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Answer of HW. Chapter 5. Answer of HW. P494 1 设将 1010101010101011 分成 x 行 y 列,则有: x*y=16 ,所以, y=16/x 需要附加的校验位数为 f=x+y+1=x+16/x+1 f 对 x 求导,得 x=4 时, f 的位数最小。即将上述位串分成 4 行 4 列,采用偶校验:. 1. 0. 1. 0. 0. 1. 0. 1. 0. 0. 1. 0. 1. 0. 0. 1. 0. 1. 1. 1. 0. 0. 0. 1. 1.
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Answer of HW Chapter 5
Answer of HW • P494 1 • 设将1010101010101011分成x行y列,则有: • x*y=16,所以,y=16/x • 需要附加的校验位数为f=x+y+1=x+16/x+1 • f对x求导,得x=4时,f的位数最小。即将上述位串分成4行4列,采用偶校验: 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 0 1 1
Answer of HW • P494 4 • If we divide 1001 into 10101010000, • we get 10010111, with a remainder of R = 001.
Answer of HW • P494 8 • The length of a polling round is • The number of bits transmitted in a polling round is .The maximum throughput therefore is
Answer of HW • P495 9 • a) b) c) see figure below:
Answer of HW • d) • Forwarding table in determines that the datagram should be routed to interface 111.111.111.002. • Host uses ARP to determine the LAN address for 111.111.111.002, namely 22-22-22-22-22. • The adapter in creates and Ethernet packet with Ethernet destination address 22-22-22-22-22-22. • The first router receives the packet and extracts the datagram. The forwarding table in this router indicates that the datagram is to be routed to 122.222.222.003. • The first router then uses ARP to obtain the associated Ethernet address, namely 55-55-55-55-55-55. • The process continues until the packet has reached Host . • e) • ARP in must now determine the LAN address of 111.111.111.002. Host sends out an ARP query packet within a broadcast Ethernet frame. The first router receives the query packet and sends to Host an ARP response packet. This ARP response packet is carried by an Ethernet frame with Ethernet destination address 00-00-00-00-00-00.