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1、Answer of HW Chapter 4R12: What is the 32-bit binary equivalent of the address 223.1.3.27?11011111 00000001 00000011 00011100 R13: Do routers have IP addresses? If so, how many?Yes. They have one address for each interfaceR18:Yes, because the entire IPv6 datagram (including header fields) is encapsu
2、lated in an IPv4 datagramR21:Link state algorithms: Computes the least-cost path between source and destination using complete, global knowledge about the network. Distance-vector routing: The calculation of the least-cost path is carried out in an iterative, distributed manner.A node only knows the
3、 neighbor to which it should forward a packet in order toreach given destination along the least-cost path, and the cost of that path from itself to the destination.Answer of HW P447 P8a) Prefix MatchLink Interface 11100000 00000000 0 11100000 00000001 1 11100000 2 11100001 3 otherwise 4b) Prefix ma
4、tch for first address is 4th entry: link interface 4 Prefix match for second address is 2nd entry: link interface 2 Prefix match for first address is 3rd entry: link interface 3Answer of HW P448 P9Destination Address Range Link Interface 0000 0011 0 0100 0111 1 1000 1011 2 1100 1111 3 number of addr
5、esses in each range = Answer of HWP448 10 Address interface 10000000 through 10111111 (64 addresses) 011000000 through 11011111 (32addresses) 111100000 through 11111111 (32 addresses) 200000000 through 01111111 (128 addresses) 3Prefix MatchInterface101111112otherwise3P448 P11Subnet 1 :2000 211 = 204
6、8Subnet 2: 1000 210 = 1024Subnet 3: 1000 210 = 1024220.2.11110000.00000000220.2.11110000.00000000 through subnet 1220.2.11110111.11111111220.2.11111000.00000000 through subnet 2220.2.11111011.11111111220.2.11111100.00000000 through subnet 3220.2.11111111.11111111P448 P12Destination Address Link Inte
7、rface 200.23.16/21 0 200.23.24/24 1 200.23.24/21 2 otherwise 3P448 P13Destination address range interface224.0/16 0224.0/16 1224/8 2225/8 3otherwise 4P448 P14101.101.101.64/26 101.101.101.01000000 through 101.101.101.01111111 Any IP address in range 101.101.101.64 to 101.101.101.127 in subnet with p
8、refix 101.101.101.64/26 Four equal size subnets from the block of addresses of the form 101.101.128.0/17 101.101.10000000.00000000 22 = 4 subnets 101.101.10000000.00000000 101.101.128/19 101.101.10100000.00000000 101.101.160/19 101.101.11000000.00000000 101.101.192/19 101.101.11100000.00000000 101.1
9、01.224/19 P449 Problem 15214.97.254/23 214.97.11111110.00000000 Subnet A: 214.97.254.0/24 (28 = 256 addresses) 214.97.11111110.00000000 Subnet B: 214.97.255.0/25-214.97.255.120/29 (27 = 128 addresses-8=120) 214.97.11111111.00000000 214.97.11111111.01111000 through - through 214.97.11111111.01111111
10、214.97.11111111.01111111Subnet C: 214.97.255.128/25 (27 = 128 addresses) 214.97.11111111.10000000Subnet D: 214.97.255.120/31 (2 addresses)214.97.11111111.01111000Subnet E: 214.97.255.122/31 (2 addresses)214.97.11111111.01111010Subnet F: 214.97.255.124/30 (4 addresses)214.97.11111111.01111100 To simp
11、lify the solution, assume that no datagrams have router interfaces as ultimate destinations. Also, label D, E, F for the upper-right, bottom, and upper-left interior subnets, respectively. For the second schedule, we have: Router 1 Longest Prefix MatchOutgoing Interface 11010110 01100001 11111111 Su
12、bnet A 11010110 01100001 11111110 0000000 Subnet D 11010110 01100001 11111110 000001 Subnet F Router 2 Longest Prefix MatchOutgoing Interface 11010110 01100001 11111111 000001 Subnet F 11010110 01100001 11111110 0000001 Subnet E 11010110 01100001 11111110 1 Subnet C Router 3Longest Prefix MatchOutgo
13、ing Interface 11010110 01100001 11111111 0000000 Subnet D 11010110 01100001 11111110 0 Subnet B 11010110 01100001 11111110 0000001 Subnet EP449 Problem 18a) Home addresses: 192.168.0.1, 192.168.0.2, 192.168.0.3 with the router interface being 192.168.0.4 b) NAT Translation Table WAN Side LAN Side 12
14、8.119.40.86, 4000 192.168.0.1, 3345 128.119.40.86, 4001 192.168.0.1, 3346 128.119.40.86, 4002 192.168.0.2, 3445 128.119.40.86, 4003 192.168.0.2, 3446 128.119.40.86, 4004 192.168.0.3, 3545 128.119.40.86, 4005 192.168.0.3, 3546 P449 Problem 19 It is not possible to devise such a technique. In order to
15、 establish a direct TCP connection between Arnold and Bernard, either Arnold or Bob must initiate a connection to the other. But the NATs covering Arnold and Bob drop SYN packets arriving from the WAN side. Thus neither Arnold nor Bob can initiate a TCP connection to the other if they are both behin
16、d NATs. P22 StepND(s),p(s) D(t),p(t) D(u),p(u)D(v),p(v)D(w),p(w) D(y),p(y)D(z),p(z) 1 x 8,x 6,x 6,x 2 xy 15,y 7,y 6,x 18,y 3 xyw 15,y 14,w 7,y 18,y 4 xywv 11,v 10,v 18,y 5 xywvu 14,u 11,v 18,y 6 xywvut 12,t 16,t 7 xywvuts 16,tS: tvyx xyvtsZ: tvyx xyvtzP24 cost tofrom u v x y z v x y z 5 2 10 0 cost tofrom u v x y z v 2 0 7 5 x 12 0 1 2 y 7 1 0 10 z 7 5 2 3 0 cost tofrom u v x y z v 2 0 7 7 5 x 9 7 0 1 2 y 9 7 1 0 3 z 7 5 2 3 0destvxuyxvy5273Next hopzs routing tableyxz681
