1

Transport Protocols (continued)

Relates to Lab 5. UDP and TCP

2

Flow Control

Congestion ControlTCP:

3

What is Flow/Congestion Control ?

• Flow Control: Algorithms to prevent that the sender overruns the receiver buffer

• Congestion Control: Algorithms to prevent that the sender overloads the network

Sliding window implements both control mechanisms.

4

Flow Control

TCP

Application

1. write 100 bytes2. write 20 bytes

queue ofbytes to betransmitted TCP

queue ofbytes thathave beenreceived

Application1. read 40 bytes2. read 40 bytes3. read 40 bytes

Segments

5

Congestion Control

TCP connection 1bottleneck

routercapacity R

6

TCP Flow Control

7

TCP Flow Control

• TCP uses the sliding window algorithm to implement flow control

8

Sliding Window Flow Control

1 2 3 4 5 6 7 8 9 10 11

Advertised window

sent but notacknowledged can be sent

USABLEWINDOW

sent andacknowledged

can't sent

• Sliding Window Protocol is performed at the byte level:

•Here: Sender can transmit sequence numbers 6,7,8.

9

Sliding Window: “Window Opens”

• Acknowledgement is received that enlarges the window to the right (AckNo = 5, Win=6):

• A receiver opens a window when TCP buffer empties (meaning that data is delivered to the application).

•1 •2 •3 •4 •5 •6 •7 •8 •9 •10 •11

•1 •2 •3 •4 •5 •6 •7 •8 •9 •10 •11

•AckNo = 5, Win = 6•is received

10

Window Management in TCP

• The receiver is returning two parameters to the sender

• The interpretation is:• I am ready to receive new data with

SeqNo= AckNo, AckNo+1, …., AckNo+Win-1

• Receiver can acknowledge data without opening the window• Receiver can change the window size without acknowledging

data

AckNowindow size

(win)32 bits 16 bits

11

IP header TCP header TCP data

Sequence number (32 bits)

DATA

20 bytes 20 bytes

0 15 16 31

Source Port Number Destination Port Number

Acknowledgement number (32 bits)

window sizeheaderlength

0 Flags

Options (if any)

TCP checksum urgent pointer

20 bytes

12

Sliding Window: Example

3K

ReceiverBuffer

0 4KSendersends 2Kof data

2K

Sendersends 2Kof data

4K

Sender blocked

13

TCP Congestion Control

14

TCP Congestion Control

• Keep a sender from congesting the network.

15

TCP Congestion Control

• Keep a sender from congesting the network.• The sender has two internal parameters:

– Congestion Window (cwnd)– Slow-start threshhold Value (ssthresh)

• Sliding window size is set to the minimum of (cwnd, receiver advertised win)

• Congestion control works in two modes:– slow start (cwnd < ssthresh)

• Probe the available bandwidth– congestion avoidance (cwnd >= ssthresh)

• Try not to overload the network.

16

Slow Start

• Initial value: Set cwnd = 1 • Note: Unit is a segment size. TCP actually is based on bytes and

increments by 1 MSS (maximum segment size)

• Modern TCP implementation may set initial cwnd to 2

• Each time an ACK is received by the sender, the congestion window is increased by 1 segment:

cwnd = cwnd + 1

• If an ACK acknowledges two segments, cwnd is still increased by only 1 segment.

• Even if ACK acknowledges a segment that is smaller than MSS bytes long, cwnd is increased by 1 maximum segment size.

• Question: how can you accelerate your TCP download?

17

Slow Start Example

• Not really slow: The congestion window size grows very rapidly– For every ACK, we

increase cwnd by 1 irrespective of the number of segments ACK’ed

• TCP slows down the increase of cwnd when cwnd > ssthresh

cwnd = 1

cwnd = 2

cwnd = 4

cwnd = 7

18

Congestion Avoidance

• Congestion avoidance phase is started if cwnd has reached the slow-start threshold value

• If cwnd >= ssthresh then each time an ACK is received, increment cwnd as follows:

• cwnd = cwnd + 1/ cwnd• So cwnd is increased by one only if all cwnd segments have

been acknowledged.

19

Example of Slow Start/Congestion Avoidance

Assume that ssthresh = 8 cwnd = 1

cwnd = 2

cwnd = 4

cwnd = 8

cwnd = 9

cwnd = 10

0

2

4

6

8

10

12

14

Roundtrip times

Cw

nd

(in

seg

men

ts)

ssthresh

20

Responses to Congestion

• TCP uses packet loss as congestion signal• A TCP sender can detect lost packets via:

• Receipt of a duplicate ACK• Timeout of a retransmission timer

21

Response to Timeout

• TCP interprets a Timeout as a severe congestion signal. When a timeout occurs, the sender performs: – cwnd is reset to one:

cwnd = 1– ssthresh is set to half of the current size of the congestion

window:

ssthressh = cwnd / 2

– and slow-start is entered

22

Reaction to Duplicate ACKs

• Fast retransmit– Three duplicate ACKs indicate a packet loss– Retransmit without timeout

• Fast recovery– Avoid slow start– Retransmit “lost packet”– ssthresh = cwnd/2– cwnd = cwnd+3– Increment cwnd by one for each additional duplicate ACK

• When ACK arrives that acknowledges “new data” set:

cwnd=ssthresh

enter congestion avoidance

23

Duplicate ACK example

1. duplicate

2. duplicate

3. duplicate

24

Flavors of TCP Congestion Control

• TCP Tahoe (1988, FreeBSD 4.3 Tahoe)– Slow Start– Congestion Avoidance– Fast Retransmit

• TCP Reno (1990, FreeBSD 4.3 Reno)– Fast Recovery– Modern TCP implementation

• New Reno (1996)• SACK (1996)

25

TCP Tahoe

Thi

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e is

co

pied

fro

m s

omew

her

e

26

TCP Reno (Jacobson 1990)

CASS

Fast retransmission/fast recovery

Thi

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27

TCP III – Retransmission and Timeout

28

Retransmissions in TCP

• A TCP sender retransmits a segment when it assumes that the segment has been lost:

1. No ACK has been received and a timeout occurs

2. Multiple ACKs have been received for the same segment

29

Retransmission Timer

• TCP sender maintains one retransmission timer for each connection

• When the timer reaches the retransmission timeout (RTO) value, the sender retransmits the first segment that has not been acknowledged

• The timer is started when 1. When a packet with payload is transmitted and timer is not running2. When an ACK arrives that acknowledges new data, 3. When a segment is retransmitted

• The timer is stopped when – All segments are acknowledged

30

How to set the timer

• Retransmission Timer:– The setting of the retransmission timer is crucial for good

performance of TCP– Timeout value too small results in unnecessary

retransmissions– Timeout value too large long waiting time before

a retransmission can be issued

– A problem is that the delays in the network are not fixed – Therefore, the retransmission timers must be adaptive

31

Setting the value of RTO:

• The RTO value is set based on round-trip time (RTT) measurements that each TCP performs

RTT #1

RTT #2

RTT #3

• Each TCP connection measures the time difference between the transmission of a segment and the receipt of the corresponding ACK

• There is only one measurement ongoing at any time (i.e., measurements do not overlap)

• Figure on the right shows three RTT measurements

32

Setting the RTO value

• RTO is calculated based on the RTT measurements– Uses an exponential moving average to estimate RTT (srtt)

and variance of RTT (rttvar) from – The influence of past samples decrease exponentially

• The RTT measurements are smoothed by the following estimators srtt and rttvar:

srttn+1 = RTT + (1- ) srttn

rttvarn+1 = ( | RTT - srttn | ) + (1- ) rttvarn

RTOn+1 = srttn+1 + 4 rttvarn+1

– The gains are set to =1/4 and =1/8

33

Setting the RTO value (cont’d)

• Initial value for RTO:– Sender should set the initial value of RTO to

RTO0 = 3 seconds

• RTO calculation after first RTT measurements arrived

srtt1 = RTT rttvar1 = RTT / 2

RTO1 = srtt1 + 4 rttvarn+1

• When a timeout occurs , the RTO value is doubled

RTOn+1 = max ( 2 RTOn, 64) seconds

This is called an exponential backoff

34

Karn’s Algorithm

Timeout !

RT

T ? R

TT

?

Karn’s Algorithm:• Don’t update RTT on any segments that have been retransmitted

If an ACK for a retransmitted segment is received, the sender cannot tell if the ACK belongs to the original or the retransmission.

RTT measurements is ambiguous in this case

35

Summary

• UDP: connectionless, unreliable, datagram service• TCP: reliable, connection-oriented, byte stream service

– TCP header– Connection management– Delayed ACKs and nagle’s algorithm– TCP flow control– TCP congestion control– TCP retransmission and timeout

• References– TCP/IP illustrated vol. 1, chapter11, 17-24– RFC793 (Transmission Control Protocol)– RFC768 (User Datagram Protocol)– RFC2581 (TCP Congestion control)– RFC2988 (Computing TCP’s Retransmission Timer)– RFC3390 (Increasing TCP’s Initial Window)

36

Lab 7 Problems

• Exercise 3(B), Part 4-5• TCP Path MTU Discovery

– How does it works?

1. TCP uses the minimum of the MSS announced by the 2 hosts in connection establishment

2. All TCP segment has the DF bit set in IP header

3. If a router along the path has a smaller MTU than the TCP segment size, it returns an “ICMP unreachable” error report with its MTU, e.g. MTUr

4. TCP set the MSS to (MTUr - 40), and try again.

5. The process continues until no “ICMP unreachable” error message is received.

37

LAB 5

38

TCP Path MTU Discovery Example

MTU 1500

MTU 1500

MTU 500

MTU 1500

10.0.1.11/24 10.0.2.22/2410.0.1.33/24 10.0.2.33/24

PC1 PC2PC3/Router

ttcp –ts –1024 –n2 –p4444 –D 10.0.2.22 ttcp –rs –1024 –n2 –p4444

SYN, MSS=1460

SYN, ACK, MSS=1460

ACK

PUSH, LEN=1460

ICMP Unreachable, MTU=500

PUSH, LEN=460

ACK

39

Exercise 6A: TCP Bulk Data transfer (fast link)

eth1 eth1

10.0.5.11/24 10.0.5.22/24

PC1 PC2

ttcp –ts –l1000 –n500 –p4444 –D 10.0.5.22 ttcp –rs –l1000 –n500 –p4444

Step 4(5):

Determine whether or not the TCP sender generally transmits the maximum amount of data allowed by the advertised window. Explain your answer.

40

Exercise 6B: TCP Bulk Data transfer (slow link)

eth0 eth0

10.0.1.11/24 10.0.2.22/24

PC2

ttcp –ts –l1000 –n500 –p4444 –D 10.0.2.22 ttcp –rs –l1000 –n500 –p4444

Step 4(4):

Does TCP sender generally transmits the maximum amount of data allowed by the advertised window? Explain your answer.

PC1

Router 1 Router 2

41

Trouble Shooting

I did everything by following the instructions, but it does not work and I can’t ping other hosts/routers. Why?

Well, something must be wrong. Let’s figure it out.

42

Trouble Shooting Techniques

• The key to trouble shooting is to isolate the problem, i.e. what is the problem and where it happened in the network.– What is the output of the ping command?

“Network unreachable” Check local routing table

“PING x.x.x.x (x.x.x.x) 56(84) bytes of data”, and it hangs there

run “traceroute x.x.x.x” to see where the ICMP Echo Request packet failed.