1 Internet Networking Spring 2004 Tutorial 9 Max-Min Fairness.

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1 Internet Networking Spring 2004 Tutorial 9 Max-Min Fairness

2 Motivation Given a network and a set of sessions we would like to find a maximal flow that it is fair We will see different definitions for max-min fairness and will learn a flow control algorithm The tutorial will give understanding what is max-min fairness

3 How define fairness? Any session is entitled to as much network use as is any other. Allocating the same share to all.

4 Example Maximal fair flow division will be to give for the sessions 0,1,2 a flow rate of 1/3 and for the session 3 a flow rate of 2/3 C=1 Session 1 Session 2 Session 3 Session 0

5 Max-Min Flow Control Rule The rule is maximizing the network use allocated to the sessions with the minimum allocation An alternative definition: is to maximize the allocation of each session i under constrain that an increase in i’s allocation doesn’t cause a decrease in some other session allocation with the same or smaller rate than i

6 Notation G=(N,A) - Directed network graph (N is set of vertexes and A is set of edges) C a – the capacity of a link a F a – the flow on a link a P – a set of the sessions r p – the rate of a session p We assume a fixed, single-path routing method

7 Definitions We have following constraints on the vector r= {r p | p Є P} A vector r satisfying these constraints is said to be feasible

8 Definitions A vector of rates r is said to be max-min fair if is a feasible and for each p Є P, r p can not be increased while maintaining feasibility without decreasing r p’ for some session p’ for which r p’ ≤ r p We want to find a rate vector that is max-min fair

9 Bottleneck Link for a Session Given some feasible flow r, we say that a is a bottleneck link with respect to r, for a session p crossing a if F a = C a and r p ≥ r p’ for all sessions p’ crossing link a :1/3 3:1/3 5:1/3 1:2/3 4:1 a b c d All link capacity is 1. Bottlenecks for 1,2,3,4,5 respectively are: c,a,a,d,a Note: c is not a bottleneck for 5 and b is not a bottleneck for 1

10 Max-Min Fairness Definition Using Bottleneck Theorem: A feasible rate vector r is max-min fair if and only if each session has a bottleneck link with respect to r Proof: Exercise

11 Algorithm for Computing Max-Min Fair Rate Vectors The idea of the algorithm: Bring all the sessions to the state that they have a bottleneck link and then according to theorem it will be the maximal fair flow We start with all-zero rate vector and to increase rates on all paths together until F a = C a for one or more links a. At this point, each session using a saturated link has the same rate as every other session using this link. Thus, these saturated links serve as bottleneck links for all sessions using them

12 Algorithm for Computing Max-Min Fair Rate Vectors At the next step, all sessions not using the saturated links are incremented equally in rate until one or more new links become saturated Note that the sessions using the previously saturated links might also be using these newly saturated links (at a lower rate) The algorithm continues from step to step, always equally incrementing all sessions not passing through any saturated link until all session pass through at least one such link

13 Algorithm for Computing Max-Min Fair Rate Vectors Init: k=1, F a 0 =0, r p 0 =0, P 1 =P and A 1 =A 1. For all a  A, n a k := num of sessions p  P k crossing link a 2. Δr=min a  A k (C a -F a k-1 )/n a k (find inc size) 3. For all p  P k, r p k :=r p k-1 + Δr, (increment) for other p, r p k :=r p k-1 4. F a k :=Σ p crossing a r p k (Update flow) 5. A k+1 := The set of unsaturated links. 6. P k+1 :=all p’s, such that p cross only links in A k+1 7. k:=k+1 8. If P k is empty then stop, else goto 1

14 Example of Algorithm Running Step 1: All sessions get a rate of 1/3, because of a and the link a is saturated. Step 2: Sessions 1 and 4 get an additional rate increment of 1/3 for a total of 2/3. Link c is saturated now. Step 3: Session 4 gets an additional rate increment of 1/3 for a total of 1. Link d is saturated. End :1/3 3:1/3 5:1/3 1:2/3 4:1 a b c d All link capacity is 1