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Packet Tracer : IP Overlap

Kaveesh
Level 1
Level 1

I'm working on my very first Cisco Packet Tracer projector for an assignment.

I'm in the part where I have to create sub-interfaces for the VLANs I created. I am getting the error "% 192.168.20.0 overlaps with GigabitEthernet0/0.10" when i try to assign the IPs for the 2nd Sub Interface. I tried it with the 3rd VLAN and got the same error as well.
- for Vlan 10
Router(config)#int gig0/0.10
Router(config-subif)#ip address 192.168.20.1 255.255.255.248Routing, Routing Protocols, Other Community Feedback

- for Vlan 12
Router(config-subif)#int gig0/0.12
Router(config-subif)#ip address 192.168.20.9 255.255.255.240

I changed the IP range to 192.168.30.9 255.255.255.240 - and it got assigned with no errors.

Below is my subnetting table 

Kaveesh_0-1738426804605.png

Kaveesh_1-1738427014211.png

Any help or advice would be appreciated. Thanks again

 

 



2 Accepted Solutions

Accepted Solutions

M02@rt37
VIP
VIP

Hello @Kaveesh 

Your VLAN 12 subnet /28 ncorrectly starts at 192.168.20.9, even though a /28 network must start at a multiple of 16.

Since 192.168.20.9/28 is part of 192.168.20.0/28, it conflicts with VLAN 10...

You resolved the problem by assigning VLAN 12 to a different network, which is the right approach. If you want to stay within the same range, you should adjust your subnets so each VLAN has a properly aligned and separate block.

 

Best regards
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View solution in original post

Joseph W. Doherty
Hall of Fame
Hall of Fame

BTW, all your networks must start on their correct address space boundaries, they cannot just be allocated back-to-back, sequentially, unless you allocate them in descending block sizes.  I.e. /26, /28, /28, /28, /29 . . .

View solution in original post

6 Replies 6

Joseph W. Doherty
Hall of Fame
Hall of Fame

The problem is, networks need to align on specific binary multiples.  For example your 192.168.20.9/28 is actually 192.168.20.0..15.

 

M02@rt37
VIP
VIP

Hello @Kaveesh 

Your VLAN 12 subnet /28 ncorrectly starts at 192.168.20.9, even though a /28 network must start at a multiple of 16.

Since 192.168.20.9/28 is part of 192.168.20.0/28, it conflicts with VLAN 10...

You resolved the problem by assigning VLAN 12 to a different network, which is the right approach. If you want to stay within the same range, you should adjust your subnets so each VLAN has a properly aligned and separate block.

 

Best regards
.ı|ı.ı|ı. If This Helps, Please Rate .ı|ı.ı|ı.

M02@rt37 Thanks so much for replying!
I will need to stay within the IP Range as that's a condition in this assignment.

Could you please give me an example? for VLAN 12?
Like on how to adjust the subnet to make sure this does not happen. and what ip should the VLAN 12 start with?

Sorry for the dumb question

 


@Kaveesh wrote:

M02@rt37 Thanks so much for replying!
I will need to stay within the IP Range as that's a condition in this assignment.

Could you please give me an example? for VLAN 12?
Like on how to adjust the subnet to make sure this does not happen. and what ip should the VLAN 12 start with?


M02@rt37 already provided that.

I.e.:

"Your VLAN 12 subnet /28 ncorrectly starts at 192.168.20.9, even though a /28 network must start at a multiple of 16."

Also, in my reply, I too identified network boundaries are important: "The problem is, networks need to align on specific binary multiples." and further described the address block for "your 192.168.20.9/28 is actually 192.168.20.0..15."  I.e. for your 192.168.20.9/28 address, would start with .0, not .8.  After .0, the next /28 would be 192.168.20.16/28, and the first usable address 192.168.20.17/28.

Oh, if you're wondering, can 192.168.20.8..15 be used?  It can, but not as a /28.

Hello @Kaveesh 

Vlan 12 with 12 users, regarding your table... So /28 needed !

For  vlan id 12 =>192.168.20.16/28

Best regards
.ı|ı.ı|ı. If This Helps, Please Rate .ı|ı.ı|ı.

Joseph W. Doherty
Hall of Fame
Hall of Fame

BTW, all your networks must start on their correct address space boundaries, they cannot just be allocated back-to-back, sequentially, unless you allocate them in descending block sizes.  I.e. /26, /28, /28, /28, /29 . . .