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Lecture 4
Two-stage stochastic linear programs
Optimality Conditions
February 2, 2015
Uday V. Shanbhag Lecture 4
Discrete random variables
Consider the two-stage stochastic linear program with discrete random
variables:
SLP minimizex
cTx+K∑k=1
pkQ(x, ξk)
subject to Ax = b, x ≥ 0,
where Q(x, ξk) is the optimal value of SecStage(x, ξk), defined as follows:
SecStage(x, ξk) minimizey
d(ξk)Ty
subject to W (ξk)y = h(ξk)− T (ξk)x, y ≥ 0.
Stochastic Optimization 1
Uday V. Shanbhag Lecture 4
From results in the previous lecture, suppose that Q(•) has a finite valuefor at least one x̃ ∈ Rn. Then from the Moreau-Rockafellar theorem, wehave that for every x0 ∈ dom Q,
∂Q(x0) = −K∑k=1
pkTTKD(x0, ξk),
where
D(x0, ξk) , arg maxπ
{πT(hk − Tkx0) : W Tk π ≤ qk
}.
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Uday V. Shanbhag Lecture 4
Optimality conditions
Consider the optimization problem
minx∈X
f(x),
where X ⊂ Rn and f : Rn → R̄ is an extended real-valued function.Convex case: Suppose that the function f is convex. If f(x̄) is finite at
some point x̄ ∈ Rn then the following holds:
f(x) ≥ f(x̄) for all x ∈ Rn ⇔ 0 ∈ ∂f(x̄).
Suppose that X is closed and convex and f is proper∗ and convex. Consider
a point x̄ ∈ X ∩ dom f . It follows that f̄(x) , f(x) + 1lX(x) is convex∗A function f is proper if f(x) > −∞ for all x ∈ Rn and its domain domQ is nonempty, where domQ = {x : f(x) < +∞}.
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Uday V. Shanbhag Lecture 4
where
1lX(x) =
0, x ∈ X+∞, x 6∈ X.Then x̄ is a minimizer of
minx∈X
f(x),
if and only if x̄ is a minimizer of f̄(x).
Suppose that
ri(X) ∩ ri(domf) 6= ∅.
Then by the Moreau-Rockafellar theorem, we have that
∂f̄(x̄) = ∂f(x̄) + ∂1lX(x̄).
From convex analysis, ∂1lX(x) = NX(x̄). Consequently, x̄ is an optimal
Stochastic Optimization 4
Uday V. Shanbhag Lecture 4
solution of the constrained problem, given by
minx∈X
f(x),
if and only if
0 ∈ ∂f(x̄) +NX(x̄).
Stochastic Optimization 5
Uday V. Shanbhag Lecture 4
Back to the optimality conditions of SLP
Theorem 1 (Optimality conditions of SLP) Let x̄ be a point such that
x̄ ∈ X and Q(x̄) < +∞. Then x̄ is an optimal solution of (SLP) if andonly if there exist πk ∈ D(x̄, ξk), k = 1, . . . ,K and µ ∈ Rm such that
K∑k=1
pkTTk πk +A
Tµ ≤ c (1)
x̄T
c− K∑k=1
pkTTk πk −ATµ
= 0. (2)Proof: The necessary and sufficient conditions of optimality for minimizing
this convex program are given by
0 ∈ ∂Q(x̄) +NX(x̄),
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Uday V. Shanbhag Lecture 4
where
NX(x) ,{z : zT(x′ − x) ≤ 0, ∀x′ ∈ X
}.
No need for additional regularity conditions, since Q(•) and X are convexand polyhedral. It follows that
0 ∈ c−K∑k=1
pkTTk πk +NX(x̄).
Next, we characterize the normal cone of this polyhedral set. Recall that
NX(x̄) can be represented as
NX(x̄) ,{z : zT(x− x̄) ≤ 0, Ax = b, x ≥ 0
}.
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Uday V. Shanbhag Lecture 4
But this implies that NX(x̄) can be written as
NX(x̄) ,{z : zT(x̄− x) ≥ 0, Ax = b, x ≥ 0
}.
In other words, z ∈ NX(x̄) if and only if
zT(x− x̄) ≤ 0, ∀x ∈ X.
When X , {x : Ax = b}, we have the following sequence of equivalence
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Uday V. Shanbhag Lecture 4
statements:
z ∈ NX(x̄)⇔ zT(x− x̄) ≤ 0, ∀x ∈ X⇔ zT(x− x̄) ≤ 0, ∀x ∈ {x : Ax = b, x ≥ 0},⇔ zT(x− x̄) ≤ 0, Ax = b, x ≥ 0⇔ zT(x̄− x) ≥ 0, Ax = b, x ≥ 0.
Since zT(x̄− x) ≥ 0 for all x satisfying Ax = b, x ≥ 0, it follows that theminimal value of the following LP, denoted by (LP), is nonnegative.
(LP) minx
zT(x̄− x)
Ax = b, (µ)
x ≥ 0. (h)
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Uday V. Shanbhag Lecture 4
Claim: This LP has optimal value equal to zero.
Proof: By the earlier sequence of arguments, f(x) , zT(x− x̄) ≥ 0 for allx ∈ X , {x : Ax = bx ≥ 0}. But there exists a feasible x = x̄ such thatf(x̄) = 0. Consequently, the optimal value of (LP) is zero and its solution
is x̄.
By LP duality theory, we have that there exists a (µ, h) such that (µ, h)
satisfies
−z −ATµ− h = 0, h ≥ 0, hT x̄ = 0.
In effect, z = −ATµ− h and NX(x̄) can be defined as
NX(x̄) , {−ATµ− h : h ≥ 0, hT x̄ = 0}.
It follows that
0 ∈ c−K∑k=1
pkTTk πk +NX(x̄).
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Uday V. Shanbhag Lecture 4
can be represented as
0 = c−K∑k=1
pkTTk πk −ATµ− h,
where hT x̄ = 0 and h ≥ 0. However, h can be expressed as
h = c−K∑k=1
pkTTk πk −ATµ,
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Uday V. Shanbhag Lecture 4
and the resulting optimality conditions reduce to
h ≥ 0 =⇒
c− K∑k=1
pkTTk πk −ATµ
≥ 0,x̄Th = 0 =⇒ x̄T
c− K∑k=1
pkTTk πk −ATµ
= 0.
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Uday V. Shanbhag Lecture 4
A scenario-based deterministic equivalent
We may also obtain these conditions for the large-scale linear programming
formulation:
SLP minimizex,y1,...,yK
cTx+∑Kk=1 pkq
Tk yk
subject to
Tkx+Wkyk = hk, (pkπk) k = 1, . . . ,K
Ax = b, (µ)
x ≥ 0,yk ≥ 0, k = 1, . . . ,K.
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Uday V. Shanbhag Lecture 4
The dual of this problem is given by
D-SLP minimizeµ,π1,...,πK
bTµ+∑Kk=1 pkh
Tkπk
subject toc−ATµ−
∑Kk=1 pkT
Tk πk ≥ 0,
qk −W Tk πk ≥ 0, k = 1, . . . ,K.
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Uday V. Shanbhag Lecture 4
The optimality conditions prescribed by the earlier Theorem can be stated
in this equivalent form:
K∑k=1
pkTTk πk +A
Tµ ≤ c,
x̄
c− K∑k=1
pkTTk πk +A
Tµ
= 0,qk −W Tk πk ≥ 0, k = 1, . . . ,K
ȳTk(qk −W Tk πk
)= 0, k = 1, . . . ,K
Stochastic Optimization 15
Uday V. Shanbhag Lecture 4
Optimality conditions for continuous random variables
We begin this subsection with a specification of the subdifferential of Q(x)when ξ is a continuous random variable.
Proposition 2 (∂Q(x) when ξ is a continuous r.v.) Suppose that the ex-pectation function Q(x) is a proper function and its domain has a nonemptyinterior. Then for any x0 ∈ dom Q,
∂Q(x0) = −E[T TD(x0, ξ)] +Ndom Q(x0),
where D(x, ξ) = arg maxπ∈Π(q) πT(h − Tx). Furthermore, Q is differ-entiable at x0 if and only if x0 ∈ int(dom Q) and D(x0, ξ) is a singletonw.p.1.
We are now prepared to derive the optimality conditions of (SLP) when ξ
is a continuous random variable.
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Uday V. Shanbhag Lecture 4
Theorem 3 (Optimality conditions of SLP) Let x̄ be a feasible solution
of SLP. Suppose that the recourse function Q(•) is proper, int(dom Q)∩X is nonempty. Then X∩int(domQ) 3 x̄ is an optimal solution of (SLP)if and only if there exists a measurable function† π(ω) ∈ D(x̄, ξ(ω)) forevery ω ∈ Ω and a vector µ ∈ Rm such that
E[T Tπ] +ATµ ≤ c (3)x̄T(c− E[T Tπ]−ATµ
)= 0. (4)
†Let (X,Σ) and (Y, ) be measurable spaces, meaning that X and Y are sets equipped with respective sigma algebras Σ andT . A function f : X → Y is said to be measurable if for every E ∈ T the preimage of E under f is in Σ; ie
f−1(E) := {x ∈ X : f(x) ∈ E} ∈ Σ, ∀E ∈ T.
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Proof: Recall that (SLP) requires the solution of
SLP minimizex
cTx+
,Q(x)︷ ︸︸ ︷E[Q(x, ξk)]
subject toAx = b,
x ≥ 0.
This problem can be equivalently stated as
minx∈X
[cTx+Q(x) + 1lX(x)].
The necessary and sufficient optimality conditions of this problem are given
by
0 ∈ ∂[cT x̄+Q(x̄) + 1lX(x̄)].
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Uday V. Shanbhag Lecture 4
Since int(dom Q) ∩ X is nonempty, by the Moreau-Rockafellar theorem,we have that
∂[cT x̄+Q(x̄) + 1lX(x̄)] = c+ ∂Q(x̄) + ∂[1lX(x̄)].
Recall, that ∂1lX(x̄) = NX(x̄). Furthermore, from Prop. 2, the optimalityof x̄ is equivalent to the existence of a measurable function π(ω) ∈D(x̄, ξ(ω)) such that
0 ∈ c+ ∂Q(x̄) +NX(x̄), where ∂Q(x) = −E[T Tπ] +Ndom Q(x).
It follows that
0 ∈ c− E[T Tπ] +Ndom Q(x̄) +NX(x̄).
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Uday V. Shanbhag Lecture 4
Note that since x̄ ∈ int(dom Q), we have that NdomQ(x̄) = {0}.Therefore, the optimality conditions reduce to
0 ∈ c− E[T Tπ] +NX(x̄).
The remainder of the proof follows from employing the structure of NX(x̄).
Stochastic Optimization 20