Can someone look at this subnet?

Posts: 238Inactive Imported Users
I'm a bit fuzzy on this custom subnet example - if you have a network address of 195.223.50.0 and you need 1 usable subnet...

1. What would the 2nd usable subnet range be?
2. What is the subnet bradcast address for the 1st usable subnet?
3. What is the 3rd usable host on the 2nd usable subnet?

The handout my class got had some errors, so I need to get these things clear in my mind. Thanks.

• Posts: 148Member ■■■□□□□□□□
Tricon7 wrote:
I'm a bit fuzzy on this custom subnet example - if you have a network address of 195.223.50.0 and you need 1 usable subnet...

1. What would the 2nd usable subnet range be?
2. What is the subnet bradcast address for the 1st usable subnet?
3. What is the 3rd usable host on the 2nd usable subnet?

The handout my class got had some errors, so I need to get these things clear in my mind. Thanks.

I'm still fairly new to this, but here's my opinion.

Firstly let's assume ip-subnet zero is off (unless it was mentioned in your class).

You've got a class C address which means your subnet is going to be 255.255.255.192 (2 usable subnets .64 & .128. .0 and .192 are not valid because of ip-subnet zero).

1. Your second usable subnet range is: 195.223.50.129 - 195.223.50.190
2. Your broadcast address for the first subnet is: 195.223.50.127
3. The third usable host on the second subnet is: 195.223.50.131

Do you understand how I got these answers (assuming they are correct - can someone check my work please!)?

Edit: typo
Regards,

CCNA R&S; CCNP R&S
• You can calculate over the formula 2^n - 2. n = number of needed subnets. you need 2 subnet as and it is 2^2 - 2 = 2 so subnetmask is 255.255.255.192. ip-subnet zero is available at ver 12.0 ISO and later as I now.

SO:

[0I]000000 = 64 is network address of first subnet
[0I]IIIIII = 127 is broadcast address of first subnet

[I0]000000 = 128 is network address of second subnet
[I0]IIIIII = 191 is broadcast of second subnet

Range for useable address:
1st subnet = 195.223.50.65-127
2nd subnet = 195.223.50.129-190
• Aquilla is correct apart from the obvious ommission of the third octet in answer 3 which should be 192.223.50.131 (not 192.223.131 )

Additionally subnet-zero is not going to affect the use of the last subnet one way or the other, this is the broadcast subnet which CISCO in the past has advised against using, but it can still be used whether or not subnet zero is enabled or disabled.

If Subnet Zero is anabled, the mask would be 255.255.255.128, where you would be using both Subnet Zero and the Broadcast subnet for IP addresses... I am sure you can figure out the IP addresses and ranges from this.

James.
• Posts: 148Member ■■■□□□□□□□
james_ wrote:
Aquilla is correct apart from the obvious ommission of the third octet in answer 3 which should be 192.223.50.131 (not 192.223.131 )

James.
Corrected. Thanks. Regards,

CCNA R&S; CCNP R&S
• Posts: 238Inactive Imported Users
aquilla wrote:
Tricon7 wrote:
I'm a bit fuzzy on this custom subnet example - if you have a network address of 195.223.50.0 and you need 1 usable subnet...

1. What would the 2nd usable subnet range be?
2. What is the subnet bradcast address for the 1st usable subnet?
3. What is the 3rd usable host on the 2nd usable subnet?

The handout my class got had some errors, so I need to get these things clear in my mind. Thanks.

I'm still fairly new to this, but here's my opinion.

Firstly let's assume ip-subnet zero is off (unless it was mentioned in your class).

You've got a class C address which means your subnet is going to be 255.255.255.192 (2 usable subnets .64 & .128. .0 and .192 are not valid because of ip-subnet zero).

1. Your second usable subnet range is: 195.223.50.128 - 195.223.50.191
2. Your broadcast address for the first subnet is: 195.223.50.127
3. The third usable host on the second subnet is: 195.223.50.131

I don't get the answer to number three. If the range on the second subnet is 128-191, wouldn't you count "128 is number one, 129 is number two, and 130 is number three." So why wouldn't 130 be the 3rd usable host on the 2nd usable subnet? And why wouldn't 128 be the network address (unusable)?
• Posts: 148Member ■■■□□□□□□□
Tricon7 wrote:
aquilla wrote:
Tricon7 wrote:
I'm a bit fuzzy on this custom subnet example - if you have a network address of 195.223.50.0 and you need 1 usable subnet...

1. What would the 2nd usable subnet range be?
2. What is the subnet bradcast address for the 1st usable subnet?
3. What is the 3rd usable host on the 2nd usable subnet?

The handout my class got had some errors, so I need to get these things clear in my mind. Thanks.

I'm still fairly new to this, but here's my opinion.

Firstly let's assume ip-subnet zero is off (unless it was mentioned in your class).

You've got a class C address which means your subnet is going to be 255.255.255.192 (2 usable subnets .64 & .128. .0 and .192 are not valid because of ip-subnet zero).

1. Your second usable subnet range is: 195.223.50.128 - 195.223.50.191
2. Your broadcast address for the first subnet is: 195.223.50.127
3. The third usable host on the second subnet is: 195.223.50.131

I don't get the answer to number three. If the range on the second subnet is 128-191, wouldn't you count "128 is number one, 129 is number two, and 130 is number three." So why wouldn't 130 be the 3rd usable host on the 2nd usable subnet? And why wouldn't 128 be the network address (unusable)?

195.223.50.128 is the network address (and therefore unusable by a host). The same way 195.223.50.191 is the broadcast address for the subnet and unusable by a host. Therefore the first IP usable by a host is 195.223.50.129.

Remember the rule of 2^x-2. You took the first two 'bits' of the fourth octet for the subnet mask (2^2 = 4 subnets, 2 of which we deemed unusable to ip-subnet zero). That left you with 6 'bits' for the host portion (2^6 = 64). The first and last address within these subnets are unusable by hosts, therefore 2^6-2 = 62 hosts per subnet.
Regards,

CCNA R&S; CCNP R&S
• Posts: 238Inactive Imported Users
aquilla wrote:
Tricon7 wrote:
aquilla wrote:
Tricon7 wrote:
I'm a bit fuzzy on this custom subnet example - if you have a network address of 195.223.50.0 and you need 1 usable subnet...

1. What would the 2nd usable subnet range be?
2. What is the subnet bradcast address for the 1st usable subnet?
3. What is the 3rd usable host on the 2nd usable subnet?

The handout my class got had some errors, so I need to get these things clear in my mind. Thanks.

I'm still fairly new to this, but here's my opinion.

Firstly let's assume ip-subnet zero is off (unless it was mentioned in your class).

You've got a class C address which means your subnet is going to be 255.255.255.192 (2 usable subnets .64 & .128. .0 and .192 are not valid because of ip-subnet zero).

1. Your second usable subnet range is: 195.223.50.128 - 195.223.50.191
2. Your broadcast address for the first subnet is: 195.223.50.127
3. The third usable host on the second subnet is: 195.223.50.131

I don't get the answer to number three. If the range on the second subnet is 128-191, wouldn't you count "128 is number one, 129 is number two, and 130 is number three." So why wouldn't 130 be the 3rd usable host on the 2nd usable subnet? And why wouldn't 128 be the network address (unusable)?

195.223.50.128 is the network address (and therefore unusable by a host). The same way 195.223.50.191 is the broadcast address for the subnet and unusable by a host. Therefore the first IP usable by a host is 195.223.50.129.

Remember the rule of 2^x-2. You took the first two 'bits' of the fourth octet for the subnet mask (2^2 = 4 subnets, 2 of which we deemed unusable to ip-subnet zero). That left you with 6 'bits' for the host portion (2^6 = 64). The first and last address within these subnets are unusable by hosts, therefore 2^6-2 = 62 hosts per subnet.

Ok. But if I don't use the 128, as it is part of the network address, then the answer to my original #2 question would be 195.223.50.129 - 195.223.50.191, wouldn't it? It wouldn't start with 195.223.50.128, as the .128 is part of the network address. Or not? This is what's confusing me.
• Posts: 148Member ■■■□□□□□□□
Tricon7 wrote:
Ok. But if I don't use the 128, as it is part of the network address, then the answer to my original #2 question would be 195.223.50.129 - 195.223.50.191, wouldn't it? It wouldn't start with 195.223.50.128, as the .128 is part of the network address. Or not? This is what's confusing me.
With every subnet you will have a network address and a broadcast address. The network address will be the first IP in the block and the broadcast will be the last IP in the block. Everything else between these is usable by your hosts.

For example:

192.168.10.0 255.255.255.0

192.168.10.0 <- network address
192.168.10.1 - 254 <- usable by hosts

195.223.50.128 255.255.255.192

195.223.50.128 <- network address
195.223.50.129 - 190 <- usable by hosts

Through my ISP, I have a /28 subnet (14 usable hosts). My setup is:

217.xxx.xxx.96 <- network address
217.xxx.xxx.97 - 110 <- usable hosts.
Regards,

CCNA R&S; CCNP R&S
• In regards to questions on the CCNA, I think its safe to assume that IP subnet zero is enabled unless otherwise stated. In the more recent IOS versions (Can't remember exact IOS #), it is enabled by default. Therefore, when calculating your subnets, you can use all available.

195.223.50.0 - 1st. Subnet
195.223.50.1 - 126 - Usable Host Addresses