37
1 CS 4396 Computer Networks Lab TCP – Part II

CS 4396 Computer Networks Lab

  • Upload
    zazu

  • View
    40

  • Download
    0

Embed Size (px)

DESCRIPTION

CS 4396 Computer Networks Lab. TCP – Part II. TCP:. Flow Control Congestion Control Retransmission Timeout. What is Flow & Congestion Control ?. Flow Control: Algorithms to prevent that the sender overruns the receiver with information - PowerPoint PPT Presentation

Citation preview

Page 1: CS 4396  Computer Networks Lab

1

CS 4396 Computer Networks Lab

TCP – Part II

Page 2: CS 4396  Computer Networks Lab

2

Flow Control

Congestion ControlRetransmission Timeout

TCP:

Page 3: CS 4396  Computer Networks Lab

3

What is Flow & Congestion Control ?

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

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

The goal of each of the control mechanisms are different.

But the implementation is combined

Page 4: CS 4396  Computer Networks Lab

4

TCP Flow Control

Page 5: CS 4396  Computer Networks Lab

5

TCP Flow Control

• TCP implements sliding window flow control

• Sending acknowledgements is separated from setting the window size at sender.

• Acknowledgements do not automatically increase the window size

• Acknowledgements are cumulative

Page 6: CS 4396  Computer Networks Lab

6

TCP Format

• TCP segments have a 20 byte header with >= 0 bytes of data.

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

Page 7: CS 4396  Computer Networks Lab

7

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

Page 8: CS 4396  Computer Networks Lab

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.

Page 9: CS 4396  Computer Networks Lab

9

Sliding Window: “Window Closes”

1 2 3 4 5 6 7 8 9 10 11

1 2 3 4 5 6 7 8 9 10 11

Transmit Byte 6

1 2 3 4 5 6 7 8 9 10 11

AckNo = 5, Win = 4is received

• Transmission of a single byte (with SeqNo = 6) and acknowledgement is received (AckNo = 5, Win=4):

Page 10: CS 4396  Computer Networks Lab

10

Sliding Window: “Window Opens”

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 = 6is received

• 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).

Page 11: CS 4396  Computer Networks Lab

11

Sliding Window: “Window Shrinks”

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 = 3is received

• Acknowledgement is received that reduces the window from the right (AckNo = 5, Win=3):

• Shrinking a window should not be used; why?

Page 12: CS 4396  Computer Networks Lab

12

Sliding Window: Example

3K

ReceiverBuffer

0 4KSendersends 2Kof data

2K

Sendersends 2Kof data

4K

Sender blocked

Page 13: CS 4396  Computer Networks Lab

TCP Flow control

• Host sends 1 byte in a segment(In-efficient usage of bandwidth)

– 40 bytes TCP+IP headers – 1 byte application data

Nagle’s algorithm Send the first byte in a segment and buffer the rest until first segment is ACKed

• Rcvr reads 1 byte at a time(RcvWindow = 1 byte to the sender)

Silly window syndrome (Clark) Rcvr does not send a window update of 1 but waits until it has a buffer space of

min(MSS, half empty buffer)

Page 14: CS 4396  Computer Networks Lab

14

TCP Congestion Control

Page 15: CS 4396  Computer Networks Lab

15

TCP Congestion Control

• TCP has a mechanism for congestion control. The mechanism is implemented at the sender

• The sender has two parameters:– Congestion Window (cwnd)

– Slow-start threshhold Value (ssthresh)Initial value is the advertised window size

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

Page 16: CS 4396  Computer Networks Lab

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)

• The receiver sends an acknowledgement (ACK) for each packet » Note: Generally, a TCP receiver sends an ACK for every other

segment.

• 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.

• Does Slow Start increment slowly? Not really. In fact, the increase of cwnd is exponential (why?)

Page 17: CS 4396  Computer Networks Lab

17

Slow Start Example

• 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

segment 1

ACK for segment 1

cwnd = 1

cwnd = 2 segment 2segment 3

ACK for segments 2

cwnd = 4 segment 4segment 5segment 6

ACK for segments 4

cwnd = 8

ACK for segments 3

ACK for segments 5

ACK for segments 6

segment 7

ACK for segments 7

Page 18: CS 4396  Computer Networks Lab

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.

Page 19: CS 4396  Computer Networks Lab

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

Page 20: CS 4396  Computer Networks Lab

20

Responses to Congestion

• So, TCP assumes there is congestion if it detects a packet loss• A TCP sender can detect lost packets via:

• Timeout of a retransmission timer• Receipt of a duplicate ACK (why?)

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

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

ssthressh = cwnd / 2

– and slow-start is entered

Page 21: CS 4396  Computer Networks Lab

21

Summary of TCP congestion control

Initially:cwnd = 1;ssthresh =

advertised window size;New Ack received:

if (cwnd < ssthresh) /* Slow Start*/ cwnd = cwnd + 1;else /* Congestion Avoidance */ cwnd = cwnd + 1/cwnd;

Timeout:/* Multiplicative decrease */ssthresh = cwnd/2;cwnd = 1;

Page 22: CS 4396  Computer Networks Lab

22

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

• New Reno (1996)• SACK (1996)

• RED (Floyd and Jacobson 1993)

Page 23: CS 4396  Computer Networks Lab

23

Acknowledgments in TCP

• Receiver sends ACK to sender

– ACK is used for flow control, error control, and congestion control

• ACK number sent is the next sequence number expected

• Delayed ACK: TCP receiver normally delays transmission of an ACK (for about 200ms)

– Why?

• ACKs are not delayed when packets are received out of sequence

– Why?

Lost segment

Page 24: CS 4396  Computer Networks Lab

24

Acknowledgments in TCP

• Receiver sends ACK to sender

– ACK is used for flow control, error control, and congestion control

• ACK number sent is the next sequence number expected

• Delayed ACK: TCP receiver normally delays transmission of an ACK (for about 200ms)

– Why?

• ACKs are not delayed when packets are received out of sequence

– Why?

Out-of-order arrivals

1K SeqNo=0

AckNo=1024

1K SeqNo=1024

AckNo=2048

SeqNo=3072

AckNo=2048

SeqNo=2048 1K1K

Page 25: CS 4396  Computer Networks Lab

25

Fast Retransmit

• If three or more duplicate ACKs are received in a row, the TCP sender believes that a segment has been lost.

• Then TCP performs a retransmission of what seems to be the missing segment, without waiting for a timeout to happen.

• Enter slow start:

ssthresh = cwnd/2

cwnd = 1

duplicate

duplicate

Page 26: CS 4396  Computer Networks Lab

26

TCP Reno

• Duplicate ACKs:• Fast retransmit• Fast recovery

Fast Recovery avoids slow start

• Timeout:• Retransmit • Slow Start

• TCP Reno improves upon TCP Tahoe when a single packet is dropped in a round-trip time.

Page 27: CS 4396  Computer Networks Lab

27

Fast Recovery

• Fast recovery avoids slow start after a fast retransmit

• Intuition: Duplicate ACKs indicate that data is getting through

• After three duplicate ACKs set:– Retransmit “lost packet”

• When ACK arrives that acknowledges “new data” (here: AckNo=2048), set:– ssthresh = cwnd/2– cwnd=ssthresh

enter congestion avoidance

1K SeqNo=0

AckNo=1024

AckNo=1024

1K SeqNo=1024

SeqNo=20481K

AckNo=1024

SeqNo=30721K

SeqNo=10241K

SeqNo=40961K

AckNo=5120

AckNo=2048

Page 28: CS 4396  Computer Networks Lab

28

TCP Tahoe and TCP Reno(for single segment losses)

Reno

time

cwn

d

time

cwn

d

Tahoe

Page 29: CS 4396  Computer Networks Lab

31

TCP Retransmission Timeout

Page 30: CS 4396  Computer Networks Lab

32

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

Page 31: CS 4396  Computer Networks Lab

33

Receiving duplicate ACKs

• If three or more duplicate ACKs are received in a row, the TCP sender believes that a segment has been lost.

• Then TCP performs a retransmission of what seems to be the missing segment, without waiting for a timeout to happen.

• This can fix losses of single segments

1. duplicate

2. duplicate

3. duplicate

Page 32: CS 4396  Computer Networks Lab

34

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 1. When a packet with payload is transmitted and timer is not running2. When an ACK arrives that acknowledges new data and there are

other segments pending ACKs, 3. When a segment is retransmitted

• The timer is stopped when – All segments are acknowledged

Page 33: CS 4396  Computer Networks Lab

35

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

Page 34: CS 4396  Computer Networks Lab

36

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

Page 35: CS 4396  Computer Networks Lab

37

Setting the RTO value

• RTO is calculated based on the RTT measurements– Uses an exponential moving average to calculate

estimators for delay (srtt) and variance of delay (rttvar)

• 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

Page 36: CS 4396  Computer Networks Lab

38

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 rttvar1

• When a timeout occurs, the RTO value is doubled

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

This is called an exponential backoff

Page 37: CS 4396  Computer Networks Lab

39

Karn’s Algorithm

Timeout !

RT

T ? R

TT

?

Karn’s Algorithm: Don’t update RTT on any segments that have been retransmitted Restart RTT measurements only after an ACK is received for a segment that is not 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