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Question 1182

SIMULATION

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Guidelines

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This is a lab item in which tasks will be performed on virtual devices

• Refer to the Tasks tab to view the tasks for this lab item.

• Refer to the Topology tab to access the device console(s) and perform the tasks

• Console access is available for all required devices by clicking the device icon or using the tab(s) above the console window

• All necessary preconfigurations have been applied

• Do not change the enable password or hostname for any device

• Save your configurations to NVRAM before moving to the next item

• Click Next at the bottom of the screen to submit this lab and move to the next question

• When Next is clicked the lab closes and cannot be reopened

Topology

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Tasks

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All physical cabling is in place. A company plans to deploy 16 new sites. The sites will utilize both IPv4 and IPv6 networks.

1. Subnet 172.16.0.0/16 to meet the subnet requirements and maximize the number of hosts

• Using the second subnet

- Assign the first usable IP address to e0/0 on Sw101

- Assign the last usable IP address to e0/0 on Sw102

2. Subnet 2001:DB8::/50 to meet the subnet requirements and maximize the number of hosts

• Using the second subnet

- Assign an IPv6 GUA using a unique 64-Bit interface identifier on e0/0 on Sw101

- Assign an IPv6 GUA using a unique 64-Bit interface identifier on e0/0 on Sw102

    Correct Answer:

Discussion
matteodd90

trying to give an aswer, correct me if I am wrong 16 sites=2^4 /16 --> n°16 /20 172.16.0.0 - 172.16.15.255 172.16.16.0 - 172.16.31.255 and so on /50 --> n°16 /54 2001:0db8:0000:0000:0000:0000:0000:0000 - 2001:0db8:0000:03ff:ffff:ffff:ffff:ffff 2001:0db8:0000:0400:0000:0000:0000:0000 - 2001:0db8:0000:03ff:ffff:ffff:ffff:ffff and so on (every field separated by colons represent 16 bit, (every hexadecimal digits is 4 bit) IPv6 Global unicast address subnet minimum prefix lenght is 64 thus the two address are 2001:db8:0:400::1/64 2001:db8:0:400::2/64 sw101 int e0/0 ip address 172.16.16.1 255.255.240.0 ipv6 address 2001:db80:400::1/64 sw102 int e0/0 ip address 172.16.31.254 255.255.240.0 ipv6 address 2011:db8:0:400::2/64

5eba813

Shouldt be last useable 30 ? Iam confused tbh

PyoJH

Using the second subnet, .17 , .30

CertBuster

The given answer is wrong. I'll break it down: We need to subnet 172.16.0.0/16. We need 16 subnets (16 sites). 2^n = 16 -> log2(16) -> 4 bits. So our subnets are the highest 4 bytes of the third octet. The second subnet would therefore start at. 172.16.[00010000].[00000000] -> 172.16.16.0 The last usable address would be 172.16.[00011111].[11111110] -> 172.16.31.254 The prefix length is calulated as network_bits + subnet_bits -> 16 + 4 -> 20. As a netmask -> 255.255.240.0 Before we apply any configurations, lets calculate the subnet for the IPv6 part: We're still deploying subnets for 16 sites, so again we need 4 host bits. Our network is 2001:DB8::/50. Remember that each hex character is 4 bytes, meaning each portion is 16 bits. The first 48 bits are therefore 2001:0DB8:0000. We then need to break down the individual bits, because a /50 prefix length crosses byte boundaries: 2001:0DB8:0000:[00xxxx00]::/54 where x -> subnet bits The second subnet means the lower-most bit is set to 1, giving us the address 2001:0DB8:0000:[00000100]::/54

CertBuster

Converting back to hexadecimal, and shortening as appropriate, we get the address of the second subnet: 2001:DB8:0:4::/54 Configuration First note that we're dealing with an L3 switch. The question says that all necessary configuratins are applied, but you'll want to check the running config to see if this is indeed the case. Particularly: 1. Ensure that ip routing is enabled 2. Ensure that ipv6 is enabled 3. Ensure that the switchports are disabled 4. Ensure that the routing interfaces are not shut down For the sake of my example, I'll assume these configurations are *not* applied. First, configure Sw101: SW101(config)#ip routing SW101(config)#ipv6 unicast-routing SW101(config)#interface e0/0 SW101(config-if)#no switchport SW101(config-if)#ip address 172.16.16.1 255.255.240.0 SW101(config-if)#ipv6 address 2001:DB8:0:4::/54 eui-64 SW101(config-if)#no shutdown SW101(config-if)#do write memory

CertBuster

Then, configure SW102: SW102(config)#ip routing SW102(config)#ipv6 unicast-routing SW102(config)#interface e0/0 SW102(config-if)#no switchport SW102(config-if)#ip address 172.16.31.254 255.255.240.0 SW102(config-if)#ipv6 address 2001:DB8:0:4::/54 eui-64 SW102(config-if)#no shutdown SW102(config-if)#do write memory Note: 1. I used eui-64 for the IPv6 addresses; the question asks for a "unique 64-Bit interface identifier", so this is appropriate. 2. I wrote the running config to startup config, as is instructed.

baanyan

SW102(config-if)#ipv6 address 2001:DB8:0:400::/54 eui-64 we cannot omit the trailing 0s, please be aware.

Jessi2302

When convert back to Hex, the second subnet is 2001:db8:0:400::/54

bymrdas

Sw101 interface ethernet 0/0 ip address 172.16.16.1 255.255.240.0 ipv6 address 2001:db8:0:400::/54 eui-64 no shutdown wr Sw102 interface ethernet 0/0 ip address 172.16.31.254 255.255.240.0 ipv6 address 2001:db8:0:400::/54 eui-64 no shutdown wr

picho707

16 sites = 2^4 11110000 = /20 = 240 SW101(config)#ipv6 unicast-routing SW101(config)#interface gigabitEthernet 0/0 SW101(config-if)#no switchport SW101(config-if)#ip address 172.16.16.1 255.255.240.0 SW101(config-if)#no shutdown SW102(config)#ipv6 unicast-routing SW102(config)#interface gigabitEthernet 0/0 SW102(config-if)#no switchport SW102(config-if)#ip address 172.16.31.254 255.255.240.0 SW102(config-if)#no shutdown The IPV6 subnet will be: 2001:DB8:0:4::/54 SW101(config)#interface gigabitEthernet 0/0 SW101(config-if)#ipv6 address 2001:DB8:0:4::1/64 SW102(config-if)#interface gigabitEthernet 0/0 SW102(config-if)#ipv6 address 2001:DB8:0:4::2/64

na4o

Address: 172.16.8.0 10101100.00010000.00001 000.00000000 Netmask: 255.255.248.0 = 21 11111111.11111111.11111 000.00000000 Wildcard: 0.0.7.255 00000000.00000000.00000 111.11111111 => Network: 172.16.8.0/21 10101100.00010000.00001 000.00000000 (Class B) Broadcast: 172.16.15.255 10101100.00010000.00001 111.11111111 HostMin: 172.16.8.1 10101100.00010000.00001 000.00000001 HostMax: 172.16.15.254 10101100.00010000.00001 111.11111110 Hosts/Net: 2046 (Private Internet)

picho707

To subnet the IPv6 address 2001:DB8::/50 into 16 subnets, you need to add 4 bits to the prefix length. This is because 2^4 equals 16, which gives us the number of subnets we need. So, the new prefix length will be 50 + 4 = 54. Therefore, the subnets will have a prefix of /54. The subnets will be: 1. 2001:DB8:0:0::/54 2. 2001:DB8:0:4::/54 3. 2001:DB8:0:8::/54 and so on. IPv6 Global unicast address subnet minimum prefix length is 64 thus the two addresses are 2001:db8:0:400::1/64 2001:db8:0:400::2/64

picho707

Correction the 1sta dn 2nd IP address of the second subnet are: 2001:DB8:0:4::1/64 and 2001:DB8:0:4::2/64

5eba813

Can u kindly explain what happend after step 3 2001:db8:0:8::/54 cause iam lost after

picho707

1. 2001:DB8:0:0::/54 2. 2001:DB8:0:4::/54 3. 2001:DB8:0:8::/54 4. 2001:DB8:0:12::/54 5. 2001:DB8:0:16::/54 6. 2001:DB8:0:20::/54 7. 2001:DB8:0:24::/54 8. 2001:DB8:0:28::/54 9. 2001:DB8:0:32::/54 10. 2001:DB8:0:36::/54 11. 2001:DB8:0:40::/54 12. 2001:DB8:0:44::/54 13. 2001:DB8:0:48::/54 14. 2001:DB8:0:52::/54 15. 2001:DB8:0:60::/54 16. 2001:DB8:0:64::/54

bymrdas

Sw101 interface ethernet 0/0 ip address 172.16.16.1 255.255.240.0 ipv6 address 2001:db8:0:400::/54 eui-64 no shutdown wr Sw102 interface ethernet 0/0 ip address 172.16.31.254 255.255.240.0 ipv6 address 2001:db8:0:400::/54 eui-64 no shutdown wr

bymrdas

We need to subnet 172.16.0.0/16. We need 16 subnets (16 sites). 2^n = 16 -> log2(16) -> 4 bits. So our subnets are the highest 4 bytes of the third octet. The second subnet would therefore start at. 172.16.[00010000].[00000000] -> 172.16.16.0 The last usable address would be 172.16.[00011111].[11111110] -> 172.16.31.254 The prefix length is calulated as network_bits + subnet_bits -> 16 + 4 -> 20. As a netmask -> 255.255.240.0 Before we apply any configurations, lets calculate the subnet for the IPv6 part: We're still deploying subnets for 16 sites, so again we need 4 host bits. Our network is 2001:DB8::/50. Remember that each hex character is 4 bytes, meaning each portion is 16 bits. The first 48 bits are therefore 2001:0DB8:0000. We then need to break down the individual bits, because a /50 prefix length crosses byte boundaries: 2001:0DB8:0000:[00xxxx00]::/54 where x -> subnet bits The second subnet means the lower-most bit is set to 1, giving us the address 2001:0DB8:0000:[00000100]::/54

Rolfer

Is ipv6 subneting CCNA scope?

d5c03de

From my understanding with IPV6, bits 49 - 64 represent the subnet. @CertBuster 2001:0DB8:0000:[00xxxx00]::/54 where x -> subnet bit Why go past the giving /50? I can understand /52 as it represents the first hexadecimal, but I'm struggling wrapping my head around /54. Why include the second hexadecimal? 1. 2001:DB8:0:0::/50 2. 2001:DB8:0:4::/50 3. 2001:DB8:0:8::/50 4. 2001:DB8:0:12::/50 Makes sense to me. Please correct me if I'm wrong.

d5c03de

Disregard my question. It's because of the "16" new sites needed thus borrow 4 host bits from 2001:DB8::/50 -> 2001:DB8::/54 -> 2001:DB8:0:400::/54 = the second subnet.

d5c03de

Shout out to CerBuster, here's another way to look at the Ipv6 subnetting, [ ] = Bianary representation 2001:0DB8:0000:[00xx xx00 0000 0000]::/54 where x -> subnet bits 2001:0DB8:0000:[0000 0000 0000 0000]::/54 second subnet = 2001:0DB8:0000:[0000 0100 0000 0000]::/54 Back to hex = 2001:0DB8:0000:0400::/54 = 2001:DB8:0:400::/54

thinqtanklearningDOTcom

Why wouldnt you just put ipv6 address 2001:db8::/64 eui-64 on the interface?