convex relaxations in mixed-integer optimization methods ... · part i { optimization of large...

Convex Relaxations in Mixed-Integer Optimization–

Methods and Control Applications

Robin Vujanic

Institut fur Automatik (IfA)Department of Electrical Engineering

Swiss Federal Institute of Technology (ETHZ)

August 20, 2014

1 / 19

Outline

1 Introduction

2 Optimization of Large Scale Systems

3 Example: Electric Vehicles Charging Coordination

2 / 19

What are Mixed–Integer Optimization Problems /Why do we look into them

• many practical and industrial systems entail continuous quantitiesI physical measurements of voltagesI concentrationsI and positions in space

as well as discrete componentsI on/off decisionsI switchesI and logic reasoning (if, or, . . . )

• when the associated control/operation tasks are tackled withoptimization, Mixed-Integer Optimization Problems (MIPs) arise

• but the additional flexibility has a price ...

3 / 19

Computational Issues – a Practical Perspective

• experience with instances from supply chain problem

• modest size, yet memory blew up

# cust. # prod. CPLEX time (sec)

100 25 Min: 10.3 Avg: 63.5 Max: 202.5300 75 out of memory

• computational requirements depend not only on structure and size ofthe problem, but also the data

I bad in a control context

The thesis focuses on particular model structures that are of practicalinterest. For these, we derived computationally attractiveapproximation schemes, equipped with guarantees.

4 / 19

Thesis Content

Part I – Optimization of Large Scale Systems.

• Lagrangian Duality

• Applications:I Power SystemsI Supply Chain Optimization

Part II – Robust Optimization of Uncertain Systems.

• Linear Programming Relaxations

• Applications:I Scheduling under UncertaintyI PWM Systems

5 / 19

• [EEM-13]

• [MED-14]

• [CDC-14]

• [Submitted, Math. Prog.-13]

• [ACC-12]

• [ECC-13]

• [CDC-13]

Outline

1 Introduction



6 / 19

Large Scale Optimization – Problem Structure Considered

• we consider the problem

P :

min∑

i cixis.t.

∑i Hixi ≤ b

xi ∈ Xi i = 1, . . . , I

where

Xi = {xi ∈ Rri × Zzi | Aixi ≤ di}

• large collection of subsystemsI subsystem model is Xi (mixed–integer)I coupled by shared resources → coupling constraints

∑i Hixi ≤ b

• # of subsystems I � # of coupling constraints m

7 / 19

. . .Substation

...

...... ...

...

aggregated flow

Rest of theNetwork

criticalbranch

flow

Problem’s Decomposition

Obtain decomposition using duality:

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b

xi ∈ Xi

⇒ minx∑

i∈I cixi + λ′(∑

i∈I Hixi − b)s.t. xi ∈ Xi

⇒∑

i∈I

minxi∈Xi

{cixi + λ′(Hixi )

}− λ′b

︸︷︷︸.=d(λ)

• Lagrangian dual (or outer) problem:

D :

{max d(λ)s.t. λ ≥ 0

8 / 19

Solutions to the inner problem

d(λ) =∑

i∈Iminxi∈Xi

{cixi + λ′(Hixi )

}− λ′b

Consider

xi (λ?) ∈ arg min

xi∈Xi

{cixi + λ?(Hixi )

}

as candidate solution to P

Properties of xi (λ?):

• satisfy Xi constraints

• obtained “for free” as by-product of methods that solve D• distributed computations• generally infeasible in the MIP case !

I violate coupling constraints9 / 19

Primal Recovery Scheme• we show that in x(λ?) only m subsystems may be “problematic”

I technique based on Shapley–Folkman–Starr theorem [CDC ’14]I or simplex tableaux argument [MED ’13]I these arguments are used to show bounded duality gap [Ekeland ’76,

Bertsekas ’83]

• so we propose to consider instead

P :

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b − ρ

xi ∈ Xi ∀i ∈ I ,

where

ρ = m ·maxi∈I

(maxxi∈Xi

Hixi − minxi∈Xi

Hixi

)

Theorem

Then x(λ?) is feasible for P. [under some uniqueness assumptions]

10 / 19

Performance of the Recovered Solutions

• under some technical assumption, ...

Theorem

The recovered solution x(λ?) is feasible and satisfies

JP(x(λ?))− J?P ≤ (m + ||ρ||∞/ζ) ·(

maxxi∈Xi

cixi − minxi∈Xi

cixi

)

• if J?P grows linearly with |I |, and Xi uniformly bounded

J(x(λ?))− J?PJ?P

→ 0 as |I | → ∞

11 / 19

Improving Approximations – Conservatism Reduction

ρ scales with m but not with I – want to keep it as small as possible

• when couplings are determined by certain network topologies

A

B

C

D

E

F

12

3

45

6 7

8

910

11

12

13

15

1617 18

19

1420

21

222325

2627

24

28

Subsystems

A

B

C

D

E

F

7 13 14 19 2320 24 28− − − −

k-th row

[Hi]i∈Ik

1 6−

A1

H1 H2 HI

A2

AI

. . .

. . .

. . .

. . . . . . . . .

• can safely use rank([Hi ]i∈Ik ) instead of m

• generally possible to use rank(H) instead of m

12 / 19

Outline

1 Introduction



13 / 19

Electric Vehicle (EV) Charging Coordination [CDC ’14]

• expected increase of EV presence

• substantial additional stress onnetwork & equipment⇒ need charging coordination

• network administrator (DSO)can’t tackle each single unit⇒ EV aggregator

14 / 19

Aggregator’s Role

• aggregator’s control task is toassign to each EV the time slotswhen charging can occur

• compatibly with...

• local requirementsI required final State of ChargeI fixed charge ratesI battery capacity limits

• global objectivesI network congestion avoidance

(limits set by DSO)I “valley fill”, cost min., . . .

15 / 19

from [Lopes ’11]

0:00 1:30 3:00 4:30 6:00 7:300

10

20

State

ofCharge(kWh)

SoC

desired final SoC

0:00 1:30 3:00 4:30 6:00 7:30

0

0.5

1

ChargeControl

Time of the Day

Computational Experiments

• cast as large optimization problem

• solve using proposed method: duality+contractionI support extensions (e.g., vehicle–to–grid “V2G”)

• population up to 10’000 EVs

• computation times ≤ 10 sec (charge only)I greedy subproblem structure

16 / 19

Solutions – Charge and V2G

0:00 1:30 3:00 4:30 6:00 7:300

2

4

6

8

10

Time of the Day

Pow

erProfiles(M

W)

reference

iteration #80

final iteration

(a) reference tracking

14:00 17:00 20:00 23:00 2:00 5:00 8:00 11:0010

20

30

40

50

60

70

Time of the Day

TotalPow

erFlow

(MW)

base load

base + EVs load

(b) resulting “valley fill”

0:00 1:30 3:00 4:30 6:00 7:300

0.5

1

1.5

2

2.5

3

Time of the Day

Branch

Pow

erFlow

(MW)

line capacity

contracted line cap.

flow iteration #80

final iteration

(c) network limits

0:00 1:30 3:00 4:30 6:00 7:300

10

20

State

ofCharge(kWh)

SoC

desired final SoC

0:00 1:30 3:00 4:30 6:00 7:30

−1

0

1ChargeControl

Time of the Day

(d) local requirements

17 / 19

Other Examples or Applications

• supply chains optimization – partial shipments [MED ’14]

• power systems operationI control of TCLsI large fleet of generators

• portfolio optimization for small investors

• . . .

18 / 19

Questions?

• Prof. Manfred Morari

• Dr. Paul Goulart

• Dr. Sebastien Mariethoz

• Dr. Peyman MohajerinEsfahani

• Dr. Joe Warrington

• Dr. Stefan Richter

• Dr. Sumit Mitra

• Gregory Ledva

• Isik Ilber Sirmatel

• Marius Schmitt

• Prof. Debasish Chatterjee

• Prof. Federico Ramponi

• Dr. Peter Hokayem

• Dr. Apostolos Fertis

• Dr. Utz-Uwe Haus

• Dr. Alexander Fuchs

• Dr. Martin Herceg

• Robert Nguyen

19 / 19

References• [Ekeland ’76]J. P. Aubin and I. Ekeland, Estimates of the duality gap

in nonconvex optimization, Mathematics of Operations Research 1(1976), no. 3, 225-–245.• [Bertsekas ’83] Dimitri P. Bertsekas, G. Lauer, N. Sandell, and T.

Posbergh, Optimal short-term scheduling of large-scale powersystems, IEEE Transactions on Automatic Control 28 (1983), no. 1,1– 11.• [Dawande ’06] Milind Dawande, Srinagesh Gavirneni, and Sridhar

Tayur, Effective heuristics for multiproduct partial shipment models,Operations Research 54 (2006), no. 2, 337–352 (en).• [Vujanic ’13] R. Vujanic, P. Mohajerin Esfahani, P. Goulart, S.

Mariethoz and M. Morari, Vanishing Duality Gap in Large ScaleMixed-Integer Optimization: a Solution Method with Power SystemApplications, submitted to Mathematical Programming (2013).• [Lopes ’11] J. Lopes, F. Soares, and P. Almeida, Integration of

Electric Vehicles in the Electric Power System, Proceedings of theIEEE, 2011, p.168-183.

19 / 19

BACKUP SLIDES

19 / 19

Convex Relaxations of MIPs

• our methods are based on convex relaxations• given pure binary problem

P :

minx cᵀxs.t. a1x ≤ b1

a2x ≤ b2

x ∈ {0, 1}n

• if you had to solve it, which relaxation would you pick?

P1 :


a2x ≤ b2

conv(x ∈ {0, 1}n)

P2 :


conv

(a2x ≤ b2

x ∈ {0, 1}n)

or

P3 :

minx cᵀx

s.t. conv

a1x ≤ b1

a2x ≤ b2

x ∈ {0, 1}n

19 / 19

Strength of the Relaxations

X1

X2min .

∑i∈I

cixi

s.t.∑i∈I

Hixi ≤ b

Aixi ≤ dixi ∈ {0, 1}ni

X1

X2min .

∑i∈I

cixi

s.t.∑i∈I

Hixi ≤ b

Aixi ≤ diconv

(xi ∈ {0, 1}ni

)

X1

X2min .

∑i∈I

cixi

s.t.∑i∈I

Hixi ≤ b

conv


X1

X2min .

∑i∈I

cixi

s.t. conv

∑i∈I

Hixi ≤ b


19 / 19

Issues affecting inner solutions

x1

x2

1

1

f

conv(Xi)

min −x1

s.t. x1 − x2 ≤ 0.5x1 + x2 ≤ 1.5x1, x2 ∈ {0, 1}

19 / 19


x1

x2

1

1

f

conv(Xi)

min −x1

s.t. x1 − x2 ≤ 0.5x1 + x2 ≤ 1.5x1, x2 ∈ {0, 1}x?

19 / 19


x1

x2

1

1

f

conv(Xi)

min −x1

s.t. x1 − x2 ≤ 0.5x1 + x2 ≤ 1.5x1, x2 ∈ {0, 1}x?

⇒ maxλ≥0

minx1,x2∈{0,1}

−x1 + λ1(x1 − x2 − 0.5) + λ2(x1 + x2 − 1.5)

19 / 19


⇒ d? = −1 λ? = (0.25, 0.25)

x1

x2

1

1

f

conv(Xi)

min −x1

s.t. x1 − x2 ≤ 0.5x1 + x2 ≤ 1.5x1, x2 ∈ {0, 1}x?

⇒ maxλ≥0

minx1,x2∈{0,1}

−x1 + λ1(x1 − x2 − 0.5) + λ2(x1 + x2 − 1.5)

19 / 19


⇒ d? = −1 λ? = (0.25, 0.25)

x1

x2

1

1

f

conv(Xi)

min −x1

s.t. x1 − x2 ≤ 0.5x1 + x2 ≤ 1.5x1, x2 ∈ {0, 1}x?

⇒ maxλ≥0

minx1,x2∈{0,1}

−x1 + λ1(x1 − x2 − 0.5) + λ2(x1 + x2 − 1.5)

x(λ?)

19 / 19


⇒ d? = −1 λ? = (0.25, 0.25)

x1

x2

1

1

f

conv(Xi)

min −x1

s.t. x1 − x2 ≤ 0.5x1 + x2 ≤ 1.5x1, x2 ∈ {0, 1}x?

⇒ maxλ≥0

minx1,x2∈{0,1}

−x1 + λ1(x1 − x2 − 0.5) + λ2(x1 + x2 − 1.5)

x(λ?)

x?LP

19 / 19

Illustration of the Problems with the L-Relaxation

x1

x2

3x1 − x2

x1 − x2 ≥ −1− x1 + 2x2 ≤ 53x1 + x2 ≥ 36x1 + x2 ≤ 15x1, x2 ≥ 0x1, x2 ∈ Z

minxs.t.

1 2 3

1

2

3

19 / 19


x1

x2

3x1 − x2

x1 − x2 ≥ −1− x1 + 2x2 ≤ 53x1 + x2 ≥ 36x1 + x2 ≤ 15x1, x2 ≥ 0x1, x2 ∈ Z

minxs.t.

1 2 3

1

2

3

x∗

19 / 19


x1

x2

3x1 − x2x1 − x2 ≥ −1− x1 + 2x2 ≤ 53x1 + x2 ≥ 36x1 + x2 ≤ 15x1, x2 ≥ 0x1, x2 ∈ Z

minxs.t.

1 2 3

1

2

3

x∗

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

19 / 19


x1

x2


minxs.t.

1 2 3

1

2

3

x∗

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

⇒ maxλ≥0

d(λ)

19 / 19


x1

x2


minxs.t.

1 2 3

1

2

3

x∗

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

⇒ maxλ≥0

d(λ) ⇒ λ∗ = 5/3

19 / 19


x1

x2


minxs.t.

1 2 3

1

2

3

x∗

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

⇒ maxλ≥0

d(λ) ⇒ λ∗ = 5/3

⇒ X (λ∗) ={(

10

),

(02

)}

19 / 19


x∗

x1

x2


minxs.t.

1 2 3

1

2

3

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

⇒ maxλ≥0

d(λ) ⇒ λ∗ = 5/3

⇒ X (λ∗) ={(

10

),

(02

)}

conv(Xi)

19 / 19


x∗

x1

x2


minxs.t.

1 2 3

1

2

3

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

⇒ maxλ≥0

d(λ) ⇒ λ∗ = 5/3

⇒ X (λ∗) ={(

10

),

(02

)}

conv(Xi)

x(λ∗)

19 / 19


x∗

x1

x2


minxs.t.

1 2 3

1

2

3

d(λ) = minx∈X

(3x1 − x2 + λ(−1− x1 + x2)

)

⇒ maxλ≥0

d(λ) ⇒ λ∗ = 5/3

⇒ X (λ∗) ={(

10

),

(02

)}

conv(Xi)

x(λ∗)

”p∗”

19 / 19

Main Idea

The following convexification of P is strongly related to D

(PLP) :

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b

xi ∈ conv(Xi ) ∀i ∈ I

Known that:

• J?PLP= J?D [Geoffrion ’74]

We show that:

1 an optimal solution x?LP to PLP has nice propertiesI feasible w.r.t. the complicating constraintsI returns integral solutions for |I | −m subproblems

2 x?LP and x(λ?) differ in at most m subproblems

3 thus we need limited compensation to make x(λ?) feasible

19 / 19

Main Idea

The following convexification of P is strongly related to D

(PLP) :

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b


Known that:

• J?PLP= J?D (surprising!)

We show that:

1 an optimal solution x?LP to PLP has nice propertiesI feasible w.r.t. the complicating constraintsI returns integral solutions for |I | −m subproblems

2 x?LP and x(λ?) differ in at most m subproblems

3 thus we need limited compensation to make x(λ?) feasible

19 / 19

The Shapley-Folkman Theorem

Theorem (Shapley-Folkman)

Let Si , i = 1, . . . , |I | be nonempty subsets of Rm, with |I | > m, and letS = S1 + · · ·+ S|I |. Then every vector s ∈ conv(S) can be represented ass = s1 + · · ·+ s|I |, where si ∈ conv(Si ) for all i = 1, . . . , |I |, and si /∈ Si forat most m indices i .

• relatively known in economics

• Lloyd Shapley won the Nobel Prize in Economic Sciences (in ’12)

• in our case

Si = HiXi

S = ⊕i∈ISi ={∑

i∈I Hixi | xi ∈ Xi

}

19 / 19

Visualization of the Theorem

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b

xi ∈ Xi ∀i ∈ I

0 0.5 1

0

0.5

1

X1

0 0.5 1

X2

00.5

1 0

0.50

1

2

X3

0 0.5 1

0

0.5

1

X4

(a) subsystems Xi

−2 0 20

1

2

3

S1 = H1X1

0 0.5 1−3

−2

−1

0

S2 = H2X2

−1 0 1−1

0

1

2

S3 = H3X3

0 1 20

0.5

1

1.5

S4 = H4X4

(b) budget consumptionHiXi

−2 0 2 4 6−3

−2

−1

0

1

2

3

4

5

6

H1(x)

H2 (

x)

S = iHiXi

○+

b1

b2

(c) aggregated budget con-sumption

19 / 19

Visualization of the Theorem

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b


0 0.5 1

0

0.5

1

X1

0 0.5 1

X2

00.5

1 0

0.50

1

2

X3

0 0.5 1

0

0.5

1

X4

(a) subsystems Xi

−2 0 20

1

2

3

S1 = H1X1

0 0.5 1−3

−2

−1

0

S2 = H2X2

−1 0 1−1

0

1

2

S3 = H3X3

0 1 20

0.5

1

1.5

S4 = H4X4

(b) budget consumptionHiXi

−2 0 2 4 6−3

−2

−1

0

1

2

3

4

5

6

H 1(x)

H2 (

x)

S and conv(S)

b1

b2

(c) aggregated budget con-sumption

19 / 19

Result #1: Bounded Duality Gap

• Duality gap for programs structured as P is bounded:

Theorem (bound on duality gap (Ekeland ’76, Bertsekas ’83))

Assume that the sets Xi are non-empty and compact, and that for anygiven xi ∈ conv(Xi ), there exists an xi ∈ Xi such that Hi xi ≤ Hixi . Then

J?P − J?D ≤ m ·maxi∈I

(maxxi∈Xi

cixi − minxi∈Xi

cixi

)

• non-convexities do not compound each other indefinitely in P, i.e. thedistance between P and PLP does not grow

• Thus, if J?P increases linearly with |I |,

J?P − J?DJ?P

→ 0 as |I | → ∞.

19 / 19

Example #2: Partial Shipments

• distributor has to supply I customers

• often, available inventory < total demandI demand uncertainty and high storage costsI restrictions on manufacturing capacity

• thusI EITHER fully supply a smaller set of customersI OR partially supply a larger set

• problem is to decide which customer gets what product in thepresence of shipping restrictions

19 / 19

Optimization Problem Model

i = customer, j = productwi = ship to i (yes/no), S j

i = fraction shipped

max∑i

reward for wi +∑i ,j

reward for S ji

s.t.∑iS ji ≤ inventory of prod. j

if wi = 1 ⇒ ∑j S

ji ≥ min. shipment

and S ji ≤ demand of j

if wi = 0 ⇒ S ji = 0

• compare with generic model P:

P :

minx∑

i∈I cixis.t.

∑i∈I Hixi ≤ b

xi ∈ Xi ∀i ∈ I

19 / 19

Results using Proposed Method

• partial shipment problem is NP-Hard

• greedy strategies perform poorly• tailored MINTO setup also not good (see [Dawande ’06])

I at N = 300, M = 75avg opt. gap = 6.2% and computation time 6h

• generate instances of industrial size

• solve with adapted method

Proposed Method CPLEXcust. prod. gap (%)† time (s)† time (s)

100 25 0.38 18.5 Min: 10.3 Avg: 63.5 Max: 202.5300 75 0.03 62.5 –500 50 0.06 101.0 –600 50 0.06 116.6 –1000 100 0.03 245.2 –

† average over 10 instances

19 / 19

x?LP can be computed

1 solve D using a subgradient method

λ[1] = 0

λ(k+1) = P+

(λ(k) + s(k) · γ(k)

) (1)

I s(k) steplength, pick s(k) = α/k .I γ(k) a subgradient, pick γ(k) =

∑i∈I Hixi (λ

(k))− b.

2 while iterating (1), calculate average

x (k) =1

k

k∑

j=1

x(λ(j))

Theorem (primal + dual convergence)

• λ(k) → λ? ∈ Λ?,

• if x?LP is the unique solution to PLP, x (k) → x?LP.

19 / 19

Computing approximate solutions to PAssume Hixi ≥ 0, and 0 ∈ Xi (relaxed in [Vujanic ’13]).

• choices of xi always “consume” budget

Procedure to compute approx. x? to P (distributed)

1 compute x?LP using averaging

2 find I1 ⊆ I such that (x?LP)i ∈ vert(Xi )I |I1| ≥ |I | −m − 1

3 for i ∈ I1, set x?i = (x?LP)i4 for the remaining i ∈ I \ I1,

I EITHER x?i = 0I OR set b → b −∑i∈I1 Hi x

?i , and solve smaller dimensional problem

Then, x? is feasible for P, and satisfies

JP(x?I

)− J?P ≤ (m + 1) maxi∈I

maxxi∈Xi

cᵀi xi .

19 / 19

Control Strategy

• Optimal Control + Lagrangian Duality

• Optimization Problem Model

minimizee,u,v

∑i∈I

Pi (C u · ui − C v · vi )subject to Pmin ≤∑

i∈IPi (ui − vi ) ≤ Pmax

(ei , ui , vi ) ∈ Xi

with local (private) model

Xi =

eiuivi

∈ RN × Z2N

∣∣∣∣∣∣∣∣

ei = E initi + B

(E ini ui − E out

i vi)

Emini ≤ ei ≤ Emax

i

ei [N] ≥ E refi

0 ≤ ui + vi ≤ 1

• ui [k], vi [k] ∈ {0, 1} : charge/discharge decision at step k for EV i ∈ I

• Pi : charge rate

• C u[k],C v [k]: prices for charging/discharging

19 / 19

Solution Method

1 set

ρcharge = N ·maxi∈I Pi

ρV2G = 2N ·maxi∈I Pi

}⇒ Pmin = Pmin + ρ

Pmax = Pmax − ρ

2 add δui [k], δvi [k] perturbations to cost function

3 dual (outer) problemI solved using a subgradient methodI constant stepsize, decreased every 20–30 iterations

4 inner problemI charge only: optimal local solution is greedy (easy to show)I charge+V2G: deploy DP (local problems are 1D)

19 / 19

Results (5000 EVs) (1/2)

0 2 4 6 8

−15

−10

−5

0

5

10

15

time (h)

∑i∈Ipi:pow

erflow

(MW)

Iteration # 20

19 / 19


0 2 4 6 8

−15

−10

−5

0

5

10

15

time (h)

∑i∈Ipi:pow

erflow

(MW)

Iteration # 40

19 / 19


0 2 4 6 8

−15

−10

−5

0

5

10

15

time (h)

∑i∈Ipi:pow

erflow

(MW)

Iteration # 70

19 / 19


0 2 4 6 8

−15

−10

−5

0

5

10

15

time (h)

∑i∈Ipi:pow

erflow

(MW)

Iteration # 100

19 / 19


0 2 4 6 8

−15

−10

−5

0

5

10

15

time (h)

∑i∈Ipi:pow

erflow

(MW)

Iteration # 150 (last)

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0 50 100 1501.5

2

2.5

3

3.5

4

4.5

5

iteration #

dualobjective(k

euro)

0 50 100 1501

2

3

4

5

iteration #

dual

objective(k

euro)

0 50 100 1500

20

40

60

80

100

iteration #

∣ ∣

∣ ∣

(∑i∈IH

ixi−

b)+∣ ∣

∣ ∣

1(M

W)

lower bound infeas.

upper bound infeas.

(c) with disturbances

0 50 100 1500

20

40

60

80

100

iteration #

∣ ∣

∣ ∣

(∑i∈IH

ixi−

b)+∣ ∣

∣ ∣

1(M

W)

lower bound infeas.

upper bound infeas.

(d) without disturbances

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Results: Charge Only

Proposed Method CPLEXOpt. Gap (%) Solve time (sec) Solve time (sec)

# of EVs Min Avg Max Min Avg Max Min Avg Max

200 3.24 3.32 3.41 * * * 1.97 2.16 3.74350 2.21 2.44 2.58 * * * 1.13 1.79 2.30500 1.40 1.46 1.54 * * * 1.02 1.24 1.52700 1.01 1.05 1.10 0.31 0.31 0.31 1.27 1.29 1.311000 0.68 0.72 0.76 0.44 0.44 0.44 1.68 1.70 1.731500 0.46 0.47 0.49 0.67 0.70 0.70 2.39 2.42 2.452000 0.33 0.35 0.36 0.88 0.88 0.89 3.22 3.30 3.415000 0.13 0.14 0.14 2.17 2.21 2.23 8.00 8.18 8.437000 0.05 0.05 0.06 3.14 3.15 3.16 11.40 11.74 13.2510000 0.03 0.03 0.04 4.45 4.51 4.52 17.41 17.77 18.43

(*) ≤ 0.3 sec (imprecise measurements).

• all solutions recovered are feasible

• fast computation times due to greedy subproblem structure

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Results: Charge + V2G

Proposed Method CPLEXOpt. Gap (%) Solve time (min) Solve time (min)

# of EVs Min Avg Max Min Avg Max Min Avg Max

200 8.82 10.51 12.37 1.05 1.06 1.08 0.06 0.07 0.07350 2.93 3.24 3.51 1.48 1.49 1.52 1.56 6.89 15.81500 2.05 2.15 2.24 1.85 1.93 2.62 15.21(*) 65.10(*) 262.81(*)700 1.48 1.54 1.61 2.43 2.44 2.48 – – –1000 1.01 1.05 1.10 3.24 3.26 3.28 – – –1500 0.65 0.68 0.72 4.72 4.74 4.81 – – –2000 0.45 0.50 0.53 6.19 6.21 6.23 – – –5000 0.12 0.15 0.20 14.88 14.90 14.95 – – –7000 0.09 0.10 0.12 20.59 20.77 21.94 – – –10000 0.06 0.07 0.07 29.34 29.39 29.58 – – –

(*) failed to solve two instances (out of memory)(–) out of memory before attaining the desired optimality gap

• 500 EVs: CPLEX times vary from 15 min to 4.5h (6+ when unsolved)

• 10k EVs: with 500k binaries and 250k continuous vars, these are among thelargest MIPs publicly available

• CPLEX does not even finish preprocessing on those

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Solutions Performance

0 2000 4000 6000 8000 100000

2

4

6

8

10

12

14

# of EVs

optimality gap (%) (avg, min, max)

• “1/|I |” rate of decrease of optimality gap verified in both cases

19 / 19

Intuitive explanation of proof that 1) x?LP is structured and2) that x?LP and xi(λ) are connected• PLP is equivalent to

min∑

i cixis.t.

∑i Hixi ≤ b

xi ∈ conv(Xi )⇔

minp∑

i∈I∑

j∈Ji pij(cixji )

s.t.∑

i∈I∑

j∈Ji pij(Hixji ) ≤ b∑

j∈Ji pji = 1 i ∈ I

pji ≥ 0 i ∈ I , j ∈ Ji

• at most |I |+ m entries of p are > 0I p is dimension K (huge), hence requires K equalities to be determinedI in PLP, only |I |+ m equalities, so K − (|I |+ m) equalities have to be

picked form p ji ≥ 0

I hence at most |I |+ m entries p ji can be > 0 (“at most” comes from

slack for coupling constr.)

• for at least |I | −m subproblems (p?) ii = 1

• inner problem can return x ji only if its “probability” is (p?)ji > 0

• we show this using strict complementarity, which requiresuniqueness

19 / 19

Robust Schedules

Is robustification of the schedule equivalent to, e.g., increasing tasklength?

19 / 19

Robust Schedules


encode possible delay

of 1 unit by assuming

task A is length 2

19 / 19

Robust Schedules


encode possible delay

of 1 unit by assuming

task A is length 2

the algorithm produces...

19 / 19

Lagrangian Duality → Divide & Conquer

• use duality to decompose P in smaller problems

Duality D

c1 H1 A1

λ1

λ2

λI

...

xD1

xD2

xDI

xD

P1

P2

PI

b

A1

H1 H2 HI

A2

AI

. . .

. . .

. . .

. . . . . . . . .

c1 c2 cI. . . . . . . . .

b

dI

...

...

...

...

c2 H2 A2

cI HI AI

d1

d2

dI

d1

d2

• but...

xD is generally suboptimal or even infeasible in the MIP case

• solutions may violate the relaxed constraints∑

i Hixi ≤ b

19 / 19

Robust PWM

ROBUST OPEN LOOP/CLOSED LOOP

19 / 19

Linear Robust MPCFinite horizon, optimal control problem formulation

min

[N−1∑k=0

(xk − xref )TP(xk − xref ) + uTk Quk

]

s.t. x0 = x0,xk+1 = Axk + Bukxk ∈ Xk ,uk ∈ Uk

• ideally, obtain optimal control policy using e.g. DPI ”optimal decision is taken at each stage in the horizon”I uk = π(xk)I often intractable

• conservative approximation: open-loop control policyI ”decide now for the entire horizon” (plan can’t be modifed)I uk = vkI poor performance, infeasibility issues

• middle ground: affine recourseI ”decisions are affinely adjusted once disturb. are meas.”I

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Linear Robust MPCFinite horizon, robust optimal control problem formulation

min E

[N−1∑k=0

(xk − xref )TP(xk − xref ) + uTk Quk

]

s.t. x0 = x0,xk+1 = Axk + Buk + Gwk ,xk ∈ Xk ,uk ∈ Uk

∀w ∈ W




19 / 19

Linear Robust MPC

Finite horizon, optimal control problem formulation

min E[(x− xref)TP(x− xref) + uTQu

]

s.t.x = Ax0 + Bu + Gw,Exx + Euu ≤ e

}∀w ∈ W




19 / 19

Linear Robust MPC



]


}∀w ∈ W



• middle ground: affine recourseI ”decisions are affinely adjusted once disturb. are meas.”I uk = vk +

∑kj=0 Kkjxj

19 / 19

Linear Robust MPC



]


}∀w ∈ W



• middle ground: affine recourseI ”decisions are affinely adjusted once disturb. are meas.”I uk = vk +

∑kj=0 Mkjwj

19 / 19

Explicit Robust Counterpart

• can show that the robust counterpart is equivalent to

minv,M,Λ E[(x− xref)

TP(x− xref) + uTQu]

s.t. ExAx0 + (ExB + Eu)v − e ≤ −ΛT · hΛT · S = ExG + ExBM + EuMΛ ≥ 0

(R-MPCavg )

where wk is bounded by S · wk ≤ h (Goulart 2006)

• finite convex QP

19 / 19

Extension to Hybrid Systems

19 / 19

Robust Control of Hybrid Systems

• hybrid system dynamics (MLD)

xk+1 = Axk + Buk + B2δk + B3zk + Gwk

Exxk + Euuk + Eδδk + Ezzk ≤ ekδk ∈ {0, 1}nδ , zk ∈ Rnz

• δk , zk characterize hybrid behavior,I in the dynamics, e.g. switching between modesI in the constr., e.g. logic conditions on the inputs

• wish to obtain a solution (with some recourse) to

minx,u,δ,z

E[(x− xref)TP(x− xref) + uTQu

]

s.t. x = Ax0 + Bu + B2δδδ + B3z + Gw,Exx + Euu + Eδδδδδδ + Ezz ≤ e , ∀w ∈ Wδδδ ∈ {0, 1}N·nδ

(RHOCP)

19 / 19

Affine recourse on continuous inputs - R-MPChybMain Idea

Proposed idea:

• we need recourse, but affine functions cannot easily provide binaryinputs

• split the inputs into continuous inputs u and binary inputs d

xk+1 = A · xk + Bcontuk + Bbindk + B2 · δk + B3zk + Gwk

• introduce affine recourse on the continuous inputs

u := M ·w + v

Assumption on G :

• disturbances only affect the continuous dynamics

19 / 19

Affine recourse on continuous inputs - R-MPChyb

• can show that the robust counterpart is

minv,M,d,δδδ,z,Λ

f + trace(D · Cw)

s.t. e ≤ −ΛTh ,ΛTS = ExBuM + ExG + EuMΛ ≥ 0 element-wiseδδδ ∈ {0, 1}N·nδ , d ∈ {0, 1}N·nd

(R-MPChyb)

• withx.

= Ax0 + Buv + Bdd + B2δδδ + B3ze.

= Exx + Euv + Edd + Eδδδδδδ + Ezz− e

(variables in case of zero disturbance)

• and appropriate D

19 / 19

Example: DC-DC Buck Converter

Linear Model

19 / 19

Plant: the Buck Converter (BC) (1/2)

RL

RC

id(t)

L

C

RoutVin

iL(t)

VC(t)Vo(t)

Figure: DC-DC buck converter circuit

• BC regulates input voltage Vin down to desired Vo,ref

• operated by switch (controlled input)

• disturbances: |id | ≤ 0.5 · iL,ref

• state constraints: iL ≤ 2 · iL,ref

• sampling frequency: 10 kHz

19 / 19

Averaged Model (standard method)

00.20.40.60.8

1

binary inputcontrol signal

(A)

AVER

AGED

cycle

• replace δ(t) ∈ {0, 1} by duty cycle d(t) ∈ [0, 1]

• average dynamics ”when off” and ”when on” weighted by d(t), obtain

x = Ax + Bu + Gwu ∈ [0, 1]

• linear model

19 / 19

Performance of R-MPCavg

-10123

iL /iL*VC/VC

*

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

control signalbinary input

time (ms)

19 / 19

R-MPCavg – Second Experiment

0 1 2 3 4 5 60.8

1

1.2

1.4

1.6

1.8

time (ms)

VC,ref constr.VC/33V (avg)

19 / 19

Hybrid Model

Hybrid Model

19 / 19

Hybrid Model [2] (1/2)

• Closer approximation ofthe switch

• Divide time into cycles withM samples each

• New binary inputs:switch on δ+

k = 1,switch off δ−k = 1

• New continuous input: uc,k

• New auxiliary state: sk(integration of δ+ − δ−)

• switch position given bysk + uc,k

00.20.40.60.8

1

binary inputcontrol signal

00.20.40.60.8

1

time

(A)

AVER

AGED

(B)

HYBRID

cycle

Figure: PWM in the hybrid model

19 / 19

Hybrid Model [2] (2/2)

• New system equations:(

iLVCs

)k+1

=(

A11 A12 B1A21 A22 B2

0 0 1

)·(

iLVCs

)k

+(

B1 0 0B2 0 00 1 −1

)·(

ucδ+δ−

)k

+(

G1G20

)wk

• New constraints:

δ+k , δ−k ∈ {0, 1} binary inputs

0 ≤ sk ≤ 1 binary state0 ≤ sk + uc,k ≤ 1 limited input−δ−k ≤ uc,k ≤ δ+

k switching time∑M−1i=0 δ+

k+i ≤ 1∑M−1i=0 δ−k+i ≤ 1

}switching constraints

• together with iL,k ≤ 2 · iL,ref they form

xk+1 = A · xk + Bcontuk + Bbindk + B2 · δk + B3zk + Gwk

Exxk + Euuk + Eddk + Eδδk + Ezzk ≤ ek

19 / 19

Performance & Simulation results – R-MPChyb

-1

0

1

2

3 iL/iL*VC/VC

*

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

time (ms)

control signal binary inputrange of recourse

19 / 19

R-MPChyb – Second Experiment

0 1 2 3 4 5 60.8

1

1.2

1.4

1.6

1.8

time (ms)

VC/33V VC,ref constr.(hyb) VC/33V (avg)

Controller RMS deviation of Vo(t) RMS deviation of Vo(t)

1st experiment 2nd experiment

MPCavg 2.09V 15.2[V ]

MPChyb 0.88V 9.74[V ]

MPCopen loophyb 0.89V Infeasibility encountered

19 / 19

convex relaxations in mixed-integer optimization methods ... · part i { optimization of large...

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