Decomposition and Reformulation in Mixed-Integer Programming

IMA New Directions Short Course on Mathematical Optimization Jim Luedtke

Department of Industrial and Systems Engineering University of Wisconsin-Madison

August 11, 2016

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 1 / 47 2 Huge number of variables and/or constraints Just solving LP relaxation is very time-consuming Possible solution: cutting plane approach – but doesn’t address huge number of variables Another possibility: Lagrangian relaxation or column generation May be able to help with either or both of these challenges Particularly useful if enables decomposition – splitting one large problem into many smaller ones

What makes integer programs hard to solve?

1 Weak LP relaxation bounds Pruning in branch-and-bound is rare – huge search trees Possible solution: use a better formulation or add (strong) valid inequalities

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 2 / 47 Another possibility: Lagrangian relaxation or column generation May be able to help with either or both of these challenges Particularly useful if enables decomposition – splitting one large problem into many smaller ones

What makes integer programs hard to solve?

1 Weak LP relaxation bounds Pruning in branch-and-bound is rare – huge search trees Possible solution: use a better formulation or add (strong) valid inequalities

2 Huge number of variables and/or constraints Just solving LP relaxation is very time-consuming Possible solution: cutting plane approach – but doesn’t address huge number of variables

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 2 / 47 What makes integer programs hard to solve?

1 Weak LP relaxation bounds Pruning in branch-and-bound is rare – huge search trees Possible solution: use a better formulation or add (strong) valid inequalities

2 Huge number of variables and/or constraints Just solving LP relaxation is very time-consuming Possible solution: cutting plane approach – but doesn’t address huge number of variables Another possibility: Lagrangian relaxation or column generation May be able to help with either or both of these challenges Particularly useful if enables decomposition – splitting one large problem into many smaller ones

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 2 / 47 Lagrangian Relaxation Outline

1 Lagrangian Relaxation Example Lagrangian Relaxation Bounds Solving the Lagrangian Dual

2 Dantzig-Wolfe Reformulation and Column Generation

3 Branch-and-Price

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 3 / 47 If only we didn’t have those pesky Dx ≤ d constraints . . .

Lagrangian Relaxation Motivation

Consider this IP:

K IP X > z = max ck xk k=1 Dx ≤ d

Akxk ≤ bk, k = 1,...,K nK x = (x1, . . . , xK ) ∈ Z+

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 4 / 47 Lagrangian Relaxation Motivation

Consider this IP:

K IP X > z = max ck xk k=1 Dx ≤ d

Akxk ≤ bk, k = 1,...,K nK x = (x1, . . . , xK ) ∈ Z+

If only we didn’t have those pesky Dx ≤ d constraints . . .

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 4 / 47 R PK The problem decomposes: z = k=1 zk where

> n zk = max ck xk : Akxk ≤ bk, xk ∈ Z+

for k = 1,...,K

zR ≥ zIP , but dropping Dx ≤ d altogether is pretty severe

Lagrangian Relaxation Motivation

Drop the pesky constraints:

K R X > z = max ck xk k=1

Akxk ≤ bk, k = 1,...,K nK x = (x1, . . . , xK ) ∈ Z+

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 5 / 47 zR ≥ zIP , but dropping Dx ≤ d altogether is pretty severe

Lagrangian Relaxation Motivation

Drop the pesky constraints:

K R X > z = max ck xk k=1

Akxk ≤ bk, k = 1,...,K nK x = (x1, . . . , xK ) ∈ Z+

R PK The problem decomposes: z = k=1 zk where

> n zk = max ck xk : Akxk ≤ bk, xk ∈ Z+ for k = 1,...,K

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 5 / 47 Lagrangian Relaxation Motivation

Drop the pesky constraints:

K R X > z = max ck xk k=1

Akxk ≤ bk, k = 1,...,K nK x = (x1, . . . , xK ) ∈ Z+

R PK The problem decomposes: z = k=1 zk where

> n zk = max ck xk : Akxk ≤ bk, xk ∈ Z+ for k = 1,...,K zR ≥ zIP , but dropping Dx ≤ d altogether is pretty severe

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 5 / 47 Lagrangian relaxation: relax the constraints Dx ≤ d by dualizing them – adding them to the objective with a penalty for violation Problem with just constraints Ax ≤ b should be easier to solve We’ll discuss how to choose the constraints to dualize later n For simplicity we assume x ∈ Z+

Lagrangian Relaxation Lagrangian Relaxation

Simplify notation: Let A be the matrix combining all Ak submatrices, and b = (b1, . . . , bK ), c = (c1, . . . , cK )

zIP = max c>x Dx ≤ d Ax ≤ b n x ∈ Z+

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 6 / 47 Problem with just constraints Ax ≤ b should be easier to solve We’ll discuss how to choose the constraints to dualize later n For simplicity we assume x ∈ Z+

Lagrangian Relaxation Lagrangian Relaxation

Simplify notation: Let A be the matrix combining all Ak submatrices, and b = (b1, . . . , bK ), c = (c1, . . . , cK )

zIP = max c>x Dx ≤ d Ax ≤ b n x ∈ Z+

Lagrangian relaxation: relax the constraints Dx ≤ d by dualizing them – adding them to the objective with a penalty for violation

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 6 / 47 Lagrangian Relaxation Lagrangian Relaxation

Simplify notation: Let A be the matrix combining all Ak submatrices, and b = (b1, . . . , bK ), c = (c1, . . . , cK )

zIP = max c>x Dx ≤ d Ax ≤ b n x ∈ Z+

Lagrangian relaxation: relax the constraints Dx ≤ d by dualizing them – adding them to the objective with a penalty for violation Problem with just constraints Ax ≤ b should be easier to solve We’ll discuss how to choose the constraints to dualize later n For simplicity we assume x ∈ Z+

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 6 / 47 Why? Let x∗ be an optimal solution to zIP (zIP = c>x∗ and d − Dx∗ ≥ 0). z(u) ≥ c>x∗ + u>(d − Dx∗) ≥ zIP

Lagrangian Relaxation Lagrangian Relaxation n First, re-write our IP, with X := {x ∈ Z+ : Ax ≤ b}

zIP = max c>x subject to Dx ≤ d x ∈ X

m Let u ∈ R+ , and deﬁne the following Lagrangian Relaxation problem:

IP (u): z(u) = max{c>x + u>(d − Dx): x ∈ X}

Theorem m IP For any u ∈ R+ , z(u) ≥ z .

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 7 / 47 Lagrangian Relaxation Lagrangian Relaxation n First, re-write our IP, with X := {x ∈ Z+ : Ax ≤ b}

zIP = max c>x subject to Dx ≤ d x ∈ X

m Let u ∈ R+ , and deﬁne the following Lagrangian Relaxation problem:

IP (u): z(u) = max{c>x + u>(d − Dx): x ∈ X}

Theorem m IP For any u ∈ R+ , z(u) ≥ z . Why? Let x∗ be an optimal solution to zIP (zIP = c>x∗ and d − Dx∗ ≥ 0). z(u) ≥ c>x∗ + u>(d − Dx∗) ≥ zIP

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 7 / 47 Modiﬁcation with equality constraints Dx = d: variables u are free

Lagrangian Relaxation Lagrangian Dual

m For u ∈ R+ ,

z(u) = max{c>x + u>(d − Dx): x ∈ X}

Deﬁnition The problem: LD m w = min{z(u): u ∈ R+ } is called a Lagrangian dual.

Properties wLD ≥ zIP LD m w ≤ z(u) for all u ∈ R+

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 8 / 47 Lagrangian Relaxation Lagrangian Dual

m For u ∈ R+ ,

z(u) = max{c>x + u>(d − Dx): x ∈ X}

Deﬁnition The problem: LD m w = min{z(u): u ∈ R+ } is called a Lagrangian dual.

Properties wLD ≥ zIP LD m w ≤ z(u) for all u ∈ R+ Modiﬁcation with equality constraints Dx = d: variables u are free

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 8 / 47 (1) ⇒ xˆ is feasible (2) ⇒ zIP ≤ z(u) = c>xˆ + u>(d − Dxˆ) = c>xˆ

Lagrangian Relaxation Lagrangian Dual

IP (u): z(u) = max{c>x + u>(d − Dx): x ∈ X} Theorem m Let u ∈ R+ and xˆ be an optimal solution to IP (u). If, (1) Dxˆ ≤ d, and (2) u>(d − Dxˆ) = 0, then xˆ is an optimal solution to IP. Note: The second condition is necessary: xˆ could be feasible to IP, but not optimal.

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 9 / 47 Lagrangian Relaxation Lagrangian Dual

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 9 / 47 Lagrangian Relaxation Example Example: Stochastic Integer Programming

Extensive form of SIP

S SMIP > X > z = min c x + psqs ys s=1 s.t. Ax ≥ b

Tsx + Wsys = hs s = 1,...,S n1 p1 x ∈ R+ × Z+ n2 p2 ys ∈ R+ × Z+ , s = 1,...,S

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 10 / 47 Lagrangian Relaxation Example Example: Stochastic Integer Programming

Copy ﬁrst-stage variables

S SMIP X > > z = min ps(c xs + qs ys) s=1

Axs ≥ bs = 1,...,S

Tsxs + Wsys = hs s = 1,...,S S X xs = ps0 xs0 s = 1,...,S s0=1 n1 p1 xs ∈ R+ × Z+ ,s = 1,...,S n2 p2 ys ∈ R+ × Z+ , s = 1,...,S

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 10 / 47 Relax these constraints using Lagrangian Relaxation with dual vectors λ = (λ1, . . . , λS):

PS > > PS > PS L(λ) := min s=1 ps(c xs + qs ys)+ s=1 psλs xs − s0=1 ps0 xs0

Axs ≥ b s = 1,...,S

Tsxs + Wsys = hs s = 1,...,S n1 p1 xs ∈ R+ × Z+ , s = 1,...,S n2 p2 ys ∈ R+ × Z+ , s = 1,...,S

¯ PS Rewrite the objective (λ = s=1 psλs): PS ¯ > > s=1 ps c + (λs − λ) xs + qs ys

Lagrangian Relaxation Example Relax Nonanticipativity PS The constraints xs = s0=1 ps0 xs0 are called nonanticipativity constraints.

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 11 / 47 ¯ PS Rewrite the objective (λ = s=1 psλs): PS ¯ > > s=1 ps c + (λs − λ) xs + qs ys

Lagrangian Relaxation Example Relax Nonanticipativity PS The constraints xs = s0=1 ps0 xs0 are called nonanticipativity constraints. Relax these constraints using Lagrangian Relaxation with dual vectors λ = (λ1, . . . , λS):

PS > > PS > PS L(λ) := min s=1 ps(c xs + qs ys)+ s=1 psλs xs − s0=1 ps0 xs0

Axs ≥ b s = 1,...,S

Tsxs + Wsys = hs s = 1,...,S n1 p1 xs ∈ R+ × Z+ , s = 1,...,S n2 p2 ys ∈ R+ × Z+ , s = 1,...,S

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 11 / 47 Lagrangian Relaxation Example Relax Nonanticipativity PS The constraints xs = s0=1 ps0 xs0 are called nonanticipativity constraints. Relax these constraints using Lagrangian Relaxation with dual vectors λ = (λ1, . . . , λS):

PS > > PS > PS L(λ) := min s=1 ps(c xs + qs ys)+ s=1 psλs xs − s0=1 ps0 xs0

Axs ≥ b s = 1,...,S

Tsxs + Wsys = hs s = 1,...,S n1 p1 xs ∈ R+ × Z+ , s = 1,...,S n2 p2 ys ∈ R+ × Z+ , s = 1,...,S

¯ PS Rewrite the objective (λ = s=1 psλs): PS ¯ > > s=1 ps c + (λs − λ) xs + qs ys

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 11 / 47 Lagrangian Relaxation Example Relax Nonanticipativity

¯ PS Rewritten objective (λ = s=1 psλs):

S X ¯ > > ps c + (λs − λ) xs + qs ys s=1

Normalize λs so that λ¯ = 0 P Lagrangian relaxation problem decomposes: L(λ) = s psDs(λs) where

> > Ds(λs) := min (c + λs) x + qs ys s.t. Ax ≥ b, Tsx + Wsy = hs n1 p1 n2 p2 x ∈ R+ × Z+ , y ∈ R+ × Z+

Each subproblem is a deterministic mixed-integer program

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 12 / 47 Lagrangian Relaxation Example Lagrangian Dual Problem

P For any λ = (λ1, . . . , λS) with s psλs = 0,

L(λ) ≤ zSMIP

Lagrangian dual Find best lower bound:

S LD n X o w := max L(λ): psλs = 0 s=1

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 13 / 47 Theorem wLD = maxc>x : Dx ≤ d, x ∈ conv(X)

n Let P = {x ∈ R+ : Ax ≤ b} zLP = maxc>x : Dx ≤ d, x ∈ P

P ⊇ conv(X), so wLD ≤ zLP If P = conv(X), then wLD = zLP wLD < zLP is only possible if P 6= conv(X) Let’s prove it for special case: X bounded, all integer vars

Lagrangian Relaxation Lagrangian Relaxation Bounds How Good is the Bound From the Lagrangian dual? The integer program we’re trying to solve

zIP = maxc>x : Dx ≤ d, x ∈ X (IP)

n where X = {x ∈ Z+ : Ax ≤ b}

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 14 / 47 n Let P = {x ∈ R+ : Ax ≤ b} zLP = maxc>x : Dx ≤ d, x ∈ P

P ⊇ conv(X), so wLD ≤ zLP If P = conv(X), then wLD = zLP wLD < zLP is only possible if P 6= conv(X) Let’s prove it for special case: X bounded, all integer vars

Lagrangian Relaxation Lagrangian Relaxation Bounds How Good is the Bound From the Lagrangian dual? The integer program we’re trying to solve

zIP = maxc>x : Dx ≤ d, x ∈ X (IP)

n where X = {x ∈ Z+ : Ax ≤ b} Theorem wLD = maxc>x : Dx ≤ d, x ∈ conv(X)

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 14 / 47 Lagrangian Relaxation Lagrangian Relaxation Bounds How Good is the Bound From the Lagrangian dual? The integer program we’re trying to solve

zIP = maxc>x : Dx ≤ d, x ∈ X (IP)

n where X = {x ∈ Z+ : Ax ≤ b} Theorem wLD = maxc>x : Dx ≤ d, x ∈ conv(X)

n Let P = {x ∈ R+ : Ax ≤ b} zLP = maxc>x : Dx ≤ d, x ∈ P

P ⊇ conv(X), so wLD ≤ zLP If P = conv(X), then wLD = zLP wLD < zLP is only possible if P 6= conv(X) Let’s prove it for special case: X bounded, all integer vars Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 14 / 47 In general wLD < zSMIP But wLD ≥ zSLP (the usual LP relaxation) wLD at least as good as any bound obtained using cuts in single scenario subproblems In many test instances, wLD is very close to zSMIP

Lagrangian Relaxation Lagrangian Relaxation Bounds Strength of Lagrangian Dual of SMIP Theorem The Lagrangian dual bound satisﬁes

( S ) LD > X w = min c x + psys :(x, ys) ∈ conv(Xs), s = 1,...,S s=1 where for s = 1,...,S

Xs := {(x, y): Ax ≥ b, Tsx + Wsy = hs n1 p1 n2 p2 x ∈ R+ × Z+ , y ∈ R+ × Z+ }

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 15 / 47 wLD at least as good as any bound obtained using cuts in single scenario subproblems In many test instances, wLD is very close to zSMIP

Lagrangian Relaxation Lagrangian Relaxation Bounds Strength of Lagrangian Dual of SMIP Theorem The Lagrangian dual bound satisﬁes

( S ) LD > X w = min c x + psys :(x, ys) ∈ conv(Xs), s = 1,...,S s=1 where for s = 1,...,S

Xs := {(x, y): Ax ≥ b, Tsx + Wsy = hs n1 p1 n2 p2 x ∈ R+ × Z+ , y ∈ R+ × Z+ }

In general wLD < zSMIP But wLD ≥ zSLP (the usual LP relaxation)

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 15 / 47 Lagrangian Relaxation Lagrangian Relaxation Bounds Strength of Lagrangian Dual of SMIP Theorem The Lagrangian dual bound satisﬁes

( S ) LD > X w = min c x + psys :(x, ys) ∈ conv(Xs), s = 1,...,S s=1 where for s = 1,...,S

Xs := {(x, y): Ax ≥ b, Tsx + Wsy = hs n1 p1 n2 p2 x ∈ R+ × Z+ , y ∈ R+ × Z+ }

In general wLD < zSMIP But wLD ≥ zSLP (the usual LP relaxation) wLD at least as good as any bound obtained using cuts in single scenario subproblems In many test instances, wLD is very close to zSMIP Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 15 / 47 Key insight z(u) is a piecewise linear convex function of u. We already have seen

z(u) = maxc>xt + u>(d − Dxt): t = 1,...,T

where {xt : t = 1,...,T } are the ﬁnitely many points in n X = {x ∈ Z+ : Ax ≤ b}. Extends also to mixed-integer, unbounded sets (like Benders analysis)

Lagrangian Relaxation Solving the Lagrangian Dual How to solve the Lagrangian dual?

Lagrangian dual wLD = min{z(u): u ≥ 0} How to ﬁnd u∗ that solves this optimization problem?

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 16 / 47 Lagrangian Relaxation Solving the Lagrangian Dual How to solve the Lagrangian dual?

Lagrangian dual wLD = min{z(u): u ≥ 0} How to ﬁnd u∗ that solves this optimization problem? Key insight z(u) is a piecewise linear convex function of u. We already have seen

z(u) = maxc>xt + u>(d − Dxt): t = 1,...,T where {xt : t = 1,...,T } are the ﬁnitely many points in n X = {x ∈ Z+ : Ax ≤ b}. Extends also to mixed-integer, unbounded sets (like Benders analysis)

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 16 / 47 Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Lagrangian dual: A nonsmooth convex optimization problem

wLD = min{z(u): u ≥ 0}

Assume for this algorithm that X is bounded so conv(X) has no rays Recall m m Let f : R → R and u ∈ R . A vector γ(u) is called a subgradient of f at u if > m f(v) ≥ f(u) + γ(u) (v − u) ∀v ∈ R

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 17 / 47 Notes:

The values µk are step sizes that satisfy at a minimum: µk > 0 and µk → 0 as k → ∞

Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Subgradient algorithm for generic convex problem: min{f(u): u ≥ 0}

1 Initialize: u = u0 2 Iteration k ≥ 0: Calculate f(uk) and ﬁnd a subgradient γ(uk) of f at uk Step: k+1 k k u = max{u − µkγ(u ), 0}

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 18 / 47 Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Subgradient algorithm for generic convex problem: min{f(u): u ≥ 0}

1 Initialize: u = u0 2 Iteration k ≥ 0: Calculate f(uk) and ﬁnd a subgradient γ(uk) of f at uk Step: k+1 k k u = max{u − µkγ(u ), 0} Notes:

The values µk are step sizes that satisfy at a minimum: µk > 0 and µk → 0 as k → ∞

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 18 / 47 Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Subgradient algorithm for Lagrangian dual problem:

min{z(u): u ≥ 0}

1 Initialize: u = u0 2 Iteration k ≥ 0: Calculate z(uk) and ﬁnd a subgradient γ(uk) of z at uk Step: k+1 k k u = max{u − µkγ(u ), 0}

Key question: How to calculate z(uk) and γ(uk)?

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 19 / 47 z(uk) = maxc>x + (uk)>(d − Dx): x ∈ X

Subgradients of z Let xk ∈ X be an optimal solution to calculation of z(uk). Then

γ(uk) = d − Dxk

is a subgradient of z at uk.

Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Subgradient algorithm for Lagrangian dual problem:

min{z(u): u ≥ 0}

How to calculate z(uk) and γ(uk)?

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 20 / 47 Subgradients of z Let xk ∈ X be an optimal solution to calculation of z(uk). Then

γ(uk) = d − Dxk

is a subgradient of z at uk.

Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Subgradient algorithm for Lagrangian dual problem:

min{z(u): u ≥ 0}

How to calculate z(uk) and γ(uk)?

z(uk) = maxc>x + (uk)>(d − Dx): x ∈ X

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 20 / 47 Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient

Subgradient algorithm for Lagrangian dual problem:

min{z(u): u ≥ 0}

How to calculate z(uk) and γ(uk)?

z(uk) = maxc>x + (uk)>(d − Dx): x ∈ X

Subgradients of z Let xk ∈ X be an optimal solution to calculation of z(uk). Then

γ(uk) = d − Dxk is a subgradient of z at uk.

Jim Luedtke (UW-Madison) Decomposition Methods Lecture Notes 20 / 47 ≥ c>xk + v>(d − Dxk) = c>xk + (uk)>(d − Dxk) + v>(d − Dxk) − (uk)>(d − Dxk) = z(uk) + (v − uk)>(d − Dxk) = z(uk) + (v − uk)>γ(uk)

Lagrangian Relaxation Solving the Lagrangian Dual Option 1: Subgradient Subgradients of z Let xk ∈ X be an optimal solution to calculation of z(uk). Then