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1、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
2、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
3、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 match for firs
4、t 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 P9Destination Address Range Link Interface 0000 0011 0 0100 0111 1 1000 1011 2 1100 1111 3 number of addresses in each
5、range=,Answer 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)3,P448 P11Subnet 1:2000 211=2048Subnet 2:1000 210=1024Subnet 3:1000 210=10240000.000000000000.0
6、0000000 through subnet 10111.111111111000.00000000 through subnet 21011.111111111100.00000000 through subnet 31111.11111111,P448 P12,Destination Address Link Interface otherwise 3,P448 P13,P448 P14,000000 through 111111 Any IP address in range 101.101.101.64 to 101.101.101.127 0000000.00000000 22=4
7、subnets 00 01 10 11,P449 Problem 15,0.00000000 Subnet A:214.97.254.0/24(28=256 addresses)0.00000000 Subnet B:214.97.255.0/25-214.97.255.120/29(27=128 addresses-8=120)1.00000000 1.01111000 through-through 1.01111111 1.01111111Subnet C:214.97.255.128/25(27=128 addresses)1.10000000Subnet D:214.97.255.1
8、20/31(2 addresses)1.01111000Subnet E:214.97.255.122/31(2 addresses)1.01111010Subnet F:214.97.255.124/30(4 addresses)1.01111100,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
9、,respectively.For the second schedule,we have:Router 1 Longest Prefix MatchOutgoing Interface 11010110 01100001 11111111 Subnet A 11010110 01100001 11111110 0000000 Subnet D 11010110 01100001 11111110 000001 Subnet F Router 2 Longest Prefix MatchOutgoing Interface 11010110 01100001 11111111 000001 S
10、ubnet F 11010110 01100001 11111110 0000001 Subnet E 11010110 01100001 11111110 1 Subnet C Router 3Longest Prefix MatchOutgoing 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.
11、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 12
12、8.119.40.86,4005 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 th
13、e 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 xyvtz,P24,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 0,
