Why are my networks not segmented?
Hi All,
Model T10-W
Version 11.12.1.B522519
1 Trusted General Office 192.168.25.1/24
2 Trusted Point-Of-Sale 192.168.29.1/24
From a computer on 192./168.25.0/24:
zenmap: nmap -sn -T4 192.168.29.0/24
Starting Nmap 7.98 ( https://nmap.org ) at 2026-02-21 17:28 -0800
Nmap scan report for 192.168.29.1
Host is up (0.0010s latency).
Nmap scan report for 192.168.29.100
Host is up (0.0030s latency).
Nmap done: 256 IP addresses (2 hosts up) scanned in 11.57 seconds
AAAHHHHH !!!!! These are suppose to be segmented. What am I doing wrong ?????
Setting up the following firewall rule fixed it, but why was it necessary?

Yours in confusion,
-T
Comments
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If you have policies with To: and/or From: Any-trusted or Any, those will potentially allow traffic between different firewall interfaces or VLANs.
Options include:
1) reviewing your policies which may allow these undesired connections from 1 firewall interface to another - and replacing Any-trusted or Any with a different From/To interface name or alias.2) change the interface type from Trusted to something else, such as Optional or Custom on the the Point-Of-Sale interface AND make sure that traffic between your 2 interfaces is allowed as desired by new or modified policies
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Thank you!
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Hi All,
Figured it out. By default, each segment is suppose to be segmented. Watchguard even advertises that they are. They are not. They are "mostly" segmented.
This is a bug in the default "Ping" firewall rule, which allow ping to any segment. To stop "nmap -sn -T4 192.168.2xx.0/24" from finding devices on the other segments, you need to change the “To:” section reading “Any” to “Any-External”
HTH someone else,
-T0 -
This is not a bug.
The default configuration is not expected to meet the needs of any site.
It is merely a starting point.0 -
okay, not a bug, but a starting point. Watchguard should not be advertising that the come segmented by default. So not a bug, but really bad judgement on the part of marketing. Or Watchguard could ship it out segmented by default.
0
