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1. LEO (Quiz 1)

1. LEO (Quiz 1). D * 2 / C = 1500*2/300,000 = 0.01 second 10,000 s + 53 s = 10,053 s. 2. Parity Bit. ‘N’ = 78 10 = 01001110 2 Even Parity  Parity Bit = 0. 3. Hidden Node Problem in A Wireless Net. Limited range Not all stations receive all transmissions Cannot use CSMA/CD

preston-ramos
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1. LEO (Quiz 1)

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  1. 1. LEO (Quiz 1) • D * 2 / C = 1500*2/300,000 = 0.01 second • 10,000 s + 53 s = 10,053 s

  2. 2. Parity Bit • ‘N’ = 7810 = 010011102 • Even Parity  Parity Bit = 0

  3. 3. Hidden Node Problem in A Wireless Net Limited range Not all stations receive all transmissions Cannot use CSMA/CD Example in diagram Maximum transmission distance is d Stations 1 and 3 do not receive each other’s transmissions Hidden node

  4. CSMA/CA Used on wireless LANs Both sides send a small message before data transmission ‘‘X is about to send to Y’’ ‘‘Y is about to receive from X’’ Data frame sent from X to Y Purpose: inform all stations in range of X or Y before transmission Known as Collision Avoidance (CA)

  5. 4. PSM • ,  /2,  *3/2,  *2/3

  6. 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

  7. 6. Checksum • 0x4E + 0x43 + 0x4E + 0x55 + 0x0A= 78+67+78+85+10 = 318= 0x13E • There is no carry in this example.

  8. 7. Distance Vector Node 1 Node 2 Node 3 Node 4

  9. Establishing the Routing Table Node 1 Node 2 Node 3 Node 4

  10. Establishing the Routing Table Node 1 Node 2 Node 3 Node 4

  11. Establishing the Routing Table Node 1 Node 2 Node 3 Node 4

  12. 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).

  13. 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 Configuration 1 U, V, W B X, Y, Z B U,V,Z,Y,X B segment 4

  14. 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 Configuration 2 U, V, W U,V,W B Z,Y,X X, Y, Z U,V,Z,Y,X B U,V,Z,Y,X B segment 4

  15. 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 Configuration 3 U, V, W B X, Y, Z B U,V,Z,Y,X B segment 4

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