1. Stability of Hop by Hop Congestion Control Stephan Bohacek Dept. of Mathematics University of Southern California

2. End to End and Hop by Hop Congestion Control End to End Congestion Control (e.g. TCP) –The sender determines the rate at which it sends data based on estimates of congestion derived from information from the receiver. There may be a large delay between when congestion occurs and when the sender reacts. 100KBps link with 250ms delay. TCP takes 1.5 secs to reach full speed. In this time 150 KB could have been sent. Most http files are smaller than 150KB. Hence, TCP is usually under utilizing the link. A large number of packets may be dropped. Hop by Hop Congestion Control –Each router negotiates with its neighbors to decide how fast data should be sent. Little delay No estimation is necessary Few drops Burden on the routers Stability is an issue (distributed control)

3. D1 S1 Connection Oriented Approach to Hop by Hop Congestion Control Neglect all other links except the ones involved in the connection. Control information (e.g. sending rate, queue size) about this particular connection is exchanged between neighboring nodes.

4. q i,i+1 (k+1) = L (q i,i+1 (k) + r i-1,i (k) – r i,i+1 (k)) r i-1,i node i-1 q i-1,i r i,i+1 node i q i,i+1 r i+1,i+2 node i+1 q i+1,i +2 data flow from node i to node i+1 queue occupancy for flow from node i-1 to node i queue dynamics r i,i+1 (k+1) = L (r i+1,i+2 (k) + F(q * - q i+1,i+2 (k))) rate dynamics Objective: match rates, r i,i+1 =M, and make q i+1,i+2 = q * “optimal queue size” r N (k+1) = M M = bandwidth of bottleneck link / number of competing flows “bottleneck link”

5. = -F-F1 11 z -1 0 00 q r σ -1 q r q r + temporal shift spatial shift It can be shown that this closed loop system is stable (Mishra 1996). Drawbacks The optimal rate M? (Why not just feed M back directly?) Optimal queue occupancy q * ? Hence, optimal delay? The router must maintain flow control for each connection (per VC congestion control). E.g., a 10Gbps router can accommodate 200,000, 56Kbps connections. If each buffer can hold 20 packets, then the total memory is 48Gb! For this reason, the ATM steering committee did not choose per VC congestion control for ABR (Jain 1996).

6. non-connection oriented approach Pazos and Gerla (Infocom, 1999) Kulkarni, Bohacek, and Safonov (Allerton, 1998) r i,j = -  r i,j +  r i,j  The rate that data exits node i to other nodes. The rate that data enters node i from other nodes. aggregate queue occupancy in node i rate that node i sends data to node j rate that data enters/leaves the network through node i q i = -  j  i,,j r i,j +  j  j,i r j,i + u i  rate controller queue dynamics (conservation of data) average data rate optimal queue size r i,j = - (r i,j - r i,j ) – F 1 (q j – q * ) + F 2 (q i – q * )  Define  i,,j = 0 otherwise 1 if node i sends data to node j

7. q = Ar + u  The close-loop LTI system has one eigenvalue at zero and the rest have negative real parts. The eigenvalue at zero corresponds to the fact that if  i u i > 0, then more data enters the network then leaves the network. The excess data accumulates in the queues. non-connection oriented approach A link is an output from one node and it is an input to another. Hence, columns of A sum to zero.

8. Blocking A B C D E fast links fast link slow link Data flows from A to D and from B to E. Since link C, E is slow, the queue in node C fills. Both flows slow down to alleviate the filling queue.

9. input rates r i,j = L (F 1 q i,j + F 2 q i,j – F 3  k  i,j,k q j,k – F 4  k  i,j,k q j,k + F 5  h  h,i,j q h,i + F 6  h  h,i,j q h,i ) Define  i,j,k to be the ratio of data going from node i to node j that will then go to node k. q i,j = L (- r i,j +  h  h,i,j r h,i + u i,j )  queue occupancy for link i, j     increase rate if local queue is full or is filling decrease rate if downstream queue is full or is filling (Back Pressure) increase rate if upstream queue is full or is filling (Forward Pressure) Node i, j adjust its rate only when directly responsible for congestion. This should alleviate the blocking problem. No preset values.

10. Congested Router Forward Pressure Data Control Back Pressure

11. Inputs and Outputs IN1 2OUT network under control q 1,2 = L( -r 1,2 + r IN ) r 1,2 = L( F 1 q 1,2 + F 2 (-r 1,2 + r IN ) – F 3 q 2,OUT - F 4 (-r 2,OUT + r 1,2 ) ) q 2,OUT = L( -r 2,OUT + r 1,2 ) r 2,OUT = L( F 1 q 2,OUT + F 2 (-r 2,OUT + r 1,2 ) + F 5 q 1,2 + F 6 (-r 1,2 + r IN ) ) Like connection oriented Hop-by-Hop, this topology is stable for all F i > 0 but queues are empty at equilibrium. No preset values.

12.  k  i,j,k  1 the data flowing along link i,j must next either flow along some other link j, k or must exit the network. 1 3 2 u1u1 u3u3 u2u2 Suppose that  1,2,3 =  2,3,1 =  3,1,2 = 1. Then all the data that enters the network remains in the network and queues overflow. We require  1,2,3 ·  2,3,1 ·  3,1,2  1 Define  i,j,k to be the ratio of data going from node i to node j that will then go to node k

13. No – Loop Condition 1 3 2 u1u1 u2u2 5 4 u4u4 u3u3 u5u5 If  1,2,3 =  3,1,2 =  3,4,5 =  4,5,3 = 1  2,3,1 =  2,3,4 =  5,3,4 =  5,3,2 = 0.5, then data that enters the network never leaves. We require that for all sequence of tuples L and some  > 0. A static requirement is that  = 1.

14. q = L ((A – I) r + u)  Since  k  i,j,k  1 the row sums of A  1. Hence, A T is a substochastic matrix. A (i,j),(j,k)  0  A (j,k),(i,j) = 0 r = L( (F 1 I – F 3 A T + F 5 A)q + (F 2 I – F 4 A T + F 6 A)(A – I)r )  no-loop condition A is time varying  LPV L accounts for state saturation Matrix Representation a convex set

15. Define F a := F 1 = F 3 and F b := F 2 = F 4 Set F 5 = 0 The system is asymptotically stable if -F b  2 + F 6 (n – n 1/2 ) < 0 F a > 0, F b > 0, F 6  0 However, the system might not be exponentially stable! (delay) Asymptotic Stability Set R = A – I r = L( F a (I – A T )q + F b (I – A T )(A – I ) r ) · q = L( (A – I) r + u ) · Note: R varies with time = -F b R T R-F a R T R0 r q · · r q L For F 6 =0

16. Is x = (A T – I)x stable?  LPV/LMI approach find P > 0 such that P(A – I) T + (A – I)P < 0 for all A Since A T substochastic and loop-free condition Exponential Stability

17. LPV/LMI does not work 2-D illustration lines of constant cost vector field for (A 1 - I) T vector field for (A 2 – I) T 1-  0 0 A 1 = 0 1-  0 A 2 = T T

18. s = 4 Lemma: Choose s > 1 such that Define the cost

19. Exponential Stability Choose F i  0 such that there exists P 11 P 12 P 22 with then the closed loop system is exponentially stable.

20. 1 in1in2in3in4 2 4 3 6 7 in5 5 in7 in8 in9 in6 A simple example of hop-by-hop control At time 0 node in4 begins to send data to node 7.

21. 1 in1in2in3in4 2 4 3 6 7 in5 5 in7 in8 in9 in6 At time 0 node in4 begins to send data to node 7. A simple example of TCP control

22. Hop-by-Hop and TCP

23. Blocking 12 3 4 5 6 fast links slow link congested node  i,j,k,l the ratio of data that is traveling from i to j and to k which then travels to l. 12 3 5 6 7 slow link congested node 4  i,j,k,l,m  i,j,...,z connection oriented per VC control

24. Conclusions Hop-by-Hop congestion control is a classic control problem. Asymptotic stability is easy. However, exponential stability is difficult to prove and may require high gain. So far the result are not scalable (they depend on n, the number of nodes, but could be changed to fan-in). Future work Topology depend strategies should be investigated. Performance issues Min delay/min queue size (minimize L  norm of queue). Minimize transaction time (high bandwidth). End to End flow control and Fairness.