Are the following IP adresses - 168.94.254.255 & 168.94.255.254 - used for broadcast, if they belong to B Class? (The adresses were randomly chosen). I suppose 168.94.255.255 would rather be a broadcast adress, but I'm not sure. Thanks in advance!
A class B network in CIDR notation is a /16 network. A broadcast address is where all of the host bits are 1. In your case, the last 16 bits would need to be 1 to be a broadcast address.
168.94.254.255 and 168.94.255.254 each have a zero bit in the last 16 bits, so they are not broadcast addresses.
If your class B is subnetted into /24's, then the 168.94.254.255 address may be a broadcast address if 168.94.254.0/24 is the network address.
Related
An ISP has provided your company with the Class C network 192.111.2.0. Divide this into four (4) subnets.
Complete the following table; both Network addresses and Broadcast addresses should be in dotted-decimal notation.
Assume the all zero's and all one's subnets are usable.
There are a lot of material on subnetting - just google it and you'll get the result pretty easily. You can start with this subnet calculator.
As for your question:
In order to get 4 subnets from 192.111.2.0/24 you should go for the following configuration:
The subnet mask for all subnets are 255.255.255.192 (or /26) and the networks are divided as follows:
192.111.2.0-192.111.2.63
192.111.2.64-192.111.2.127
192.111.2.128-192.111.2.191
192.111.2.192-192.111.2.255
Notice that the first address is the network address and the last is the broadcast address
I am learning Computer Networks and was reading about limited and directed broadcast.
If a host with IP 200.100.1.1 wants to send packets to all host in the same network, then does the procedures A and B, to achieve our goal, differs?
Procedure A SRC- 200.100.1.1 DEST- 200.100.1.255
Procedure B SRC- 200.100.1.1 DEST- 255.255.255.255
So, can we use Procedure A to do the task required as the directed broadcast(destination) is aiming at its own network? Or Procedure A is incorrect and Procedure B is the right way to go.
IP directed broadcast packets have a destination IP address that is a valid broadcast address for the subnet that is the target of the directed broadcast (the target subnet). The intent of an IP directed broadcast is to flood the target subnet with the broadcast packets without broadcasting to the entire network. reference; emphasis added
Assume the following:
We have an IP address of 200.100.1.1 (11001000.01100100.00000001 .00000001)
The netmask is /24 (or 255.255.255.0) (11111111.11111111.11111111 .00000000)
Then:
The network address is a bitwise AND between these two:
11001000.01100100.00000001 .00000001
11111111.11111111.11111111 .00000000
=
11001000.01100100.00000001 .00000000
200 100 1 0
So we know the network is 200.100.1.0.
To determine the broadcast address, we invert the subnet mask and OR it with the network address.
Network: 11001000.01100100.00000001 .00000000
Inverted Netmask: 00000000.00000000.00000000 .11111111
11001000.01100100.00000001 .11111111
=
200 100 1 255
The broadcast for 200.100.1.0/24 will be on 200.100.1.255
We also know that this network will have 254 possible hosts in it (or 253 given a gateway address; and the assumption that .0 is not addressable so 200.100.1.1 -> 200.100.1.254).
Consider the following classless address block:
154.78.177.3/27
List the addresses from this block that would be used as:
a) the network address,
b) the direct broadcast address, and
c) the range available for hosts to use
Show the steps you took to arrive at your answers.
154.78.177.3/27 means that this ip address is having network address to be 27 bit long ( most significant bit). i.e if we consider the ip address to be 32 bit long. So, in 154.78.177.00000011 ( last 3 number is represented as binary for simplicity) upto 154.78.177.000 (8+8+8+3 = 27) is network address and remaining 5 bit (00000) is for host ip addresses.
Note-
Network address is the first ip address of the total ip's of host.
Direct broadcast address is the last ip address of total ip's of host.
So final conclusion:
1. Network address is 154.78.177.0. This is calculated by setting all 5 bits of host id to 0.
2. Direct broadcast address is 154.78.177.31. This is calculated by setting all 5 bits of host to 1.
3. Range available for host is from 1 to 30 in the last octect. I.e from 154.78.177.1 to 154.78.177.30 ( As first and last ip is reserved for network address and direct broadcast address respectively)
I tried to understand mapping multicast IP address to layer 2 ethernet frame i.e IEEE 802.3 multicast frame but didn't understand the concept behind it. I'm will appreciate the answers with examples here, of which mapping of an IP address to Ethernet frame can be explained on the scale side by side.
This is very simple:
For ipv4 :
Mac adress has to start with 01:00:5E the 01 in binary gives 00000001 the the lowest bit of the hihgest byte give multicast or not. This is the first bit received by network card so then can quickly ignore or not multicast packet
So you have 01:00:5E:XX:YY:ZZ the range of available mac adresses goes from 01:00:5e:00:00:00 to 01:00:5e:7f:ff:ff
So you have 23 bits to complete with the 23 lowest bits of the ipv4 adresses
IPV4 adresses multicast go from 224.0.0.0 to 239.255.255.255
So for example OSPF (well know routing protocol) use these ipv4 adresses :
224.0.0.5 that give 01:00:5E:00:00:05
224.0.0.6 that give 01:00:5E:00:00:06
For ipv6 :
ipv6 adress are ff00::/8
You use mac prefix 33:33:XX:YY:ZZ (remark the lowest bit of the highest bytes is 1)
and you complete by the lowest bytes of ipv6 adress
I have around 500 IP addresses. 172.45.67.1 - 172.45.67.200. How do I find the Wifi subnet for these IP addresses? If I could use a java API, that would be great. If not, any other technique to determine the subnet?
Your IP range appears to be part of a Class B IPv4 subnet based on the starting octet value 172.
http://en.wikipedia.org/wiki/IP_address#IPv4_subnetting
As such the subnet mask would be 255.255.0.0.
http://www.subnet-calculator.com/subnet.php?net_class=B
A Class B subnet allows for a maximum of 65,536 addresses.
Your building may be allocated just a slice of that subnet by the people administering that subnet. However, there is no way of knowing how much of that subnet is allocated to the building without further information (if there are 500 addresses, they cannot all be allocated from 172.45.67.* as there are only 255 addresses in that range).