1

Internet TechnologyNETW 902Tutorial 6

Mohamed Esam

2

Curriculum

• First 3 Chapters • Chapter 4: from the chapter start (page

255) up to (page 372)• Chapter 5: from page 491 to 554 

3

Main Points for Problems 1. ALOHA2. Slotted ALOHA3. CSMA/CD4. Token Ring, Token BUS & Ring Types. (last Tutorial)5. BUS Segments6. WLAN7. FDDI8. Spanning Tree9. Performance of (Aloha, CSMA & Token Ring)

10. SONET/SDH11. ATM12. Virtual Connections13. ISDN14. Dial up15. DSL

Ch’s 1&2Till page: 200

Ch’s 3Till page: 255

Ex.1:LAN-Inter working& Spanning TreeEx.2:MAC protocols Performance

Ex.3:IP Addressing, ATM& Access Networks

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16.Internet Arch. & IP Addressing.17.ARP18.MTU19.Routing Functions20.Forwarding21.Routing Principles22.Distance Vector Algor.23.Link State Algor.24.Path Vector Algor.25.Routing in Internet.26.Internet Routing Protocols27.ARQ28.UDP29.TCP30.QoS

Some of Ch’s 4&5Till page: 554

Ex.4:Distance Vector &Link State Routing

Ex.5:TCP

5

ARP (Automatic Resolution Protocol)

6

In Star Network

• Switch is the responsible device in the network for ARP.• It’s done one time after connection and then updated regularly

Router Switch

PC1PC2

PC3

PC4

PC5

10.10.10.5

Internet

Send data to 10.10.10.3

Who has IP10.10.10.3, replyWith MAC Address to MAC:8AIP:10.10.10.0

My MAC is 3BMy MAC is 3B

Data

MAC IP Port

8A 10.10.10.0 _

1F 1

2A 2

3B 10.10.10.3 3

7D 4

2C 5

10.10.10.4

10.10.10.3

10.10.10.2

10.10.10.1

7

MTU (Maximum Transmission Unit)

• Sender doesn’t send larger than MTU of the used Technology.

• Routers may divide the payload into smaller data units while forwarding.

Tech. MTU

(Bytes)

IPv4 65535

Eth.2 1500

FDDI 4352

8

Dynamic Routing

Update Routing Table periodically

9

I- Distance Vector Algorithm

• Bellman Ford Algorithm.

N7

N1

N4

N5N6

N2 7

3

8

6

51

4

2

N3

Link IDLink ID

N6 is the DestinationN6 is the Destination

10

N1 N2

-1 ∞ -1 ∞

-1 ∞ -1 ∞

Link No. Distance(No. of hops)

Steps/nodes

0

1

N3 N4 N5 N6 N7

-1 ∞ -1 ∞ -1 ∞ D 0 -1 ∞

1 1 -1 ∞ 5 1 D 0 2 1

D 0

D 0

D 0

D 0

N7

N1

N4

N5N6

N2 7

3

8

6

51

4

2

N3N6 is the DestinationN6 is the Destination

Assumes whole linkshave the same cost (=1).So Total path cost = number of hops.In case of same cost select randomly.

Not Unique Solution

11

N1 N2

-1 ∞ -1 ∞

-1 ∞ -1 ∞

3 2 6 2

Link No. Distance(No. of hops)

Steps/nodes

0

1

2

N3 N4 N5 N6 N7

-1 ∞ -1 ∞ -1 ∞ D 0 -1 ∞

1 1 -1 ∞ 5 1 D 0 2 1

1 1 8 2 5 1 D 0 2 1

D 0

D 0

D 0

N7

N1

N4

N5N6

N2 7

3

8

6

51

4

2

N3

N6 is the DestinationN6 is the Destination

Shorter than (4, 2)

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N1 N2

-1 ∞ -1 ∞

-1 ∞ -1 ∞

3 2 6 2

3 2 6 2

Steps/nodes

0

1

2

3

N3 N4 N5 N6 N7

-1 ∞ -1 ∞ -1 ∞ D 0 -1 ∞

1 1 -1 ∞ 5 1 D 0 2 1

1 1 8 2 5 1 D 0 2 1

1 1 8 2 5 1 D 0 2 1

D 0

N7

N1

N4

N5N6

N2 7

3

8

6

51

4

2

N3

N6 is the DestinationN6 is the Destination

Shorter than (4, 2)Shorter than (7,3)

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• Repeat with All destinations Get the routing Table• N7 to All Destinations

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• Count to infinity Problem:

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Exerc.4 Part 1 Dynamic Routing

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Q2-a

17

Q2-b

18

Q2-C

19

Q3-a

20

Q3-b

21

Q3-c

• Looping (Count to Infinity)

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Q3-d

23

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II- Link State Algorithm

1. Discover the neighbours: 2. Measure the cost to each

neighbour3. Flooding: Send Link State

packet to all routers saying the new information; includes:

• Sender ID• Packet Sequence No. (32 bits)• Packet age (TTL)• List of neighbours and

corresponding Cost.

4. Compute shortest path to all routers using Dijkstra algorithm.

Echo (Round Trip) packet

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Part 2- Exerc.4

EA

CB 2

5

3

16

4

D

Link IDLink ID

11

11

111212

44

11

CostCost

EA

CB 2

5

3

16

4

D11

11

111212

44

11

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A-

EA

CB 2

5

3

16

4

D11

11

111212

44

11

EA

CB 2

3

16

4

D11

11

111212

44

11

If Duplicate Messages are not avoidedIf Duplicate Messages are not avoided

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B- Link State DatabaseFrom To Link Cost

A B 1 1

A E 5 1

B C 2 4

C D 3 1

C E 6 1

E D 4 12

EA

CB 2

5

3

16

4

D

Link IDLink ID

11

11

111212

44

11

CostCost

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C-

• Using Time Stamps (Synchronized routers)

• Using Seq. Number:– 32 bits address to avoid Wrap around

problem.– Incremented with every new message

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D- Apply Dijkstra

EA

CB 2

5

3

16

4

D

Link IDLink ID

11

11

111212

44

11

CostCost

Step B C D E

0 ∞ ∞ ∞ ∞

1 1/A ∞ ∞ 1/A

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D- Apply Dijkstra

EA

CB 2

5

3

16

4

D

Link IDLink ID

11

11

111212

44

11

CostCost

Step B C D E

0 ∞ ∞ ∞ ∞

1 1/A ∞ ∞ 1/A

2 1/A* 5/B ∞ 1/A

Doesn’t Send to the Source Node (A)

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D- Apply Dijkstra

EA

CB 2

5

3

16

4

D

Link IDLink ID

11

11

111212

44

11

CostCost

Step B C D E

0 ∞ ∞ ∞ ∞

1 1/A ∞ ∞ 1/A

2 1/A* 5/B ∞ 1/A

3 1/A* 5/B 2/E 13/E 1/A*

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D- Apply Dijkstra

EA

CB 2

5

3

16

4

D

Link IDLink ID

11

11

111212

44

11

CostCost

Step B C D E

0 ∞ ∞ ∞ ∞

1 1/A ∞ ∞ 1/A

2 1/A* 5/B ∞ 1/A

3 1/A* 5/B 2/E 13/E 1/A*

4 1/A* 2/E* 13/E 3/C 1/A*

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D- Node A Routing Table

Dest. Link No. Cost Next Hop

A _ 0 _

B 1 1 B

C 5 2 E

D 5 3 E

E 5 1 E