Lecture 4
Two-stage stochastic linear programs
January 24, 2018
Uday V. Shanbhag
Lecture 4
Introduction
Consider the two-stage stochastic linear program defined as
SLP
cT x + Q(x)
minimize
x
Ax = b,
subject to
x ≥ 0,
where Q(x) is defined as follows:
Q(x) , E[Q(x, ξ)].
If ξ(ω) , (T (ω), W (ω), q(ω), b(ω)), then the integrand of this expectation, denoted by Q(x, ξ), is the optimal value of the following second-stage
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problem:
SecLP(ξ)
minimize
q(ω)T y(ω)
subject to
T (ω)x + W (ω)y(ω) = b(ω)
y(ω) ≥ 0.
y(ω)
• T (ω) is referred to as the tender
• W (ω) is called the recourse matrix
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Polyhedral functions
Definition 1 (Polyhedra) A polyhedron (or a polyhedral convex set) is the
intersection of a finite number of linear inequalities.
Definition 2 (Polyhedral function) A function f : Rn → (−∞, +∞] is
said to be polyhedral if its epigraph is a polyehdral set in Rn+1 or epi(f ) is
a polyhedral set where
epi(f ) , {(x, w) : x ∈ Rn, w ∈ R, f (x) ≤ w} .
A polyhedral function f is closed, convex, and proper since epi(f ) is a
closed, convex, and nonempty set.
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Discrete random variables
• Suppose ξ is a discrete random variable taking on values in Ξ.
• Suppose K1 and K2(ξ) (elementary feasibility set) are defined as
K1 , {x : Ax = b, x ≥ 0}
and
K2(ξ) , {x : y ≥ 0
subject to
W (ω)y = h(ω) − T (ω)x} .
• Moreover, K2 is defined as
K2 ,
\
ξ∈Ξ
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K2(ξ).
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Lecture 4
Next, we prove some useful properties of K2(ξ) but need to define the
positive hull:
Definition 3 (Positive hull) The positive hull of W is defined as
pos W , {t : W y = t, y ≥ 0} .
In fact, pos W is a finitely generated cone which is the set of nonnegative
linear combinations of finitely many vectors. Note that it is convex and
polyhedral.
Proposition 1 (Polyhedrality of K2(ξ) and K2) The following hold:
1. For a given ξ , K2(ξ) is a convex polyhedron;
2. When ξ is a finite discrete random variable, K2 is a convex polyhedron.
Proof:
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(i) Consider an x 6∈ K2(ξ) and ξ = (T (ω), W (ω), q(ω), b(ω)). Consequently there exists no nonnegative y such that W (ω)y =
h(ω) − T (ω)x; equivalently h(ω) − T (ω)x 6∈ pos W (ω), where
pos W (ω) is a finitely generated convex cone. Therefore, by the separating hyperplane theorem, there exists a hyperplane, defined as H ,
such that H = {v : uT v = 0} such that uT (h(ω) − T (ω)x)) < 0
and uT v > 0 where v ∈ pos W (ω). Since pos W (ω) is a finitely
generated cone, there can only be finitely many such hyperplanes.
It follows that K2(ξ) can be viewed as an intersection of a finite
number of halfspaces and is therefore a convex polyhedron.
(ii) Since K2 is an intersection of a finite number of polyhedral sets, K2
is a convex polyhedron.
• Next, we examine the properties of Q(x, ξ)
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Proposition 2 (Properties of Q(x, ξ)) For a given ξ , Q(x, ξ) satisfies
the following:
(i) Q(x, ξ) is a piecewise linear convex function in (h, T )
(ii) Q(x, ξ) is a piecewise linear concave function in q
(iii) Q(x, ξ) is a piecewise linear convex function in x for x ∈ K2
Proof:
(i) Suppose f (b) is defined as f (b) = min q T y : W y = b, y ≥ 0 .
We proceed to show that f (b) is convex in b. Consider b1, b2
such that bλ = λb1 + (1 − λ)b2 where λ ∈ (0, 1). Let y1 and
T
y2 denote solutions of q y : W y = b, y ≥ 0 for b = b1 and
b2 respectively. Furthermore, suppose yλ denotes the solution of
min q T y : W y = bλ, y ≥ 0 and it follows that
f (bλ) = q T yλ ≤ q T (λy1 + (1 − λ)y2),
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where λy1 + (1 − λ)y2 is feasible i.e.
W (λy1 + (1 − λ)y2) = λW y1 + (1 − λ)W y2 = b
(λy1 + (1 − λ)y2) ≥ 0.
Consequently,
f (bλ) = q T yλ ≤ q T (λy1 + (1 − λ)y2) = q T ȳ
(By optimality of yλ )
= λ q| T{zy1} +(1 − λ) q| T{zy2}
=f (b1 )
=f (b2 )
= λf (b1) + (1 − λ)f (b2),
where λy1 + (1 − λ)y2 is feasible with respect to the constraint
{y : W y = bλ}. The convexity of f in (h, T ) follows.
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(ii) To show the concavity of Q(x, ξ) in q , we may write f (q) as
n
T
T
o
f (q) = max π b : W π ≤ q .
Proceeding as earlier, it is relatively easy to show that f (q) is concave
in q .
(iii) The convexity of Q(x, ξ) in x follows from (i). Next, we show the
piecewise linearity of Q(x, ξ) in (h, T ). Suppose y is a solution of
min q T y : W y = b, y ≥ 0 . Then if yB and yN (≡ 0) denote the
basic and non-basic variables and B(ω) denotes the basis and N (ω)
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represents the remaining columns, we have
B(ω)yB + N (ω)yN = h(ω) − T (ω)x
B(ω)yB = h(ω) − T (ω)x
yB = (B(ω))−1(h(ω) − T (ω)x).
Recall that when a feasible basis is optimal, we have that
q(ω)T ỹ ≥ qB (ω)T yB
= qB (ω)T (B(ω))−1(h(ω) − T (ω)x)
= Q(x, ξ),
implying that Q(x, ξ) is linear in (h, T, q, x) on a domain prescribed
by feasibility and optimality conditions. Piecewise linearity follows
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from noting the existence of a finite number of different optimal bases
for the second-stage program.
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Support functions
The support function sC (•) of a nonempty closed and convex set C is
defined as
sC (h) , sup z T h.
z∈C
Given a convex set C , its support function, denoted by sC (•), is convex,
positively homogeneous,∗ and lower semicontinuous.†
If s1(•) and s2(•) are support functions of C1 and C2 respectively, then
s1(•) ≤ s2(•) ⇔ C1 ⊂ C2.
Furthermore,
s1(•) = s2(•) ⇔ C1 = C2.
∗ A function f (x) is positively homogeneous if f (λx) = λf (x) where λ ≥ 0 and x ∈ Rn .
† A function f (x) is lower semicontinuous at x if lim inf f (x) ≥ f (x ).
0
0
x→x0
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Lemma 1 (Support function of unit ball) Consider the unit ball B defined with the Euclidean norm as follows
B , {π : kπk2 ≤ 1}.
Then sB (.) = k.k2.
Proof: By definition, we have that
sB (χ) = sup π T χ.
π∈B
Since B is a compact set, by the continuity of the objective, the supremum
is achieved. Suppose the maximizer is denoted by π ∗. Then, kπ ∗k ≤ 1 and
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it follows that π ∗ =
χ
.
kχk
Consequently,
χT
kχk2
(π ) χ =
χ=
= kχk.
kχk
kχk
∗ T
Lemma 2 (Support function of Minkowski addition of sets) Consider a
set X = X1 + X2 + . . . + Xn where X1, X2, . . . , Xn are nonempty closed
and convex sets. Then for any y ∈ Rn, the support function sX (y) is
given by the following:
sX (y) = sX1 (y) + sX2 (y) + . . . + sXn (y).
Proof: Recall that by the definition of a support function, we have the
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following sequence of equalities:
sX (y) = sup z T y
z∈X
=
zT y
sup
z∈X1 +...+Xn
=
zT y
sup
z1 ∈X1 ,...,zn ∈Xn
!
=
sup z1T y + . . . +
z1 ∈X1
!
sup znT y
zn ∈Xn
= sX1 (y) + . . . + sXn (y).
Lemma 3 (Support function is positively homogeneous) Consider the support function of a closed and convex set X . Then given any y ∈ Rn, we
have that sX (λy) = λsX (y) when λ ≥ 0.
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Proof: This result follows by noting that for λ ≥ 0, we have that
sX (λy) = sup λy T z
z∈X
= λ sup y T z
z∈X
= λsX (y).
Other examples of support functions:
•
(
C= x:
n
X
)
xi = 1, xi ≥ 0 ,
i=1
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sC (χ) = max χj .
j∈{1,...,n}
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Lecture 4
Subgradients of Q(x, ξ)
Consider the second-stage linear program given by
SecLP(ξ)
minimize
q(ω)T y(ω)
subject to
T (ω)x + W (ω)y(ω) = b(ω)
y(ω) ≥ 0.
y(ω)
Then the associated dual problem is given by
D-SecLP(ξ)
maximize b(ω)T π(ω)
π(ω)
subject to W T (ω)π(ω) ≤ q(ω).
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The recourse function Q(x, ξ) is defined as follows:
n
T
n
T
Q(x, ξ) , inf q(ω) y : W (ω)y = h(ω) − T (ω)x, y ≥ 0
o
o
≡ inf q(ω) y : W (ω)y = χ, y ≥ 0 , sq (χ).
(Suppressing ω ) Suppose Π(q) is defined as
n
o
T
Π(q) , π : W π ≤ q .
It follows that from LP duality, we have that
n
T
o
n
T
T
o
inf q(ω) y : W (ω)y = χ, y ≥ 0 ≡ sup χ π : W (ω) π ≤ q(ω) .
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As a consequence,
sq (χ) = sup π T χ.
π∈Π(q)
This implies that sq (χ) is the support function of Π(q), a closed and convex
polyhedral set. Next we examine the subdifferential of Q(x, ξ) and need
some definitions.
Definition 4 (Differentiability) Given a mapping g : Rn → Rm. It is
said that g is directionally differentiable at a point x0 ∈ Rn in a direction
h ∈ Rn if the following limit exists:
!
g(x0 + th) − g(x0)
g (x0; h) := lim
.
t→0
t
0
If g is directionally differentiable at x0 for every h ∈ Rn, then it is said
to be directionally differentiable at x0. Note that whenever this limit
exists, g 0(x0; h) is said to be positively homogeneous in h or g 0(x0; th) =
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tg 0(x0; h) for any t ≥ 0.
If g(x) is directionally differentiable at x0 and g 0(x0; h) is linear in h,
then g(x) is said to be Gâteaux differentiable at x0. This limit can also
be written as follows:
g(x0 + h) = g(x0) + g 0(x0; h) + r(h),
where r(h) is such that
r(th)
t
→ 0 as t → 0 for any fixed h ∈ Rn.
If g 0(x0; h) is linear in h and r(h)
→ 0 as h → 0 (or r(h) = o(h)), then
khk
g(x) is said to be Fréchet differentiable at x0 or merely differentiable at
x0
Note that Fréchet implies Gâteaux differentiability and when the functions are locally Lipschitz, both notions coincide. Recall that a mapping
g is said to be locally Lipschitz on a set X ⊂ Rn, if it is Lipschitz continuous on a neighborhood of every point of X (with possibly different
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constants).
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Summary of differentiability concepts
• Directionally differentiable at x0 =⇒ Directionally differentiable for all
h at x0
• g(x) is Gauteaux differentiable at x0 =⇒ Directionally differentiable
at x0 and g 0(x0; h) is linear in h
• g(x) is Fréchet differentiable at x0 =⇒ g 0(x0; h) is linear in h and
r(h)
→ 0 as h → 0
khk
• In fact Fréchet differentiable =⇒ Gauteaux differentiable; if g is locally
Lipschitz continuous, then both notions coincide.
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Subgradients and subdifferentials
A vector z ∈ Rn is said to be a subgradient of f (x) at x0 if
f (x) − f (x0) ≥ z T (x − x0),
∀x ∈ Rn.
The set of all subgradients of f (x) at x0 is referred to as a subdifferential
and is denoted by ∂f (x0). The subdifferential ∂f (x0) is a closed and
convex subset of Rn and f is subdifferentiable at x0 if ∂f (x0) 6= ∅.
• Consider the function f (x) = |x|. Then ∂f (x) is defined as follows:
∂f (x) =
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


−1,



x<0




1,
x > 0.
[−1, 1],
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• Consider the function f (x) = max{x, 0}
∂f (x) =
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


0,



x<0




1,
x > 0.
[0, 1],
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x=0
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Convexity and polyhedrality of Q(x, ξ)
Proposition 3 (Convexity and polyhedrality of Q(x, ξ)) For any given
ξ , the function Q(., ξ) is convex. Moreover, if the set Π(q) is nonempty
and D-SecLP(ξ) is feasible for at least one x, then the function Q(., ξ) is
polyhedral.
Proof: Recall that the set Π(q) is closed, convex, and polyhedral. If Π(q)
is nonempty, then sq (χ), defined as
sq (χ) = sup χT π,
π∈Π(q)
is a polyhedral function,‡ where χ = h−T x. Finally, since Q(x, ξ) = sq (χ),
the result follows.
‡ Recall that an extended real-valued function is called polyhedral, if it is proper, convex, and lower semicontinuous, its domain
is a closed and convex polyhedron, and it is piecewise linear on its domain. A function f is said to be proper if f (x) > −∞ for
all x ∈ Rn and its domain dom(f ) is nonempty where dom(f ) , {x : f (x) < +∞}.
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Next we characterize the subdifferential of Q(x, ξ) but require the FenchelMoreau theorem:
Theorem 4 (Fenchel-Moreau) Suppose that f is a proper, convex, and
lower semicontinuous function. Then f ∗∗ = f , where f ∗, the conjugate
function of f , is defined as
f ∗(z) , sup {z T x − f (x)}.
x∈Rn
Note that the conjugate function f ∗ : Rn → R̄ is always convex and lsc.
Furthermore, we have that
n
∗
T
o
∂f (x) = arg max
z x − f (z) .
n
z∈R
Proposition 5 (Subdifferential of Q(x, ξ)) Suppose that for a given x =
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x0 and for ξ ∈ Ξ, the function Q(x0, ξ) is finite.
subdifferentiable at x0 and
Then Q(., ξ) is
∂Q(x0, ξ) = −T T D(x0, ξ),
where D(x, ξ) the set of dual solutions is defined as
D(x, ξ) := arg max π T (h − T x).
π∈Π(q)
Proof: Since Q(x0, ξ) is finite, then Π(q) is nonempty and sq (χ) is its
support function, defined as
sq (χ) , sup χT π.
π∈Π(q)
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By the definition of a conjugate function, it can be seen that sq (χ) is the
conjugate of the indicator function 1lq (π), defined as
1lq (π) :=

0,
if π ∈ Π(q)
+∞,
otherwise.
Since Π(q) is closed and convex, the function 1lq (π) is convex and lower
semicontinuous. By the Fenchel-Moreau theorem, the conjugate of sq (.) is
1lq (.). This can be easily seen by noting the following:
sq (z) = sup {z T π} = sup {z T π − 1lq (π)},
π
π∈Π(q)
implying that sq (·) is the conjugate of 1lq (·). Since 1lq (·) is convex and
lower semicontinuous, by Fenchel-Moreau, we have that 1lq (·) = 1l∗∗
q (·)
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(the biconjugate) where sq (·) = 1l∗q (.) implying that 1lq (·) = (sq (·))∗.
Then for χ0 = h − T x0,
n
o
T
∂sq (x0) = arg max π χ0 − 1lq (π) = arg max π T χ0.
π
π∈Π(q)
Since Π(q) is polyhedral and sq (χ0) is finite, it follows that ∂sq (χ0) is
nonempty. Moreover, sq (.) is piecewise linear (earlier in this lecture) and
by the chain rule
∂xQ(x, ξ) = (∂xχ0)T ∂χ0 sq (χ0) = −T T D(x0, ξ).
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Expected recourse for discrete distributions
In this section, we consider the expected cost of recourse, denoted by Q(x),
where
Q(x) , E[Q(x, ξ)].
Suppose, the distribution of ξ has finite support, implying that ξ can take
on a finite number of realizations given by ξ1, . . . , ξK with a mass function
denoted by p1, . . . , pK . For a given x, yj is given by the solution to
SecLP(ξj )
minimize
y
qjT yj
subject to
Tj x + Wj yj = bj
yj ≥ 0,
j
where (T (ξj ), W (ξj ), q(ξj ), b(ξj )) = (Tj , Wj , qj , bj ).
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Given that the
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problems in y are separable, we may specify the expected cost of recourse
as the following problem in (y1, . . . , yK ) as follows:
SecLP(ξ1, . . . , ξK ) minimize
y ,...,y
1
K
subject to
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K
X
pj qjT yj
j=1
Tj x + Wj yj = bj ,
yj ≥ 0,
31
j = 1, . . . , K
j = 1, . . . , K
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Note that the entire two-stage problem can then be specified as follows:
SLP
minimize
x,y ,...,y
1
K
T
c x+
K
X
pj qjT yj
j=1
Ax = b
subject to Tj x + Wj yj = bj ,
x, yj ≥ 0,
j = 1, . . . , K
j = 1, . . . , K
Theorem 6 (Moreau-Rockafellar) Let fi : Rn → R̄ be proper convex
P
functions for i = 1, . . . , N . Let f (.) = N
i=1 fi (.) and x0 be a point such
that fi(x0) are finite or x0 ∈ ∩N
i=1 dom fi . Then
∂f1 + . . . + ∂fN ⊂ ∂f.
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Furthermore, we have equality in this relationship or
∂f1 + . . . + ∂fN = ∂f,
if one of the following hold:
1. The set ∩m
i=1 ri(domfi ) is nonempty;
2. the functions f1, . . . , fk for k ≤ m are polyhedral and the intersection of
the sets ∩ki=1domfi and ∩m
i=k+1 ri(domfi ) is nonempty;
3. there exists a point x̄ ∈ dom(fm) such that x̄ ∈ int(domfi) for i =
1, . . . , m − 1.
The subdifferential of Q(x) is specified by the following proposition.
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Proposition 7 (Subdifferential of ∂Q(x)) Suppose that the probability
distribution function for ξ has finite support or Ξ , {ξ1, . . . , ξK } and the
expected recourse cost has a finite value for at least some x̃ ∈ Rn. Then
Q(x) is polyhedral and
∂Q(x0) =
K
X
∂Q(x, ξj ).
j=1
Proof: This follows as a consequence of the Moreau-Rockafellar theorem.
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Expected recourse for general distributions
Consider Q(x, ξ) where ξ : Ω → Rd. The cost of taking recourse given ξ is
the minimum value of q(ω)T y(ω).
Lemma 4 Q(x, ξ) is a random lower semicontinuous function.
Proof: This follows from noting that Q(x, .) is measurable§ with respect to
(.). Furthermore, Q(., ξ) is lower semicontinuous. It follows that Q(x, ξ)
is a random lower semicontinuous function.
Proposition 8 Suppose either E[Q(x, ξ)]+ or E[Q(x, ξ)]− is finite. Then,
Q(x) is well defined.
Proof: This requires verifying whether Q(x, .) is measurable with respect
to the Borel sigma algebra on Rd. This follows by directly employing
Theorem 7.37 [SDR09].
§ Given a probability space (Ω, F , P), a function ξ : Ω → [−∞, +∞] is said to be measurable if the set ξ −1 ((a, ∞]) =
{ω ∈ Ω : ξ(ω) > a} is a measurable set. Any element of F is said to be a measurable set.
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Theorem 9 Let F : Rn × Ω → R̄ is a random lower semicontinuous
function. Then the optimal value function ϑ(ω) and the optimal solution
multifunction X ∗(ω) are both measurable where
ϑ(ω) , infn F (y, ω) and X ∗(ω) , arg minn F (x, ω).
y∈R
x∈R
Next, we consider several settings where either E[Q(x, ξ)]+ or E[Q(x, ξ)]−
is finite(ii). However, this requires taking a detour in discussing various
notions of recourse.
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Notions of recourse
• The two-stage problem is said to have fixed recourse if W (ω) = W for
every ω ∈ Ω.
• The problem is said to have complete recourse if there exists a solution
y satisfying W y = χ and y ≥ 0 for every χ. In effect, for any first-stage
decision, the primal second-stage problem is feasible.
More formally, this can be stated as follows:
posW = Rm,
where h − T x ∈ Rm.
• By employing duality, the fixed recourse is said to be complete if and
only if the feasible set Π(q) of the second-stage dual problem is bounded
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(could also be empty).
Complete recourse ⇔ Primal second-stage is feasible
⇔ Dual second-stage is bounded (could be empty).
• Given a set X , then X∞ refers to the recession cone of X and is defined
as follows:
X∞ , {y : ∀x ∈ X, ∀λ ≥ 0, x + λy ∈ X, } .
• Boundedness of Π(q) implies that its recession cone denoted by
Π0 = Π(0) contains only the zero element; Π0 = {0}, provided
that Π(q) is nonempty.
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• Claim: Π(q) = {π : W T π ≤ q} is nonempty and bounded implies that
Π0 contains only the zero element.
Proof sketch: Suppose this claim is false. Then Π0 6= ∅ and suppose
u ∈ Π0. This implies that W T u ≤ 0 where u 6≡ 0. Since Π(q) is
nonempty, then it contains at least one element, say π̂ . It follows that
W T (π̂ + λu) ≤ W T π̂ ≤ q,
for all λ ≥ 0. This implies that Π(q) is unbounded since it contains a
ray given by π̂ + λu where λ ≥ 0 and π̂ ∈ Π(q).
• A subclass of problems with
recourse problems where the
shortage decisions determined
T and q are deterministic and
Stochastic Optimization
fixed and complete recourse are simple
recourse decisions are either surplus or
by the first-stage decisions. Furthermore,
q > 0.
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Uday V. Shanbhag
Lecture 4
Specifically, y(ω) = y +(ω) + y −(ω). If [h − T x]i ≥ 0, yi+(ω) =
[h−T x]i and yi−(ω) = 0. Similarly, if [h−T x]i ≤ 0, yi−(ω) = [h−T
x]
i
and yi+(ω) = 0. This is compactly captured by defining W = I −I ,
leading to the following system:
y +(ω) − y −(ω) = h(ω) − T x
y +(ω), y −(ω) ≥ 0.
• The recourse is said to be relatively complete if for every x in the set
X , {x : Ax = b, x ≥ 0}, the feasible set of the primal second-stage
problem is nonempty for a.e. ω ∈ Ω. Note that there may be events that
occur with zero probability (measure) that may have Q(x, ξ) = +∞.
However, these do not affect the overall expectation.
• A sufficient condition for relatively complete recourse is the following:
for every x ∈ X, Q(x, ξ) < +∞ for all ξ ∈ Ξ.
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Uday V. Shanbhag
Lecture 4
• Note that this condition becomes necessary and sufficient in two cases:
• The vector ξ has finite support
• Fixed recourse
• Next we consider an instance of a problem with random recourse and
discuss some concerns.
Example 1 (Random recourse) Suppose Q(x, ξ) := inf{y : ξy =
x, y ≥ 0}, with x ∈ [0, 1] and ξ having a density function given by
p(z) = 2z, 0 ≤ z ≤ 1. Then for ξ > 0, x ∈ [0, 1], Q(x, ξ) = x/ξ
and E[Q(x, ξ)] = 2x. For ξ = 0 and x > 0, Q(x, ξ) = +∞.
It can be seen that for almost every ξ ∈ [0, 1], Q(x, ξ) < +∞. –
relatively complete recourse. Furthermore, P(ξ : ξ = 0) = 0 and
E[Q(x, ξ)] < +∞ for all x ∈ [0, 1].
However, a small perturbation in the distribution may change that.
For instance, a discretization of the distribution with the first point at
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Uday V. Shanbhag
Lecture 4
ξ = 0, immediately leads to E[Q(x, ξ)] = +∞ for x > 0. In effect,
this problem is unstable and this issue cannot occur if the recourse is
fixed.
• We now consider the support function sq associated with Π(q). Recall
that this is defined as
sq (χ) = sup π T χ.
π∈Π(q)
Our goal lies in finding sufficiency conditions for the finiteness of
the expectation E[sq (χ)]. We will employ Hoffman’s Lemma [SDR09,
Theorem 7.11] which is defined as follows:
Lemma 5 (Hoffman’s lemma) Consider the multifunction M(b) :=
{x : Ax ≤ b}, where A ∈ Rm×n. Then there exists a positive constant
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42
Uday V. Shanbhag
Lecture 4
κ, depending on A, such that for any given x ∈ Rn and any b ∈ dom M,
dist(x, M(b)) ≤ κk(Ax − b)+k.
Lemma 6 Suppose there exists a constant κ depending on W such that
if for some q0, the set Π(q0) is nonempty, then for every q , the following
inclusion holds:
Π(q) ⊂ Π(q0) + κkq − q0kB,
where B := {π : kπk ≤ 1} and k.k denotes the Euclidean norm. Then
the following holds:
sq (•) ≤ sq0 (•) + κkq − q0kk • k.
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Uday V. Shanbhag
Lecture 4
If q0 = 0, then
| sq (χ1) − sq (χ2) |≤ κkqkkχ1 − χ2k,
∀χ1, χ2 ∈ dom(sq ).
Proof.
Since the support function of the unit ball is the Euclidean
norm, when Π(q0) is nonempty, and that
s1(•) ≤ s2(•) ⇔ C1 ⊂ C2.
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Uday V. Shanbhag
Lecture 4
If C1 = Π(q) and C2 = Π(q0) + κkq − q0kB , then we have that
sq (•) ≤ sΠ(q0)+κkq−q0kB (•)
= sΠ(q0)(•) + sκkq−q0kB (•)
(Support function of Minkowski sum)
= sΠ(q0)(•) + κkq − q0ksB (•)
= sq0 (•) + κkq − q0kk • k.
(Positive homogeneity of support function)
(Support function of unit ball is Euclidean norm)
(1)
• Consider q0 = 0 and the set Π(0) is defined as
n
T
o
Π(0) , π : W π ≤ 0 .
This set is a cone and suppose its support function is denoted by
s0 = sq0 .
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Uday V. Shanbhag
Lecture 4
• The support function is defined as follows (proof omitted):
s0(χ) =

0,
if χ ∈ pos W
+∞,
otherwise
• Therefore, if q0 = 0 in (1) allows one to claim that if Π(q) is nonempty,
then sq (χ) ≤ κkqkkχk for all χ ∈ pos W , and sq (χ) = +∞ if
χ 6∈ pos W .
• By the polyhedrality of Π(q), if Π(q) is nonempty, we have that sq (·)
is piecewise linear on its dom(sq (·)) which is given by posW . In fact,
sq (.) is Lipschitz continuous on its domain or
| sq (χ1) − sq (χ2) |≤ κkqkkχ1 − χ2k,
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∀χ1, χ2 ∈ dom(sq ).
Uday V. Shanbhag
Lecture 4
• We now provide a necessary and sufficient condition for a fixed recourse
problem having a finite E[Q(x, ξ)+].
Proposition 10 (Nec. and suff. condition for finiteness of E(Q(x, ξ)+))
Suppose that recourse is fixed and E[kqkkhk] < +∞ and E[kqkkT k] <
+∞. Consider a point x ∈ Rn. Then E[Q(x, ξ)+] is finite if and only
if h(ξ) − T (ξ)x ∈ pos W for almost every ξ.
Proof:
(⇒) Suppose h(ξ) − T (ξ)x 6∈ pos W with probability one. This implies
that for some ξ ∈ U ⊆ Ξ, h(ξ) − T (ξ)x 6∈ pos W where µ(U ) =
P[ξ : ξ ∈ U ] > 0. But for any such ξ , Q(x, ξ) = +∞. But
(Q(x; ξ))+ , max(Q(x; ξ), 0)) ≥ Q(x; ξ) = +∞ with positive
probability. Consequently, it follows that E[Q(x, ξ)+] = +∞.
(⇐) Suppose h(ξ) − T (ξ)x ∈ pos W with probability one.
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Then
Uday V. Shanbhag
Lecture 4
Q(x, ξ) = sq (h − T x). By (1), we have that
sq (χ) ≤ s0(χ) + κkqkkχk.
Furthermore, we recall that s0(χ) = 0 when χ ∈ pos W implying
that
sq (χ) ≤ κkqkkχk.
By using the triangle inequality, we have that
Q(x; ξ) = sq (h(ξ) − T (ξ)x)
≤ κkq(ξ)(kh(ξ)k + kT (ξ)kkxk)
(with probability one)
Since
0 ≤ κkq(ξ)(kh(ξ)k + kT (ξ)kkxk) wp1,
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(wp1)
Uday V. Shanbhag
Lecture 4
it follows that
max(Q(x; ξ), 0) ≤ κkq(ξ)(kh(ξ)k + kT (ξ)kkxk) wp1.
Taking expectations, we have the following:
E[Q(x; ξ)+] ≤ κE[kq(ξ)kh(ξ)k] + κE[kq(ξ)kkT (ξ)k]kxk.
By assumption, we have that E[kqkkhk] < +∞ and E[kqkkT k] <
+∞. As a result, E[Q(x, ξ)+] < +∞.
• Next, instead of requiring that (h(ξ) − T (ξ)x) ∈ pos(W ) wp1, we
assume that Π(q(ξ)) is nonempty wp1.
Proposition 11 Suppose that (i) the recourse is fixed, (ii) for a.e. q the
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Uday V. Shanbhag
Lecture 4
set Π(q) is nonempty, and (iii) the following holds:
E[kqkkhk] < +∞ and E[kqkkT k] < +∞.
Then the following hold: (a) the expectation function Q(x) is welldefined and Q(x) > −∞ for all x ∈ Rn; (b) Moreover, Q(x) is
convex, lower semicontinuous, and Lipschitz continuous on dom φ, and
its domain is a convex closed subset of Rn is given by
domQ = {x : h − T x ∈ pos W w.p.1.}
Proof:
(a): By (ii), Π(q) is nonempty with probability one. Consequently, for
almost every ξ ∈ Ξ, Q(x, ξ) = sq (h(ξ) − T (ξ)x) for every x and for
almost every ξ .
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Uday V. Shanbhag
Lecture 4
Suppose π(q) denotes the element of Π(q) closest to zero; by the
closedness of Π(q), it exists and by Hoffman’s lemma, there exists a
constant κ such that
kπ(q)k ≤ κkqk.
By the definition of the support function,
sq (h − T x) =
sup
π(q)T (h − T x) ≥ π(q)T (h − T x)
π(q)∈Π(q)
and for every x, the following holds with probability one:
−π(q)T (h − T x) ≤ kπ(q)k (khk + kT kkxk) (Cauchy-Schwartz and triangle inequality)
≤ κkqk (khk + kT kkxk)
=⇒ π(q)T (h − T x) ≥ −κkqk (khk + kT kkxk)
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Uday V. Shanbhag
Lecture 4
By (iii), we have that
E[kqkkhk] < +∞ and E[kqkkT k] < +∞
and it can be concluded that Q(•) is well defined and Q(x) > −∞ for
all x ∈ Rn.
(b): Since sq (•) is lower semicontinuous in (•) (since it is a support
function), it follows that Q(•) is also lower semicontinuous by Fatou’s
Lemma¶
Lemma 7 (Fatou’s Lemma) Suppose that there exists a P-integrablek
function g(ω) such that fn(•) ≥ g(•). Then
h
i
lim inf E[fn] ≥ E lim inf fn .
n→∞
n→∞
¶ Just to jog your memory, Fatou’s Lemma allows for interchanging integrals and limits under assumptions on the integrand.
k A random variable Z : Ω → Rd is P-integrable if E[Z] is well-defined and finite. Recall that E[Z(ω)] is well-defined if it
does not happen that E[Z(ω)+ ] and E[(−Z(ω))+ ] are +∞ in which case E[Z] = E[Z+ ] − E[(−Z)+ ]
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Uday V. Shanbhag
Lecture 4
• Convexity of Q(x) can be claimed as follows. From convexity of Q(x, ξ)
in x, we have that for any x, y
Q(x, ξ) ≥ Q(y, ξ) + u(ξ)T (x − y), where u(ξ) ∈ ∂Q(y, ξ).
(2)
Taking expectations on both sides and by the well-defined nature of
Q(x) for all x, we obtain the following:
E[Q(x, ξ)] ≥ E[Q(y, ξ)] + E[u(ξ)]T (x − y), where u(ξ) ∈ ∂Q(y, ξ).
(3)
But under suitable assumptions, we may interchange subdifferential and
expectation, implying that
u = E[u(ξ)] ∈ E[∂Q(x, ξ)] = ∂ E[Q(x, ξ)] = ∂Q(x).
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Uday V. Shanbhag
Lecture 4
It follows that for any x, y , we have that
Q(x) ≥ Q(y) + uT (x − y), where u ∈ ∂Q(y).
(4)
In other words, Q(x) is a convex function in x.
• Convexity and closedness of domQ is a consequence of the convexity
and lower semicontinuity of Q. By Prop. 10, Q(x) < +∞ if and only
if h(ξ) − T (ξ)x ∈ pos W with probability one. But this implies that
domQ = {x ∈ Rn : Q(x) < +∞}
= {x ∈ Rn : h(ξ) − T (ξ)x ∈ pos W, with probability one. } .
(c): In the remainder of the proof, we show the Lipschitz continuity of
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Uday V. Shanbhag
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Q(x). Consider x1, x2 ∈ domQ. Then, we have
h(ξ) − T (ξ)x1 ∈ pos W, w.p.1 and h(ξ) − T (ξ)x2 ∈ pos W, w.p.1 .
By leveraging this claim, if Π(q) is nonempty, we have that
|sq (h − T x1) − sq (h − T x2)| ≤ κkqkkT kkx1 − x2k.
Taking Expectations on both sides, we have that
|sq (h − T x1) − sq (h − T x2)| ≤ κkqkkT kkx1 − x2k.
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Uday V. Shanbhag
Lecture 4
|Q(x1) − Q(x2)| ≤ E[|sq (h − T x1) − sq (h − T x2)|]
≤ κE[kqkkT k]kx1 − x2k,
where the first inequality follows from the application of Jensen’s inequality∗∗ and the convexity of the norm. The Lipschitz continuity of
Q(x) on its domain follows.
∗∗ Recall that Jensen’s inequality implies that f (E[X]) ≤ E[f (X)], when f is a convex function and E[X] denotes the
expectation of a random variable X.
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Lecture 4
References
[SDR09] A. Shapiro, D. Dentcheva, and A. Ruszczyński. Lectures on
stochastic programming, volume 9 of MPS/SIAM Series on Optimization. Society for Industrial and Applied Mathematics (SIAM),
Philadelphia, PA, 2009. Modeling and theory.
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