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Subnet calculation - pls confirm

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17 years 11 months ago #19312 by skylimit
Hi all, could someone please verify the range of hosts i have on each subnet below pls?

IP address: 235.85.105.66 gives 6 subnets, 30 valid hosts each and 180 hosts in total

Subnets Network ID Minimum IP Maximum IP Broadcast
1 | 235.85.105.66 |235.85.105.67 -235.85.105.96|
235.85.105.97

2 |235.85.105.98 |235.85.105.99 -235.85.105.128| 235.85.105.129

3 |235.85.105.130 |235.85.105.131-235.85.105.160 |235.85.105.161

4 |235.85.105.162 |235.85.105.163 - 235.85.105.192| 235.85.105.193

5 |235.85.105.194 |235.85.105.195 - 235.85.105.224| 235.85.105.225

6 |235.85.105.226 |235.85.105.227- 235.85.105.256 |235.85.105.257


My major concern is the last subnet where i have my last host to be
235.85.105.256 and my broadcast ip to be:235.85.105.257

thanks in advance for any responses

"...you are never too old to learn" anon
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17 years 11 months ago #19313 by Dove
Hi There,

The IPs subnet should be like this

IP : 235.85.105.66
subnet: 255.255.255.224

It will give 6 subnet and 30 host on each subnets.

1) Subnet : 235.85.105.64
Broad : 235.85.105.95
Host : 235.85.105.65 - 235.85.105.94

2) Subnet : 235.85.105.96
Broad : 235.85.105.127
Host : 235.85.105.97 - 235.85.105.126

3) Subnet : 235.85.105.128
Broad : 235.85.105.159
Host : 235.85.105.129 - 235.85.105.158

4) Subnet : 235.85.105.160
Broad : 235.85.105.191
Host : 235.85.105.161 - 235.85.105.190

5) Subnet : 235.85.105.192
Broad : 235.85.105.223
Host : 235.85.105.193 - 235.85.105.222

6) Subnet : 235.85.105.224
Broad : 235.85.105.255
Host : 235.85.105.225 - 235.85.105.254


My major concern is the last subnet where i have my last host to be
235.85.105.256 and my broadcast ip to be:235.85.105.257



The max rane in IP is between 0 - 255, so you couldn't get the IP with 235.85.105.257


Dove
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17 years 11 months ago #19319 by S0lo
You can also check this in a very fast manner using IP subnet Calculator, here:

www.wildpackets.com/products/free_utilit...psubnetcalc/overview

Studying CCNP...

Ammar Muqaddas
Forum Moderator
www.firewall.cx
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17 years 11 months ago #19327 by skylimit

Hi There,

The IPs subnet should be like this

IP : 235.85.105.66
subnet: 255.255.255.224

It will give 6 subnet and 30 host on each subnets.

1) Subnet : 235.85.105.64
Broad : 235.85.105.95
Host : 235.85.105.65 - 235.85.105.94

2) Subnet : 235.85.105.96
Broad : 235.85.105.127
Host : 235.85.105.97 - 235.85.105.126

3) Subnet : 235.85.105.128
Broad : 235.85.105.159
Host : 235.85.105.129 - 235.85.105.158

4) Subnet : 235.85.105.160
Broad : 235.85.105.191
Host : 235.85.105.161 - 235.85.105.190

5) Subnet : 235.85.105.192
Broad : 235.85.105.223
Host : 235.85.105.193 - 235.85.105.222

6) Subnet : 235.85.105.224
Broad : 235.85.105.255
Host : 235.85.105.225 - 235.85.105.254


My major concern is the last subnet where i have my last host to be
235.85.105.256 and my broadcast ip to be:235.85.105.257



The max rane in IP is between 0 - 255, so you couldn't get the IP with 235.85.105.257


Hi, thanks for your reply but i think you have lost me there. I was thinking that subnetting an IP address of 235.85.105.66 would give:
1st subnet:

subnet:235.85.105.66
host range:235.85.105.67 - 235.85.105.96
broadcast:235.85.105.97

2nd subnet:
subnet:235.85.105.98
hostrange:235.85.105.99-235.85.105.128
broadcast:235.85.105.129
....
[000, 001, 010, 011,100,110, 101,111]
i.e. 000 ( 0 dec), 001 (32 dec), 010 (64 dec) are all invalid since the IP address starts from .... .66. pls correct me if im wrong.

Also given that the ip address has 66 as the last octet therefore the first host in the first subnet will have ...67 as the last byte of the first host. In other words assuming that the block of IP given is say: 235.85.105.66/27 - 235.85.105.128/27.

What will the subnets then be for 235.85.105.64?
1st subnet

subnet:235.85.105.64

range:235.85.105.65-235.85.105.94
broadcast:235.85.105.95

Here as you can see, the last byte of the first host in the 1st subnet is greater than the last octet of the IP:235.85.105.64. I may be getting things mixed up here. pls help

More explanation would be great. I'll also give the subnet calculator ago but i want to understand manual calculation perfectly first. thanks in advance.

"...you are never too old to learn" anon
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17 years 11 months ago #19331 by Smurf

Hi, thanks for your reply but i think you have lost me there. I was thinking that subnetting an IP address of 235.85.105.66 would give:
1st subnet:

subnet:235.85.105.66
host range:235.85.105.67 - 235.85.105.96
broadcast:235.85.105.97

2nd subnet:
subnet:235.85.105.98
hostrange:235.85.105.99-235.85.105.128
broadcast:235.85.105.129
....
[000, 001, 010, 011,100,110, 101,111]
i.e. 000 ( 0 dec), 001 (32 dec), 010 (64 dec) are all invalid since the IP address starts from .... .66. pls correct me if im wrong.

Also given that the ip address has 66 as the last octet therefore the first host in the first subnet will have ...67 as the last byte of the first host. In other words assuming that the block of IP given is say: 235.85.105.66/27 - 235.85.105.128/27.

What will the subnets then be for 235.85.105.64?
1st subnet

subnet:235.85.105.64

range:235.85.105.65-235.85.105.94
broadcast:235.85.105.95

Here as you can see, the last byte of the first host in the 1st subnet is greater than the last octet of the IP:235.85.105.64. I may be getting things mixed up here. pls help

More explanation would be great. I'll also give the subnet calculator ago but i want to understand manual calculation perfectly first. thanks in advance.


IP Address 235.85.105.66 isn't a subnet, its a host address.

I would take a look at the following threads which exaplains it all already (to just save repeating it all again).

If you then have further question then please feel free to update this post and we can provide further details.

www.firewall.cx/ftopict-3948.html - Good Thread

www.firewall.cx/ftopict-3953.html

Thanks

P.S. Please do come back to us if the two threads don't fully explain your issues with subnetting.

Wayne Murphy
Firewall.cx Team Member
www.firewall.cx

Now working for a Security Company called Sec-1 Ltd in the UK, for any
Penetration Testing work visit www.sec-1.com or PM me for details.
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