I do not understand one thing: does each device have a broadcast address? I am subnetting a network and I am giving each subnet one broadcast address and not each host. Is this correct?
For IPv4 networks, each subnet has a broadcast address. Whatever the highest address is for that particular subnet is the broadcast address.
Suppose I had a network at 203.0.113.0 with a netmask of 255.255.255.0. With this mask, hosts on this network can have addresses 203.0.113.1 through 203.0.113.254. Hosts on this network wishing to send a broadcast message to all hosts on the network would use 203.0.113.255.
Related
What is the exact difference between Local IP, External Ip, Static IP, Dynamic IP?
And what does it mean you should be connected on the same local network?
Local v External
static v dynamic
Every IP address has a network and host part, determined by the subnet mask.
A device has an IP and subnet mask of:
192.168.0.5;255.255.255.0 means that it has an network address of 192.168.0 and the rest of the ip address is allocated to hosts on the network (for a host to be on the same network it would need an ip in the range of 192.168.0.1 -> 192.168.0.254; 192.168.0.0 is the subnet address and 192.168.0.255 is the broadcast address)
another example:
A device has an IP and subnet mask of:
192.168.0.5;255.0.0.0 means that it has an network address of 192 and the rest of the ip address is allocated to hosts on the network (for a host to be on the same network it would need an ip in the range of 192.0.0.1 -> 192.255.255.254; 192.0.0.0 is the subnet address and 192.255.255.255 is the broadcast address)
As a final, more complicated example, A device has an IP and subnet mask of:
192.168.200.5;255.255.240.0. To find the network we convert the mask to binary
255.255.240.0
11111111.11111111.11110000.00000000
then convert the ip to binary, and apply the mask to find our ip range of the network
192.168.200.5
11000000.10101000.11001000.00000101 apply mask ->
11111111.11111111.11110000.00000000 logical AND to get our network portion:
11000000.10101000.11000000.00000000 (192.168.192.0)
to get our first and last host address on the same network, fill in the network portion plus all 0's for the first, and all 1's for the last address:
11000000.10101000.11000000.00000000(192.168.192.0 -> subnet)
11000000.10101000.11001111.11111111(192.168.207.255-> broadcast)
so to be on the same network as 192.168.200.5;255.255.240.0; a device would need an ip in the range of 192.168.192.1;255.255.240.0 -> 192.168.240.254;255.255.240.0
It has got something to the with the Local network but I am not sure. Can anyone clear this doubt?
Essentially, in the usual IPv4 + Ethernet scenario there are three different kinds of broadcasts:
local IP broadcast to 255.255.255.255 - this broadcast reaches all nodes within the local broadcast domain; it is not forwarded by a router
directed IP subnet broadcast to the subnet address with all hosts bits set to "1", e.g. 192.168.1.255 for the 192.168.1.0/24 subnet - this broadcast reaches all nodes within the (possibly remote) subnet's broadcast domain; even though it is not forwarded by default on most routers, they are often configured to do so
Ethernet broadcast to FF:FF:FF:FF:FF:FF - this broadcast also reaches all nodes on the local broadcast domain and is the method of layer 2 transport used by the former two L3 broadcasts; a local IP broadcast directly translates to an Ethernet broadcast whereby a directed IP subnet broadcast is first routed to the destination network and then wrapped in an Ethernet broadcast by the last router
I read about the rules of how hosts communicate with each other. Let's say hostA and hostB. hostA gets its own network ID by bitwise-and its ip address and its mask, then gets hostB's network ID by bitwise-and hostB's ip address and hostA's mask. If the results are the same, then they are in the same network segment.
Assuming hostA doesn't know hostB's MAC address: if they are in the same segment, hostA sends ARP request to switch and finally gets hostB's MAC address; if they are not in the same segment, hostA uses gateway MAC address as destination MAC address and hostB's ip address as destination ip address, and send the frame to gateway. That is how it works.
I can understand when the two hosts' ip addresses are like 192.168.1.1/24 and 192.168.1.2/24. But what happened when they are like the following:
If hostA is 192.168.0.1/16 and hostB is 192.168.1.1/24, hostA gets its own network ID:
192.168.0.1 & 255.255.0.0 -> 192.168.0.0
then hostA gets hostB's network ID:
192.168.1.1 & 255.255.0.0 -> 192.168.0.0
So hostA thinks they are in the same segment. But for hostB, hostB gets its own network ID:
192.168.1.1 & 255.255.255.0 -> 192.168.1.0
then hostB gets hostA's network ID:
192.168.0.1 & 255.255.255.0 -> 192.168.0.0
So hostB thinks hostA in a different segment.
How do they communicate in this case? For example, they don't know each other's MAC address, hostA uses one way to find hostB and hostB uses another way to find hostA?
Thanks,
You have a mistake because you have overlapping networks. You cannot configure that in a router. You must have consistent addressing, and you will have problems communicating until your IP addressing is corrected.
Very Much confused about assigning IP address to WAN Interface and LAN Interface on Router/Firewall.
I know the process of assigning IP address but don't know which IP address to assign. Support My default gateway for ISP is 14.141.46.129, then what IP should I assign to my WAN Interface? And what Subnet Mask I have to assign?
Now What IP address I should assign to LAN Interface? And what Subnet Mask I have to assign?
After Assigning IP, I have to get Internet access from My LAN interface to WAN Interface.....
How to find the route to WAN Interface so that I Can ping google.com??
First, 14.141.46.129 is a valid IP address. The WAN interface needs a public IP address which should be given by your ISP provider.you can covert through cyberoam interface A port for-wan, B port for local lan & new range of 192.168.2.1 to 192.168.2.255 subnet mask 255.255.255.0 as u need you have to covert in subnetmask 192.168.2.1/32 to as per your needing (You assing your local ip for Lan The LAN interface can be in the range 192.168.2.2-254 with a mask of 255.255.255.0, )
First, 14.141.46.129 is not a valid IP address.
The WAN interface needs a public IP address which should be given by your ISP provider.
The LAN interface can be in the range 192.168.1.2-254 with a mask of 255.255.255.0
I was studying about IP addresses when I came Across an address of type 127.0.0.1/8. I know that127.0.0.1 is a loop-back address but what did /8 mean in the address.what /8 (in 127.0.0.1/8) specify
the /8 in your ip is the subnet mask.
It allows you to know on which subnetwork you are, find your broadcast and unicast IP.
Take a look at http://en.wikipedia.org/wiki/Subnetwork to have more informations
/8 is a subnet mask written in CIDR notation and which is used by hosts to determine if they are on the same network as some other host and can communicate directly or they need to use a router (default gateway). In your case this is a loopback address, and actually all IP addresses from 127.0.0.0/8 network (from 127.0.0.0 to 127.255.255.255) are reserved loopback addresses as defined in rfc6890: Special-Purpose Address Registries