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Computer Networks
AIU_CN_Lecture_2
Multiple access
(Chapter 12)
Outline
1. Introduction
2. RANDOM ACCESS
2.1 CSMAlCD
2.2 CSMAlCA
3. CONTROLLED ACCESS
3.1 Reservation
3.2 Polling
3.3 Token Passing
4. CHANNELIZATION
4.1 FDMA
4.2 TDMA
4.3 CDMA 2
1. Introduction
 Data Link layer sublayers:
 Data Link Control
• Framing, flow control, error control, protocols.
 Medium Access Resolution
• Medium access control
Data Link layer
Data Link Control
Multiple-Access Resolution
3
Multiple-Access Protocols
4
2. Random Access
 Also called contention method
 no station has control over another.
 No station permits/does not permit, another station to
send.
 At each instance, a station that has data to send uses
a procedure defined by the protocol to make a
decision on whether or not to send.
 It’s a random method because the transmission
is random among the stations.
 It’s called a contention method because stations
compete with one another to access the medium
5
2.1 CSMA/CD
 Carrier Sense Multiple Access with Collision
Detection (CSMA/CD)
 Control access to the medium
 When a node wants to transmit information, it
first “listens” to the network (or medium).
 If no one is transmitting over the network, the node
begins transmission
 It is however possible for two nodes to transmit
simultaneously thinking that the network is clear
6
CSMA/CD
 When two nodes transmit at the same time, a
collision occurs
 The first station to detect the collision sends a
jamming signal into the network
 Both nodes back off, wait for a random period of
time and then re-transmit
7
CSMA/CD
8
Minimum Frame Size
 Any station must discover collision before ending
transmission, because stations don’t listen after
finishing transmission
 If a station succeeds to send the last bit of a
frame without collision, it considers the frame’s
been sent successfully and discards it from
buffer, and starts handling the next frame.
 Time required to transmit a frame, Tfr , must be
as much as at least twice the propagation time,
TP , on the medium
9
Minimum Frame Size
 Assume that the involved stations, A and B, are situated
at the end of the medium
 The signal needs to propagate from A to B in TP, and if a
collision takes place at B, another TP is required for the
signal to propagate to A
 So in order to discover the collision, A needs to be in
sending state after at least 2 x TP
10
Packet almost at
B at TP - ε
Packet starts at
time 0
Collision at time
TP
Noise burst gets back
to A at 2TP
Minimum Frame Size
 1st bit collision
 At t1: A starts sending
 At t2: C starts sending
 At t3: C detects collision and aborts
 At t3: A detects collision and aborts 11
Minimum Frame Size
 Exercise
 A network using CSMA/CD has a bandwidth of 10
Mbps. If the maximum propagation time (including all
sort of delay and ignoring the time needed to send a
jamming signal) is 25,6 s, what is the minimum size
of the frame?
 Solution
• Tf = 2 * Tp = 51.2 s
• the minimum frame length =
10 x 106 x 51.2 x 10-6 =
512 bits (64 bytes)
 This is the minimum frame length of a traditional
Ethernet frame 12
Exercise
 A network using CSMA/CD has a bandwidth of 100
Mbps. If the maximum propagation time (including
all sort of delay and ignoring the time needed to
send a jamming signal) is 20,0 s, what is the
minimum size of the frame?
13
2.2 CSMA/CA
 Carrier Sense Multiple Access with Collision
Avoidance
 Control access to the medium in wireless networks
 In wireless networks, collision detection is not
effective
 Collision avoidance is used
 Collision avoidance strategies:
 Inter-Frame Space (IFS)
 Contention Window
 Acknowledgement
14
Inter-frame space (IFS)
 A station senses the medium till it becomes free
(ie, not busy),
 next it waits for an IFS time,
 Then it waits for a contention time.
15
Contention Window
 The contention window is an amount of time divided
into slots.
 A station that is ready to send chooses a random
number of slots as its wait time.
 The number of slots in the window changes according
to the binary exponential back-off strategy.
 i.e., 1 slot first time, 2 slots 2nd time,4 slots 3rd time, etc
 Note: If the station finds the channel busy, it does not
restart the process;
 it just stops the timer and restarts it when the channel
becomes idle.
16
Acknowledgement
 With all these precautions, there still may be a
collision resulting in destroyed data.
 In addition, the data may be corrupted during
transmission.
 The positive acknowledgement and the time-out
timer can help guarantee that the receiver has
received the frame.
17
3. Controlled Access
 Stations consult one another to find which
station has the right to send
 Prior to transmission, a station get authorization
to send from other stations.
 Controlled access methods:
 Reservation
 Polling
 Token Passing
18
3.1 Reservation
 a station needs to make a reservation before
sending data
 Time is divided into intervals.
 In each interval, a reservation frame precedes
the data frames sent in that interval.
19
3.2 Polling
 Polling works with topologies in which one
device is designated as a primary station and
the other devices are secondary stations.
 Stations exchange data through the primary
20
3.3 Token Passing
 Stations are organized in logical ring
 Every station has predecessor and successor
 A station can send if it holds the token
 When finishing transmission, it passes the token to its
successor.
21
4. Channelization
 Available bandwidth of a link is shared in time,
frequency, or through code, between stations.
In this section,
 we discuss three channelization protocols:
 Frequency-Division Multiple Access (FDMA)
 Time-Division Multiple Access (TDMA)
 Code-Division Multiple Access (CDMA)
22
4.1 FDMA
23
4.2. TDMA
24
4.3. CDMA
 One channel carries all transmissions simultaneously
25
Chips
 Each station is assigned a code, which is a
sequence of numbers, called chips
 multiplication by number
2. [+1 +1-1-1]=[+2+2-2-2]
 Multiplication of two equal sequences
[+1 +1 -1 -1] • [+1 +1 +1 +1] = 1 + 1 - 1 - 1 = 0
[+1 +1-1 -1] . [+1 +1 -1 -1] = 1 + 1 + 1 + 1 = 4
 Addition of 2 sequences
[+1+1-1-1]+[+1+1+1+1]=[+2+2 0 0] 26
Data Representation
 If a station needs to send a 0 bit, it encodes it
as -1;
 if a station needs to send a 1 bit, it encodes it as
+1
 When a station is idle, it sends no signal, which
is interpreted as a 0
27
Encoding and Decoding
28
Sequence Generation
 Walsh table is used to generate chip sequences
 Walsh table is a two-dimensional table with an equal
number of rows and columns
29
Example
 Find the chips for a network with
 a. Two stations
 b. Four stations
 Solution
 We can use the rows of W2 and W4 in the precedent
Figure
a. For a two-station network, we have [+1 +1] and [+1 -
1].
b. For a four-station network we have [+1 +1 +1 +1], [+1
-1 +1 -1], [+1 +1 -1 -1], and [+1-1-1 +1].
30

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AIU_CN_2.pptx

  • 2. Outline 1. Introduction 2. RANDOM ACCESS 2.1 CSMAlCD 2.2 CSMAlCA 3. CONTROLLED ACCESS 3.1 Reservation 3.2 Polling 3.3 Token Passing 4. CHANNELIZATION 4.1 FDMA 4.2 TDMA 4.3 CDMA 2
  • 3. 1. Introduction  Data Link layer sublayers:  Data Link Control • Framing, flow control, error control, protocols.  Medium Access Resolution • Medium access control Data Link layer Data Link Control Multiple-Access Resolution 3
  • 5. 2. Random Access  Also called contention method  no station has control over another.  No station permits/does not permit, another station to send.  At each instance, a station that has data to send uses a procedure defined by the protocol to make a decision on whether or not to send.  It’s a random method because the transmission is random among the stations.  It’s called a contention method because stations compete with one another to access the medium 5
  • 6. 2.1 CSMA/CD  Carrier Sense Multiple Access with Collision Detection (CSMA/CD)  Control access to the medium  When a node wants to transmit information, it first “listens” to the network (or medium).  If no one is transmitting over the network, the node begins transmission  It is however possible for two nodes to transmit simultaneously thinking that the network is clear 6
  • 7. CSMA/CD  When two nodes transmit at the same time, a collision occurs  The first station to detect the collision sends a jamming signal into the network  Both nodes back off, wait for a random period of time and then re-transmit 7
  • 9. Minimum Frame Size  Any station must discover collision before ending transmission, because stations don’t listen after finishing transmission  If a station succeeds to send the last bit of a frame without collision, it considers the frame’s been sent successfully and discards it from buffer, and starts handling the next frame.  Time required to transmit a frame, Tfr , must be as much as at least twice the propagation time, TP , on the medium 9
  • 10. Minimum Frame Size  Assume that the involved stations, A and B, are situated at the end of the medium  The signal needs to propagate from A to B in TP, and if a collision takes place at B, another TP is required for the signal to propagate to A  So in order to discover the collision, A needs to be in sending state after at least 2 x TP 10 Packet almost at B at TP - ε Packet starts at time 0 Collision at time TP Noise burst gets back to A at 2TP
  • 11. Minimum Frame Size  1st bit collision  At t1: A starts sending  At t2: C starts sending  At t3: C detects collision and aborts  At t3: A detects collision and aborts 11
  • 12. Minimum Frame Size  Exercise  A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including all sort of delay and ignoring the time needed to send a jamming signal) is 25,6 s, what is the minimum size of the frame?  Solution • Tf = 2 * Tp = 51.2 s • the minimum frame length = 10 x 106 x 51.2 x 10-6 = 512 bits (64 bytes)  This is the minimum frame length of a traditional Ethernet frame 12
  • 13. Exercise  A network using CSMA/CD has a bandwidth of 100 Mbps. If the maximum propagation time (including all sort of delay and ignoring the time needed to send a jamming signal) is 20,0 s, what is the minimum size of the frame? 13
  • 14. 2.2 CSMA/CA  Carrier Sense Multiple Access with Collision Avoidance  Control access to the medium in wireless networks  In wireless networks, collision detection is not effective  Collision avoidance is used  Collision avoidance strategies:  Inter-Frame Space (IFS)  Contention Window  Acknowledgement 14
  • 15. Inter-frame space (IFS)  A station senses the medium till it becomes free (ie, not busy),  next it waits for an IFS time,  Then it waits for a contention time. 15
  • 16. Contention Window  The contention window is an amount of time divided into slots.  A station that is ready to send chooses a random number of slots as its wait time.  The number of slots in the window changes according to the binary exponential back-off strategy.  i.e., 1 slot first time, 2 slots 2nd time,4 slots 3rd time, etc  Note: If the station finds the channel busy, it does not restart the process;  it just stops the timer and restarts it when the channel becomes idle. 16
  • 17. Acknowledgement  With all these precautions, there still may be a collision resulting in destroyed data.  In addition, the data may be corrupted during transmission.  The positive acknowledgement and the time-out timer can help guarantee that the receiver has received the frame. 17
  • 18. 3. Controlled Access  Stations consult one another to find which station has the right to send  Prior to transmission, a station get authorization to send from other stations.  Controlled access methods:  Reservation  Polling  Token Passing 18
  • 19. 3.1 Reservation  a station needs to make a reservation before sending data  Time is divided into intervals.  In each interval, a reservation frame precedes the data frames sent in that interval. 19
  • 20. 3.2 Polling  Polling works with topologies in which one device is designated as a primary station and the other devices are secondary stations.  Stations exchange data through the primary 20
  • 21. 3.3 Token Passing  Stations are organized in logical ring  Every station has predecessor and successor  A station can send if it holds the token  When finishing transmission, it passes the token to its successor. 21
  • 22. 4. Channelization  Available bandwidth of a link is shared in time, frequency, or through code, between stations. In this section,  we discuss three channelization protocols:  Frequency-Division Multiple Access (FDMA)  Time-Division Multiple Access (TDMA)  Code-Division Multiple Access (CDMA) 22
  • 25. 4.3. CDMA  One channel carries all transmissions simultaneously 25
  • 26. Chips  Each station is assigned a code, which is a sequence of numbers, called chips  multiplication by number 2. [+1 +1-1-1]=[+2+2-2-2]  Multiplication of two equal sequences [+1 +1 -1 -1] • [+1 +1 +1 +1] = 1 + 1 - 1 - 1 = 0 [+1 +1-1 -1] . [+1 +1 -1 -1] = 1 + 1 + 1 + 1 = 4  Addition of 2 sequences [+1+1-1-1]+[+1+1+1+1]=[+2+2 0 0] 26
  • 27. Data Representation  If a station needs to send a 0 bit, it encodes it as -1;  if a station needs to send a 1 bit, it encodes it as +1  When a station is idle, it sends no signal, which is interpreted as a 0 27
  • 29. Sequence Generation  Walsh table is used to generate chip sequences  Walsh table is a two-dimensional table with an equal number of rows and columns 29
  • 30. Example  Find the chips for a network with  a. Two stations  b. Four stations  Solution  We can use the rows of W2 and W4 in the precedent Figure a. For a two-station network, we have [+1 +1] and [+1 - 1]. b. For a four-station network we have [+1 +1 +1 +1], [+1 -1 +1 -1], [+1 +1 -1 -1], and [+1-1-1 +1]. 30