ACFrOgADrA4dUBFymTmKRfSmh21uqQTorw8cNAb2SO32TbM5OlXyDt7NEwOU3AC0DhCebVN6yfJYF8ovy5as3X7Uh0-3gphEz jS67V0vcek4UTmJZ7XcPrsmaUoJLHxRRhy4wFvgkezkhnXY9ZK PDF
ACFrOgADrA4dUBFymTmKRfSmh21uqQTorw8cNAb2SO32TbM5OlXyDt7NEwOU3AC0DhCebVN6yfJYF8ovy5as3X7Uh0-3gphEz jS67V0vcek4UTmJZ7XcPrsmaUoJLHxRRhy4wFvgkezkhnXY9ZK PDF
ACFrOgADrA4dUBFymTmKRfSmh21uqQTorw8cNAb2SO32TbM5OlXyDt7NEwOU3AC0DhCebVN6yfJYF8ovy5as3X7Uh0-3gphEz jS67V0vcek4UTmJZ7XcPrsmaUoJLHxRRhy4wFvgkezkhnXY9ZK PDF
192 . 168 . 10 . 10
•Because the first 24 bits in the subnet mask are consecutive numeral
ones , the corresponding first 24 bits in the IP address in binary is
11000000. 10101000.00000001;these represent the network portion of
the address.
•The remaining 8 bits are 1100100 and represent the host portion of the
address.
Prof. Veena.Gadad, Dept of CSE, RVCE. 51
• The network address is the first address in the
network.(192.168.1.0)
• The broadcast address is the last address in the
network. (192.168.1.255)
• The first usable host address in the network is
the first address after the network
address.(192.168.1.1)
• The last usable host address is the address
prior to the broadcast address. (192.168.1.254).
• How many hosts can be connected in this
network ?(or how many usable IP addresses?)
28-2= 254
Prof. Veena.Gadad, Dept of CSE, RVCE. 52
Example:
IP address: 192.168.1.10.
Subnet mask: 255.255.248.0.
Determine
1. Network address.
2. Number of hosts in this network.
3. First host address.
4. Last host address.
5. Broadcast address.
•Because the first 21 bits in the subnet mask are consecutive numeral
ones , the corresponding first 21 bits in the IP address in binary is
11000000 10101000 00000 ;these represent the network portion of the
address.
•The remaining 11 bits are 00100001010 and represent the host portion
of the address.
Prof. Veena.Gadad, Dept of CSE, RVCE. 54
• The network number and the broadcast
address use two addresses out of the subnet.
• The number of hosts available in an IPv4
subnet is the number 2 to the power of the
number of host bits available , minus 2:
Number of available hosts = 2 (number of host bits)–2
• In the previous example, number of hosts
available in the subnet are 211-2= 2046.
• First host address: 192.168.0.1.
• Last host address: 192.168.7.254.
• Broadcast address: 192.168.7.255