The network core:
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CSE 413: Computer Network
Circuit Switching and Packet Switching Networks
Md. Kamrul Hasan09-03-2010
The network core:• mesh of interconnected
routers• the fundamental
question: how is data transferred through net?– circuit switching:
dedicated circuit per call: telephone net
– packet-switching: data sent through net in discrete “chunks” (packets) on shared media
The network core:Circuit Switching
End-to-end resources reserved for “call”
• link bandwidth, switch capacity
• dedicated resources: no sharing
• circuit-like (guaranteed) performance
• call setup required
Circuit Switching
DATA
Caller Callee
Boston Switch
LA Switch
propagation delay between caller and Boston switch
processing delay at switch
• It’s the method used by the telephone network
• A call has three phases:1. Establish circuit
from end-to-end (“dialing”),
2. Communicate,3. Close circuit (“tear
down”).• If circuit not available:
“busy signal”
(1)
(2)
(3)
Switch
Circuit Switching: Multiplexing/Demultiplexing
• Time divided into frames and frames divided into slots• Relative slot position inside a frame determines which
conversation the data belongs to – E.g., slot 0 belongs to the red conversation
• Need synchronization between sender and receiver
Frames
0 1 2 3 4 5 0 1 2 3 4 5Slots =
One way for sharing a circuit is TDM:
Lecture notes use the word “frame” for slot
The network core: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)• Consumers are charged on a per-minute basis• 2 ways of dividing the link bandwidth into
“pieces”– frequency division multiplexing (FDM)– time division multiplexing (TDM)
Circuit Switching: FDM and TDM
Frequency Division Multiplexing (FDM)
frequency
time
Time Division Multiplexing. (TDM)
frequency
time
4 users
Example:
Numerical example• How long does it take to send a file of
640,000 bits from host A to host B over a circuit-switched network?• The link’s transmission rate = 1.536 Mbps• Each link uses TDM with 24 slots/sec• 500 msec to establish end-to-end circuit
Figure it out …• Solution:
– Bandwidth of circuit = 1.536/24 = 64 kbps– Time to send: 640 kbits/64 kbps + 0.5s = 10.5s
What would be different if we use FDM instead of TDM?
Common mistake/confusion :Question:• A) Express transmission rate of 1Kbits/sec in bits/sec• B) Express the file size of 1KBytes in bitsAnswer: • A) 1000 bits/sec (in throughput, K = 103=1000)• B) 1024 Bytes = 8192 bits (in data size, K = 210=1024)
• Electronic speeds/times: K = 103, M = 106, G = 109
• Computer file/memory sizes: K = 210 , M = 220, G = 230
• Common computer notation:– b(bits) Kb, Mb, Gb– B(Bytes) KB, MB, GB
• Better computer notation:– b(bits) Kib, Mib, Gib– B(Bytes) KiB, MiB, GiB
Packet Switching• Used in the Internet• Data is sent in Packets
(header contains control info, e.g., source and destination addresses)
• Per-packet routing• At each node the entire
packet is received, stored, and then forwarded (store-and-forward networks)
• No capacity is allocated
Header Data
Packet 1
Packet 2
Packet 3
Packet 1
Packet 2
Packet 3
Packet 1
Packet 2
Packet 3
processing
delay of Packet 1 at Node 2
propagationdelay betweenHost 1 & Node 2 transmission
time of Packet 1at Host 1
Host 1 Host 2
Node 1 Node 2
Router
Packet Switching: Multiplexing/Demultiplexing
• Multiplex using a queue– Routers need memory/buffer
• Demultiplex using information in packet header– Header has destination – Router has a routing table that contains information
about which link to use to reach a destination
Queue
Packet switching also show reordering
Host A
Host BHost E
Host D
Host C
Node 1 Node 2
Node 3
Node 4
Node 5
Node 6 Node 7
Packets in a flow may not follow the same path (depends on routing as we will see later) packets may be reordered
The network core:Packet Switching
• all streams share network resources
• each packet uses full link bandwidth
• resources used as needed
Resource contention: • aggregate resource
demand can exceed amount available
• congestion: packets queue, wait for link
Bandwidth division into “pieces”
Dedicated allocation
Resource reservation
The network core:Packet switching
• Data transmitted in small, independent pieces – Source divides outgoing messages into
packets – Destination recovers original data
• Each packet travels independently – Includes enough information for delivery – May follow different paths – Can be retransmitted if lost
The network core:Functions of packet-switching
networks
• Packet construction– encode/package data at source
• Packet transmission– send packet from source to destination
• Packet interpretation– unpack/decode data from packet at destination– acknowledge receipt
statistical multiplexing Sequence of A & B packets does not have fixed pattern; shared on demand.
Compare: in TDM, each host gets same slot (periodically)in FDM, each host gets same bandwidth (continuously)
A
B
C100 Mb/sEthernet
1.5 Mb/s
D E
statistical multiplexing
queue of packetswaiting for output
link
The network core:statistical multiplexing
Differences Between Circuit & Packet Switching
Circuit-switching Packet-Switching
Guaranteed capacity No guarantees (best effort)
Capacity is wasted if data is bursty
More efficient
Before sending data establishes a path
Send data immediately
All data in a single flow follow one path
Different packets might follow different paths
No reordering; constant delay; no pkt drops
Packets may be reordered, delayed, or dropped
End-to-end delay (nodal delay) :• Total time from initiating “send” (from source) to
completed “receive” (at destination)
Throughput :• Rate (bits/sec) at which bits are actually being
transferred between sender/receiver– instantaneous: rate at given point in time– average: rate over longer period of time
Network performance metrics
Four sources of packet delay
• 1. nodal processing: – check bit errors– determine output link
A
B
propagation
transmission
nodalprocessing queueing
• 2. queueing delay– time waiting at output
link for transmission – depends on
congestion level of router
• 3. Transmission delay:– R=link bandwidth (speed
in bits per second, i.e. “bps”)
– L=packet length (in bits)– transmission delay = L/R
• 4. Propagation delay:– d = length of physical link (in
meters)
– s = propagation speed in medium (~2.5 x 108 m/sec)
– propagation delay = d/s
Note: R and s are very different quantities!
Four sources of packet delay
A
B
propagation
transmission
nodalprocessing queueing