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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