Encodings in Mixed Integer Linear...

ORC Seminar, December 2012 – Cambridge, Massachusetts

Encodings in Mixed Integer Linear Programming

Juan Pablo VielmaMassachusetts Institute of Technology

0

Mixed Integer Binary Formulations

MIP Formulations = Model Finite Alternatives

/302////302222222

f(x,y)

y

x

Pi

x ∈n⋃

i=1

Pi ⊆ Rd

0

Textbook Formulation

/303 0000000////////3333333000000033333333

4∑i=1

diλi = x,4∑

i=1

f(di)λi = z

4∑i=1

λi = 1, λi ≥ 0

3∑i=1

yi = 1, yi ∈ {0, 1}

λ1 ≤ y1, λ2 ≤ y1 + y2

λ3 ≤ y2 + y3, λ4 ≤ y3Formulation for f(x)=z

d1 d2 d3 d4

f(d4)

0

f(d1)

f(d2)

f(d3)

0

Textbook Formulation

/303 0000000////////3333333000000033333333

4∑i=1

diλi = x,4∑

i=1

f(di)λi = z

4∑i=1

λi = 1, λi ≥ 0

3∑i=1

yi = 1, yi ∈ {0, 1}

λ1 ≤ y1, λ2 ≤ y1 + y2

λ3 ≤ y2 + y3, λ4 ≤ y3Formulation for f(x)=z

d1 d2 d3 d4

f(d4)

0

f(d1)

f(d2)

f(d3)

“Weak”

0

Better Formulation

/304

Formulation for f(x)=z

d1 d2 d3 d4

f(d4)

0

f(d1)

f(d2)

f(d3)

d0 +3∑

i=1

(di+1 − di)δi = x,

f(d0) +3∑

i=1

(f(di+1)− f(di))δi = z

δ3 ≤ y2 ≤ δ2 ≤ y1 ≤ δ1

yi ∈{0, 1}

0

Better Formulation

/304

Formulation for f(x)=z

d1 d2 d3 d4

f(d4)

0

f(d1)

f(d2)

f(d3)

d0 +3∑

i=1

(di+1 − di)δi = x,

f(d0) +3∑

i=1

(f(di+1)− f(di))δi = z

δ3 ≤ y2 ≤ δ2 ≤ y1 ≤ δ1

yi ∈{0, 1}Integral

0

Solve Times in CPLEX 11

/305

Weak Strong Mystery

0/305Transportation Problems in V., Ahmed and Nemhauser ‘10.

0

Solve Times in CPLEX 11

/305

0 s

500 s

1000 s

1500 s

2000 s

4 8 1632

Weak Strong Mystery

0/305Transportation Problems in V., Ahmed and Nemhauser ‘10.

0

Outline

MIP v/s constraint branching.

“Have your cake and eat it too” formulation

Step 1: Encoding alternatives.

Step 2: Combine with strong “standard” formulation.

Summary, Extensions and More.

/306

min |x|s.t.

x ∈ {αi}ni=1

00

Formulating Discrete Alternatives

//30//3077α1 α2 α3 α4 α5

min |x|s.t.

x ∈ {αi}ni=1

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

00


//30//3077α1 α2 α3 α4 α5

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

00


//30//30888

α1 α2 α3 α4 α5

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

00


//30//30888

α1 α2 α3 α4 α5

1

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

00


//30//30888

α1 α2 α3 α4 α5IPopt = 1, LPopt = 0

1

α +

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

00


//30//30888


1

Solve by binary Branch-and-Bound:

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

Branch on λ2

00


//30//30888


1


min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

Branch on λ2

00


//30//30888


1

Solve by binary Branch-and-Bound:• λ2 = 1→ Feasible with |x| = 1

+α

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

Branch on λ2

00


//30//30888


1


• λ2 = 0→ Best Bound = 0

+α

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

Branch on λ2

Branch on λ2, λ4, λ5 → Best Bound = 000


//30//30888


1


• λ2 = 0→ Best Bound = 0

α +

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

Worst case: n/2 branches to solve

(i.e. 2n/2 B-and-B nodes!).

00


//30//30888


1


min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

/30/30



1

Solve by constraint B-and-B:

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

/30/30



1


Branch on λ1 + λ2

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

/30/30



1

α α

+Solve by constraint B-and-B:

Branch on λ1 + λ2• λ1 + λ2 = 1→ Feasible with |x| = 1

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

/30/30



1



• λ1 + λ2 = 0→ Feasible with |x| = 1

+

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1

λ ∈ {0, 1}n

/30/30



1



• λ1 + λ2 = 0→ Feasible with |x| = 1

Never more than one branch (2 nodes).

Ryan and Foster, 1981.

Discrete Alternatives: SOS1 branching of Beale and Tomlin 1970. Also SOS2 (B. and T, 70) and piecewise linear functions (Tomlin 1981).

SOS1:

Problem: Need to re-implement advanced branching selection (e.g. pseudocost ).

00

Constraint Branching is the Solution?

/30/30111010

∑t

i=1λi = 1 or

∑t

i=1λi = 0

�λi = 0 ∀i > t or λi = 0 ∀i ≤ t

0

Binary v/s Specialized Branching

/30111Weak Integer SOS2 Branching Mystery Integer

0


/30111

0 s37.5 s75 s112.5 s150 s

Med.Large

Weak Integer SOS2 Branching Mystery Integer

CPLEX 9: Basic SOS2 branching implementation (graph from Nemhauser, Keha and V. ‘08)

0


/30111

0 s37.5 s75 s112.5 s150 s

Med.Large

0 s500 s1,000 s1,500 s2,000 s

1632

Weak Integer SOS2 Branching Mystery Integer

CPLEX 9: Basic SOS2 branching implementation (graph from Nemhauser, Keha and V. ‘08)

CPLEX 11: Improved SOS2 branching implementation (graph from Nemhauser, Ahmed and V. ‘10)

Formulation Step 1:Encoding Alternatives

∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

{bi}n

i=1= {0, 1}log2 n

0

Formulation for Discrete Alternatives

Li and Lu 2009, Adams and Henry 2011, V. and Nemhauser 2008.

Sommer, TIMS 1972.

Log = Binary Encoding

Other choices of lead to standard and incremental formulations

/30113

{bi}n

i=1

Unary Encoding

000000000000000000000000/////////////////////////333333333333333333333333333000000000000000000000000011111111111111111111111111111111111111111111111111444444444444444444444444444

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 1 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1 0 00 0 0 0 0 0 1 00 0 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

λ = y,

�λi = yi

∑8

i=1λi = 1,

λ ∈ R8, y ∈ {0, 1}8

0

Binary Encoding

/30115

⎛⎝ 1 1 1 1 0 0 0 0

1 1 0 0 1 1 0 01 0 1 0 1 0 1 0

⎞⎠λ = y,

∑8

i=1λi = 1,

λ ∈ R8, y ∈ {0, 1}3

∑8

i=1λi = 1,

λ ∈ R8, y ∈ {0, 1}7

⇓

0

Incremental Encoding

/30116

y1 ≥ y2 ≥ . . . ≥ y7

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 1 1 1 1 1 1 10 0 1 1 1 1 1 10 0 0 1 1 1 1 10 0 0 0 1 1 1 10 0 0 0 0 1 1 10 0 0 0 0 0 1 10 0 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

λ = y,

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000

Example: Unary Encoding

//30117

1

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

1

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

+

1

=

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

1

=

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

y1 =

⎛ ⎞⎛0

⎞ ⎛1

⎞ ⎛0

⎞⎛0

⎞

1

=

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

1

⎛ ⎞

y1 =

⎛ ⎞⎛0

⎞⎛0

⎞ ⎛1

⎞ ⎛0

⎞

1

=

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

y1 =

⎛ ⎞⎛0

⎞ ⎛1

⎞ ⎛0

⎞⎛0

⎞

0

⎛ ⎞ ⎛ ⎞⎛ ⎞

1

=

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠

y1 = y4 = 0

⎛0⎞ ⎛

0⎞⎛ ⎞

⎝0

⎠ ⎝0

⎠ ⎝ ⎠

1

=

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

Need n/2branches

l

min |x|s.t. ∑n

i=1λiαi = x∑n

i=1λi = 1∑n

i=1biλi = y

λ ∈ Rn+

y ∈ {0, 1}m

000


//30117

⎛⎜⎜⎝1000

⎞⎟⎟⎠

⎛⎜⎜⎝0100

⎞⎟⎟⎠

⎛⎜⎜⎝0001

⎞⎟⎟⎠

⎛⎜⎜⎝0010

⎞⎟⎟⎠{ {

n/2 n/2

2k − 1

000

Example: Binary Encoding

//301188

1

⎛⎝001

⎞⎠

⎛⎝010

⎞⎠

⎛⎝000

⎞⎠ =

⎛⎝y1y2y3

⎞⎠

. . .

{

2k − 1

000


//301188

1

⎛⎝001

⎞⎠

⎛⎝010

⎞⎠

⎛⎝000

⎞⎠ =

⎛⎝y1y2y3

⎞⎠

. . .

{

y1 = y2 = 0

⎛⎝

⎞⎠

⎛⎝⎛⎛ ⎞

⎠

2k − 1

000


//301188

1

⎛⎝001

⎞⎠

⎛⎝010

⎞⎠

⎛⎝000

⎞⎠ =

⎛⎝y1y2y3

⎞⎠

. . .

{

y1 = y2 = 0

⎛⎝

⎞⎠

⎛⎝⎛⎛ ⎞

⎠

Best Bound = 0 unless:

yi = 0 ∀i

2k − 1

Need k = log2 nbranches (n nodes)

000


//301188

1

⎛⎝001

⎞⎠

⎛⎝010

⎞⎠

⎛⎝000

⎞⎠ =

⎛⎝y1y2y3

⎞⎠

. . .

{

y1 = y2 = 0

⎛⎝

⎞⎠

⎛⎝⎛⎛ ⎞

⎠

Best Bound = 0 unless:

yi = 0 ∀i

000

Example: Incremental Encoding

//301199

1

000


//301199

1

⎛⎜⎜⎝0000

⎞⎟⎟⎠

⎛⎜⎜⎝1100

⎞⎟⎟⎠

⎛⎜⎜⎝1111

⎞⎟⎟⎠

⎛⎜⎜⎝1110

⎞⎟⎟⎠ =

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

⎛⎜⎜⎝1000

⎞⎟⎟⎠

000


//301199

1

⎛⎜⎜⎝0000

⎞⎟⎟⎠

⎛⎜⎜⎝1100

⎞⎟⎟⎠

⎛⎜⎜⎝1111

⎞⎟⎟⎠

⎛⎜⎜⎝1110

⎞⎟⎟⎠ =

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

⎛⎜⎜⎝1000

⎞⎟⎟⎠

y3 = 1 ∨ y3 = 0

⎝ ⎠⎜⎝⎜⎜ ⎟⎟⎟⎟⎠ ⎜⎜⎜⎜ ⎟⎟⎟⎟ ⎜⎜⎜⎜⎝1

⎟⎟⎟⎟⎠⎜⎜⎜⎜⎝ ⎟⎟⎟⎟⎠ =⎜⎜⎜⎜⎝ ⎟⎠⎟⎟⎜⎜⎜⎜⎝ ⎟⎟⎟⎟⎠

yi∗ = 0 ∨ yi∗ = 1

Only need1 branch!

Best Bound = 1 if:

000


//301199

1

⎛⎜⎜⎝0000

⎞⎟⎟⎠

⎛⎜⎜⎝1100

⎞⎟⎟⎠

⎛⎜⎜⎝1111

⎞⎟⎟⎠

⎛⎜⎜⎝1110

⎞⎟⎟⎠ =

⎛⎜⎜⎝y1y2y3y4

⎞⎟⎟⎠

⎛⎜⎜⎝1000

⎞⎟⎟⎠

y3 = 1 ∨ y3 = 0

⎝ ⎠⎜⎝⎜⎜ ⎟⎟⎟⎟⎠ ⎜⎜⎜⎜ ⎟⎟⎟⎟ ⎜⎜⎜⎜⎝1

⎟⎟⎟⎟⎠⎜⎜⎜⎜⎝ ⎟⎟⎟⎟⎠ =⎜⎜⎜⎜⎝ ⎟⎠⎟⎟⎜⎜⎜⎜⎝ ⎟⎟⎟⎟⎠

0

Induced Constraint Branching

/30220

⎛⎝ 0 1 1 1

0 0 1 10 0 0 1

⎞⎠λ = y

Incremental

(0 0 1 10 1 0 1

)λ = y

Binary

0


/30220

⎛⎝ 0 1 1 1

0 0 1 10 0 0 1

⎞⎠λ = y

Incremental

→ λ1 =λ2 = 0

or

λ3 =λ4 = 0

SOS1 Branching

(0 0 1 10 1 0 1

)λ = y

Binary

⎛⎝⎛ ⎞

⎠⎞

0


/30220

⎛⎝ 0 1 1 1

0 0 1 10 0 0 1

⎞⎠λ = y

Incremental

→ λ1 =λ2 = 0

or

λ3 =λ4 = 0

SOS1 Branching

(0 0 1 10 1 0 1

)λ = y

Binary

→ λ1 =λ2 = 0

or

λ3 =λ4 = 0

SOS1 Branching( )

0


/30220

⎛⎝ 0 1 1 1

0 0 1 10 0 0 1

⎞⎠λ = y

Incremental

→ λ1 =λ2 = 0

or

λ3 =λ4 = 0

SOS1 Branching

(0 0 1 10 1 0 1

)λ = y

Binary

→( )Odd/Even Branching

λ1 =λ3 = 0

or

λ2 =λ4 = 0

Formulation Step 2:Combining with Strong Formulation

n∑i=1

∑v∈ext(P i)

vλiv = x

∑v∈ext(P i)

λiv =

n∑i=1

yi = 1

y ∈ {0, 1}, λiv ≥ 0

n∑i=1

yi

yi

0

Long Lost Integral Formulation

Jeroslow and Lowe 1984./30222

{P i

}n

i=1polytopes

x ∈n⋃

i=1

P i ⇔

Also for general polyhedrawith common recession cones.

1

n

n∑i=1

∑v∈ext(P i)

vλiv = x

∑v∈ext(P i)

λiv =

n∑i=1

yi = 1

y ∈ {0, 1}, λiv ≥ 0

n∑i=1

yi

yi

0

Combining with Alternative Encoding


{P i

}n

i=1polytopes

x ∈n⋃

i=1

P i ⇔


1

n

n∑i=1

∑v∈ext(P i)

vλiv = x

∑v∈ext(P i)

λiv =

n∑i=1

yi = 1

y ∈ {0, 1}, λiv ≥ 0

n∑i=1

yn∑

i=1

∑v∈ext(P i)

biλiv

0



{P i

}n

i=1polytopes

x ∈n⋃

i=1

P i ⇔


m

1

n∑i=1

∑v∈ext(P i)

vλiv = x

∑v∈ext(P i)

λiv =

n∑i=1

yi = 1

y ∈ {0, 1}, λiv ≥ 0

n∑i=1

yn∑

i=1

∑v∈ext(P i)

biλiv

0


/30224

{P i

}n

i=1polytopes

x ∈n⋃

i=1

P i ⇔


m

1

V., Ahmed and Nemhauser 2010; V. 2012.

Univariate Transportation Problems

00//3300222255

DiscontinuousPiecewise Linear

Disc. PWL+

“Semicontinuous”

0

Piecewise Linear

/30226

Unary Incremental / Strong Integer Binary / Mystery Integer

0

Piecewise Linear

/30226

0

20

40

60

80

816

32


0

Piecewise Linear

/30227

0

1000

2000

3000

4000

48

16


000

Combining Codes for Multivariate

//3022288

01

01

0 1

y1

y2

y3

000


//3022299

01

01

0 1

01

0

0 1 0 1

000


//3033330010

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


/303311

0

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


/303311

00

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//30333221

0 0

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//3033333

10 0

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//3033333

10 0 1

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//3033344

10 0 1

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//3033344

10 0 10 1 0 1

10 0 1

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//3033355

10 0 10 1 0 1

10 0 1

01

01

0 1

01

y1

y2

y3

0

0 1 0 1

000


//30333366

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0

y1

y2

y3

0 1 0 1

000


//30333366

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0000

y1

y2

y3

0 1 0 1

000


//303337

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

y1

y2

y3

0 1 0 1

000


//303337

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

11

y1

y2

y3

0 1 0 1

000


//30333388

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1

y1

y2

y3

0 1 0 1

000


//30333388

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1111

y1

y2

y3

0 1 0 1

000


//30333399

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1 111

y1

y2

y3

0 1 0 1

000


//30333399

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1 111

y1

y2

y3 y1

y2

y3

0 1 0 1

000


//30444400

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1 111

y1

y2

y3

y1

y2

y3

0 1 0 1

000


//30444400

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1 111

y1

y2

y3

y1

y2

y3

Binary

0 1 0 1

000


//30444400

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1 111

y1

y2

y3

y1

y2

y3

0

10

10

1

Binary

0 1 0 1

000


//30444400

10 0 10 1 0 1

10 0 1

01

01

0 1

01

0 000

1

1 111

y1

y2

y3

y1

y2

y3

0

10

10

1

Binary

01

11

00

11

00

01

01

01 →

Incremental Diagonal

0

Incremental “Delta” Formulation

P1

P2

P3

v22

v21

v23 v24

v25

v13

v31

v11

v12

Yıldız and V. ‘12 (Generalization of Lee and Wilson 1999)

x = v11 +k∑

i=1

ri∑j=2

(vij − vi1

)δij

+

k∑i=2

(vi1 − vi−1

ri

)yi

δij ∈ [0, 1], yi ∈ {0, 1}

/309229

0


P1

P2

P3

v22

v21

v23 v24

v25

v13

v31

v11

v12

ri∑j=2

δij ≤ 1, δij ≥ 0


x = v11 +k∑

i=1

ri∑j=2

(vij − vi1

)δij

+

k∑i=2

(vi1 − vi−1

ri

)yi

/309229

0


P1

P2

P3

v22

v21

v23 v24

v25

v13

v31

v11

v12

ri∑j=2

δij ≤ yi

2


x = v11 +k∑

i=1

ri∑j=2

(vij − vi1

)δij

+

k∑i=2

(vi1 − vi−1

ri

)yi

/309229

0


P1

P2

P3

v22

v21

v23 v24

v25

v13

v31

v11

v12yi+1 ≤ δiri , yi+1 ≤ yi

2

1


x = v11 +k∑

i=1

ri∑j=2

(vij − vi1

)δij

+

k∑i=2

(vi1 − vi−1

ri

)yi

/309229

0

Summary, Extensions and More. Effective formulations: Encode and Formulate

Best encoding? Why not try a few.

Smaller formulations for shared vertex case

Need encodings with special structure.

Where to find more:

15.099 Spring ’13: Theory / Practice.

Survey: V., “MIP Formulation Techniques”./30330

Encodings in Mixed Integer Linear...

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