minimum cost flows: introduction and basic algorithms

MINIMUM COST FLOWS: INTRODUCTION AND BASIC ALGORITHMS

Ben Klein – 11/03/13

IntroductionApplicationsOptimality ConditionsPrimal Dual in LPAlgorithms


Max flow – A reminder

The graph: G=(V,E) is a directed graph Capacity u(v,w) > 0 for every Two distinguished vertices s and t.

s

u

v

w

z

t

3

2

4

2

5

12


The flow: Flow is a function on the edges. A feasible flow is a flow that satisfies:

The value of a flow f is

The max flow problem is to find a feasible flow f, with maximum value


Minimum Cost Flows - Definition

The graph: G=(V,E) is a directed graph Capacity u(v,w) for every Balances: For every we will have a number b(v) Cost c(v,w) for every

v1 v2

v3 v4

v5

5

-2

-3

0 04,1

3,4

5,1

1,1

3,3



Assumptions:

All arc costs are nonnegative – No loss of generality – due to a known transformation which converts a min cost flow problem with negative costs to a one with non-negatives costs.

If then



Assumptions:


If then


v1 v24,1

2,3


Assumptions:


If then


v1 v2

u

w

4,1 4,0

2,32,0


The flow: Flow is a function on the edges. A feasible flow is a flow that satisfies:

The cost of a feasible flow f: The min cost flow problem: is to find a feasible flow f, with the minimum cost.



For a feasible flow x, G(x) is the residual network that corresponds to the flow x.We replace each arc by two arcs and .

The arc has cost and residual capacity .The arc has cost and residual capacity .

The residual network consists only of arcs with positive residual capacity.



As a linear program:

s.t.


Minimum Cost Flows – Finding a Feasible Solution

How to find a feasible solution?

A simple observation:

But what is the value of the following expression:




is not enough:

v1 v2

v3 v4

v5

5

-2

0 04,1

3,4

5,1

1,1

1,3-3



v1 v2

v3 v4

v5

5

-2

-3

0 04,1

3,4

5,1

1,1

3,3


A feasible flow is a flow that satisfies:

A simple observation: A feasible flow does not depend on the cost values.



v1 v2

v3 v4

v5

5

-2

-3

0 04

3

5

1

3


A feasible flow is a flow that satisfies:

A simple observation: A feasible flow does not depend on the cost values.



v1 v2

v3 v4

v5

5

-2

-3

0 04

3

5

1

3


Reduction to max flow problem.

Add vertices s and t. For every v such that b(v)>0:

Add an edge (s,v) such that u(s,v)=b(v) For every v such that b(v)<0:

Add an edge (v,t) such that u(v,t)= -b(v)

There exists a feasible solution iff all the edges from s are saturated.

s

t

3

2

5



v1 v2

v3 v4

v5

4

3

5

1

3


Reduction to max flow problem.

Add vertices s and t. For every v such that b(v)>0:

Add an edge (s,v) such that u(s,v)=b(v) For every v such that b(v)<0:

Add an edge (v,t) such that u(v,t)= -b(v)

There exists a feasible solution iff all the edges from s are saturated.

s

t

3

2

5


Minimum Cost Maximum Flow

A variation of the Min Cost Flow problem is to find a flow which is maximum, but has the lowest cost among the maximums.

How do we solve it?

s

v

3,1

2,2

4,0

2,1

5,3

1,12,1

u w

z

t



Forget about the costs. Use your favorite Max Flow algorithm to find the value of the max flow, f*.

s

v

3

2

4

2

5

12

u w

z

t



Reduction to Min Cost Flow:

Define:

s

v

3,1

2,2

4,0

2,1

5,3

1,12,1

u w

z

tf* -f*

0 0

0 0


Minimum Cost Flow – Application #1

We have T tasks and M machines ( We shall assume that every machine will execute only one task.

Let be the cost/time of running the task on the machine .

We want to find the assignment A*, of tasks to machines, with the minimum total cost/running time.



A

t1

t2

m1

m3

m2 B

2 -2

0 0

0 0

0

All the capacities are 1.

0

0

0

0

0

c1,1

c1,2

c1,3

c2,3

c2,2

c2,1



We have a bus that can accommodate P passengers at a given time.

The bus stops at L stations (First at station #1, then at station #2 and finally at station #L).

At every station i, we know in advance how many passengers are willing to take the bus from station i to station j (j > i). We shall denote this number by pij.

The price of a single ticket for taking the bus from station i to station j is c ij.

Our goal is to find a plan which brings the profit to maximum.



If one desires to find the maximum, M, of a function g(x), he/she can instead find the minimum, m, of the function and then we have that



As a linear program:

s.t.



1 2 3 4

1-4

2-4

3-4

1-3

2-3

1-2

P,0 P,0 P,0

∞,-c34∞,-c24

∞,-c23

∞,-c14

∞,-c13

∞,-c12

Red Edges have ∞capacity

0 cost

p1,4 p2,4 p3,4

p1,3 p2,3

p1,2

-p1,2 -p1,3 –p2,3 -p1,4 –p2,4 –p3,4


Minimum Cost Flow – Optimality Conditions

We’ll consider three different optimality conditions:

Negative cycle Reduced cost Complementary slackness


MCF - Optimality Conditions – Negative Cycle

Theorem: A feasible solution x* is an optimal solution of the MCF problem if and only if it satisfies the negative cycle optimality conditions: namely, the residual network G(x*) contains no negative cost (directed) cycle.



Proof:

If x is feasible and there is a negative cycle in G(x), then x is not an optimal solution

Suppose that x is a feasible flow and that G(x) contains a negative cycle, C. Then by simply augmenting flow through C, we got a new solution x’ which is feasible and has a lower cost.Therefore x is not an optimal solution to the MCF.

a

b

c

d

4-5

1 -2

Residual network



Proof:

If x is feasible and there is no negative cycle in G(x), then x is an optimal solution



Help is needed:

Theorem (Augmenting Cycle Theorem): Let x and y be any two feasible solutions of a network flow problem. Then the cost of x equals the cost of y plus the cost of the flow on at most m directed cycles in G(y).

Now we are ready:

Let x be a feasible flow and there is no negative cycle in G(x). Let x* be the optimal flow. Then . But x and x* are both feasible solutions of the network. Therefore:.

Therefore and x is an optimal solution.


MCF - Optimality Conditions – Reduced cost

Node potentials:

Let be a labeling of the nodes in V.

We shall call , the potential of the node v.For a given labeling , we define the reduced cost of an arc (u,v) as:



Node potentials:

Property 1:For any directed path P from node u to node v:

v0 v1 … vk-1 vk

=



Node potentials:

Property 2:For any directed cycle C:

v0 v1 … vk-1 v0

Using property 1:



Theorem: A feasible solution x* is an optimal solution of the minimum cost flow problem if and only if some set of node potentials satisfy the following reduced cost optimality conditions:



Proof:

if for a feasible solution x*, some set of node potentials satisfy the following reduced cost optimality conditions: then x* is an optimal solution

Let C be a cycle in G(x*).

Therefore there is no negative cycle in G(x*) x* is optimal according to the negative cycle optimality conditions.



Proof:

if x* is an optimal solution, then some set of node potentials satisfy the following reduced cost optimality conditions:

x* is an optimal solution and therefore there is no negative cycle in G(x*).No negative cycle shortest paths are well defined in G(x*)

We shall choose an arbitrary node s.Let denote the shortest path from s to v.Therefore:

We shall denote Then:

s

u

v


MCF - Optimality Conditions – Complementary Slackness

Theorem: A feasible solution x* is an optimal solution of the MCF problem if and only if for some set of node potentials , the reduced costs and flow values satisfy the following complementary slackness conditions for every edge (u,v) in the network:

If then If then If then



A feasible solution x* is optimal, if some set of reduced costs and the flow values satisfy the following complementary slackness conditions for every edge (u,v) in the network:


Proof: We shall prove that every edge (v,w) in G(x*) has a non-negative value. By contradiction, let (v,w) be an edge in G(x*) with .Then according to the third condition (v,w) is not in G(x*). Contradiction.



If a feasible solution x* is optimal, then some set of node potentials satisfy the following the reduced costs and flow values satisfy the following complementary slackness conditions for every edge (u,v) in the network:


Proof:Let x* be an optimal solution. Therefore there exist s.t. every edge (v,w) in G(x*) has .

Case 1: Assume that and Then (j,i) is in G(x*) and contradiction.



If a feasible solution x* is optimal, then some set of node potentials satisfy the following the reduced costs and flow values satisfy the following complementary slackness conditions for every edge (u,v) in the network:



Case 2: If . Therefore the residual network contains both (i,j) and (j,i).If then (i,j) or (j,i) has a negative reduced cost and that is a contradiction.



If a feasible solution x* is optimal, then some set of node potentials satisfy the following the reduced costs and flow values satisfy the following complementary slackness conditions for every edge (u,v) in the original network:



Case 3: If and , then (i,j) is in the residual graph. A contradiction.


PRIMAL DUAL IN LP


Finding an Upper Bound on OPT

min 9x1 + 2x2 + 6x3

s.t.4x1-x2+2x3 ≥ 22x1+x2+x3≥ 10x1,x2,x3 ≥0

How can one find an upper bound on OPT?Any feasible solution to this problem is an upper bound!Look at (5,2,3) As an example.

4*5 – 2 + 2*3 = 24 ≥ 225+2+3 = 10 ≥ 10

Therefore, 9*5+2*2+6*3 = 67, is an upper bound to OPT


Finding a Lower Bound on OPT

min 9x1 + 2x2 + 6x3

s.t.4x1-x2+2x3 ≥ 22x1+x2+x3≥ 10x1,x2,x3 ≥0

How can one find a lower bound on OPT?We can “play” with the constraints.

1) 9x1+2x2+6x3 ≥ 4x1 – x2 + 2x3 ≥ 222) 9x1+2x2+6x3 ≥ 1*(4x1-x2+2x3) + 3*( x1+x2+x3) = 7x1+2x2+5x3 ≥ 22 + 30 = 52

An optimal solution to our problem is (1,0,9) with the value = 63


Finding a Lower Bound on OPT - Generalizationmin 9x1 + 2x2 + 6x3

s.t.4x1-x2+2x3 ≥ 22 (y1)x1+x2+x3≥ 10 (y2)x1,x2,x3 ≥0

For which y=(y1,y2) we can say that 9x1+2x2+6x3 ≥ y1(4x1-x2+2x3)+y2(x1+x2+x3)≥10y1+22y2?9x1 + 2x2 + 6x3 ≥ y1(4x1-x2+2x3) + y2(x1+x2+x3) = (4y1+y2)x1 + (-y1+y2)x2 + (2y1+y2)x3

Any y s.t.4y1+y2 ≤ 9-y1+y2 ≤ 22y1+y2 ≤ 6y1,y2≥0

And of course that we want the greatest lower bound, so we want max 10y1 + 22y2


The Dual ProblemSo while searching for a good lower bound, we created a new LP problem. We will call to the original problem – The Primal, and to the new one – The Dual.

The Primal:

min 9x1 + 2x2 + 6x3

s.t.4x1-x2+2x3 ≥ 22 x1+x2+x3≥ 10 x1,x2,x3 ≥0

The Dual:

max 22y1 + 10y2

s.t. 4y1 +y2 ≤ 9 -y1 + y2 ≤ 22y1 + y2 ≤ 6y1,y2 ≥ 0

min cTxs.t.Ax ≥ bx ≥ 0

max bTys.t.ATy ≤ cy ≥ 0

The dual of the dual is the primal


Weak Duality - TheoremNote: In this lecture, Primal is a min problem Dual is a max problem. We will refer to the Primal as (P) and to the Dual as (D).

Theorem (Weak Duality): For any x feasible in (P) and y feasible in (D) we have.

Proof:

x is feasible for the

primal.y is feasible for the dual.




Weak Duality - Corollaries Let x* denote an optimal solution to (P) and let y* denote an optimal solution to (D).

Corollary 1: for any feasible y; that is, yb, for any feasible y is a lower bound on the optimal solution of (P).

Corollary 2: for any feasible x.

Corollary 3: If for any x and y, feasible for (P) and (D), respectively, then x and y are both optimal.


Strong Duality - Theorem Theorem (Strong Duality): Let x and y be feasible for (P) and (D) respectively, thenx is optimal to primal and y is optimal to dual if and only if .


Complementary Slackness - TheoremTheorem (Complementary Slackness): Let x and y be feasible for (P) and (D) respectively, thenx is optimal to primal and y is optimal to dual if and only if the conditions of complementary slackness hold:

Primal complementary slackness conditions

For each j, j = 1,...,n we have xj = 0 or

Dual complementary slackness conditions

For each i, i= 1,...,m we have yi = 0 or


Complementary Slackness - ProofLet x and y be feasible for (P) and (D) respectively, then x is optimal to primal and y is optimal to dual if and only if .

We know that:

Therefore, cx=yb if and only if:

and




MCF – Linear Programming

s.t.



s.t.

𝜋 (𝑖)

𝛼 𝑖𝑗

primalmin cTxs.t.Ax ≥ bx ≥ 0

dualmax bTys.t.ATy ≤ cy ≥ 0



s.t. s.t.



s.t.



s.t.

Define :



s.t.



s.t.

We know that . Therefore we want to be small.We can only control. Therefore



Theorem: Let z(x) denote the objective function value of some feasible solution x of the MCF problem and let denote the objective function value of some feasible solution of its dual. Then .



Proof: Weak Duality Theorem.



Theorem: Let x* denote a feasible solution of the primal, and denote a feasible solution of the dual. Then x* is optimal for the primal and is optimal for the dual iff



Proof: Strong Duality Theorem.



Theorem: Let x* denote an optimal solution of the primal, and denote a optimal solution of the dual. Then the pair satisfies the complementary slackness optimality conditions:




Proof: Complementary Slackness Theorem.

s.t. s.t.

Case 1: If then and therefore .




s.t. s.t.

Case 2: If then can’t be greater than 0, because then (Case 1). We know that for the optimal solution . Therefore it must be zero.




s.t. s.t.

Case 3: If then . Therefore


Computing Optimal Node Potentials from Optimal Solution

Let be an optimal solution to the MCF problem. How one can find the optimal node potentials?



Solution: If is an optimal solution then there are no negative cycles in G(). Therefore shortest paths are well defined in G(). Choose an arbitrary node s. Compute d(i) – shortest distance from s to i in G(). And denote .


Computing Optimal Flow from Optimal Node Potentials

Given the optimal node potentials, how can one find the optimal flow?



Solution: If are the optimal node potentials, then we shall first compute for every edge (i,j) in the network.

Examine every If then (Case 1) and delete (i,j) from the network. If then (Case 3) and delete (i,j) from the network and decrease b(i) and increase b(j) by . If then we’ll demand: .

Let G’=(V,E’) the resulting network and b’ denote the modified supplies/demands.

Now the problem reduces to finding a feasible flow in the network G’ (And we already know how to do this).


Cycle Canceling Algorithm

The negative cycle optimality conditions suggest one simple algorithmic approach:

algorithm cycle-canceling:1. Establish a feasible flow x in the network2. While G(x) contains a negative cycle do:

1. Use some algorithm to identify a negative cycle W

2. augment units of flow in the cycle W and update G(X)

Finding negative cycle in a graph: Bellmand-Ford O(nm).



Theorem: If all arc capacities and supplies/demands of nodes are integer, then the minimum flow problem always has an integer minimum cost flow.



Running Time:

Let U be an upper bound to the capacities and C an upper bound to the costs.

mCU is an upper bound on the initial cost.-mCU is a lower bound on the initial cost.

Each iteration the algorithm changes the objective function value by an amount of . Therefore we have at most O(mCU) iterations.

With Bellman-Ford we get running time.


THANK YOU


minimum cost flows: introduction and basic algorithms

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