daniel bienstock abhinav verma - columbia universitydano/talks/latk.pdf · a conceptual game we are...

Multi-stage integer programs in power grids

Daniel BienstockAbhinav Verma

Columbia University, New York

Daniel Bienstock Abhinav Verma ( Columbia University, New York)Multi-stage integer programs in power grids 1 / 43

A conceptual game

We are given a graph with

Multicommodity demands: for 1 ≤ h ≤ K , demand dh betweenvertices sh and th

For every edge {i, j}, a “weight” wij and a “threshold” τij .

We will play T steps:

A conceptual game

Step i < T of the game

For 1 ≤ h ≤ K , we route an amount d ′h ≤ dh of commodity h.

The routing must use minimum-weight paths between sh andth, for 1 ≤ h ≤ K .

Ater routing all commodities, if, on any edge {i, j} the total flowexceeds the threshold τij , then the edge is deleted (from nowon).

Last step of the game

For 1 ≤ h ≤ K , we route an amount dTh ≤ dh of commodity h.

The total flow on any edge must not exceed the threshold.

The value of game is∑

h dTh ; which we want to maximize.

Gist of the game: use early iterations to get rid of small-capacity edges.

WARNING: FLUFF

Some recent national-scale blackouts

August 2003: North America. 50 million people affected duringtwo days; New York City loses power

September 2003: Switzerland-France-Italy. 57 million peopleaffected during one day; Italy loses power

Why?Too much consumption? No. Inadequate generation? No.

→ The transmission network failed

U.S.-Canada joint commission: The leading cause of the blackoutwas “inadequate system understanding”

A power grid has three components

TRANSMISSION

GENERATION

DISTRIBUTION

END OF FLUFF

Linear power flow model

We are given a network G with:

A set of S of supply nodes (the “generators”); for each generatori an “operating range” 0 ≤ SL

i ≤ SUi ,

A set D of demand nodes (the “loads”); for each load i a“maximum demand” 0 ≤ Dmax

For each arc (i, j) values xij and uij .

i ≤ SUi ,

Feasible power flows

A power flow is a solution f , θ to:∑ij fij −

∑ij fji = bi , for all i , where

SLi ≤ bi ≤ SU

i OR bi = 0 for i ∈ S,

0 ≤ −bi ≤ Dmaxi for i ∈ D,

and bi = 0, otherwise.

xij fij − θi + θj = 0 for all (i, j). (Ohm’s equation)

Lemma Given a choice for b, the system has a unique solution.

The solution is feasible if |fij | ≤ uij for every (i, j).

Its throughput is∑

i∈D −bi .

SLi ≤ bi ≤ SU

i∈D −bi .

SLi ≤ bi ≤ SU

i∈D −bi .

SLi ≤ bi ≤ SU

i∈D −bi .

SLi ≤ bi ≤ SU

i∈D −bi .

Cascades

An initial fault (= edge deletion) or set of faults occurs

We now have a new network

In the new network, some of the power flows may exceed theedge capacities

This may trigger a new set of faults, etc.

Cascades

FAULTINITIAL

NEWTOPOLOGY

POWER FLOWSNEW

MOREFAULTS

This talk: the attack problem

Delete a small set of arcs, such that in the resulting network allfeasible flows have small throughput

Used to model “natural” blackouts

“Small” throughput: we satisfy less than some amount Dmin oftotal demand

“Small” set of arcs = very small

Delete 1 arc = the “N-1” problem

Of interest: delete k = 2, 3, 4, . . . edges

Naive enumeration blows up

Two types of successful attacks

Type 1: Network becomes disconnected with a mismatch of supplyand demand.

Two types of successful attacks

Type 2: Uniqueness of power flows means exceeded capacities orinsufficient supply.

A little game:

The controller’s problem: Given a set A of arcs that has beendeleted by the attacker, choose a set G of generators to operate, so asto feasibly meet demand (at least) Dmin.

The attacker’s problem: Choose a set A of arcs to delete, so as todefeat the controller, no matter how the controller chooses G.

A little game:

The controller’s problem for a given choice of generators

Given a set A of arcs that has been deleted by the attacker, and achoice G of which generators to operate, set demands and suppliesso as to feasibly meet total demand (at least) Dmin.

This a linear program:

The controller’s problem for a given choice of generators

Given a set A of arcs that has been deleted by the attacker, and achoice G of which generators to operate, set demands and suppliesso as to feasibly meet total demand (at least) Dmin.

This a linear program:

tA(G).= min t

Subject to:∑ij fij −

∑ij fji − bi = 0, for all nodes i ,

Smini ≤ bi ≤ Smax

i for i ∈ G, 0 ≤ −bi ≤ Dmaxi for i ∈ D

bi = 0 otherwise.

xij fij − θi + θj = 0 for all (i, j) /∈ A

−∑

i∈D bi + Dmin t ≥ 2Dmin

uij t ≥ |fij | for all (i, j) /∈ A

fij = 0 for all (i, j) ∈ A

Lemma: tA(G) > 1 iff the attack is successful against the choice G.

tA(G).= min t

bi = 0 otherwise.

−∑

tA(G).= min t

bi = 0 otherwise.

−∑

tA(G).= min t

bi = 0 otherwise.

−∑

tA(G).= min t

bi = 0 otherwise.

−∑

tA(G).= min t

bi = 0 otherwise.

−∑

tA(G).= min t

bi = 0 otherwise.

−∑

tA(G).= min t

bi = 0 otherwise.

−∑

for all (i, j) ∈ A, t ≥ 1 + |fij |/uij

Attack problem

min∑

ij zij

Subject to:

zij = 0 or 1, for all arcs (i, j), (choose which arcs to delete)

tsuppt(z)(G) > 1, for every subset G of generators.

[ suppt(v) = support of v]

→ Use dual to represent tsuppt(z)(G)

Attack problem

min∑

ij zij

Subject to:

[ suppt(v) = support of v]

→ Use dual to represent tsuppt(z)(G)

Building the dual

tA(G).= min t

∑ij fji − bi = 0, for all nodes i , (αi)

i for i ∈ G,

0 ≤ −bi ≤ Dmaxi for i ∈ D

bi = 0 otherwise.

xij fij − θi + θj = 0 for all (i, j) /∈ A (βij)

−(∑

i∈D bi)/Dmin + t ≥ 2

uij t ≥ |fij | for all (i, j) /∈ A (pij , qij)

uij t ≥ uij + |fij | for all (i, j) ∈ A (r+ij , r−

Building the dual

∑ij fij −

ij )∑ij βij −

∑ji βji = 0 ∀i

αi − αj + xijβij = pij − qij + r+ij − r−

ij ∀(i, j)

Building the dual

∑ij fij −

ij )∑ij βij −

∑ji βji = 0 ∀i

ij ∀(i, j)

Building the dual

∑ij fij −

ij )∑ij βij −

∑ji βji = 0 ∀i

ij ∀(i, j)

Again:

∑ij βij −

∑ji βji = 0 ∀i

ij ∀(i, j)

0-1 -ify: form mip-dual

pij + qij ≤ Mij(1 − zij)

r+ij + r−

ij ≤ M ′ijzij

→ “big M” formulation: what’s the problem

Again:

∑ij βij −

∑ji βji = 0 ∀i

ij ∀(i, j)

r+ij + r−

ij ≤ M ′ijzij

Again:

∑ij βij −

∑ji βji = 0 ∀i

ij ∀(i, j)

r+ij + r−

ij ≤ M ′ijzij

The dual has structure

Nβ = 0

X β + NT α = p − q + r+ − r−

N = node-arc incidence matrix; X = diag(xij).

Lemma - Given p, q, r+, r−, the above system has a unique solutionin αi − αj , β.

The dual has structure

Nβ = 0

X β + NT α = p − q + r+ − r−

Lemma - Given p, q, r+, r−, the above system has a unique solutionin αi − αj , β.

Connected network:

Nβ = 0

X β + NT α = p − q + r+ − r−

‖X−1/2NT α‖ = ‖X−1/2NT (NX−1NT )−1NX−1(p − q + r+ − r−)‖

( N = N minus one row)

Mij =√xij max(k ,l) (

√xkl ukl)

Connected network:

Nβ = 0

X β + NT α = p − q + r+ − r−

‖X−1/2NT α‖ = ‖X−1/2NT (NX−1NT )−1NX−1(p − q + r+ − r−)‖

( N = N minus one row)

Mij =√xij max(k ,l) (

√xkl ukl)

A formulation for the attack problem

min∑

ij zij

Subject to:

min∑

ij zij

Subject to:

value of dual mip (G) > 1, for every subset G of generators.

→ very large

min∑

ij zij

Subject to:

value of dual mip (G) > 1, for every subset G of generators.

→ very large

Cutting planes

If z∗ is not a successful attack, then there is a “witness” set G ofgenerators such that tsupp(z∗)(G) ≤ 1.

So we can try to separate from the convex hull of successful attacks:

Combinatorial cuts? Don’t know of any yet

Conic cuts: they exist, but the ones we know are not very strong?

→ Farkas’ Lemma cuts, i.e. Benders’ cuts

Cutting planes

Benders’ cuts

For a given 0 − 1 vector z, and a set of generators G,

tsuppt(z)(G) = max µT y

Ay ≤ bz

y ∈ P

for some vectors µ, b, matrix A and polyhedron P,(all dependent on G, but not z).

→ If tsuppt(z)(G) ≤ 1, use LP duality to separate z,

getting a cut αtz ≥ β violated by z.

Benders’ cuts

For a given 0 − 1 vector z, and a set of generators G,

tsuppt(z)(G) = max µT y

Ay ≤ bz

y ∈ P

for some vectors µ, b, matrix A and polyhedron P,(all dependent on G, but not z).

→ If tsuppt(z)(G) ≤ 1, use LP duality to separate z,

getting a cut αtz ≥ β violated by z.

Cutting planes

Could set up a pure cutting-plane algorithm, or even branch-and-cut,but ...

Cutting planes

Could set up a pure cutting-plane algorithm, or even branch-and-cut,but ...

Algorithm outline

→ Maintain a “master (attacker) MIP”:

Made up of valid inequalities (for the attacker)Initially empty

Iterate:

1. Solve master MIP, obtain 0 − 1 vector z∗.

2. Solve controller problem to test whether supp(z∗) is a successfulattack:

If successful, then z∗ is an optimal solution

If not, then for some set of generators G, tsupp(z∗)(G) ≤ 1.

3. Add to the master MIP the entire formulation for the dual mipcorresponding to G, and go to 1.

Algorithm outline

Iterate:

Algorithm outline

Iterate:

Algorithm outline

Iterate:

Algorithm outline

Iterate:

Algorithm outline

Iterate:

Algorithm outline

Iterate:

Amended algorithm

3a. Augment the formulation for the master MIP by adding all theformulation for the dual mip corresponding to G, and go to 1. – do this“at the beginning”, else:3b. Separate z∗ using a Benders’ cut, go to 1.

Amended algorithm

3a. Augment the formulation for the master MIP by adding all theformulation for the dual mip corresponding to G, and go to 1. – do this“at the beginning”, else:3b. Separate z∗ using a Benders’ cut, go to 1.

98 nodes 203 arcs, 15 generators

Surviving |A|Demand Fraction 2 3 4

(2) 223 (11) 6540.94 Not Enough Kill

(2) 212 (16) 9817 (21) 135400.92 Not Enough Not Enough Kill

(2) 181 (11) 6912 –0.90 Not Enough Not Enough

98 nodes 203 arcs, 10 generators

Surviving |A|Demand Fraction 2 3 4

(2) 177 (30) 555 –0.89 Not Enough Kill

(2) 233 (13) 5931 (73) 82730.86 Not Enough Not Enough Kill

(2) 152 (41) 15211 –0.84 Not Enough Not Enough

PART II

A better game

We are given a power grid with generators, demands, capacities,impedances, etc.

... and something bad has happened: some flows exceed capacities

We will play T steps of a one-player game:

A better game

Step 1, 2, · · · , T − 1 of the game

1 If all flows are within capacities, do nothing

2 Otherwise, we reduce all demands by a fixed factor0 < λ < 1.

3 Ater the new flows are instantiated, any edge whose flow exceedsthe edge’s capacity is deleted.

(more complex versions of the game ...)

Step T of the game

In the current network, we compute a feasible flow that satisfies amaximum amount of total demand.

Step 1, 2, · · · , T − 1 of the game

Step T of the game

Step 1, 2, · · · , T − 1 of the game

Step T of the game

Step 1, 2, · · · , T − 1 of the game

Step T of the game

Step 1, 2, · · · , T − 1 of the game

Step T of the game

600 nodes, 1268 edges, 25 generators, 344 demands, 6 rounds

0 0.2 0.4 0.6 0.8 1

throughput

AC Power flows

Voltage at a node k is of the form Uk ejθk , where j =√

Power flowing on edge {k , m} equals pkm + jqkm, where

(A) pkm = U2k gkm − Uk Um gkm cos θkm − Uk Um bkm sin θkm

(B) qkm = −U2k (bkm + bsh

km) + Uk Um bkm cos θkm− Uk Um gkm sin θkm

Here, θkm.= θk − θm

gkm, bkm, bshkm are known parameters (series conductance, series

reactance, shunt susceptance)Daniel Bienstock Abhinav Verma ( Columbia University, New York)Multi-stage integer programs in power grids 41 / 43

AC Power flows

Net power injected at node k equals Pk + jQk , where

(C) Pk = Σ{k ,m}pkm (active power)

(D) Qk = Σ{k ,m}qkm (reactive power)

Node k is a generator ⇒ Pk > 0; if k is a load ⇒ Pk < 0.Similarly with the reactive powers.

Power flow problem: given known loads, find values for all thevariables Uk and θk so that equations (A) - (D) hold

Boundary condition: the angles θk can be fixed at some nodes

Linearization – DC “model”

Power flowing on edge {k , m} equals pkm + jqkm, where

(A) pkm = U2k gkm − Uk Um gkm cos θkm − Uk Um bkm sin θkm,

(θkm = θk − θm)

Approximation:θkm is small, and hence sin θkm ≈ θkm.

Uk ≈ 1, for each node k ; gkm ≈ 0(A’) xkm pkm − θk + θm = 0 ( xkm = 1/bkm )

(C’) Σ{k ,m}pkm = net supply of power at k(power) flow conservation

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