33 hosts : A 192.168.72.0/24 B 88 hosts
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There is a router and the two attached subnets below (A and B). The number of hosts is also shown below. The subnets share the 24 high-order bits of the address space: 192.168.72.0/24
- Is the address space public or private?
- How many hosts can there be in this address space?
- What is the subnet address of subnet A? (CIDR notation)
- What is the broadcast address of subnet A?
- What is the starting address of subnet A?
- What is the ending address of subnet A?
- What is the subnet address of subnet B? (CIDR notation)
- What is the broadcast address of subnet B?
- What is the starting address of subnet B?
- What is the ending address of subnet B?
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- R1 1500 Hosts 1000 Hosts 172.16.0.0/16 800 Hosts 400 Hosts R2 R3 200 Hosts 150 Hosts 1. Get the Network Ranges for the following Host Requirementsfind network id, broadcast id and host range, dont use binary methodDetermine whether or not the following IPv6 address notations are correct:•::0F53:6382:AB00:67DB:BB27:7332•7803:42F2:::88EC:D4BA:B75D:11CD•::4BA8:95CC::DB97:4EAB•74DC::02BA•::00FF:128.112.92.116
- Show abbreviations for the following IPv6 addresses: 0000 : FFFF : FFFF : 0000:0000 : 0000 : 0000 :0000 1234 : 2346 : 3456: 0000:0000 : 0000 : 0000 :FFFF 0000 : 0001 : 0000: 0000:0000 : FFFF : 1200 :1000Q1: Compress the following IPv6 addresses 0000:0000:0000:0000:0000:0000:0000:0000 2001:0000:0000:0000:0000:0000:0000:0001 2FFF:0000:1111:0000:0000:ABCD:0000:0025 3000:0000:0000:101A:0000:0000:0000:0001 20FF:3756:0005:0001:ACAD:0000:0000:0025 3FFF:0000:0000:0000:ACAD:25FF:0001:0127 20FF:ACAD:0000:0BCD:FFFF:0000:0000:0001 3ABC:0001:ACAD:0000:0000:0000:0000:0005 FE80:0000:0000:0000:03E0:1275:0000:0034 FE80:0000:0540:0000:0000:0000:0000:9800 Q2: Decompress the following IPv6 Addresses FE80::1 ::1 2F00:0:0:0:0:BCD:0:127 3E80:70::00FF:0:0001 2001:380:F:0:ACAD::5 3001::ACAD:205:0:100 2002:ACAD:0:1BCD:FFFF::4 3F0:25:ACAD::ABCD:0:5 FF::4E00:235:0:FF 3E01:6C:400::FF00 Q3: SLACC – eui-64Calculate the IPv6 addresses for the hosts with the following network prefix and MAC addresses A:Network prefix: 2001:4400:0FFF:1234::/64MAC Address: 90-61-AE-7B-A4-6B B:Network prefix: 2603:8080:2A40:76D::/64MAC Address: FE-AA-54-66-AC-11 C:Network Prefix: 2000:1:A:B:C::/64MAC Address:…what is the highest host address in 128.211.0.16/28
- The subnet mask in binary format: ● 255.255.255.240 is 111 111.11111000 11111.1 11111.11110000 11111111 111 1111100 11111111.11111111.111111 .11111110 * The system that mimics exact behavior of :another system is Emulator Simulator SOI IOS : The non-intelligent network device is Hub O Network cloud *Figure out how many hosts are available given the following network addresses: 116.244.0.0/20IPV6 TUNNELING AND ENCAPSULATION Consider the network shown below which contains four IPV6 subnets, connected by a mix of IPV6-only routers(shaded blue), IPV4-only routers (shaded red) and dual-capable IPV6/IPV4 routers (shaded blue with red interfaces to IPV4 routers). Suppose that a host of subnet D wants to send an IPV6 datagram to a host on subnet A. Assume that the forwarding between these two hosts goes along the path: D --> E -> d--> c --> a --> C -->A 4FB4:297e:962F:179: 46a6:0:ACe:6F28 IPV6 subnet F 31.148.34. 160 67.74. 20. a b F 16.224,99.241 62.15. 168. 10L A D IPV6/v4 IPV4 IPV6/v4 IPV6 subnet D IPV4 IPV6 subnet A IPV6 101.20.229.229 E IPV6 135. 11.212.12 IPV6 IPV6/v4 IPV4 IPV4 IPV6/v4 subnet B 4.42.20T.201 C314:2371ISR31IE4C6:BEBF1351BI 958TIFIED What is the destination address of the D to E datagram? 4985:E8B9:6716:3F35:BF52:5ADE:113A:F5A3 C314:2371:5A31:E4CE:BEBF:351B:95B7:F4ED C911:AA28:691D:57DE:4117:3619:5867:8848 O 8AB4:568F:BDF5:3B61:CFD8:AE58:8562:6969
- Just answer please without explanation"liu.edu.lb" is considered as: IP address MAC address Transport layer address Application layer addressShort Answer Questions Apply Rule 1 and Rule for the following IPv6 Addresses a) 3FFF:FF00:0000:0000:ACAD:0000:0000:0127 b) 3ABC:0001:ACAD:0000:0000:1234:0000:0005 c) 3FFF:FF00:0000:0000:ACAD:0000:0000:0127 Apply EUI-64 Process and generate interface id. d) 00:00:5E:00:53:AF Your answer