Optimization Problems with EquilibriumConstraints

GIAN Short Course on Optimization:Applications, Algorithms, and Computation

Sven Leyffer

Argonne National Laboratory

September 12-24, 2016

Outline

1 Introduction: Stackelberg Games

2 Difficulties with MPECs

3 Stationarity Conditions for MPECsBouligand and Strong StationarityAlphabet Soup of Spurious Stationarity

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Introduction: Nash Games

ISOgameNash producers

Nash Game: non-cooperative equilibrium of several producers

z∗i ∈

argmin

zibi (z)

subject to ci (zi ) ≥ 0zi ≥ 0

producer i

Producer i optimizes own zi , given other producers choices

All producers z = (z∗1 , . . . , z∗i−1, zi , z

∗i+1, . . . , z

∗l )

No shared constraints (otherwise called Nash-Gournot)

All producers/players are equal

Definition (Nash Equilibrium)

No producer i can improve objective, if other producer’s variables,zi , ∀j 6= i , remain unchanged.

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Solution of Nash Games

Form first-order optimality conditions for each player ...

(NCP)

{0 ≤ µ ⊥ ∇b(z)−∇c(z)λ ≥ 0

0 ≤ λ ⊥ c(z) ≥ 0

where

b(z) = (b1(z), . . . , bk(z)) & c(z) = (c1(z), . . . , ck(z))

⊥ means λT c(z) = 0, either λi > 0 or ci (z) > 0

Called a nonlinear complementarity problem (NCP)

Robust large scale solvers exist: e.g. PATH

Setting y = (z , λ, µ)T and F (y) = (b(z)−∇c(z)λ, c(z))T , wecan rewrite (NCP) equivalently as

0 ≤ y ⊥ F (y) ≥ 0

... change of notation: y both variables and multipliers!

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Stackelberg Games & Bilevel Optimization

Single dominant producer & Nash followers

minimize

x≥0,yf (x , y)

subject to c(x , y) = 00 ≤ y ⊥ F (x , y) ≥ 0

ISOgameNash producers

LARGEproducer # 1

Nash game (0 ≤ y ⊥ F (x , y) = 0)... parameterized in leader’s variables x

Mathematical Program with Equilibrium Constraints (MPEC)

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Bilevel Optimization as MPECs

Single dominant producer & Nash followers equivalent to

minimize

x≥0,yf (x , y)

subject to c(x , y) = 0{miny

b(y)

s.t. d(y , x) ≥ 0ISO

gameNash producers

LARGEproducer # 1

Lower-level problem (min b(y) s.t. d(y , x) ≥ 0)... parameterized in leader’s variables x

Mathematical Program with Equilibrium Constraints (MPEC)

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Example: Optimal Taxation Model

Government sets tax rates, tg , for certain goods to maximizerevenue

Consumers buy goods to maximize own utility function

Consumers react to tax rates by changing purchase behavior

Government is leader ... knows how consumers will react

Assume we have seven goods:

G ={

Beer, Pizza, Movie, Wine, Cheese, Ballet, Leisure}

... and two classes of consumers

C ={

Students, Professors}

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Example: Optimal Taxation Model

Consumer c buys quantities qc,g ≥ 0 of goods, g ∈ G tomaximize

qUc(q) =

∏g∈G

qαc,gc,g utility function

subject to∑g∈G

pg (1 + tg )qc,g ≤ bc budget constraint

where∑αc,g = 1, with prices, pg , and tax-rates, tg of good g ∈ G

KKT conditions of consumer c are:

−αc,gq(αc,g−1)c,g

∏g ′∈G:g ′ 6=g

qαc,g′

c,g ′ + πcpg (1 + tg )− ξc,g = 0 ∀g ∈ G

∑g∈G

pg (1 + tg )qc,g ≤ bc ⊥ πc ≥ 0 and 0 ≤ qc,g ⊥ ξc,g ≥ 0

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Example: Optimal Taxation Model

Government maximizes tax revenue subject to consumer actions

maxt

∑c∈C

∑g∈G

tgqc,gNc

s.t. −αc,gq(αc,g−1)c,g

∏g ′∈G:g ′ 6=g

qαc,g′

c,g ′ + πcpg (1 + tg )− ξc,g = 0 ∀g ∈ G

∑g∈G

pg (1 + tg )qc,g ≤ bc ⊥ πc ≥ 0

0 ≤ qc,g ⊥ ξc,g ≥ 0, ∀c ∈ C, ∀g ∈ G

where Nc is the number of consumers in class c ∈ C

So who gets taxed the most???

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Example: Optimal Taxation Model

Government maximizes tax revenue subject to consumer actions

maxt

∑c∈C

∑g∈G

tgqc,gNc

s.t. −αc,gq(αc,g−1)c,g

∏g ′∈G:g ′ 6=g

qαc,g′

c,g ′ + πcpg (1 + tg )− ξc,g = 0 ∀g ∈ G

∑g∈G

pg (1 + tg )qc,g ≤ bc ⊥ πc ≥ 0

0 ≤ qc,g ⊥ ξc,g ≥ 0, ∀c ∈ C, ∀g ∈ G

where Nc is the number of consumers in class c ∈ C

So who gets taxed the most???

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The Problem for the Rest of the Day

Mathematical Program with Equilibrium Constraints (MPEC)minimize

x ,yf (x , y)

subject to c(x , y) ≥ 00 ≤ y ⊥ F (x , y) ≥ 0

f : Rp × Rq → R, and c : Rp × Rq → Rm smooth

Complementarity constraint: F : Rp × Rq → Rq smoothyi = 0 or Fi (x , y) = 0 ... yTF (x , y) = 0

more general l ≤ c(x , y) ≤ u: no problem

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MPEC: Economic Applications

Stackelberg games [Stackelberg, 1952]

modeling of competition in deregulated electricity markets[Pieper, 2001, Hobbs et al., 2000]

volatility estimation in American option pricing[Huang and Pang, 1999]

transportation network design:1 toll road pricing: how to set toll levels leader2 consumers move optimally (Wardrop’s principle) followers

[Hearn and Ramana, 1997, Ferris et al., 1999]

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MPEC: Engineering Applications

design of structures involving friction[Ferris and Tin-Loi, 1999a]

brittle fracture identification [Tin-Loi and Que, 2002]

problems in elastoplasticity [Ferris and Tin-Loi, 1999b]

process engineering models[Rico-Ramirez and Westerberg, 1999,Raghunathan and Biegler, 2002]

floor planning (design of semi-conductors)[Anjos and Vanelli, 2002]

obstacle problems (PDE); packaging problems[Outrata et al., 1998]

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Outline

1 Introduction: Stackelberg Games

2 Difficulties with MPECs

3 Stationarity Conditions for MPECsBouligand and Strong StationarityAlphabet Soup of Spurious Stationarity

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Why Not Simply Solve MPECs as NLPs?

Mathematical Program with Equilibrium Constraints (MPEC)minimize

x ,yf (x , y)

subject to c(x , y) ≥ 00 ≤ y ⊥ F (x , y) ≥ 0

Equivalent smooth nonlinear program (NLP):minimize

x ,yf (x , y)

subject to c(x , y) ≥ 0F (x , y) ≥ 0 and y ≥ 0yTF (x , y) = 0

NLP solvers converge slowly, and sometimes fail completely!

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Why Not Simply Solve MPECs as NLPs?

Mathematical Program with Equilibrium Constraints (MPEC)minimize

x ,yf (x , y)

subject to c(x , y) ≥ 00 ≤ y ⊥ F (x , y) ≥ 0

Equivalent smooth nonlinear program (NLP):minimize

x ,yf (x , y)

subject to c(x , y) ≥ 0F (x , y) ≥ 0 and y ≥ 0yTF (x , y) = 0

NLP solvers converge slowly, and sometimes fail completely!

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Example of Linear Convergence of SQP

Consider

minimizex ,y

(x − 1)2 + (y − 1)2 subject to 0 ≤ x ⊥ y ≥ 0

SQP method:

Start at (1, 1)

(x2, y2) = (1/2, 1/2)

(x3, y3) = (1/2k , 1/2k)

... linear convergence to (0, 0)

... multipliers →∞

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y

x

2(x−1) + (y−1)

2

... not even stationary! s = (0, 1) s = (1, 0) descend!

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Example of Linear Convergence of SQP

Consider

minimizex ,y

(x − 1)2 + (y − 1)2 subject to 0 ≤ x ⊥ y ≥ 0

SQP method:

Start at (1, 1)

(x2, y2) = (1/2, 1/2)

(x3, y3) = (1/2k , 1/2k)

... linear convergence to (0, 0)

... multipliers →∞

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y

x

2(x−1) + (y−1)

2

... not even stationary! s = (0, 1) s = (1, 0) descend!

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Example of Linear Convergence of SQP

Consider

minimizex ,y

(x − 1)2 + (y − 1)2 subject to 0 ≤ x ⊥ y ≥ 0

SQP method:

Start at (1, 1)

(x2, y2) = (1/2, 1/2)

(x3, y3) = (1/2k , 1/2k)

... linear convergence to (0, 0)

... multipliers →∞

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y

x

2(x−1) + (y−1)

2

... not even stationary! s = (0, 1) s = (1, 0) descend!

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A Nonlinear Programming Approach

Replace equilibrium 0 ≤ x1 ⊥ x2 ≥ 0 by X1x2 ≤ 0 or xT1 x2 ≤ 0

⇒ standard nonlinear program (NLP)

(NLP)

minimize

xf (x)

subject to c(x) ≥ 0x1, x2 ≥ 0

X1x2 ≤ 0x

x

1

2

Advantage: standard (?) NLP; use large-scale solvers ...Snag: nonlinear program (NLP) violates standard assumptions!

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Mangasarian Fromowitz CQ fails

Mangasarian Fromowitz Constraint Qualification at feasible x :

x1 = 0, x2 > 0

⇒ x1 ≥ 0, and x2x1 ≤ 0 active

⇒ MFCQ: s1 > 0, and x2s1 < 0

1 2

x1

x2

x1

x x

> 0

< 0

MFCQ is important (minimalist) stability assumption for NLP

Failure of MFCQ implies:

1 Lagrange multiplier set unbounded ... ∇2L may blow up

2 Constraint gradients linearly dependent ... ill-conditioned steps

3 Central path does not exist ... IPMs may not work at all!

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Dependent Constraints and Unbounded Multiplier Sets

Consider the two QPECs{minimize

zfi (x , y)

subject to 0 ≤ y ⊥ y − x ≥ 0

with f1(z) = (x − 1)2 + y2 and f2(z) = x2 + (y − 1)2

Solution at (x , y)∗ = (1/2, 1/2)T

z1

f (z)1

f (z)2 z

1

z2

z2

1

1

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Dependent Constraints and Unbounded Multiplier Sets

Equivalent NLP of QPECs isminimize

zfi (z) multiplier

subject to y ≥ 0 ν ≥ 0y − x ≥ 0 λ ≥ 0y (y − x) ≤ 0 ξ ≥ 0.

with KKT conditions:(−1

1

)or

(1−1

)= λ∗

(−1

1

)− ξ∗

(−1

212

).

... active constraint normals are clearly dependent!

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Dependent Constraints and Unbounded Multiplier Sets

Since y∗ = 12 > 0 we see ν∗ = 0, and multiplier sets ...

M1 ={

(λ, ξ) | ξ ≥ 0, λ + 12ξ = 1

}M2 =

{(λ, ξ) | λ ≥ 0, −λ+ 1

2ξ = 1},

... are unbounded

λ

ξ

1

−2

1

2

1

M2

λ

ξM

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Inconsistent Linearizations

MPECs can have inconsistent linearizations arbitrarily close tostationary point

minimizez

x + y

subject to y2 ≥ 10 ≤ x ⊥ y ≥ 0.

Nice solution: (x , y)∗ = (0, 1)T multipliers λ∗ = 0.5Linearize at (x , y) = (ε, 1− δ)T with ε, δ > 0:

(1− δ)2 + 2(1− δ)(y − (1− δ)) ≥ 1 ⇒ y ≥ 1 + (1− δ)2

2(1− δ)> 1

and

(1− δ)ε+ (1− δ)(x − ε) + ε(y − (1− δ)) ≤ 0 ⇒ y ≤ 1− δ < 1

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How Else Can We Solve MPECs?

minimize

x ,yf (x , y)

subject to c(x , y) ≥ 0F (x , y) ≥ 0 and y ≥ 0yTF (x , y) = 0

Goal

Want to use the good NLP solvers, such as IPM, SQP, SLQP, ...Trouble caused by too many dependent active constraints:F (x , y) = 0 and y = 0 and yTF (x , y) = 0 ... remove one!

Two alternative approaches that use NLP solvers:

1 Relax the complementarity constraint

2 Penalize the complementarity constraint

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NLP-Based Relaxation Approach to MPECs

Formulate a relaxed NLP

(R-NLP(ρ))

minimize

x ,yf (x , y)

subject to c(x , y) ≥ 0F (x , y) ≥ 0 and y ≥ 0yTF (x , y) = ρ

... for ρ↘ 0

Given initial ρ > 0repeat

Solve (R-NLP(ρ)) for (xρ, yρ)

Reduce ρ := ρ/4until (xρ, yρ) is solution of MPEC ;

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NLP-Based Penalization Approach to MPECs

Formulate a penalized NLP

(P-NLP(ρ))

minimize

x ,yf (x , y) + π‖yTF (x , y)‖

subject to c(x , y) ≥ 0F (x , y) ≥ 0 and y ≥ 0

... for π ↗ 0 ... problem satisfies MFCQ!

Given initial π > 0repeat

Solve (P-NLP(π)) for (xπ, yπ)

Reduce π := 4πuntil (xπ, yπ) is solution of MPEC ;

Relaxation and penalization are loosely related ...

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An Even Simpler Trick Seems to WorkConsider an alternative (lazy) reformulation of MPEC

minimizex ,y

f (x , y)

subject to c(x , y) ≥ 00 ≤ y ⊥ F (x , y) ≥ 0

Introduce slack variables s:

Write F (x , y) = s as nonlinear equation

Simplify the complementarity to bilinear inequality yT s ≤ 0

Equivalent, because s, y ≥ 0 ... solvers satisfy bounds easily

Equivalent smooth nonlinear program (NLP):minimize

x ,yf (x , y)

subject to c(x , y) ≥ 0F (x , y) = s, s ≥ 0, y ≥ 0 and yT s ≤ 0

... more in the next lecture!25 / 32

Outline

1 Introduction: Stackelberg Games

2 Difficulties with MPECs

3 Stationarity Conditions for MPECsBouligand and Strong StationarityAlphabet Soup of Spurious Stationarity

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MPEC Bouligand-Stationarity

Definition (MPEC B-Stationarity)

(x∗, y∗) is B-stationary , iff d = 0 solves LPEC

minimized

g∗Td

subject to c∗ + A∗Td ≥ 0,

0 ≤ y∗ + dy ⊥ F ∗ + B∗Td ≥ 0,

where g∗ = ∇f (x∗, y∗), A∗ = ∇c(x∗, y∗), B∗ = ∇F (x∗, y∗)

B-stationarity is a structural stationarity condition

Applies stationarity to nonlinear functions

Retains structure of the problem ⇒ strong result

Absence of feasible descend directions!... similar to LP being stationary for NLP

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MPEC Strong-Stationarity

(x∗, y∗) is weakly-stationary, iff ∃ λ, µ, and ν:

g∗ − A∗λ− B∗µ−(

0ν

)= 0,

0 ≤ c∗ ⊥ λ ≥ 0,0 ≤ y∗ ⊥ F ∗ ≥ 0.

where ν ⊥ y∗ and µ ⊥ F (x , y) ... µ, ν unrestricted

Degenerate complementarity conditions:

D(z) :={i : yi = 0 = Fi (z)

}(x∗, y∗) is strongly-stationary iff

µi ≥ 0, νi ≥ 0, ∀i ∈ D∗

... equivalent to KKT conditions of equivalent NLP

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Alphabet Soup of Spurious Stationarity

(x∗, y∗) is weakly-stationary, iff ∃ λ, µ, and ν:

g∗ − A∗λ− B∗µ−(

0ν

)= 0,

0 ≤ c∗ ⊥ λ ≥ 0,0 ≤ y∗ ⊥ F ∗ ≥ 0.

where ν ⊥ y∗ and µ ⊥ F (x , y)

Degenerate complementarity: D(z) :={i : yi = 0 = Fi (z)

}A-stationary, iff µi ≥ 0 or νi ≥ 0, ∀i ∈ D∗

C-stationary, iff µiνi ≥ 0 ∀i ∈ D∗

M-stationary, iff(µi > 0 and νi > 0

)or µiνi = 0, ∀i ∈ D∗

all have trivial descend directions

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Spuriousness of C-Stationarity

Consider min (x − 1)2 + (y − 1)2 subject to 0 ≤ y ⊥ x ≥ 0:

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y

x

2(x−1) + (y−1)

2

strongly−stationary

C−stationary

(0, 0) C-stationary: µ = ν = −2 < 0!!!⇒ ∃ descend directions

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Spuriousness of A/M-Stationarity

Consider min (x − 1)2 + y3 + y2 subject to 0 ≤ y ⊥ x ≥ 0

(0, 0) M/A-stationarity: µ = 0, ν = −2 < 0!!!⇒ exists descend directions

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Alphabet Soup of Stationarity

trivial descend direction

B−stationary

strongly−stationary

C−stationary M−stationary

A−stationary

A/B/C/M/S-stationarity equivalent, iff D∗ = ∅

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What Have We Learned?

Complementarity constraints are important class of problems

Arise in many applications ... useful modeling paradigm

Students should pay more taxes than their professors

MPECs are a challenging class of problems

Violate MFCQ ⇒ unbounded multipliers, infeasiblelinearizations

NLP solvers can fail

Extended optimality conditions

B-stationarity is the best ... and most difficult

Strong stationarity is good ... but does not always hold

Many useless stationarity concepts: A-, C-, L-, M-, W- ...

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