Chapter4-new answer

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 interfaceAnswer of HW P447 P8 a) Prefix MatchLink Interface 11100000 00000000 0 1110000

2、0 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 3Answer of HW P448 P9Destination Address Range Link Interface 000

3、0 0011 0 0100 0111 1 1000 1011 2 1100 1111 3 number of addresses in each range = 422Answer of HW P448 10 Address interface 10000000 through 10111111 (64 addresses) 011000000 through 11011111 (32addresses) 111100000 through 11111111 (32 addresses) 200000000 through 01111111 (128 addresses) 3Prefix Ma

4、tchInterface101111112otherwise3 P448 P11Subnet 1 :2000 211 = 2048Subnet 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 subn

5、et 3220.2.11111111.11111111P448 P13Destination address range interface224.0/16 0224.0/16 1224/8 2225/8 3otherwise 4P448 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

6、 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 15214.97.25

7、4/23 214.97.11111110.00000000 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 214.97.11111111.01111111Subnet C:

8、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 simplify the solution, assume that no

9、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 Subnet A 11010110 01100001 11111110

10、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 MatchOutgoing Interface 11010110 01100001 11

11、111111 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 128.119.40.86, 4000 192.168.0.1, 334

12、5 128.119.40.86, 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

13、 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 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 xyvtzP