ms&e 211, lecture 11 the dual of min-cost flow ashish goel

MS&E 211, Lecture 11The dual of Min-Cost Flow

Ashish Goel

Taking stock

• Today: Max-flow Min-Cut• Next eight lectures (with possible order changes):

– Zero sum games– Interior Point Method– Introduction to convex and quadratic optimization– Two examples from finance– Support vector machines– Gittins index and dynamic programming– Lagrange multipliers– Semi-definite and conic programming

Recap: The MCF

The MCF: A Useful Trick

Assume that the sum of all the demands = 0

(Necessary for the problem to be feasible)


≤


≤

- Maximize -


NOW, THE LP IS IN A STANDARD PRODUCTION FORM

Adding Labels to the MCF

The Dual of MCF

PRIMAL

DUAL

Coefficient of variable z(p) in constraint x(v,w) in Dual?

PRIMAL

The Transpose rule of thumb: Same as the coefficient of variable x(v,w) in constraint z(p) in primal

• p=v : x(v,w) appears as part of OUTx(p), i.e. with coefficient 1

• p=w : x(v,w) appears as part of INx(p), i.e. with coefficient -1

• p≠v, p≠w : x(v,w) does not appear in z(p), i.e. occurs with coefficient 0

The Dual of MCF

PRIMAL

DUAL

Coefficient of variable y(q,r) in constraint x(v,w) in Dual?

PRIMAL

The Transpose rule of thumb: Same as the coefficient of variable x(v,w) in constraint y(q,r) in primal

• (q,r)=(v,w) : x(v,w) appears in y(q,r) with coefficient 1

• (q,r)≠ (v,w): x(v,w) does not appear in y(q,r), i.e. occurs with coefficient 0

The Dual of MCF

PRIMAL

DUAL

The Dual of MCF

PRIMAL

DUAL

FLIP SIGNS

The Dual of MCF

PRIMAL

DUAL

A Lot of Work!!

Why?

A Lot of Work!!

Why?

We can now apply this to many problems

Next module: Dual of Shortest Path Problem

The Dual of MCF

PRIMAL

DUAL

Applying to Shortest Path Problem

• The constraints x(v,w) ≤ u(v,w) don’t exist in the primal !

=> No variables y(v,w) in the Dual !!

• d(s) = -1; d(t) = 1; all other d(v)’s are 0 !!!

DUAL

Dual for the Shortest Path Problem

Really Simple.

Informal Interpretation:z(v) – z(s) = Shortest path distance

from s to v

An Algorithmic Interpretation

• Ford’s Algorithm– Initialize: Set z(s) = 0; z(v) = 1 for all v s – Repeat: While there exists an edge (v,w) which violates the

constraint, set z(w) = z(v) + c(v,w) [Known as relaxation]• Intuition: z(v) is the shortest path distance from s to v• Most (perhaps all) popular shortest path algorithms

are special cases of Ford’s algorithm (if multiple edges violate the constraint, which one to choose)

Dual Complementary Slackness for Shortest Path Example

Let x,z be optimum solutions to primal, dual respectively

Either z(w) = z(v) + c(v,w) or x(v,w) = 0

Informal Interpretation: Either there exists shortest path to w that goes via v or the flow on edge (v,w) is zero

Example

6

2

4

st

p

qr

12

3

4 What is the value of z(t) – z(s) in any optimum solution of the dual for the shortest path problem?

Example

6

2

4

st

p

qr

12

3

4

One optimum solution to the dual is

z(s) = 0z(p) = 4z(q) = 3z(r) = 5z(t) = 6

Using dual complementary slackness, for which of the following edges can we conclude that the flow in an optimum solution to the primal shortest path is 0?

(a) (s,p) (b) (p,s)

(c) (s, q) (d) (r,t)

Example

6

2

4

st

p

qr

12

3

4


z(s) = 0z(p) = 4z(q) = 3z(r) = 5z(t) = 6

In fact, we can conclude that edges (p,s), (p,t), and (q,s) can not have positive flow in the primal

Example

6

2

4

st

p

qr

12

3

4


z(s) = 0z(p) = 4z(q) = 3z(r) = 5z(t) = 6

In fact, we can conclude that edges (p,s), (p,t), and (q,s) can not have positive flow in the primal

INTRIGUING FACT: The remaining edges form a shortest path tree!

Not totally a coincidence, since this corresponds to our informal interpretation, but we will not develop this further in this class.

Applying to Max-Flow Problem

The values d(v) and c(v,w) are 0 for max-flow (except for the fake edge, which does not have a capacity constraint)

DUAL

Applying to Max-Flow Problem

The values d(v) and c(v,w) are 0 for max-flow (except for the fake edge, which does not have a capacity constraint and does not need xt,s ¸ 0)

DUAL

The Dual of Max-Flow

The values y(v,w) must be 0 or 1 (increasing the capacity of an edge by a small amount ± increases the max-flow by 0 or ±

The Dual of Max-Flow

The values y(v,w) must be 0 or 1 (increasing the capacity of an edge by a small amount ± increases the max-flow by 0 or ± Similarly, z must be either 0 or 1 [Proof omitted] We have the “min-cut” problem: Find the set of edges of

smallest capacity which “cuts” the source from the sink, i.e., removes all paths from source to sink

Theorem: Max-flow = Min-Cut

The maximum flow you can send from s to t in a network = The minimum s-t in the same network

[Minimum s-t cut is the smallest total capacity of any s-t cut]

[An s-t cut is a set of edges such that removing these edges from the network will result in no path remaining from s to t]

2

1

Sink

Source

6

34

5

3

4

4

2

3

73

Yinyu Ye, Stanford, MS&E211 Lecture Notes #8 36

Look at Another Max-Flow Problem

3

Let xij be the flow rate from node i to node j. Then the problem

can be formulated as

max x41

x21 + x31 + x41 − x12 − x13

x12 + x32 + x42 − x21 − x23 − x24 x13 +

x23 + x43 − x31 − x32 − x34 x24 + x34 −

x41 − x42 − x43

xij

xij

z1

z2

z3

z4

y ij

s.t. = 0,

= 0,

= 0,

= 0,

≤ uij , ∀(i, j) ∈ E,

≥ 0, ∀(i, j) ∈ E.

37

The Primal Formulation

Corresponding Dual variables

38

The Dual of Max-Flow: the Min-Cut Problem

min ∑ (i,j) A ∈ uij yij

s.t. z4 − z1 = 1,

−z2 + z1 + y12 ≥ 0,

−z3 + z1 + y13 ≥ 0,

...

−z4 + z2 + y24 ≥ 0,

−z4 + z3 + y34 ≥ 0,

y,z ≥ 0.

x41

x12

x13

…

x 24

x34

1 if i is on the sink side 1, if zi=0 and zj=1

zi = and yij=

0 if i is on the source side 0 otherwise

CorrespondingPrimal variables

Yinyu Ye, Stanford, MS&E211 Lecture Notes #8 39

The Min-Cut Solution: Min-Cut Value=8

y24 = 1

1

Sink

Source

3 4

2

3

3

Z4=1

Z3=1

Z2=0

z1=0

y23 = 1

y13 = 1

2

All other yij=0

THANK YOU!!!

ms&e 211, lecture 11 the dual of min-cost flow ashish goel

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