maximum flow problem - github pagesmaximum flow problem what is the greatest amount of ow that can...

Maximum flow problem

CE 367R

MAXIMUM FLOWPROBLEM

Maximum Flow Problem

What is the greatest amount of flow that can be shipped between a givensource and sink without exceeding link capacities?

Maximum flow problem Maximum flow problem

Applications

Determining the capacity of a network

Identifying critical links in a network

Rounding a matrix


Network capacity

Assume we are trying to transport something from one point to another(fluid in pipes, people to evacuate, etc.)

Network links represent the physical system and its capacities; set theorigin as source and destination as sink.

The solution gives the way to transport goods from the source to sink atthe fastest possible rate.


Bottleneck analysis

By examining the solution to the network capacity problem, you canidentify the “critical links” where any capacity reduction will affect theoverall rate of flow from source to sink.

You can also identify “noncritical links,” where the capacity can bereduced slightly without affecting the overall flow rate.


Matrix rounding problem

Assume we are given a matrix, and want to round its entries to integers.The row and column sums should also be rounded, and give the correctvalues for the rounded entries.

3.1 6.8 7.3 17.29.6 2.4 0.7 12.73.6 1.2 6.5 11.3

16.3 10.4 14.5

If we can round any entry up or down to the nearest integer, we can solvethis as a max flow problem.


Create a column of nodes for each row, and a column of nodes for eachcolumn. Also create two artificial nodes, the source and the sink.

Links have lower and upper flow bounds representing the range ofrounding possibilities. Any feasible integer flow in this network (includingthe maximum) is a valid rounding.


FORMULATION

In the maximum flow problem, we are given...

A directed network G = (N,A) with a specified “source” node s and“sink” node t.

Each link (i , j) has a capacity uij . (Costs are irrelevant in the maximumflow problem.)

The objective is to ship the maximum amount of flow from s to t whilerespecting the link capacities.

Maximum flow problem Formulation

What do we mean by amount of flow?

Think about shipping one unit of flow from s to t along a particular path.

Because links have capacities, you can probably send more flow if you usemultiple paths (keeping in mind where they intersect.)

The xij values show the total amount of flow on link (i , j), which is notallowed to exceed the capacity uij .


A feasible flow satisfies conservation: except for the source and the sink,the flow entering a link must be the same as the flow leaving.

We can express this as ∑(i ,j)∈A(i)

xij −∑

(h,i)∈B(i)

xhi = 0

for all nodes i except for s and t.

Furthermore the net flow leaving the source should be the same amountentering the sink:∑

(s,j)∈A(s)

xsj −∑

(h,s)∈B(s)

xhs =∑

(h,t)∈B(t)

xht −∑

(t,j)∈A(t)

xtj

and this is what we are trying to maximize. We can call this b forconvenience.


Formulation

maxb,x

b

s.t.∑

(h,i)∈A

xhi −∑

(i ,j)∈A

xij =

−b i = s

+b i = t

0 otherwise

0 ≤ xij ≤ uij ∀(i , j) ∈ A

A feasible solution x to this problem is called a flow. The amount of flowshipped is given by b.


AUGMENTING PATHALGORITHM

An augmenting path is an undirected path from the source to the sink,satisfying the following properties:

1 Every link in the forward direction has flow less than capacity,xij < uij .

2 Every link in the reverse direction has flow greater than zero, xij > 0.

The residual capacity of this path is the smallest gap between the left andright hand sides of the above inequality (that is, find the smallest value of

uij − xij for forward links, the smallest value of xij for reverse link; the residual

capacity is the smaller of those two numbers.)

The idea: we will increase flow on forward links and decrease flow onreverse links. This maintains flow conservation.

Maximum flow problem Augmenting path algorithm

Example

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CapacityMaximum flow problem Augmenting path algorithm

Algorithm

1 Initialize the flow with x = 0, b ← 0.

2 Identify an undirected path π connecting s and t with positiveresidual capacity. (Terminate if no such path exists.)

3 Find the residual capacity u∗ of π.

4 For each forward arc (i , j) in π, set xij = xij + u∗. For each reverse arc(j , i) in π, set xij = xij − u∗.

5 Increase b by u∗, and return to step 2.

If you were writing code to do this, how would you implement step 2?

Maximum flow problem Augmenting path algorithm

MAX-FLOW/MIN-CUTDUALITY

To prove correctness of the augmenting path algorithm, we’ll take a shortdetour.

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Capacity u(S,T) = 7

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52 3

In a network, a cut is a partitioning of the nodes into two sets S and T ,where the source is in S and the sink is in T .A cut link is a link whose tail is in S and whose head is in T .

Maximum flow problem Max-flow/min-cut duality

The capacity of a cut u(R, S) is the sum of the capacities in the cut links.

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Capacity u(S,T) = 7

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Capacity u(S,T) = 5

7

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Capacity u(S,T) = 10

7

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63 2

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52 3


Weak max-flow/min-cut theorem

.

If (b, x) is a feasible solution to the max flow problem, and if (S ,T ) is acut, then b ≤ u(S ,T )

Proof sketch: Any unit of flow shipped from s to t must cross one ofthe cut links. So, the total amount of flow moving from s to t cannotexceed the sum of the capacities in the cut links.


Strong max-flow/min-cut theorem

.

If (b∗, x∗) is an optimal solution to the max flow problem, and if u∗ =min{u(S ,T )} among all possible cuts, then b∗ = u∗

This is a “duality” result: finding the maximum flow in a network isequivalent to finding the cut with minimum capacity.

We will prove this theorem in conjunction with the correctness of theaugmenting path algorithm.


Assume for now that the augmenting path algorithm terminates with someflow x. When it does so, there is no undirected path from s to t withpositive residual capacity.

Let S be the set of nodes for which undirected paths with positive residualcapacity do exist (including s itself), and let T be all other nodes.


Since s ∈ S and t ∈ T , (S ,T ) is a valid cut. Furthermore, by definitionevery cut link has zero residual capacity.

Therefore, the capacity of the cut is exactly equal to the sum of the flowon the cut arcs, so b = u(S ,T ).

By the weak max-flow/min-cut theorem, this means b must be maximaland u(S ,T ) must be minimal. (Strong theorem has been proved.)

Therefore, the augmenting path algorithm terminates with the correctanswer.


COMPLEXITY

The other issue is proving that the augmenting path algorithm mustterminate.

This is easy to do if we assume that the capacities are integers.

(If they are fractional, multiply by a sufficiently large number.)

Each iteration adds at least 1 to b, so b will reach its maximum value b∗

in a finite number of iterations.

Maximum flow problem Complexity

If U is the largest capacity on any link, then the capacity of the min cut isbounded by nU. (Look at a cut where S is just the source.)

So, the augmenting path algorithm requires O(nU) iterations.

The basic search algorithm for finding a path also has O(n2) steps.

So, the augmenting path algorithm in all requires O(n3U) steps.

Maximum flow problem Complexity

CAPACITY SCALINGALGORITHM

The augmenting path algorithm can take a long time if U is large, and ifthe path is chosen poorly.

Can we do better?

Maximum flow problem Capacity Scaling Algorithm

The capacity scaling algorithm overcomes this drawback, by finding anaugmenting path with sufficiently large capacity.

One option is to find the augmenting path with maximum availablecapacity.

This works, but involves more computation at each iteration. There is afaster approach...


For a given value of ∆, we will only search for undirected paths whoseresidual capacity is at least ∆.

The idea: start with a large ∆, augment flows along paths with residualcapacity at least ∆ until no more exist. Reduce ∆ and continue.

Let bxc mean rounding x down to the nearest integer.


Capacity scaling algorithm

1 Initialize x = 0, b = 0, and ∆ = 2blog2 Uc.

2 If ∆ < 1 then terminate.

3 Try to find an augmenting path π from s to t with residual capacityat least ∆.

4 If no such path exists, set ∆ = ∆/2 and return to step 2.

5 Augment flow along that path as in the augmenting flow algorithm,and return to step 3.


We can show that the capacity scaling algorithm also terminates with thecorrect solution, and requires O(n4 logU) steps; a slightly moresophisticated version of capacity scaling can reduce this to O(n3 logU).

If capacity is large, this is much better than O(n3U).


maximum flow problem - github pagesmaximum flow problem what is the greatest amount of ow that can...

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