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TOPIC: Subnet question

Subnet question 7 years 1 week ago #32759

  • lav_plsb1
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Wan consists of 3 networks.
1st network 192.168.10.0/26
2nd network 192.168.10.64/27
3rd network 12 hosts ?
A new subnet with 12 hosts has been added to the certkiller network shown above. which subnet address should this network use to provide enough useable addresses, while wasting the fewest number of ip addresses?
the options are
a. 192.168.10.80/29
b. 192.168.10.80/28
c. 192.168.10.96/28
d. 192.168.10.96/29
e. none of the above.
plz guide me for the correct answer.
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Re: Subnet question 7 years 1 week ago #32762

  • novembre
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cccc
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Re: Subnet question 7 years 1 week ago #32763

  • lav_plsb1
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hi novembre,

could you plz explain to select answer C.
thnks.
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Re: Subnet question 7 years 1 week ago #32764

  • novembre
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What part did you get stuck on when you tried to answer it?
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Re: Subnet question 7 years 1 week ago #32769

  • katzebnt
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Wan consists of 3 networks.
1st network 192.168.10.0/26
2nd network 192.168.10.64/27
3rd network 12 hosts ?
A new subnet with 12 hosts has been added to the certkiller network shown above. which subnet address should this network use to provide enough useable addresses, while wasting the fewest number of ip addresses?
the options are
a. 192.168.10.80/29
b. 192.168.10.80/28
c. 192.168.10.96/28
d. 192.168.10.96/29
e. none of the above.
plz guide me for the correct answer.

C is the correct answer, lemme break it down VLSM style...

Subnet 1
192.168.10.0/26 <-- Net ID
Usable IPs 1 - 62
192.168.10.63/26 <-- Bcst ID
Subnet 2
192.168.10.64/27 <-- Net ID
Usable IPs 65 - 94
192.168.10.95/27 <-- Bcst ID
Subnet 3
192.168.10.96/28 <-- Net ID
Usable IPs 97 - 110
192.168.10.111/28 <-- Bcst ID

For the third subnet the answer must be either C or D, as the Net ID must be "x.x.x.96 / something". If we borrow 4 host bits from the fourth octect (giving you a /28 CIDR notation) that gives this subnet 16 IPs, 14 of which are usable. This satisfies the need for 12 hosts with a tiny bit of expansion room left over.

If you were to borrow 5 host bits from the fourth octect that would give you a CIDR notation of /29, giving your subnet 8 IPs total, only 6 of which are usable. This does not satisfy the requirement of 12 hosts. I always tell my students "you can borrow for more hosts than you need, but you can never borrow for less".

Hope this helps clear things up! :)
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Re: Subnet question 7 years 1 week ago #32779

  • Losh
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katzebnt wrote:

For the third subnet the answer must be either C or D, as the Net ID must be "x.x.x.96 / something".
Hope this helps clear things up! :)

That can't be true!

The correct answer can ONLY be C.

This is because a CIDR of /29 will only provide a total of 8 hosts with 6 of them usable. The correct CIDR is /28 to give 16 hosts, 14 of which are usable.

Well explained answer though.
~ Networking :- Just when u think its starting to make sense......... ~
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