1. Generalizing The Network Performance Interference Problem
Jiyong Park
Real-Time Operating Systems Laboratory,
Seoul National University, Korea
12, Dec. 2005

2. Contents Overview Background Knowledge Network Model
Analysis of The Model
Solution Mechanism
Future Work
Conclusion

3. Overview (1/3) Goal JEC paper
Generalizing the network performance interference problem.
Proposing a universal solution mechanism for the problem.
JEC paper
Submitted at Aug. 2005
Analysis and solution only for a specific network configuration
Queue size = 100, upload speed = 800 Kbps, ...

4. Number of ACK packets for opposite direction
Overview (2/3)
Result of Analysis
Factors that affects the throughput
Delay of ACK should be as short as possible.
Number of ACK in the queue should be minimized.
Number of ACK packets for opposite direction
in the queue
delay of ACK packet

5. Overview (3/3) Basic Idea Solution Mechanism
Delay of ACK should be as short as possible.
Solution: Priority of ACK > priority of data
Number of ACK in the queue should be minimized.
Solution: Drop probability of ACK > drop probability of data
Solution Mechanism
Idea from RED (Random Early Detection) queue
Separate queues for ACK and non-ACK packets

6. Background Knowledge

7. Bandwidth-Delay Product
Bandwidth-delay product is
the number of packets that are in-flight in a link.
Example of a conveyor belt.
For a network link
Number of people in a conveyor belt = 2 * 2 = 4 peoples
2 peoples/minute
It takes 2 minute.
Maximum number of packets = μ t / L
Transmission speed = μ (bytes/second)
Number of packets = λ t / L
Packet size = L(byte)
Propagation delay = t (seconds)
Packet speed = λ (byte/second) (λ ≤μ)

8. Capacity of a Path Capacity of a Path (C)
Maximum number of packets (of bytes) that can be in-flight in the path.
Sum of bandwidth-delay product of all links and capacity of buffers
If we send more packets than C, excessive packets will be dropped at the queue. B2 λc tc λa ta λb tb B1

9. TCP Overview – 1. Flow Control (1/3)
Window
Number of data packets that can be sent without being acknowledged.
Represented as variable wnd.
Sliding window
Upon receiving ACK, window slides.

10. TCP Overview – 1. Flow Control (2/3)
Operation on Sender’s Side
If an ACK is received,
Transmit k data packets immediately (slide the window by k).
k = seq. number of current ACK – seq. number of previous ACK
Sent & ACKed
Sent & Not ACKed
Not sent
window (wnd = 5)
λ data packets/sec
Sender
user data (infinite) 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 12 13 14
λa ACK packets/sec
ACK received: 5 6 9
Data sent: 11 12 13 14

11. TCP Overview – 1. Flow Control (3/3)
Operation on Receiver’s Side
If a data packet is received,
Send one ACK packet immediately.
Seq. number of the ACK := seq. number of last data packet that has been received in-order
λ data packets/sec
Receiver
Data received 1 2 3 4 5 6 7 8 9 10 11
ACK sent: 6 7 7 7 7 11
λ ACK packets/sec
Rate of incoming ACK packets = Rate of outgoing data packets (packets/sec)

12. TCP Overview – 2. Congestion Avoidance (1/2)
Window size (wnd) can be vary.
wnd = min(cwnd, maxwnd)
maxwnd: buffer size of the receiver (determined when establishing a connection)
cwnd: size of the congestion window (variable)
cwnd is updated as follows.
If an ACK is received, cwnd is increased.
 Network is not congested. Put more packets in the link.
If data loss is detected, cwnd is decreased.
 Network is congested. Put less packets in the link.

13. TCP Overview – 2. Congestion Avoidance (2/2)
cwnd increment algorithm has two phases.
Slow start phase (aggressive phase)
when cwnd < ssthresh
cwnd += 1 (upon receiving 1 ACK)
Congestion avoid phase (cautious phase)
when cwnd > ssthresh
cwnd += 1/cwnd (upon receiving 1 ACK)
cwnd
ssthresh
logarithmic increment
exponential increment
time
slow start
congestion avoid

14. TCP Overview – 3. Loss Recovery (1/2)
Loss of a data packet can be detected by
Timer expiration,
Three duplicated ACKs,
or Selective ACK.
If packet loss is detected,
cwnd is decreased.
ssthresh is updated.
...
Exact mechanisms are different among various TCP implementations.
TCP tahoe, TCP reno, TCP new-reno, TCP-SACK, ...
Good mechanisms are less affected by the data packet loss.

15. TCP Overview – 3. Loss Recovery (2/2)
Figure taken from Anurag Kumar, “Comparative Performance Analysis of Versions of TCP in a Local Network with a Lossy Link”, In IEEE Transactions on Network, 1998

16. Performance of TCP (1/2) Assumption
Data packet is not lost.
If you want to calculate the throughput with data packet loss, refer to the previous slide.
Throughput (λ) (unit: bytes per second)
Average number of bytes sent during a certain period of time.
A certain period of time: RTT (round-trip-time)
During RTT, TCP sender transmits wnd data packets.

17. wnd = number of data packets + number of ACK packets
Performance of TCP (2/2)
During RTT, TCP sender transmits wnd data packets.
data packet
Sender
Receiver
1st
2nd
RTT
wnd-th
ACK packet
wnd = number of data packets + number of ACK packets
The sender has been just transmitted a data packet.
That packet is wnd-th packet currently in the link.
It takes RTT for the sender to receive ACK packet for the data packet. (by the definition of RTT)
If the sender receives the ACK packet, it is the wnd-th ACK packet since 1.
The sender need wnd ACK packets to transmit wnd data packets.
So, the sender should have been transmitted wnd data packets during RTT.

18. Network Model

19. Network Model (1/2) BU Node A Node B uplink μU tU downlink WU μD WD tDBD
Subscript D and U mean downlink and uplink respectively.
B: queue capacity (in packets)
μ: link speed (in byte per seconds)
t: link delay (in seconds)
D: size of a TCP data packet (in byte)
A: size of a TCP ACK packet (in byte)
W: maximum window size
Links are loss free.
Constraints
μD > μU (downlink is faster than uplink)
D > A

20. Network Model (2/2) Data is sent and received greedily.
There are always packets to send.
There are always rooms to receive packets.
Two simultaneous TCP connections
Connection d: node B  node A (download)
Connection u: node A  node B (upload)

21. Generality of the Model
Chain of links can be simplified as a single link.
bottleneck link μ B
total delay = t t μ B

22. Analysis of The Model

23. Throughput To calculate throughput, we need Average window size: wnd
Round-trip-time: RTT

24. Calculation of Average Window Size (1/2)
Assumption
maxwnd should be larger than P = C-BU-BD
If not, link is always underutilized.
Average window size can be calculated.
wnd
wnd
maxwnd C C
average wnd
maxwnd
average wnd P P
time
time
Case 1: maxwnd > C
Case 2: P < maxwnd < C

25. Calculation of Average Window Size (2/2)
From now on, let
wndD: average window size for download traffic
wndU: average window size for upload traffic

26. Calculation of RTT RTT = dD+dU dU : delay on downlinkBU
Node A
Node B
uplink μU tU
downlink WU μD WD tD BD
dD : delay on downlink

27. Calculation of RTT - Delay on Downlink
dD = tD + queuing delay
downlink μD tD BD
dD : delay on downlink tD
queuing delay

28. Calculation of RTT – Queuing Delay (1/5)
Queuing delay consists of
Transmission time of ACK packets in the queue.
Transmission time of data packets in the queue. μD
DDQ: number of data packets in this queue
AUQ: number of ACK packets in this queue

29. Calculation of RTT – Queuing Delay (2/5)
There are total wndD packets (ACK+data) in the path.
Data packets on downlink (DDL)
Data packets on downlink queue (DDQ)
ACK packets on uplink (ADL)
ACK packets on uplink queue (ADQ)
ADQ
ADL
Node A
Node B μU WU μD WD
DDL
DDQ

30. Calculation of RTT – Queuing Delay (3/5)
DDL: number of data packets on downlink
Use bandwidth-delay product
Let λD as download throughput
λD/D data packets per second tD

31. Calculation of RTT – Queuing Delay (4/5)
ADL + ADQ = number of ACK packets for download traffic
For the node B, uplink is seen as a queue-less link with delay dU
Use bandwidth-delay product again
λD/D ACK packets per second dD

32. Calculation of RTT – Queuing Delay (5/5)
ADQ
ADL
Node A
Node B μU WU μD WD
DDL
DDQ

33. Calculation of RTT

34. Calculation of Throughput
Non-zero
Delay of ACK
Number of ACK packets in queue (from opposite direction)

35. Factors That Effects The Throughput
Delay of ACK in reverse link
Number of ACK packets from opposite direction in queue in forward link
We should minimize these two values!

36. Solution Mechanism

37. Basic Idea Minimize delay of ACK packets.
Give high priority to ACK packets.
(ACK packets are processed prior to data packets)
Minimize number of ACK packets in queue.
Drop ACK packets more than data packets.
This does not affect to the performance since ACK is cumulative.

38. Solution Mechanism (1/2)
Minimize delay of ACK packets
Give high priority to ACK packets
(ACK packets are processed prior to data packets)
 ACK-separation queuing mechanism
Output queue for data packets (100 packets)
low
data
Priority-
based
Packet
Scheduler
Packet
Classifier
ACK
high
Output queue for ACK packets (100 packets)

39. Solution Mechanism (2/2)
Minimize number of ACK packets in queue
Drop ACK packets more than data packets.
This does not affect to the performance since ACK is cumulative.
Use modified RED queue (Random Early Detection)

40. time-averaged queue size
RED Queue
S. Floyd and V. Jacobson, “Random early detection gateways for TCP congestion avoidance,” IEEE/ACM Transactions on Networking vol. 1, no. 4, pp , August 1993.
When queue is almost full: drop many packets
When queue is almost empty: don’t drop packets
time-averaged queue size
incoming packet
queue S
drop probability
drop probability
drop S

41. Modified RED Queue for Our Problem
Give high drop probability (Pa) to ACK packets
Give low drop probability (Pd) to data packets
Sa = time-averaged queue size of ACK queue
Sd = time-averaged queue size of data queue Pa Pd 1 Sd

42. Future Work Devise better drop probability equations.
Summarize related work.
Verify the proposed solution using Simulator (NS-2)
Modification of NS-2 is required.
It is a very time-consuming job.

43. Thank You!!!