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Answer of HW

Answer of HW. Chapter 4. Answer of HW. p404 4 No VC number can assigned to the new VC; thus the new VC can not be established in the network. Each link has two available VC numbers. There are four links. So the number of combinations is 2 4 = 16. One example combination is (10,00,00,10).

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Answer of HW

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  1. Answer of HW Chapter 4

  2. Answer of HW • p404 4 • No VC number can assigned to the new VC; thus the new VC can not be established in the network. • Each link has two available VC numbers. There are four links. So the number of combinations is 24 = 16. One example combination is (10,00,00,10).

  3. Answer of HW • p404 5 In a virtual circuit network, there is an end-to-end connection in the sense that each router along the path must maintain state for the connection; hence the terminology connection service. In a connection-oriented transport service over a connectionless network layer, such as TCP over IP, the end systems maintain connection state; however the routers have no notion of any connections; hence the terminology connection-oriented service

  4. Answer of HW • P405 7 • a) Prefix Match Link Interface 11100000 0 11100001 00000000 1 11100001 2 otherwise 3 • b) • Prefix match for first address is 4th entry: link interface 3 • Prefix match for second address is 2nd entry: link interface 1 • Prefix match for first address is 3rd entry: link interface 2

  5. Answer of HW • P405 8 Destination Address Range Link Interface 00000000 ~~ 00111111 0 01000000 ~~ 01111111 1 10000000 ~~10111111 2 11000000 ~~ 11111111 3 number of addresses in each range =

  6. Answer of HW • P405 9 Address interface 10000000 through 10111111 (64 addresses) 0 11000000 through 11011111 (32addresses) 1 11100000 through 11111111 (32 addresses) 2 00000000 through 01111111 (128 addresses) 3

  7. Answer of HW • P406 Problem 10. • 223.1.17.0/25 • 223.1.17.128/26 • 223.1.17.192/26 • P407 Problem 15 • The maximum size of data field in each fragment = 480 (20 bytes IP header). Thus the number of required fragments • Each fragment will have Identification number 422. Each fragment except the last one will be of size 500 bytes (including IP header). The last datagram will be of size 120 bytes (including IP header). The offsets of the 7 fragments will be 0, 60, 120, 180, 240, 300, 360. Each of the first 6 fragments will have flag=1; the last fragment will have flag=0.

  8. Answer of HW • P407 16 • MP3 file size = 4 million bytes. Assume the data is carried in TCP segments, with each TCP segment also having 20 bytes of header. Then each datagram can carry 1500-40=1460 bytes of the MP3 file • Number of fragments required All but the last fragment will be 1,500 bytes; the last fragment will be 1060+40 = 1100 bytes. The offsets will be multiples of 185.

  9. Answer of HW • P408 21

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