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i have to construct a subnet in cisco packet tracer using this ip that was supposedly given by the IPS:

79.48.0.0/16

But when i try to configurate the routing using RIP, and put for example 79.48.2.0 i get in the RIP network addresses 79.0.0.0 (255.0.0.0)

In normal cases, when using for example: 192.168.1.0 it appears correctly, so what it's going on? I'm using a 24 bit mask for 4 subnet

(79.48.0.0, 79.48.1.0, 79.48.2.0, 79.48.3.0)(all using this mask 255.255.255.0)

and a serial cable (10.0.0.0 255.0.0.0)

(here's a example)

enter image description here

  • If you want four subnets for a /16 network, the correct mask length is /18, not /24. – Ron Maupin Jul 06 '16 at 17:13
  • Why? I thought it was: for 256 host it's 2^8, so 32 - 8 = 24 so i put 79.48.0.1 255.255.255.0, then for the other 256 host i do the same but i change it to 79.48.1.1 255.255.255.0 – Sebastian Tare B. Jul 06 '16 at 17:25
  • Four subnets requires extending a network mask by two bits (2^2=4). Yes, 254 hosts (you can't use the first, network, address, or the last, broadcast, address) only requires a /24, but your question says you need four subnets, and that is a very different requirement. – Ron Maupin Jul 06 '16 at 17:28
  • So you say i should configure it like this? ip add 79.48.0.1 255.255.248.0 – Sebastian Tare B. Jul 06 '16 at 17:32
  • Actually, 79.48.0.0/18 has a mask of 255.255.192.0. You really need to do this in binary, then it becomes obvious. – Ron Maupin Jul 06 '16 at 17:46
  • To be more specific, i have 2 routers, 2 switch connected to each router, each switch connected to a FastEthernet port – Sebastian Tare B. Jul 06 '16 at 17:47
  • You need to determine the subnet requirements. Is it by the number of subnets, or is it by the number of hosts? This answer provides and excellent explanation of how to manipulate IP addressing. – Ron Maupin Jul 06 '16 at 17:51
  • It's by number of hosts, i have 2 floors and in each floor a laboratory with 31 hosts, so i'm using 64 ip, and also 2 classrooms per floor, for that i'm using 4 access point, that's why i just put it like i have 254 hosts, so i can connect a AP to the same switch – Sebastian Tare B. Jul 06 '16 at 18:03
  • Do whatever meets your requirements. I was merely pointing out that what you wrote in your question, "I'm using a 24 bit mask for 4 subnet," will actually give you 256 possible /24 subnets for a /16 network, not just four subnets. If the requirement is to subnet the /16 network into four subnets, then you use a /18. Your question is unclear, but it seemed to imply that the requirement was by subnet, not host. – Ron Maupin Jul 06 '16 at 18:13

2 Answers2

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This is correct, the reason why you are seeing this is because you are using RIPv1 which can only deal with classful addresses and does not support VLSM. The address range you are trying to configure is a Class A address which has a /8 (255.0.0.0) subnet mask and you can't use a /16 mask as it belongs to Class B addresses in a classful process.

If you want to use classless addresses and use VLSM so that you can divide your subnet into smaller bits, you need to change the RIP version to RIPv2.

To change this, you can do something like:

router rip

version 2

network 79.48.2.0

no auto-summary

We also turn off auto-summary so that it doesn't summarize the network into a classful network as you are seeing now.

Please also remember that RIPv1 is very old and you probably shouldn't be using it. If you must use RIP, make sure it is version 2 as it has many improvements over version 1, and using classless addresses is one of them :)

Hope this helps you on your quest.

SleepyMan

SleepyMan
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Unless, of course, he is being required to use RIP in a class or something.

Per subnetting and finding the right CIDR (/xx)...

In the 3rd octet for your four subnets, you only need the first two bits to express 0, 1, 2, 3. The value of those two bits is/are 128 + 64 --> 192, thus 10.48.192.0/18. This would leave you 6 bits of the 3rd octet and 8 in the 4th octet to address specific devices (a lot of them, 8K I think).