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

}.

Stochastic Optimization 2

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) < +∞}.

Stochastic Optimization 3

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),

Stochastic Optimization 6

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

}.

Stochastic Optimization 7

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

Stochastic Optimization 8

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)

Stochastic Optimization 9

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).

Stochastic Optimization 10

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,

Stochastic Optimization 11

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.

Stochastic Optimization 12

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.

Stochastic Optimization 13

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.

Stochastic Optimization 14

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.

Stochastic Optimization 16

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.

Stochastic Optimization 17

Uday V. Shanbhag Lecture 4

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)].

Stochastic Optimization 18

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).

Stochastic Optimization 19

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