Unit 3 – Data Link Layer

and Medium Access Sub

Layer

Chapter 12 – Forouzan
Multiple Access Mechanisms

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ALOHA - Additive Links On-line Hawaii Area

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Packet Transmission Time
 Packet Transmission Time: It is amount of
time from beginning until the end of packet
transmission.
 Packet transmission Time= Packet Size/Bit Rate
Assuming, 100Mbits/sec. Ethernet
Maximum packet size of 1526 Bytes Result in,
 Packet Transmission Time =
(1526*8)/(100,000,000) = 122usec.

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Propagation Time
 Propagation Time= Distance/Propagation Speed
 It is nothing but the time it takes for first bit to
travel from the sender to receiver.
 Propagation speed depends on physical
medium of the link.
 Propagation speed for fiber, twisted pair, copper
wire is in the range of 2*10^8m/sec. and 3*10^8
m/sec for wireless communications.

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ALOHA: Vulnerable Time

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Example:
A pure ALOHA n/w transmit 200 bit frames on a shared
channel of 200 Kbps. What is the requirement to make
this frame collision-free?

Frame transmission time Tfr=200 bits/200 Kbps = 1 ms

The vulnerable time = 2*1 ms = 2ms

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Throughput:

How much data can be transferred

from one location to another in given
time.

Throughput is the rate

of successful message delivery over
a communication channel
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ALOHA: Throughput
 The throughput for pure ALOHA is
S = G × e−2G
where G is the average number of frames
generated during one frame transmission
time.
 The maximum throughput
 Smax = 0.184 when G= 1/2

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Example
 A pure ALOHA network transmits 200-bit
frames on a shared channel of 200 kbps.
What is the throughput if the system (all
stations together) produces
 1000 frames per second
 500 frames per second
 250 frames per second

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 If the system creates 1000 frames per
seconds, that is 1 frame per millisec. The
load is 1. In this case S=G*e^-2G =
0.135(13.5%)
 This means that throughput is
1000*0.135=135 frames.
 Only 135 frames out of 1000 will probably
survive.

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 If the system creates 500 frames per
seconds, that is 1/2 frame per millisec. The
load is 1/2. In this case S=G*e^-2G =
0.184(18.4%)
 This means that throughput is 500*0.184=92
frames.
 Only 92 frames out of 500 will probably
survive.

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 If the system creates 250 frames per
seconds, that is 1/4 frame per millisec. The
load is 1/4. In this case S=G*e^-2G =
0.152(15.2%)
 This means that throughput is 250*0.152=38
frames.
 Only 38 frames out of 250 will probably
survive.

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Slotted ALOHA

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Slotted ALOHA: Vulnerable Time

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Slotted ALOHA: Throughput
 The throughput for Slotted ALOHA is

S = G × e−G

where G is the average number of frames

requested per frame-time
 The maximum throughput
 Smax = 0.368 when G= 1

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Comparison

S=0.368

S=0.18
Example
 A Slotted ALOHA network transmits 200-bit
frames on a shared channel of 200 kbps.
What is the throughput if the system (all
stations together) produces
 1000 frames per second
 500 frames per second
 250 frames per second

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Solution
 The frame transmission time is 200/200 kbps
or 1 ms.
 a. If the system creates 1000 frames per
second, this is 1 frame per millisecond. The
load is 1. In this case S = G× e −G or S =
0.368 (36.8 percent). This means that the
throughput is 1000 × 0.368 = 368 frames.
Only 368 frames out of 1000 will probably
survive.
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 b. If the system creates 500 frames per
second, this is (1/2) frame per millisecond.
The load is (1/2). In this case S = G × e −G or
S = 0.303 (30.3 percent). This means that the
throughput is 500 × 0.0303 = 151. Only 151
frames out of 500 will probably survive.

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 c. If the system creates 250 frames per
second, this is (1/4) frame per millisecond.
The load is (1/4). In this case S = G × e −G or
S = 0.195 (19.5 percent). This means that the
throughput is 250 × 0.195 = 49. Only 49
frames out of 250 will probably survive.

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Space/time model of the collision
in CSMA

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 Example 12.5

A network using CSMA/CD has a bandwidth of 10

Mbps. If the maximum propagation time (including the
delays in the devices and ignoring the time needed to
send a jamming signal, as we see later) is 25.6 μs,
what is the minimum size of the frame?
Solution
The frame transmission time is Tfr = 2 × Tp = 51.2 μs.
This means, in the worst case, a station needs to
transmit for a period of 51.2 μs to detect the collision.
The minimum size of the frame is 10 Mbps × 51.2 μs
= 512 bits or 64 bytes. This is actually the minimum
size of the frame for Standard Ethernet.
12.42
Flow diagram for the CSMA/CD

12.43
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CSMA/CA
 Carrier Sense Multiple Access with Collision
Avoidance
 Used in a network where collision cannot be
detected
 E.g., wireless LAN

IFS – Interframe Space 46

 When a frame is ready, the transmitting station checks
whether the channel is idle or busy.
 If the channel is busy, the station waits until the channel
becomes idle.
 If the channel is idle, the station waits for an Inter-frame
gap (IFG) amount of time and then sends the frame.
 After sending the frame, it sets a timer.
 The station then waits for acknowledgement from the
receiver. If it receives the acknowledgement before
expiry of timer, it marks a successful transmission.
 Otherwise, it waits for a back-off time period and
restarts.
CSMA/CA: Flow Diagram

contention window
size is 2K-1

After each slot:

- If idle, continue counting
- If busy, stop counting

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