Chapter 2: Direct Link
Networks (continued)
Slide Set 5
Reminder + Announcements
• Quiz January 27th.
• Chapters 1 and 2.1 to 2.6 (Upto
this set of slides)
• Bring Scantron sheets, calculators
In this set..
• Reliable Transmissions -retransmissions.
• Ethernet
Why retransmissions ?
• Error correction although feasible, is not
enough to handle all kinds of errors -especially burst errors.
• Corrupt frames cannot be deciphered and
are therefore dropped.
• Retransmissions needed to provide
reliability.
ACKs and Time-outs
• When frames are sent piggyback an
acknowledgement (ACK) for received packets
onto sent packets.
• If no ACK received up to a preset time-out,
resend frame.
• Called ARQ -- Automatic Repeat request.
1
Ack 1
100
Stop and Wait
• Allow only one outstanding packet at any given time.
• If ACK not received within time-out, send again.
Receiver
Sender
Fram
Receiver
Sender
Fram
e
e
ACK
ACK
Fram
e
ACK
(c)
(a)
Receiver
Sender
Fram
e
Receiver
Sender
Fram
e
ACK
Fram
Fram
e
e
ACK
ACK
(b)
(d)
How efficient is Stop and
Wait ?
• Consider a 1.5 Mbps link with 45 ms
RTT.
• BW - Delay product = 67.5 Kb = 8KB.
• You can fill 8 Kilo bytes of data prior to
receiving an ACK.
• However, if your frame size is 1 KB, you
are using only 1/8 of the capacity.
– Inefficient.
Sliding Window
• What we really like is that the
9th frame be transmitted when
ACK for the first frame arrives
:).
• Say we do this -- A window of
packets sent -- as ACKs are
received window slides i.e.,
more packets sent.
• Now what do we need in
addition ?
• Need to know which packets
have been received and which
have not.
– Packets labeled using
sequence numbers.
Sender
Receiver
Some definitions
• We have a window of packets sent
-- Send Window Size or SWS.
• Last Acknowledgement received is
denoted LAR.
• LFS represents the last frame
sent.
• NOTE: LFS - LAR <= SWS.
Sender Functions
<─ SWS
■■■
■■■
LAR
LFS
• When an ACK is received, the LAR moves to the
right.
• This allows for the transmission of an additional
frame.
Receiver functions
<
─
Notation:
RWS
■■■
■■■
RWS -- Receive Window Size
LAF -- Largest Acceptable Frame.
LFR
•
LAF
LFR -- Last Frame Received.
When frame with Seq_Num arrives:
1.
2.
3.
4.
5.
If Seq_Num <= LFR or Seq Num > LAF, discard.
–
Frame is outside window.
–
Frame within window.
If LFR < Seq_Num < LAF, accept frame.
Let Seq_Num_to_Ack be the largest sequence number yet
to be acked. This implies, all frames <= Seq_Num_to_Ack
have been received.
LFR = Seq_Num_to_Ack
Adjust LAF = LFR + RWS.
An Example
• Let LFR =5 and RWS = 4.
• This implies LAF = 9
• If packets 7 and 8 arrive (not 6), they
are buffered.
– Note that they are out of order.
• Typically, Receiver will resend an ACk for
packet 5.
• When 6 arrives, it can cumulatively ACK
all buffered packets i.e., it ACKs 8 and
moves LFR to 8 and LAF to 12.
Other possibilities
• Send NAK (negative acknowledgement)
for lost packets -- example for 6, when
7 is received.
• Duplicate ACKs -- send an ACK for 5
again when 7 is received to trigger
retransmission of 6.
• Selective ACKs : Explicitly ACK frames
that are received -- more complex.
Setting the Window Size
• SWS -- Set considering the BW
product.
• RWS -- Receiver can set it to
something appropriate -- may
depend on buffering resources.
– If RWS = 1 what happens to out of
order frames ?
Sequence number wrapping
• Sequence numbers are finite -thus there is a need to reuse -called wrap around.
• What is the relationship between
the SWS and MaxSeqNum ?
Maximum Sequence Number
and SWS
•
•
•
Should it not be SWS <= MaxSeqNum + 1 ?
Let us consider an example:
– Sender has eight Seq Nums from 0 to 7.
– SWS = RWS = 7.
– Sender transmits frames 0-6
– Receiver gets them, ACKs but ACKs get lost.
– Receiver expects 7 and next 0...5. But sender sends the
previous 0..5.
– When the receiver gets these, he cannot distinguish.
Thus, when RWS = SWS, SWS < (MaxSeqNum+1)/2.
(Verify that system works).
Sequence Numbers and
SWS
• For other cases i.e., when RWS is not
equal to SWS, other rules may apply.
• Depends on the specific case.
• A different way with TCP -- we will see
later.
• Easy solution -- large Sequence number
space.
Flow Control +
• Feedback control that allows the receiver
to throttle the sender.
• Informs sender not only about what
frames it has received but also how many
more frames it can receive.
• Need this in order to ensure that
receiver buffer does not overflow.
• Read rest of the parts on Sliding Window
-- implementation etc.
Multiple Access
• Nodes send and receive frames over a
shared “direct link” network.
• One way -- contention based i.e., nodes
contend to send -- example Ethernet.
• A second way is scheduled access -everyone knows when and how to send -i.e., they take turns -- example Token
Ring.
• Decentralized operations needed.
CSMA - CD
• Stands for Carrier Sense Multiple Access
with Collision Detection.
• Carrier sensing -- nodes can distinguish
between an idle and a busy line.
• Collision detection -- nodes can detect
when a transmission is “interfered” with.
• CSMA-CD can work on a bus architecture
as an example.
• Forms the basis for Ethernet.
The Ethernet Standard
• Specified by IEEE.
• Referred to as the 802.3 standard.
• Initially started out as a standard for a
bus topology -- 10 Mbps Ethernet.
• Today, we have switched Ethernet -Fast Ethernet with 100 Mbps and
Gigabit Ethernet with 1000 Mbps.
Ethernet on a Coax
• We consider the 10 Mbps Ethernet.
• Ethernet implemented on a coax cable of up to
500m.
• Hosts tap into this cable and must be at least 2.5
meters apart.
• Transceiver which is attached detects idle line and
drives signal.
Transceiver
Ethernet cable
Adaptor
Host
Repeaters
• Multiple Ethernet segments
can be joined using
“repeaters”.
• Repeaters amplify and
forward signals.
• No more than four
repeaters between any two
hosts.
– This ensures that
maximum reach of
Ethernet = 2500
meters.
– We will see why this is
needed later.
É
É
É
É
Repeater
Host
Ethernet -- Signal
broadcasts
• Signal broadcast over the multiple-access
link
– Broadcast means, sent to everyone.
• Each node propagates signal in both
directions.
• Repeaters forward on all outgoing
segments.
• Note: Ethernet uses Manchester encoding.
Cables
• Use of a 10base2 cable -- 10
Mbps, base --> it is baseband -frequency shifting -- No more than
200 m (actually only 185 m).
• 10base T -- T stands for twisted
pair.
Ethernet Frame
64
48
48
16
Preamble
Dest
addr
Src
addr
Type
32
Body
CRC
• Nuggets: (Other details refer book)
• 64 bit preamble for synchronization.
• 48 bit hardware or Ethernet address.
• Ethernet frame size: up to 1500 bytes of data and “at
least” 46 bytes --> padding may be necessary.
• Note -- Body carries IP datagram -- length of IP
header specified packet size -- can determine padding.
Ethernet Address
• Ethernet address unique for every host.
• Technically address belongs to adaptor
and not host -- burned into ROM.
• 6 bytes -- represented as six numbers
separated by colons: 8:0:2b:e4:b:1:2.
• Each number corresponds to a pair of
hexadecimal units -- one for each of the
nibbles in the byte.
Addressing (cont)
• Uniqueness: Each manufacturer (e.g. AMD) is
allocated a different prefix -- 8:0:2b --> 24
bits.
• Every frame transmitted on the Ethernet is
received by every adaptor on the network -- the
address is recognized and only the particular
adaptor recovers frame.
• All 1’s implies broadcast -- every adaptor picks it
up.
• An address that has the 1st bit set to 1 but is not
the broadcast address is used for multicast.
Transmitter Algorithm
•
•
Intelligence mainly on transmitter side - receiver dumb.
Algorithm:
1. If line idle, transmit frame immediately; no
negotiation with other adaptors.
2. If line busy, wait for the line to be idle and
either
a. Transmit immediately (1-persistent version)
b. Transmit with a probability p (p-persistent
version). With a probability q=1-p, defer the
transmission for a later time.
Why p-persistency?
• Deference is needed since multiple
transmitters may be waiting to transmit.
• If for example, p = 0.25, one could have
up to 4 transmitters on average -- only
one would transmit when the line is free.
• If multiple transmitters send
simultaneously, a collision will result.
Waiting time
• Each node looks at a “slot” time.
• Each slot is of a duration equal to that of
a frame.
• When node has something to send, in each
idle slot, it sends with a probability p, and
defers to the next slot with a probability q
and so on.
• Just a note: 1-persistency is good if nodes
are lightly loaded -- has been found
effective.
Upon Collision Detection
• When an adaptor detects a collision, it
transmits a 32 bit jamming sequence and
stops transmission.
• Minimal amount of data sent 64 bit
preamble + 32 jamming sequence = 96
bits.
• How does a node detect a collision ? As
nodes send, they monitor the line for
other transmissions.
Runt frame
• If two hosts that simultaneously
transmit are very close to each
other, they can detect a collision
quickly.
• In this case, enough only the 96
bits are sent -- such a frame is
called a runt frame.
Effects of distance on CD
•
•
•
•
•
•
•
•
What if nodes are far apart ? Let us
consider the case where propagation time is
d.
Let A transmit at time t.
A’s transmission jams that of node B which
just begins transmission at “t+d”.
It then sends out a jamming sequence -this takes another d seconds to reach A.
Thus, A knows of the collision at “t+2d”.
A has to be transmitting at this time in
order to realize the collision i.e., it has to
be transmitting for 2d seconds.
Note that 2d = RTT.
So transmission time should be at least
equal to RTT.
A
B
A
B
A
B
A
B
(a)
(b)
(c)
(d)
Transmission time and RTT
• To repeat : Transmission time at least equal
to RTT.
• Now since RTT is at most 51.2 ms given that
maximum reach is 2500 meters.
• Thus, Transmission time at least 51.2 ms. On
a 10 Mbps link, transmitter needs to send at
least 512 bits.
• Thus, we have a minimum size of 46 bytes
for the body (header makes frame 512 bits).
Ethernet back-off
• Once adaptor detects collision, it waits before
trying again.
• First delay -- selected randomly from the pair 0,
(21-1)51.2 ms = {0, 51.2} ms.
• If a second collision is experienced, choose backoff time from 0, (22-1)51.2 ms i.e, from {0, 51.2,
102.4 and 153.6} ms.
• Thus, after N collisions choose a back-off time
from {0.. (2N-1)51.2} ms.
• In reality, Ethernet gives up after a maximum
number of attempts which is 16. However, maximum
value of N is fixed; typically N = 10.
Key properties
• Ethernet works best when loads are light -small number of active nodes.
• Typically utilization is less than 30 %.
• Easy to administer and maintain; inexpensive.
• Currently, switched Ethernet.
Hub
Hub
Next time
• Token Ring,
• FDDI,
• Wireless LANs.