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So your gateway IP is 192.168.1.120. What is its subnet mask? If it is 255.255.255.0, in no way you can use a 192.168.2.x IP and hope that it works. Client IP and network gateway must be on the same network, and this is the very same reason Windows is getting you an error. EDIT: maybe you can assign two IP addresses on you Windows machine, ...


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I guess the gateway is something like 192.168.1.X since it worked while the server was in that range? Then you need another subnet mask, the smallest mask to be usable here that has both the server and its gateway in the same network ist 255.255.252.0. But the networks also needs to be configured in a way that allows that mask to be used. You should ...


5

It can introduce performance problems since packets to the broadcast IP address are broadcast to all hosts on the segment. This additional traffic which is intended for only one host but send to every host can negatively impact the performance of those hosts needlessly receiving that traffic. Some switches can rate limit broadcast traffic. If your switch ...


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That means that the 192.168.1.1 address is associated with a 240.0.0.0 mask. You have to think binary to understand that. In binary your IP address would be: 11000000.10101000.00000001.00000001 In the same way /4 represent the number of "1" bits in the mask. So here the mask is 11110000.00000000.00000000.00000000 (which is 240.0.0.0 in decimal) /8 would ...


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No, you can't. In router 192.168.1.254 it need a route back to talk to 192.168.10.X, you can't have two route to the same network. As the router 192.168.10.254 will have an IP in the WAN (aka 192.168.1.X), thus two distinct IP there for the two router. The only trick you can do, is to make a lower subnet on 192.168.1.254's side. (/25) For 1. .128 it ...


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cannot comment :( no, sorry that is not possible. But you can "cut" the main network into two halfs. e.g. 192.168.10.1-192.168.10.127 and 192.168.10.128-192.168.10.254 so it will be 192.168.10.1 and 192.168.10.128 for the two routers Maybe tell us a littlebit more, I will then update this answer



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