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2 Private IP Ranges

Posted on 2010-09-08
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Last Modified: 2012-05-10
Ok I will explain this the best I can.  Our network, that was set up long before me.  Our Ip range is 172.16.x.x and now we also have 172.17.x.x.  Our DC has our dhcp server on it and works fine for the 172.16.x.x address.  The 17 address range cannot see any assets on the network that carry 172.16 range and cannot access the internet.  Both ranges are setup in the router.  Our 172.16 range for uses a 255.255.255.0 mask.  Our server is 2003.  Can anyone tell me how to get the two ranges to see each other?
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Question by:MaximusTech
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by:rfc1180
ID: 33628842
you are either running into a routing issue are IP filtering issue. What type of router do you have.
What is the default gateway and subnet masks of the hosts that are in the 172.17.x.x?
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by:Kvistofta
ID: 33628853
What hardware is the router? Can you post the configuration here?

/Kvistofta
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by:MaximusTech
ID: 33628867
the default gateway is 172.17.1.1
and the subnet mask 255.255.0.0

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by:rfc1180
ID: 33628898
your subnet mask is incorrect, change it to 255.255.255.0
Also please post the router config
Billy
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by:Kvistofta
ID: 33628943
rfc1180: Why is the mask incorrect?? 172.17 is a B-class network and /16 is perfectly fine.

/Kvistofta
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by:rfc1180
ID: 33629020
his hosts in the 172.16 are using 255.255.255.0
hosts in the 172.17.x.x that are using 255.255.0.0 that are trying to communicate with 172.16 will never send a packet to the default gateway as the host(s) will assume that 172.16 is local

Also, we have asked for a posting of his config which will verify if the 2 networks are separate broadcast domains or if they are in fact local, if that is the case, then possible the hosts in 172.16.0.0 will need to change the mask to 255.255.0.0

Billy
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by:Kvistofta
ID: 33629097
"hosts in the 172.17.x.x that are using 255.255.0.0 that are trying to communicate with 172.16 will never send a packet to the default gateway as the host(s) will assume that 172.16 is local"

What? Why would a properly configured host in 172.17.0.0/16 believe that anything addressed 172.16.x.x is local???

/Kvistofta
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by:rfc1180
ID: 33629170
>What? Why would a properly configured host in 172.17.0.0/16 believe that anything addressed 172.16.x.x is local???

yeah, you are right! My bad, I am thinking /8 not sure why; you are correct, the 172.17.x.x and 255.255.0.0 is the correct mask.

at any rate, we still need the config of the router

Billy
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by:Fred Marshall
ID: 33629186
It's a bit unclear if the objective is to have two separate subnets that are interconnected or to have a single subnet.    That's because the author says: "and now we have ...."  So, it would help the Author to have this clear and it would help us to know so we can provide better advice.  Anyway, I strongly suspect that someone set up "17" addresses because they should have been available with 255.255.0.0

If that's the case then it's a matter of subnet masks matching the intent.  255.255.255.0 seems incorrect under *any* circumstance here - or there's not enough information to go on.
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by:Kvistofta
ID: 33629335
Yes, we are all guessing. The author needs to give more information.

/Kvistofta
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by:goyal_251
ID: 33634521
you donot need to add router to connect 172.16.x.x and 172.17.x.x.change the subnet mask 255.255.254.0 in both the network.both network will communicate each other without router(will become same network actually).
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by:Kvistofta
ID: 33635475
goyal_251: That is totally wrong. Maybe you mean subnet mask 255.254.0.0?

/Kvistofta
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by:goyal_251
ID: 33635989
no its 255.255.254.0 .......I am taking about CIDR(Classless Inter-Domain Routing)
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Kvistofta earned 100 total points
ID: 33636042
Sure, but with the mask 255.255.254.0 you can never have anything addressed 172.16.x.x and 172.17.y.y in the same network. Sorry, get back to the books.

172.16.0.0/23 contains 172.16.0.0 - 172.16.1.255
172.17.98.0/23 contains 172.17.98.0 - 172.17.99.255.

You sure talk about CIDR becaus this is what all this is about, without that you would have the entire 172.16.0.0/16 as a classful network and 172.17.0.0/16 as another. But in your mask 255.255.254.0 you have the network-host-boundary in the third octet, which means that everything in the second octet defines the network, not host. Hence 172.16.whatever.whatever with 255.255.254.as mask contains nothing more than 172.16.whatever.0 to 172.16.whatever+1.255. Never 172.17, because then your second octet in the netmask needs to be something else than 255.

/Kvistofta
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Assisted Solution

by:chris-burns
chris-burns earned 25 total points
ID: 33640269
Kvistofta is right, he could change his subnet to 255.254.0.0.

that would have all addresses from 172.16.0.0 to 172.17.255.255.
By doing this you would have a single subnet, but by far is the easiest fix.

http://www.subnet-calculator.com/cidr.php Is worth having to hand.
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by:goyal_251
ID: 33644357
Yes.. Kvistofta you are right,atucally we need to make the change in second octet.But the calculation I had given was for third octact.thanks for notifying the error.
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by:Kvistofta
ID: 33644507
You are welcome. ;)

/Kvistofta
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