1. Understanding Congestion Control Mohammad Alizadeh Fall 2018
6.829 Computer Networks
Understanding Congestion Control
Mohammad Alizadeh
Fall 2018

2. What Problem is TCP really solving?
Max-min rate allocation?
1 Mb/s
x1 = ?
x2 = ?
x3 = ? x1 x2 x3
Max-min
1/3
TCP
Assuming equal RTTs

3. What Problem is TCP really solving?
Max-min rate allocation?
1 Mb/s
1 Mb/s
x1 = ?
x3 = ?
x2 = ? x1 x2 x3
Max-min
1/2
TCP
~0.4
~0.6

4. What Problem is TCP really solving?
1 Mb/s
1 Mb/s
1 Mb/s
x1 = ?
x3 = ?
x2 = ?
What is the difference between these links?

5. What Problem is TCP really solving?

6. Network Utility Maximization
TCP is solving an optimization problem!
Spurred a lot of research on analyzing and designing network protocols from the lens of optimization

7. Is My Protocol “Stable”?
Example: ns2 simulations, 50 AIMD TCP sources, capacity = 9000 pkt, RED AQM
Credit: Low et al. “Dynamics of TCP/RED and a Scalable Control”, INFOCOM 2002.

8. Control Theoretic Analysis of TCP
Model TCP as a dynamical system (“fluid model”)
Analyze stability properties via control theory

9. Outline for Rest of Lecture
Network Utility Maximization (NUM) Distributed Algorithm for NUM TCP’s NUM Problem TCP Fluid Models

10. Utility Function The benefit derived from sending at rate x
Good model for elastic flows
e.g. file downloads
The benefit derived from sending at rate x
We’ll assume U(.) is increasing & concave

11. Examples of Concave Utility Functions

12. Elastic vs. Inelastic Flows
utility,
U(x)
rate, x
utility,
U(x)
rate, x
elastic traffic
(e.g., file download) delay-tolerant, no intrinsic rate
prefers to share
inelastic traffic
(e.g., real-time video)
delay-sensitive, intrinsic rate prefers to randomize
Credit: Frank Kelly (University of Cambridge)

13. Network Utility Maximization (NUM)
c1 = 1 Mb/s
c2 = 1 Mb/s x1 x3 x2
maximize log(x1) + log(x2) + log(x3) subject to: x1 + x2 ≤ 1 x1 + x3 ≤ 1

14. Network Utility Maximization (NUM)
c1 = 1 Mb/s
c2 = 1 Mb/s x1 x3 x2
maximize log(x1) + log(x2) + log(x3) subject to: x1 x2 x3 1

15. RLxN NUM: General Case maximize subject to: routing matrix N flows
L links
maximize
subject to: x1 x2 xN c1 c2 cL
RLxN
routing matrix
If Ui(.) strictly concave, the problem has unique solution

16. NUM Example 1: Throughput Maximization
c1 = 1 Mb/s
c2 = 1 Mb/s x1 x3 x2
maximize x1 + x2 + x3 subject to: x1 + x2 ≤ 1 x1 + x3 ≤ 1
x1★ = 0 Mb/s
x2★ = 1 Mb/s
x3★ = 1 Mb/s

17. NUM Example 2: Proportional Fairness
c1 = 1 Mb/s
c2 = 1 Mb/s x1 x3 x2
maximize log(x1) + log(x2) + log(x3) subject to: x1 + x2 ≤ 1 x1 + x3 ≤ 1
x1★ = 1/3 Mb/s
x2★ = 2/3 Mb/s
x3★ = 2/3 Mb/s

18. NUM Example 3: “α-fairness”
c1 = 1 Mb/s
c2 = 1 Mb/s x1 x3 x2
α ≥ 0 (a constant)
maximize subject to: x1 + x2 ≤ 1 x1 + x3 ≤ 1
alpha
Objective
α = 0
Thrput Maximization
α → 1
Proportional Fairness
α →∞
Max-min Fairness

19. Outline
The Network Utility Maximization (NUM) Problem Distributed Algorithm for NUM TCP’s NUM Problem TCP Fluid Models

20. Proportional Fairness Example
c1 = 1 Mb/s
c2 = 1 Mb/s x1 x3 x2
maximize log(x1) + log(x2) + log(x3) subject to: x1 + x2 ≤ 1 x1 + x3 ≤ 1

21. Distributed Algorithm
c1 = 1 Mb/s
c2 = 1 Mb/s x1 p1 p2 x3 x2
pl = “congestion price” for link l
price per unit of bandwidth (pl ≥ 0)
qi = total price for source i
Sum of prices on source i’s path

22. This is done independently at each source.
The Source Problem
c1 = 1 Mb/s
c2 = 1 Mb/s x1 p1 p2 x3 x2
Pick rate xi to maximize utility: log(xi) – qi xi
benefit
cost
xi = 1/qi
This is done independently at each source.

23. The Network Problem p1 p2 pl(t+1) = [ pl(t) + κ(yl(t) – cl) ]+
c1 = 1 Mb/s
c2 = 1 Mb/s x1 p1 p2 x3 x2
Adapt link prices periodically based on congestion
pl(t+1) = [ pl(t) + κ(yl(t) – cl) ]+
total traffic at link l
[.]+ = max(., 0)

24. This is done independently at each link.
The Network Problem
c1 = 1 Mb/s
c2 = 1 Mb/s x1 p1 p2 x3 x2
Adapt link prices periodically based on congestion
pl(t+1) = [ pl(t) + κ(yl(t) – cl) ]+
This is done independently at each link.

25. Putting it All Together
c1 = 1 Mb/s
c2 = 1 Mb/s x1 p1 p2 x3 x2
Sources
Links
x1(t) = 1 / (p1(t) + p2(t))
x2(t) = 1 / p1(t)
x3(t) = 1 / p2(t)
p1(t+1) = [p1(t) + κ(y1(t) – c1)]+
p2(t+1) = [p2(t) + κ(y2(t) – c2)]+

26. Example x1 x2 x3

27. Outline
The Network Utility Maximization (NUM) Problem NUM Distributed Algorithm TCP’s NUM Problem TCP Fluid Models

28. TCP + PI p1 p2 Sources Links p1(t+1) = [p1(t) + κ(y1(t) – c1)]+
c1 = 1 Mb/s
c2 = 1 Mb/s x1 p1 p2 x3 x2
Sources
Links
x1(t) = 1 / RTT1
x2(t) = 1 / RTT2
x3(t) = 1 / RTT3
p1(t+1) = [p1(t) + κ(y1(t) – c1)]+
p2(t+1) = [p2(t) + κ(y2(t) – c2)]+

29. What is TCP’s Utility Function?

30. Outline
The Network Utility Maximization (NUM) Problem NUM Distributed Algorithm TCP’s NUM Problem TCP Fluid Models

31. Is My Protocol “Stable”?
Example: ns2 simulations, 50 AIMD TCP sources, capacity = 9000 pkt, RED AQM
Credit: Low et al. “Dynamics of TCP/RED and a Scalable Control”, INFOCOM 2002.

32. TCP/RED Fluid Model - Setup
p(t)
Bi(t)
q(t)
Receiver
Sender
Receiver
Sender C
Receiver i
Sender i
One bottlenecked RED router
capacity {C (packets/sec) }
queue length q(t)
drop prob. p(t)
N TCP flows
window sizes Wi(t)
round trip time
Ri(t) = Ai+q(t)/C
throughputs
Bi (t) = Wi(t)/Ri(t)

33. Fluid Model vs. Simulation
Flow set 1
Flow set 2
Flow set 3
Flow set 4
Flow set 5
RED router 1
RED router 2
# flows per flow set: 40 -> 10 (t=75) -> 40 (t=150)
Credit: Vishal Misra (Columbia)

34. Fluid Model vs. Simulation
Queue 1
Queue 2

36. System of Differential Equations
All quantities are average values. Timeouts and slow start ignored
dWi dt ^ = 1
Ri(q(t))
Additive
increase
Wi(t-t) ^
Ri (q(t-t))
p(t-t)
Loss arrival
rate - Wi 2 ^
Mult.
decrease
Window Size:
Queue length: dq dt = ^
+ S
Ri(q(t)) ^
Wi(t)
Incoming
traffic
-1[q(t) > 0]C ^
Outgoing
traffic

37. System of Differential Equations (cont.)
Average queue length:
q(t) ^ dx dt =
ln (1-a) d -
x(t)
Where
a = averaging parameter of RED(wth)
d = sampling interval ~ 1/C
Loss probability: dp dt = dx ^
Where dp is obtained from the marking profile