AMS 550.666: Combinatorial Optimization
Homework Problems - Week V
For the following problems, A ∈ Rm×n will be m×n matrices, and b ∈ Rm . An affine subspace
is the set of solutions to a a system of linear equations, i.e., {x ∈ Rn : Ax = b} is an
affine subspace of Rn . A polytope is a bounded polyhedron.
1. Show the following:
(i) The intersection of an affine subspace with a polyhedron is a polyhedron.
Solution: Express the original polyhedron as P = {x : Ax ≤ b} and the sffine subspace
as L = {x : A0 x = b0 }. The intersection P ∩ L = {x : A0 x = b0 , Ax ≤ b} = {x : A0 x ≤
b0 , −A0 x ≤ −b0 , Ax ≤ b} which is by definition a polyhedron.
(ii) Let P1 , P2 be two polytopes. Define C = {x + y : x ∈ P1 , y ∈ P2 }. Show that C is a
polytope.
Solution: Express P1 = conv{x1 , . . . , xp } and P2 = conv{y1 , . . . , yq }. We claim that
C = conv{xi + yj : i ∈ {1, . . . , p}, j ∈ {1, .P
. . , q}}. Consider zP
∈ C, then z = x + y for
p
q
some
x
∈
P
and
y
∈
P
.
We
can
write
x
=
λ
x
and
y
=
1
2
i
i
i=1
j=1 µj yj , where λi ≥ 0,
Pp
Pq
P
λ
=
1,
µ
≥
0
and
µ
=
1.
Observe
that
λ
µj
=
1. Also,
j
i,j i
i=1 i
j=1 j
X
λi µj (xi + yj ) =
i,j
p
X
λi xi +
i=1
q
X
µj yj = x + y = z.
j=1
P
j ∈ {1, . . . , q}}, we have
Since
i,j λi µj (xi + yj ) ∈ conv{xi + yj : i ∈ {1, . . . , p},
z ∈ conv{xi + yj : i ∈ {1, . . . , p}, j ∈ {1, . . . , q}}. Therefore, C ⊆ conv{xi + yj : i ∈
{1, . . . , p}, j ∈ {1, . . . , q}}. To show the reverse inclusion,
consider z ∈ conv{xi + yj :
P
iP∈ {1, . . . , p}, j ∈ {1, . . . , q}}; so we can write z = i,j σij (xi + yj ) where σij ≥ 0 and
i,j σij = 1. We now simply this expression:
X
σij (xi +yj ) =
i,j
where λi =
X
i,j
Pq
j=1 σij
p X
q
q X
p
p
q
X
X
X
X
X
σij xi +
σij yj =
(
σij )xi + (
σij )yj =
λ i xi +
µj yj ,
i,j
and µj =
i=1 j=1
Pp
i=1 σij .
p
X
λi =
i=1
j=1 i=1
i=1
j=1
Observe that
p X
q
X
σij = 1.
i=1 j=1
Pq
Pp
Pq
Pp
Similarly,
j=1 µj = 1. Thus,
i=1 λi xi ∈ P1 and
j=1 µj yj ∈ P2 . z =
i=1 λi xi +
Pq
µ
y
∈
P
+
P
.
1
2
j=1 j j
Thus, we express C as the convex hull of a finite set of points; by the Minkowski-Weyl
theorem it is a polytope.
(iii) The intersection of two polyhedra is a polyhedron. Thus, show that the intersection of
a polyhedron with a polytope is a polytope.
Solution: Let one polyhedron be represented by A1 x ≤ b1 and the second polyhedron
be A2 x ≤ b2 . The intersection is given by A1 x ≤ b1 , A2 ≤ b2 which is another polyhedron. Since the intersection of a bounded set set with another set is also bounded, the
intersection of a polytope with another polyhedron is a polytope.
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(iv) Let T : Rn → Rd be a linear map. Show that if P ⊆ Rn is a polytope, then T (P ) is a
polytope in Rd .
Solution: We represent P as the convex hull conv{v1 , . . . , vt }. We claim that T (P ) =
conv{T v1 , . . . , T vt } because of the following sequence of equivalences.
y
⇔ y
⇔ y
⇔ y
∈ conv{T v1 , . . . , T vt }
= λ1 T v1 + . . . + λt T vt
= T (λ1 v1 + . . . + λt vt )
∈ T (P )
Thus, we express T (P ) as the convex hull of a finite set of points; by the Minkowski-Weyl
theorem it is a polytope.
2. Show that
max{cT x : Ax ≤ b, x ≥ 0} = min{y T b : y T A ≥ cT , y ≥ 0}
provided that both values are finite. [Adapt the standard LP duality result from class to
incorporate the constraints x ≥ 0 and then show that the right hand side above is equivalent
to the dual LP]
A
b
T
0
0
0
0
Solution: The primal problem is max{c x : A x ≤ b }, where A =
and b =
. By
I
0
LP duality, max{cT x : A0 x ≤ b0 } = miny0 ∈Rm+n {y 0T b0 : y 0T A0 = cT , y 0 ≥ 0}. Decompose y 0 ∈
y
m+n
0
where y ∈ Rm and s ∈ Rn . Rewriting the dual, we get miny0 ∈Rm+n {y 0T b0 :
R
as y =
s
y 0T A0 = cT , y 0 ≥ 0} = miny∈Rm ,s∈Rn {y T b+sT 0 : y T A+s = cT , y ≥ 0, s ≥ 0}. The constraint
s ≥ 0 implies that miny∈Rm ,s∈Rn {y T b + sT 0 : y T A + s = cT , y ≥ 0, s ≥ 0} = min{y T b : y T A ≥
cT , y ≥ 0}. Thus, we are done.
3. (Complementary slackness) Let x∗ ∈ Rn be an optimal solution to the problem max{cT x :
Ax ≤ b} and y ∗ ∈ Rm be an optimal solution to the problem max{y T b : y T A = cT , y ≥ 0}.
Show that for every i = 1 . . . , m either ai · x∗ = bi or yi∗ = 0 (or both). (Here ai denotes the
i-th row of A and bi is the i-th component of b.) [Hint: consider the vector (y ∗ )T (Ax∗ − b)]
The theorem is saying that in an optimal primal-dual pair of solutions for an LP, either a
constraint is tight at the optimal primal solution or the corresponding dual multiplier is 0.
Solution: We will in fact show that two feasible solutions x∗ (for the primal) and y ∗ (for
the dual) are optimal solutions if and only if for every i = 1 . . . , m either ai · x∗ = bi or
yi∗ = 0 (or both). Observe that since x∗ and y ∗ are feasible for the primal and dual problems
respectively,
cT x∗ − (y ∗ )T b = ((y ∗ )T A)x∗ − (y ∗ )T b = (y ∗ )T (Ax∗ − b) ≤ 0
Moreover, since each component of y ∗ is nonnegative and each component of Ax∗ − b is also
nonnegative, cT x∗ − (y ∗ )T b = 0 if and only if for every i = 1 . . . , m either ai · x∗ = bi or yi∗ = 0
(or both). Finally, cT x∗ − (y ∗ )T b = 0 is equivalent to x∗ and y ∗ being optimal solutions to
the primal and dual problems respectively.
4. Find an example of a pair of primal and dual linear programs such that both problems are
infeasible. [Hint: There is an example with 2 variables in each problem]
Solution:
2
Dual:
Primal:
max
subject to
max
x1
subject to −x1 + x2 ≤ 1
x2
≤ 0
−x2
≤ −1
y1 − y3
−y1
= 1
y1 + y2 − y3 = 0
y1 , y2 , y3
≥ 0
5. Let P = {x ∈ Rn : Ax ≤ b} be a nonempty polyhedron. Let C = {r ∈ Rn : x + λr ∈
P for all x ∈ P, λ ∈ R+ } where R+ is the set of nonnegative real numbers. Show that
(i) C is a convex cone. [This cone is called the recession cone of the polyhedron P ]
(ii) C = {r ∈ Rn : Ar ≤ 0}.
Solution:
(i) We need to first verify that C is convex. Consider any r1 , r2 ∈ C and any µ ∈ [0, 1] and
let r̄ = µr1 + (1 − µ)r2 . For any x ∈ P, λ ≥ 0 x + λr̄ = x + λµr1 + λ(1 − µ)r2 . Since r1 ∈ C,
x + λµr1 ∈ P and since r2 ∈ C, (x + λµr1 ) + λ(1 − µ)r2 ∈ P . A very similar argument shows
that C is a cone.
(ii) Consider any r such that Ar ≤ 0. Then A(x + λr) = Ax + λAr ≤ Ax for λ ≥ 0
and Ax ≤ b if x ∈ P . Thus, A(x + λr) ≤ b for all x ∈ P and λ ≥ 0. This shows that
{r ∈ Rn : Ar ≤ 0} ⊆ C. Consider any r such that Ar 6≤ b. This means, for some i = 1, . . . , m,
ai · r > 0. Since P is nonempty, let x ∈ P . Since ai · r > 0, there exists λ ≥ 0 such that
ai (x + λr) = ai · x + λ(ai · r) > bi . This shows that r 6∈ C. Hence C ⊆ {r ∈ Rn : Ar ≤ 0}.
6. Let P = {x ∈ Rn : Ax ≤ b} be a nonempty polyhedron. Let L = {r ∈ Rn : x + λr ∈
P for all x ∈ P, λ ∈ R}. Show that
(i) L is a linear subspace of Rn . [This is called the lineality space of the polyhedron P ]
(ii) L = {r ∈ Rn : Ar = 0}.
Solution:
(i) For any r1 , r2 ∈ L, x + λ(r1 + r2 ) = x + λr1 + λr2 . Since r1 ∈ L, x + λr1 ∈ P and
since r2 ∈ L, (x + λr1 ) + λr2 ∈ P . Thus r1 + r2 ∈ L. Similarly, for any r ∈ L and µ ∈ R,
x + λ(µr) = x + (λµ)r ∈ P for any lambda ∈ R since r ∈ L. Thus, µr ∈ L. Therefore, L is a
linear subspace.
(ii) Proof is similar to the case of the cone.
7. Using the notation from Problems 5 and 6, show that L = C ∩(−C). We say that P is pointed
if L = {0}. Suppose P 6= ∅; show that P has at least one vertex if and only if P is pointed.
Solution:
C ∩ (−C) =
=
=
=
{r ∈ Rn : Ar ≤ 0} ∩ {r ∈ Rn : A(−r) ≤ 0}
{r ∈ Rn : Ar ≤ 0, Ar ≥ 0}
{r ∈ Rn : Ar = 0}
L.
If P has a vertex z, then Az has rank n, thus A has rank n. Therefore Ar = 0 only has the
0 solution, and so L = {0}. So if P has a vertex, it is pointed.
Assume P is pointed. So Ar = 0 has only the 0 solution and so A has rank n. Let z ∈ P
since P is nonempty. If Az has rank n, then z is a vertex and we are done. Else, consider
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any r such that Az r = 0. We know that A has rank n, and thus there exists a row ai such
that ai · r 6= 0. By choosing r or −r, we can make sure ai · r > 0. Then let
=
min
{
i∈{1,...,m}
bi − ai · z
: i 6∈ Tz , ai · r > 0}
ai · r
(1)
(recall that Tz is the index set of the rows in Az .) Let z 0 = z + r. Observe that Az z 0 =
Az (z + r) = Az z = bz . Moreover, let i∗ be the index which achieves the minimum in (1).
∗
∗
i∗
∗
∗
z
)(ai · r) = bi∗ . Thus, Tz 0 ⊇ Tz ∪ {i∗ }. Since
Thus, ai · z 0 = ai · (z + r) = ai z + ( bi∗a−a
i∗ ·r
∗
ai · r > 0, rank(Az 0 ) > rank(Az ). Continuing in this manner, we can find a sequence of
points v in P with strictly increasing rank(Av ). Hence, we will at some point find a point
v ∈ P where Av has rank n. At that point, v would be a vertex of P .
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