+1
r
= x r
1
f (xr )
f (x )
2
r
assuming that the Newton dir

Solutions Chapter 4
SECTION 4.2
4.2.4
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Problem correction: Assume that Q is symmetric and invertible. (This correction has been made
in the 2nd printing.)
Solution:
We have
1
x Qx
2
subject to Ax = b.
minimize f (x) =
Since x is an optimal solution of

Lecture 6: Optimization over a Convex Set
Optimality conditions
Projection theorem
Feasible direction methods
Conditional gradient method
Gradient projection methods
Lecture 6
Zhi-Quan Luo
Optimality Conditions
minimize f (x)
subject to x X,
where f

Nonlinear Programming
2nd Edition
Solutions Manual
Dimitri P. Bertsekas
Massachusetts Institute of Technology
Athena Scientic, Belmont, Massachusetts
1
NOTE
This solutions manual is continuously updated and improved. Portions of the manual, involving
prim

Section 6.3
Solutions Chapter 6
SECTION 6.3
6.3.1
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(a) Let be a dual optimal solution. Similar to the proof of Prop. 6.3.1, we obtain
|k+1 |2 |k |2 2sk q q(k ) + (sk )2 |g k |2 ,
where q = q( ). Since sk =
q q(k )
,
|g k |2
we have
q(k ) 2
q
|k+1 |2 |