ALL CHAPTERS COVERED
SOLUTIONS MANUAL
, Solutions to selected exercises
These solutions are intended for course instructors and course graders only. Please
do not distribute these solutions to students, either electronically or in hard-copy
form.
These solutions have benefitted from the suggestions of many students at The
University of Texas at Austin, particularly including Seung Jun Baek, David Burn-
ham, Chia-Chih Chen, Ye Chen, Seyeong Choi, Kuldeep Dixit, Jerome Froment-
Curtil, Philippe Girolami, Sergey Gorinsky, Bing Huang, Hyun-Moo Kim, Aditya
Lele, Caleb Lo, Pandarinath Murali, Ted Junseok Song, Paul Tan, Ping Xia, Lin
Xu, Chi-Ming Yeh, and Minqi Zhong. Any remaining errors are, of course, my
responsibility alone. If you come across any errors or have any comments about
the solutions, please send me email, .
Copyright c Ross Baldick, Austin, Texas, June 2018.
iii
, 2 Problems, algorithms, and solutions 1
2 Problems, algorithms, and solutions
2.1
(i) {1},
(ii) /
0,
(iii) R.
2.2 Suppose that there is another solution x⋆⋆⋆ , say, to g(x) = 0 that is different
to x⋆ = −3 and x⋆⋆ = 1. There are three cases, depending on the relationship of
x⋆⋆⋆ to the solutions x⋆ = −3 and x⋆⋆ = 1. We consider each case in turn:
(i) x⋆⋆⋆ < −3. Then:
g(x⋆⋆⋆ ) = (x⋆⋆⋆ )2 + 2x⋆⋆⋆ − 3,
= x⋆⋆⋆ (x⋆⋆⋆ + 2) − 3,
> (−3)(−1) − 3, since x⋆⋆⋆ < −3, x⋆⋆⋆ + 2 < −1,
= 0.
(ii) −3 < x⋆⋆⋆ < 1. Then:
g(x⋆⋆⋆ ) = (x⋆⋆⋆ )2 + 2x⋆⋆⋆ − 3,
= (x⋆⋆⋆ + 1)2 − 4,
< (2)2 − 4, since |x⋆⋆⋆ + 1| < 2,
= 0.
(iii) x⋆⋆⋆ > 1. Then:
g(x⋆⋆⋆ ) = (x⋆⋆⋆ )2 + 2x⋆⋆⋆ − 3,
= x⋆⋆⋆ (x⋆⋆⋆ + 2) − 3,
> (1)(3) − 3, since x⋆⋆⋆ > 1, x⋆⋆⋆ + 2 > 3,
= 0.
In each case, g(x⋆⋆⋆ ) 6= 0, so no such solution x⋆⋆⋆ exists that is different to x⋆ = −3
and x⋆⋆ = 1.
2.3
(i) 1,
, 2 Solutions to selected exercises
(ii) {2}.
2.4 Suppose that f ≤ 1. Then:
f ≤ 1,
≤ (x − 2)2 + 1, ∀x ∈ R, since (x − 2)2 ≥ 0.
So, f is a lower bound for the problem minx∈S f (x) according to Definition 2.2.
2.5 Suppose that f ≤ f ⋆ . Then:
f ≤ f ⋆,
≤ f (x), ∀x ∈ S,
by definition of minimum. That is, f is a lower bound for minx∈S f (x) according
to Definition 2.2.
2.6
Part x⋆ x⋆⋆ x⋆⋆⋆
(i) h1 (x) ≤ 0 active? Yes Yes No
(ii) h2 (x) ≤ 0 active? Yes No No
(iii) Active set? {1, 2} {1} 0/
(iv) Strictly feasible for h1 (x) ≤ 0? No No Yes
(v) Strictly feasible for h2 (x) ≤ 0? No Yes Yes
(vi) Strictly feasible for h(x) ≤ 0? No No Yes
(vii) On boundary of {x ∈ R2 |h(x) ≤ 0}? Yes Yes No
2.7
(i) The contour set is defined by:
C f ( f˜) = x ∈ S f (x) = f˜ ,
= x ∈ S (x1 )2 + (x2 + 1)2 − 4 = f˜ ,
p
x
which is the set of points 1 on the circle of radius f˜ + 4 and center
x2
SOLUTIONS MANUAL
, Solutions to selected exercises
These solutions are intended for course instructors and course graders only. Please
do not distribute these solutions to students, either electronically or in hard-copy
form.
These solutions have benefitted from the suggestions of many students at The
University of Texas at Austin, particularly including Seung Jun Baek, David Burn-
ham, Chia-Chih Chen, Ye Chen, Seyeong Choi, Kuldeep Dixit, Jerome Froment-
Curtil, Philippe Girolami, Sergey Gorinsky, Bing Huang, Hyun-Moo Kim, Aditya
Lele, Caleb Lo, Pandarinath Murali, Ted Junseok Song, Paul Tan, Ping Xia, Lin
Xu, Chi-Ming Yeh, and Minqi Zhong. Any remaining errors are, of course, my
responsibility alone. If you come across any errors or have any comments about
the solutions, please send me email, .
Copyright c Ross Baldick, Austin, Texas, June 2018.
iii
, 2 Problems, algorithms, and solutions 1
2 Problems, algorithms, and solutions
2.1
(i) {1},
(ii) /
0,
(iii) R.
2.2 Suppose that there is another solution x⋆⋆⋆ , say, to g(x) = 0 that is different
to x⋆ = −3 and x⋆⋆ = 1. There are three cases, depending on the relationship of
x⋆⋆⋆ to the solutions x⋆ = −3 and x⋆⋆ = 1. We consider each case in turn:
(i) x⋆⋆⋆ < −3. Then:
g(x⋆⋆⋆ ) = (x⋆⋆⋆ )2 + 2x⋆⋆⋆ − 3,
= x⋆⋆⋆ (x⋆⋆⋆ + 2) − 3,
> (−3)(−1) − 3, since x⋆⋆⋆ < −3, x⋆⋆⋆ + 2 < −1,
= 0.
(ii) −3 < x⋆⋆⋆ < 1. Then:
g(x⋆⋆⋆ ) = (x⋆⋆⋆ )2 + 2x⋆⋆⋆ − 3,
= (x⋆⋆⋆ + 1)2 − 4,
< (2)2 − 4, since |x⋆⋆⋆ + 1| < 2,
= 0.
(iii) x⋆⋆⋆ > 1. Then:
g(x⋆⋆⋆ ) = (x⋆⋆⋆ )2 + 2x⋆⋆⋆ − 3,
= x⋆⋆⋆ (x⋆⋆⋆ + 2) − 3,
> (1)(3) − 3, since x⋆⋆⋆ > 1, x⋆⋆⋆ + 2 > 3,
= 0.
In each case, g(x⋆⋆⋆ ) 6= 0, so no such solution x⋆⋆⋆ exists that is different to x⋆ = −3
and x⋆⋆ = 1.
2.3
(i) 1,
, 2 Solutions to selected exercises
(ii) {2}.
2.4 Suppose that f ≤ 1. Then:
f ≤ 1,
≤ (x − 2)2 + 1, ∀x ∈ R, since (x − 2)2 ≥ 0.
So, f is a lower bound for the problem minx∈S f (x) according to Definition 2.2.
2.5 Suppose that f ≤ f ⋆ . Then:
f ≤ f ⋆,
≤ f (x), ∀x ∈ S,
by definition of minimum. That is, f is a lower bound for minx∈S f (x) according
to Definition 2.2.
2.6
Part x⋆ x⋆⋆ x⋆⋆⋆
(i) h1 (x) ≤ 0 active? Yes Yes No
(ii) h2 (x) ≤ 0 active? Yes No No
(iii) Active set? {1, 2} {1} 0/
(iv) Strictly feasible for h1 (x) ≤ 0? No No Yes
(v) Strictly feasible for h2 (x) ≤ 0? No Yes Yes
(vi) Strictly feasible for h(x) ≤ 0? No No Yes
(vii) On boundary of {x ∈ R2 |h(x) ≤ 0}? Yes Yes No
2.7
(i) The contour set is defined by:
C f ( f˜) = x ∈ S f (x) = f˜ ,
= x ∈ S (x1 )2 + (x2 + 1)2 − 4 = f˜ ,
p
x
which is the set of points 1 on the circle of radius f˜ + 4 and center
x2
