Computer Communication &
Networks
Lecture 4
Circuit Switching, Packet Switching, Delays
http://web.uettaxila.edu.pk/CMS/coeCCNbsSp09/index.asp
Waleed Ejaz
waleed.ejaz@uettaxila.edu.pk 1
Communication Network
Communication networks
Switched networks Broadcast networks
End nodes send to one (or more) end nodes End nodes share a common channel
(TV, radio…)
Circuit switching Packet switching
Dedicated circuit per call Data sent in discrete portions
(telephone, ISDN) (the Internet)
(physical)
2
Communication Network
Communication networks
Switched networks Broadcast networks
End nodes send to one (or more) end nodes End nodes share a common channel
(TV, radio…)
Circuit switching Packet switching
Dedicated circuit per call Data sent in discrete portions
(telephone, ISDN) (the Internet)
(physical)
3
Circuit switching
A dedicated communication path (sequence of links-
circuit) is established between the two end nodes
through the nodes of the network
Bandwidth: A circuit occupies a fixed capacity of
each link for the entire lifetime of the connection.
Capacity unused by the circuit cannot be used by
other circuits.
Latency: Data is not delayed at switches
4
Circuit switching (cnt’d)
Three phases involved in the communication process:
1. Establish the circuit
2. Transmit data
3. Terminate the circuit
If circuit not available: busy signal (congestion)
5
Time diagram of circuit switching
switch
host 1 node 1 node 2 host 2
Delay
host 1- node 1
Processing
circuit delay node 1
establishment
Delay
host 2- host 1
data
transmission DATA
time
6
Circuit Switching
Network resources (e.g., bandwidth)
divided into “pieces”
pieces allocated to calls
resource piece idle if not used by owning call
(no sharing)
dividing link bandwidth into “pieces”
frequency division
time division
7
Circuit Switching: FDM and TDM
Example:
FDM
4 users
frequency
time
TDM
frequency
time
8
Example
Assume that a voice channel occupies a bandwidth of 4
kHz. We need to combine three voice channels into a link
with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the
configuration, using the frequency domain. Assume there
are no guard bands.
Solution
We shift (modulate) each of the three voice channels to a
different bandwidth, as shown in Figure on next Slide. We
use the 20- to 24-kHz bandwidth for the first channel, the
24- to 28-kHz bandwidth for the second channel, and the
28- to 32-kHz bandwidth for the third one. Then we
combine them.
9
Example (contd.)
10
Example
Five channels, each with a 100-kHz bandwidth, are to be
multiplexed together. What is the minimum bandwidth of
the link if there is a need for a guard band of 10 kHz
between the channels to prevent interference?
Solution
For five channels, we need at least four guard bands. This
means that the required bandwidth is at least
5 × 100 + 4 × 10 = 540 kHz
11
Applications
AM Radio
Band 530-1700KHz
Each AM Station needs 10KHz
FM Radio
Band 88-108MHz
Each FM Station needs 200KHz
TV
Each Channel needs 6MHz
AMPS
12
Synchronous TDM
In synchronous TDM, the data rate
of the link is n times faster, and the unit
duration is n times shorter.
13
Example
In Figure on Last Slide, the data rate for each input connection is 1
kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the
duration of (a) each input slot, (b) each output slot, and (c) each
frame?
Solution
We can answer the questions as follows:
a. The data rate of each input connection is 1 kbps. This means
that the bit duration is 1/1000 s or 1 ms. The duration of the input
time slot is 1 ms (same as bit duration).
b. The duration of each output time slot is one-third of the input time
slot. This means that the duration of the output time slot is 1/3
ms.
c. Each frame carries three output time slots. So the duration of a
frame is 3 × 1/3 ms, or 1 ms. The duration of a frame is the same
as the duration of an input unit.
14
Example
Figure below shows synchronous TDM with a data stream
for each input and one data stream for the output. The unit
of data is 1 bit. Find (a) the input bit duration, (b) the
output bit duration, (c) the output bit rate, and (d) the
output frame rate.
15
Disadvantages of Sync. TDM
16
Statistical Multiplexing
On-demand time-division
Schedule link on a per-packet basis
Packets from different sources interleaved on link
Buffer packets in switches that are contending for the link
Do you see any problem ?
17
Statistical Multiplexing
An application needs to break-up its message in
packets, and re-assemble at the receiver
Fair allocation of link capacity: FIFO or QoS
Buffer may overflow – congestion at the switch
18
TDM slot comparison
•Slot Size
•No Synchronization
Bit
•Bandwidth
19
Communication networks
Communication networks
Switched networks Broadcast networks
end nodes send to one (or more) end nodes End nodes share a common channel
(TV, radio…)
Circuit switching Packet switching
Dedicated circuit per call Data sent in discrete portions
(telephone, ISDN) (the Internet)
20
Packet Switching
each end-end data stream resource contention:
divided into packets aggregate resource
user A, B packets share
network resources
demand can exceed
amount available
each packet uses full link
bandwidth congestion: packets
resources used as needed queue, wait for link use
store and forward:
Bandwidth division into “pieces” packets move one hop
Dedicated allocation at a time
Resource reservation
Node receives complete
packet before forwarding
21
Packet switching
- Why not message switching?-
host 1 node 1 node 2
host 2
propagation delay
host 1 – node1
message
processing &
set-up delay
of a message at
node 1
message
time message
Store-and-Forward
22
Message switching
EXAMPLE
node 1 node 2
host 1 host 2
for simplicity: ignore processing and propagation delays
M=7.5 Mb
transmission delay:
R=1.5 Mbps
M
3 15 [s]
R
Store complete message and than forward
23
Message switching versus packet
switching
Example
host 1 node 1 host 2
node 2
R=1.5 Mbps R=1.5 Mbps R=1.5 Mbps
For simplicity ignore processing and propagation delays
Split the message into packets each with1500 bits long
Store only 1 packet and then forward it
1 ms to transmit packet on 1 link
Pipelining: each link works in parallel
Delay reduced from 15 s to 5.002 s!!!
24
Packet switching
25
Packet Switching
router
router
router
Sequence of A & B packets does not have fixed pattern statistical multiplexing.
26
Packet switching versus circuit switching
Packet switching allows more users to use network!
1 Mb/s link
each user:
100 kb/s when “active”
active 10% of time
circuit-switching: N users
10 users 1 Mbps link
packet switching:
with 35 users, probability that
there are 11 or more
simultaneously active users
is approximately .0004
27
Packet switching versus circuit switching
Is packet switching a “winner?”
Great for bursty data
resource sharing
simpler, no call setup
Excessive congestion: packet delay and loss
protocols needed for reliable data transfer, congestion
control
Q: How to provide circuit-like behavior?
bandwidth guarantees needed for audio/video apps
still an unsolved problem
28
Packet switching versus circuit switching
(cnt’d)
Advantages of packet switching over circuit switching
Statistical multiplexing, and therefore efficient bandwidth usage
Simple to implement
Disadvantages of packet switching over circ. switching
Excessive congestion: packet delay and high loss
Protocols needed for reliable data transfer, congestion control
Packet header overhead
Provides no transparency to a user
Analogy: a road versus a railroad
29
How do loss and delay occur?
packets queue in router buffers
packet arrival rate to link exceeds output link capacity
packets queue, wait for turn
packet being transmitted (delay)
B
packets queueing (delay)
free (available) buffers: arriving packets
dropped (loss) if no free buffers
30
Four sources of packet delay
1. Nodal processing: 2. Queueing
check bit errors time waiting at output
determine output link link for transmission
depends on congestion
level of router
transmission
A propagation
B
nodal
processing queueing
31
Delay in packet-switched networks
3. Transmission delay: 4. Propagation delay:
R=link bandwidth (bps) d = length of physical link
L=packet length (bits) s = propagation speed in
time to send bits into link medium (~2x108 m/sec)
= L/R propagation delay = d/s
Note: s and R are very
transmission different quantities!
A propagation
B
nodal
processing queueing
32
Nodal delay
d nodal d proc d queue d trans d prop
dproc = processing delay
typically a few microsecs or less
dqueue = queuing delay
depends on congestion
dtrans = transmission delay
= L/R, significant for low-speed links
dprop = propagation delay
a few microsecs to hundreds of msecs
33
Queueing delay (revisited)
R=link bandwidth (bps)
L=packet length (bits)
a=average packet arrival
rate
traffic intensity = La/R
La/R ~ 0: average queueing delay small
La/R -> 1: delays become large
La/R > 1: more “work” arriving than can be serviced,
average delay infinite!
34
Packet loss
queue preceding link in buffer has finite
capacity
when packet arrives to full queue, packet is
dropped
lost packet may be retransmitted by previous
node, by source end system, or not
retransmitted at all
35
Assignment 1
You can find Assignment 1 from course web.
Due Date: First class of Next Week
Quiz 1
On the day of submission of Assignment
related with topics covered in Assignment 1.
36
Readings
Computer Networking, a top-down approach
featuring the Internet (3rd edition), J.K.Kurose,
K.W.Ross
Chapter 1: Section 1.3, 1.6
37
38
