Specific subnetting quetions from 2 practice questions...

Nunya

I'm practicing subnetting and have hit 2 snags that make me want to question the source (usually very reliable) based on the answers provided by the question generator; Here the questions and answers:

1. Given address of 172.21.0.0 /23 how many subnets and host per subnet?
My answer is 2^7-2 subnets =126, 2^9-2 =510 hosts. The answer provided (by the question generator) is 2^7 subnets = 128, 2^9-2 = 510 hosts...so I'm confused now as to when the subnets include a network ID and host reservation and when it wouldn't.

2. Given address of 192.168.24.0 /28, assign a server the last valid address from the 4th subnet.
My answer given the subnet mask would be 255.255.255.240 makes the increment 16...so the subnets would be x.x.x.0-15, x.x.x.16-31, x.x.x.32-63, x.x.x.64-79, making the 192.168.24.64-79 the 4th subnet, .78 being the broadcast and .78 the last valid host of the 4th subnet.

The answer provided by the question generator has the last valid host of the 4th subnet to be 192.168.24.62, but by my understanding this would be the last valid address of the 192.168.24.32-63 range which I belive to be the 3rd subnet.

Could anyone help me to understand where I'm going wrong/right?

Thanks,

Nunya

Forsaken_GA

Nunya wrote: »

1. Given address of 172.21.0.0 /23 how many subnets and host per subnet?
My answer is 2^7-2 subnets =126, 2^9-2 =510 hosts. The answer provided (by the question generator) is 2^7 subnets = 128, 2^9-2 = 510 hosts...so I'm confused now as to when the subnets include a network ID and host reservation and when it wouldn't.

The restrictions on not being able to use subnet zero or the all-ones subnet went away a long time ago. I have no idea which way the CCNA is actually teaching these days, but you don't lose entire subnets anymore.

For hosts, different ballgame. You still can't assign the prefix address or the broadcast address to a host, so you do lose 2 host IP's from each subnet.

2. Given address of 192.168.24.0 /28, assign a server the last valid address from the 4th subnet.
My answer given the subnet mask would be 255.255.255.240 makes the increment 16...so the subnets would be x.x.x.0-15, x.x.x.16-31, x.x.x.32-63, x.x.x.64-79, making the 192.168.24.64-79 the 4th subnet, .78 being the broadcast and .78 the last valid host of the 4th subnet.

The answer provided by the question generator has the last valid host of the 4th subnet to be 192.168.24.62, but by my understanding this would be the last valid address of the 192.168.24.32-63 range which I belive to be the 3rd subnet.

Could anyone help me to understand where I'm going wrong/right?

You need to check your math. You went from a range of 16 in the first two subnets to a range of 32 in the last (the difference between 32 and 63 is greater than 16)

0-15, 16-31, 32-47, 48-63, so the 4th subnet is 192.168.24.48/28, so the last addressable host in that subnet is 192.168.24.62, with .63 being the broadcast.

Nunya

1. Thanks for the clarification on the 1/0's in relation to subnets/hosts.
2. F* me...math. Thanks for pulling my head out.

bigmantenor

I don't know if you're using the Odom book or not, but that may be where the confusion about ip-subnet zero came from. That particular book was published in 2007 (so possibly written in more like '06), so it is slightly antiquated in very few places.

onesaint

On a side note, this video may be a good suppliment and add to your understanding. It was a go to for me.

Subnetting and Binary Math for IPv4 and IPv6 - Nova Datacom

hbbnt

for your exam, only subtract 2 when dealing with host bits not subnet bits. don't get confused