Network Switching and Layered Models: Circuit Switching vs. Packet Switching, Lecture notes of Data Communication Systems and Computer Networks

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Chapter 3 – End-to-End Switching and

Layered Models

3.1 – Network Switching

• (^) Few computers and small distance  direct

connection (LAN)

• (^) Many computers or large distance 

switching

• (^) Two fundamental approaches:
  • (^) Circuit Switching
  • (^) Packet SWitching

Example:

• (^) Utilization is defined as the ratio of the Arrival rate (L.) to the Transmission rate (R)
• (^) Consider the following:

Given:

L = 1000 bytes (8000 bits)

λ = 1 message per second

Td < 0.1 seconds (delay

requirement)

Thus, T

d

= L/R

And hence, R > 8000/0.1 = 80,000 bps, and

Utilization = 8000/80000 = 10%

• (^) Message Switching:
  • (^) No physical path is established in advance
  • (^) Store and forward, in its entirety
• (^) Packet Switching:
  • (^) Messages are broken into discrete units

called packets

• (^) Each packet contains data and headers
• (^) Store and forward, on a packet basis
• (^) Two approaches:
  • (^) Datagram– mainly used in the Network Layer
  • (^) Virtual Circuit– a Data Link Layer technology

• (^) Virtual Circuit packet switching
  • (^) It is connection-oriented
  • (^) Merges datagram packet switching and circuit

switching to extract advantages of both

• (^) Packets flow on logical circuits; no physical

resources are allocated

• (^) Each packet carries a circuit identifier which is

local to a link and updated by each switch on the

path of the packet from its source to its destination

• (^) A virtual circuit: sequence of mappings

(established during connection setup) between a

link taken by packets and the circuit identifier

packets carry on this link.

• (^) Advantages of VC Packet Switching:
  • (^) In order delivery of packets
  • (^) Smaller overhead in each packet
  • (^) More reliable connection
  • (^) Billing is easier (billing per call and not per packet)
• (^) Disadvantages of VC Packet Switching:
  • (^) The switching equipment needs to be more powerful, since each switch needs to store details of all the calls that are passing through it and to allocate capacity for any traffic that each call could generate;
  • (^) Resilience to the loss of a trunk is more difficult, since if there is a failure all the calls must be dynamically reestablished over a different route.

• (^) Packet Switching Vs Circuit Switching
  • (^) Packet-switching is not suitable for real-time

services

• (^) However, Packet switching :
  • (^) offers better sharing of bandwidth, and
  • (^) is simpler, more efficient, and less costly to

implement than circuit switching

• (^) is great for bursty data.

Examples

• (^) 1 Mbps link, each user sends data at

100kbps when busy, busy for 10% of time

• (^) Circuit switching with TDM:
  • (^) Only 10 users can connect at a time
• (^) Packet switching with statistical multiplexing:
  • (^) With 35 users, probability that more than 10 users

are active at the same time is less than .0004;

thus 35 users can connect at a time

Delay and Loss

• Packet Delay or Latency

• Processing Delay (Tproc) – time for checking errors

and determining next link

• (^) Queuing Delay (Tque) – time between when the

processing delay is over and start of transmission

• (^) Transmission delay ( Ttrans ) - time required to push all

of the packet's bits into the wire (L/R)

• (^) Propagation delay ( Tprop ) - time a signal takes to

reach destination node once transmitted by transmitter

(d/s)

• Thus, total delay (Ttot) is given by:
  • Ttot= Tproc + Tque + Ttrans + Tprop

• (^) Packet Loss
  • (^) Failure of one or more transmitted packets to arrive at

their destination

• (^) Effects of packet loss:
  • (^) In text and data: errors.
  • (^) In videoconference: jitter.
  • (^) In audio communications: jitter and frequent gaps
  • (^) In the worst cases:
    • (^) severe mutilation of received data
    • (^) broken-up images
    • (^) unintelligible speech, or
    • (^) even the complete absence of a received signal
• (^) Cause of packet loss:
  • (^) inadequate signal strength at the destination natural or

human-made interference, excessive system noise,

hardware failure, software corruption or congestion

• (^) Suppose the cars “propagate” at the rate of

1000 km/hr and petrol station takes 1 min to

service a car. Will cars arrive at 2nd station

before all cars are serviced at 1st station?

Solution

• (^) After 7 min ( ), 1st^ car will be at

2 nd^ station and 3 cars are still at 1st^ station.

km h hr km 1 60 min

1000 / 100  1 min

Delay Analysis

Circuit Switching:

Assume: Number of hops = M Per-hop processing delay = P Link propagation delay = L Transmission speed = R bit/s Message size = S bits Total Delay = total propagation + total transmission + total processing = 4ML + S/R + (M-1)P Total Delay excluding time for acknowledgment: = 3ML + S/R + (M-1)P

Virtual Circuit Packet Switching:

Assume: Number of hops = M Per-hop processing delay = P Link propagation delay = L Packet transmission delay = T Message size = N packets Total Delay = total propagation

• total transmission
• total store & forward
• total processing = 4ML + NT + (M-1)T + 4(M-1)P Excluding time for acknowledgement: = 3ML + NT + (M-1)T + 3(M-1)P

Q1. Almaz and Belay are 4 hops apart on a datagram packet-switched network where each link is 160 km long. Per-hop processing delay is 5s. Packets are 1500 bytes long. All links have a transmission speed of 56kbps (kbit/s). The speed of light in the wire is approximately 200,000 km/s. If Belay sends a 10-packet message to Almaz, how long will it take Almaz to receive the message up to the last bit (measured from the time Belay starts sending)? Solution: We know the following:

• (^) Number of hops M=4,
• (^) Number of packets N=10,
• (^) Per-hop processing delay P=5  s=0.000005s,
• (^) Link propagation delay L = distance/speed of light = 160/200,000 = 0.0008s,
• (^) Packet size = 1500 bytes = 1500*8=12,000 bits,
• (^) Packet transmission delay T = packet size/transmission speed = 12,000/ =0.214s. Delay = ML + NT + (M-1)T + (M-1)P =0.0032 + 2.14 + 0.642 + 0.000015 = 2.785s. Note that the total delay is dominated by the transmission delay which depends on link speed. A link with a higher transmission speed can reduce the delay dramatically.