1

1. LEO (Quiz 1)

• D * 2 / C = 1500*2/300,000 = 0.01 second

• 10,000 s + 53 s = 10,053 s

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2. Parity Bit

• ‘N’ = 7810 = 010011102

• Even Parity Parity Bit = 0

3

3. Hidden Node Problem in A Wireless Net

• Limited rangeLimited range– Not all stations receive all transmissionsNot all stations receive all transmissions– Cannot use CSMA/CDCannot use CSMA/CD

• Example in diagramExample in diagram– Maximum transmission distance is Maximum transmission distance is dd– Stations Stations 1 and 3 do not receive each other’s1 and 3 do not receive each other’s transmissionstransmissions

• Hidden nodeHidden node

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CSMA/CA

• Used on wireless LANsUsed on wireless LANs

• Both sides send a small message before Both sides send a small message before data transmissiondata transmission– ‘‘‘‘X is about to send to Y’’X is about to send to Y’’– ‘‘‘‘Y is about to receive from X’’Y is about to receive from X’’– Data frame sent from Data frame sent from X to YX to Y

• Purpose: inform all stations in range of Purpose: inform all stations in range of X or X or Y before Y before transmissiontransmission

• Known as Known as Collision Avoidance (CA)Collision Avoidance (CA)

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4. PSM

, /2, *3/2, *2/3

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5. Unix commands

• ls - list directory contents

• pwd – print working directory name

• less – similar to more; display the contents of a file and pause at each page, but allow backward movement in the file as well as forward movement.

• grep - print lines matching a pattern

• gcc - GNU C and C++ compiler

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6. Checksum

N C N U \n

4E 43 4E 55 0A

• 0x4E + 0x43 + 0x4E + 0x55 + 0x0A= 78+67+78+85+10 = 318= 0x13E

• There is no carry in this example.

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7. Distance Vector

Dst Next Cost

1 - 0

Node 3

Dst Next Cost

2 - 0

Dst Next Cost

3 - 0

Dst Next Cost

4 - 0

Node 4Node 2Node 1

9

Establishing the Routing Table

Dst Next Cost

1 - 0

3 3 3

Node 3

Dst Next Cost

2 - 0

3 3 1

4 4 2

Dst Next Cost

3 - 0

1 1 3

2 2 1

4 4 5

Dst Next Cost

4 - 0

2 2 2

3 3 5

Node 4Node 2Node 1

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Establishing the Routing Table

Dst Next Cost

1 - 0

3 3 3

2 3 4

4 3 8

Node 3

Dst Next Cost

2 - 0

3 3 1

4 4 2

1 3 4

Dst Next Cost

3 - 0

1 1 3

2 2 1

4 2 3

Dst Next Cost

4 - 0

2 2 2

3 2 3

1 3 8

Node 4Node 2Node 1

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Establishing the Routing Table

Dst Next Cost

1 - 0

3 3 3

2 3 4

4 3 6

Node 3

Dst Next Cost

2 - 0

3 3 1

4 4 2

1 3 4

Dst Next Cost

3 - 0

1 1 3

2 2 1

4 2 3

Dst Next Cost

4 - 0

2 2 2

3 2 3

1 2 6

Node 4Node 2Node 1

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8. Switch

• Note that there are cycles in the network topology.– To prevent the broadcast storm, the spanning

tree algorithm must be performed to disable some links (bridges).

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Configuration 1

• The bridge learns the locations of U,V,Z,Y,X.

• For frames destined to V, U, Z, V, X, W, Z,– only frames 1,3,5,6

are forwarded to segment 4

B

B

B

U, V, W

X, Y, Z

U,V,Z,Y,X

segment 4

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Configuration 2

• The bridge learns the locations of U,V,Z,Y,X.

• For frames destined to V, U, Z, V, X, W, Z,– only frames

1,3,5,6 are forwarded to segment 4

B

B

B

U, V, W

X, Y, Z

U,V,Z,Y,X

segment 4

U,V,Z,Y,X

U,V,W

Z,Y,X

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Configuration 3

• The bridge learns the locations of U,V,Z,Y,X.

• For frames destined to V, U, Z, V, X, W, Z,– only frames 1,3,5,6

are forwarded to segment 4

B

B

B

U, V, W

X, Y, Z

U,V,Z,Y,X

segment 4