计算机网络双语课件第4章答案.ppt

Answer of HW,Chapter 4,R12: 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 interface,R18:Yes, because the entire IPv6 datagram (including header fields) is encapsulated in an IPv4 datagram,R21: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 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 P8 a) Prefix Match Link Interface 11100000 00000000 0 11100000 00000001 1 11100000 2 11100001 3 otherwise 4 b) Prefix match 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 3,Answer of HW,P448 P9 Destination Address Range Link Interface 0000 0011 0 0100 0111 1 1000 1011 2 1100 1111 3 number of addresses in each range =,Answer of HW,P448 10 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,P448 P11 Subnet 1 :2000 211 = 2048 Subnet 2: 1000 210 = 1024 Subnet 3: 1000 210 = 1024 220.2.11110000.00000000 220.2.11110000.00000000 through subnet 1 220.2.11110111.11111111 220.2.11111000.00000000 through subnet 2 220.2.11111011.11111111 220.2.11111100.00000000 through subnet 3 220.2.11111111.11111111,P448 P12,Destination Address Link Interface 200.23.16/21 0 200.23.24/24 1 200.23.24/21 2 otherwise 3,P448 P13,P448 P14,101.101.101.64/26 101.101.101.01000000 through 101.101.101.01111111 Any IP address in range 101.101.101.64 to 101.101.101.127 in subnet with prefix 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 101.101.224/19,P449 Problem 15,214.97.254/23 214.97.11111110.00000000 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 214.97.11111111.01111111 Subnet C: 214.97.255.128/25 (27 = 128 addresses) 214.97.11111111.10000000 Subnet D: 214.97.255.120/31 (2 addresses) 214.97.11111111.01111000 Subnet E: 214.97.255.122/31 (2 addresses) 214.97.11111111.01111010 Subnet F: 214.97.255.124/30 (4 addresses) 214.97.11111111.01111100,To simplify 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 Match Outgoing Interface 11010110 01100001 11111111 Subnet A 11010110 01100001 11111110 0000000 Subnet D 11010110 01100001 11111110 000001 Subnet F Router 2 Longest Prefix Match Outgoing Interface 11010110 01100001 11111111 000001 Subnet F 11010110 01100001 11111110 0000001 Subnet E 11010110 01100001 11111110 1 Subnet C Router 3Longest Prefix Match Outgoing Interface 11010110 01100001 11111111 0000000 Subnet D 11010110 01100001 11111110 0 Subnet B 11010110 01100001 11111110 0000001 Subnet E,P449 Problem 18,a) 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 128.119.40.86, 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 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 behind NATs.,P22,Step N D(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