Question 2. Assume a host with global unicast address (2000:A45B:0ACD:1:1BA9:77FF:FE89:F7D1/128)16 Find: Link local address. Subnet ID....
Question 2.
Assume a host with global unicast address (2000:A45B:0ACD:1:1BA9:77FF:FE89:F7D1/128)16
Find:
- Link local address.
- Subnet ID.
- Physical (MAC) address of this host.
Homework Answers
1. Link Local Address: Link Local address in IPv6 is formed as 64-bit(FE80::) reserved + next 64-bit ( 64-bit MAC Address from IPv6). So, in above case Link Local Address will be FE80:0:0:0:77FF:FE89:F7D1
2. Subnet ID: It is the fourth (16 bit block) in IPv6 address. From the given IPv6 address 2000:A45B:0ACD:1:1BA9:77FF:FE89:F7D1. "1 is the subnet ID"
3. Physical (MAC Address): It is extracted from the last 64-bits of IPv6 address. If you look at the format. 48-bit MAC address is converted to last 64-bit IP address by dividing 48-bits into 24-bits and 24-bits. and inserting FFFE in between them. Apart from that, In IEEE format, we invert bit 7 to finally convert 48-bit into 64-bit. So, with this process, the MAC address will come as 19:A9:77:89:F7:D1
Add Answer to:
Question 2.
Assume a host with
global unicast address
(2000:A45B:0ACD:1:1BA9:77FF:FE89:F7D1/128)16
Find:
Link local address.
Subnet ID....
Addressing Table IPv4 Address Subnet Mask Device Interface Default Gateway IPv6 Address/Prefix IP...
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Addressing Table IPv4 Address Subnet Mask Device Interface Default Gateway IPv6 Address/Prefix IPv6 Link-local N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A G0/O 2001 DB8:ACAD:: 1/64 FE80::1 G0/1 2001:DB8:ACAD:1:1/64FE80::1 R1 G0/2 2001:DB8:ACAD:21/64FE80::1 172.16.1.2 2001:DB8:2::1/64 209.165.200.226 2001:DB8:1::1164 172.16.1.1 2001:DB8:2:2/64 255.255.255.252 FE80::1 255.255.255.252 FE80::2 255.255.255.252 FE80:2 S0/0/1 S0/0/0 Central S0/0/1 S1 S2 S3 VLAN 1 VLAN 1 VLAN 1 Staff NIC 2001:DB8:ACAD::2/64FE80::2 FE80::1 Sales 2001 DB8:ACAD:1:2/64FE80::2 FE80::1 IT 2001 DB8:ACAD:2: 2/64 FE80::2 64.100.0.3 2001 DB8 CAFE::3/64 192.168.0.196 2001:DB8...
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