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

timeTime 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 allocationResource 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-SwitchingGuaranteed 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

propagationtransmission

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

propagationtransmission

nodalprocessing queueing