07-02-2010 10:01 AM - edited 03-03-2019 05:59 AM
Hi,
I am studying for CCNA and right now going through Subnetting section. Came across this question : What is the default subnet mask for ip address 192.168.10.7 / 10? If my requirement is to connect 1000 hosts then how many subnets, hosts per subnet and host range can i get using this?
Using 10 mask bits, we get dsm as 255.192.0.0. but 192.168.10.7 is class c address then how can we hav 10 mask bits?
Thanx
Mahima
07-04-2010 08:29 AM
Hi,
I am studying for CCNA and right now going through Subnetting section. Came across this question : What is the default subnet mask for ip address 192.168.10.7 / 10? If my requirement is to connect 1000 hosts then how many subnets, hosts per subnet and host range can i get using this?
Using 10 mask bits, we get dsm as 255.192.0.0. but 192.168.10.7 is class c address then how can we hav 10 mask bits?
Thanx
Mahima
Hi Mahima,
Supernetting acts to bridge the gap between a Class C network that is limited to 254 addresses and a Class B network that is too large, with over 65,000 addresses.Supernetting achieves this by making a single network that has your specified number of hosts and corresponding supernet (like a subnet mask). A supernetted address will look like any other TCP/IP address in dotted decimal format (XXX.XXX.XXX.XXX), but it will have a supernetted subnet mask.
In the your example 192.168.10.7/10 which is class range so subnet mask 255.192.0.0 you will have subnet with 192.128.0.0 with 192.128.0.1 to 192.191.255.254 as the host range and next subnet will start with 192.192.0.0 etc
Hope to Help !!
Ganesh.H
Remember to rate the helpful post
07-05-2010 01:32 AM
Hello Ganesh,
So here, 192.168.10.7 is class C address with class A subnet mask and this is possible because of supernetting.Is this what you mean to say?
I got your last part that there will be four subnets with 64 block size starting from 0, 64, 128 and 192.
thanx
mahima
07-05-2010 11:42 PM
Hello Ganesh,
So here, 192.168.10.7 is class C address with class A subnet mask and this is possible because of supernetting.Is this what you mean to say?
I got your last part that there will be four subnets with 64 block size starting from 0, 64, 128 and 192.
thanx
mahima
Hi Mahima,
Actually if you see with the use CIDR and classless routing the networks and the subnet mask make for a pure bit-wise comparison to find the network range but there is a difference between the default subnet mask and the mask /10(255.192.0.0) would basically give you 64 networks each with over 1000 hosts. But default subnet mask in Class C is 255.255.255.0 or /24.But if you want a network that has 1000 hosts then a /10.
Hope to help !!
Ganesh.H
Remember to rate the helpful post
07-06-2010 11:59 AM
hello Ganesh
u are making me confused. A /10 cidr means i have 2 subnet bits which give 2^2 = 4 networks and remainings bits are the host bits (22) which definitely give more than 1000 hosts. According to Todd calculation, there will be four network, 64 will be the block size so we will hav four network of 192.0.0.0 , 192.64.0.0 , 192.128.0.0 , 192.192.0.0 .Isn't it?
How u arrived at 64 networks?
thanx
mahima
07-06-2010 10:34 PM
hello Ganesh
u are making me confused. A /10 cidr means i have 2 subnet bits which give 2^2 = 4 networks and remainings bits are the host bits (22) which definitely give more than 1000 hosts. According to Todd calculation, there will be four network, 64 will be the block size so we will hav four network of 192.0.0.0 , 192.64.0.0 , 192.128.0.0 , 192.192.0.0 .Isn't it?
How u arrived at 64 networks?
thanx
mahima
Hello Mahima,
You are right 64 will be the block size and following will be the subnetting for 192.168.10.7/10
192.168.10.7/10
current network 192.128.0.0 mask 255.192.0.0
current host range 192.128.0.1 to 198.191.255.254
Next subnet will be 192.192.0.0/10
current host range 192.192.0.1 to 192.255.255.254
And following will be the subnet
Subnet , Valid Hosts , Broadcast
192.128.0.0 , 192.128.0.1 to 192.191.255.254 , 192.191.255.255
192.192.0.0 , 192.192.0.1 to 192.255.255.254 , 192.255.255.255
193.0.0.0 , 193.0.0.1 to 193.63.255.254 , 193.63.255.255
193.64.0.0 , 193.64.0.1 to 193.127.255.254 , 193.127.255.255
193.128.0.0 , 193.128.0.1 to 193.191.255.254 , 193.191.255.255
193.192.0.0 , 193.192.0.1 to 193.255.255.254 , 193.255.255.255
194.0.0.0 , 194.0.0.1 to 194.63.255.254 , 194.63.255.255
194.64.0.0 , 194.64.0.1 to 194.127.255.254 , 194.127.255.255
194.128.0.0 , 194.128.0.1 to 194.191.255.254 , 194.191.255.255
194.192.0.0 , 194.192.0.1 to 194.255.255.254 , 194.255.255.255
195.0.0.0 , 195.0.0.1 to 195.63.255.254 , 195.63.255.255
195.64.0.0 , 195.64.0.1 to 195.127.255.254 , 195.127.255.255
195.128.0.0 , 195.128.0.1 to 195.191.255.254 , 195.191.255.255
195.192.0.0 , 195.192.0.1 to 195.255.255.254 , 195.255.255.255
196.0.0.0 , 196.0.0.1 to 196.63.255.254 , 196.63.255.255
196.64.0.0 , 196.64.0.1 to 196.127.255.254 , 196.127.255.255
196.128.0.0 , 196.128.0.1 to 196.191.255.254 , 196.191.255.255
196.192.0.0 , 196.192.0.1 to 196.255.255.254 , 196.255.255.255
197.0.0.0 , 197.0.0.1 to 197.63.255.254 , 197.63.255.255
197.64.0.0 , 197.64.0.1 to 197.127.255.254 , 197.127.255.255
Hope to Help !!
Ganesh.H
07-07-2010 01:44 AM
Hello Ganesh,
Isn't 192.168.0.0 - 192.168.255.255 a range of private IP addresses? Then how come the range of IP that you are showing in that example is all the way up to 198.191.255.254 if the CIDR is /10? The first octet can not change. The broadcast address of 192.0.0.0 is 192.255.255.255 and the last IP address is 192.191.255.254, how did u get to 198.191.255.254? So basically the default mask is 255.192.0.0 with 4.194.304 hosts a little bit too much what the question is mentioning.
I do not even think that 192.168.10.7/10 is used in the real world.
Thanks,
Marius.
07-08-2010 12:51 AM
Hi,
How hav you arrived at change in first octet from 192 to 197?
the calculation for 192.168.10.7/10 will be like :
Ist Subnet : 192.0.0.0
Range : 192.0.0.1 to 192.63.255.254
Broadcast : 192.63.255.255
2nd subnet : 192.64.0.0
Range : 192.64.0.1 to 192.127.255.254
Braodcast : 192.127.255.255
3rd subnet 192.128.0.0
range: 192.128.0.1 to 192.191.255.254
broadcast : 192.191.255.255
4th subnet : 192.192.0.0
range: 192.192.0.1 to 192.255.255.254
broadcast : 192.255.255.255
thanx
mahima
07-06-2010 01:35 AM
Hi Mahima,
Hope you will find this attachment helpful.
Happy Networking !!!
Regards,
Narendrakumar B
07-06-2010 01:23 PM
Hello Narendrakumar B,
That was indeed helpful.
Do you know whether this technique works with cidr also?
ex. 150.250.150.250/23 is the ip address.requirement is total 3 networks?
Thanx
Mahima
07-07-2010 02:33 AM
Yeah sure.
150.250.150.250/23 is the ip address.requirement is total 3 networks?
So since you need 3 networks, and as the rule says in the file, you have to calculate for 3-1 = 2. And to get 2 youo need 2 bits. you have to turn on 2 additional network bits!!!
Do rate the helpful posts !!!
Regards,
Narendrakumar B
07-08-2010 01:35 AM
For 150.250.150.250/23 my subnet mask is 255.255.254.0. Now i borrow 2 more network bits from host bits so new mask is
255.255.255.128. Right?
Here increment bit give 128 as the decimal value.
the network will be then :
150.250.150.0 and 150.250.150.128
but that is only 2 networks.
thanx
Mahima
07-09-2010 03:50 PM
Hello,
You are right. It is basically Hex number system. The decimal weights are (128, 64, 32, 16, 8, 4, 2, 1) for each byte. It follows powers of 2. As you increase the network bits, you start adding corresponding weight in decimal.
Here are few useful links for starters.
http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a00800a67f5.shtml
http://www.ralphb.net/IPSubnet/
Hope this helps.
Regards,
NT
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