Decimal to Binary Conversion
Step 1: Divide the given number by 2 and find the remainder.
Step 2: Divide the given continuously until we get zero as the quotient.
Step 3: Write the remainders from bottom to the top, thus binary conversion of the given
number is obtained.
Binary to Decimal Conversion :
Step 1: Write the binary number and count the powers of 2 from right to left (starting from
0)
Step 2 : Write each binary digit (right to left) with corresponding powers of 2 from right to
left , such that MSB or the first binary digit will be multiplied by the greatest power of 2.
Step 3 :Add all the products in the step 2
Step 4: The answer is our decimal number.
Classes High Range of IP Addresses Network Hosts per Subnet Mask /Slash
Order and Host Network Notati
Bits Portion on
A 0 1.0.0.0 to 126.255.255.255 NHHH 224-2=16777214 255.0.0.0 /8
B 10 128.0.0.0 to 191.255.255.255 NNHH 216-2=65534 255.255.0.0 /16
C 110 192.0.0.0 to 223.255.255.255 NNNH 28-2=254 255.255.255.0 /24
D 1110 224.0.0.0 to 239.255.255.255 - - - -
E 11110 240.0.0.0 to 254.255.255.255 - - - -
No of Hosts = 2n-2
where n= no of host bits
Subnet Mask : A subnet mask is a number that distinguishes the network address and the host address
within an IP address. In subnet mask Network bits represented with 1’s and host bits-0’s
192.168.10.0/24
192.168.10.00000000
Number of host IDs: - ?
Subnet mask:- ?
Range: ?
Network id: ?
Broadcast id: ?
Subnet mask for class C network :
192.168.10.0/24
255 255 255 0
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 00000000
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks = 2n
Where n indicates the total no of bits borrowed from the host.
Number of IP addresses on each network=2n
Here n indicates number of host bits.
192.168.10.0/25
255 255 255 128
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10000000
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =21
The first network is 192.168.10.0 - 192.168.10.127
The second network is 192.168.10.128 - 192.168.10.255
Number of IP addresses on each network=27 =128
192.168.10.0/26
255 255 255 192
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11000000
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =22=4
Number of IP addresses on each network=26 =64
192.168.10.0/27
255 255 255 224
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 111 00000
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =23=8
Number of IP addresses on each network=25=32
192.168.10.0/28
255 255 255 ?
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ?
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =?
Number of IP addresses on each network=?
192.168.10.0/29
255 255 255 248
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11111000
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =25=32
Number of IP addresses on each network=23=8
192.168.10.0/30
255 255 255 252
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1111 1 1 0 0
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 242322 21 20
Number of networks =26=64
Number of IP addresses on each network=22=4
192.168.10.0/31
255 255 255 254
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1111111 0
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =27=128
Number of IP addresses on each network=21=2
192.168.10.0/32
255 255 255 255
1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 111111111
27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20 27 26 25 24232221 20
Number of networks =28=256
Number of IP addresses on each network=20=1
For class---- C
CIDR Subnet mask value
/25 128
/26 192
/27 224
/28 240
/29 248
/30 252
/31 254
/32 255
VLSM : 100 hosts
bbA
20 hosts D
bbr
B 10 hosts
ab
40 hosts
Hosts FLSM VLSM
100 128 128
40 128 64
20 128 32
10 128 16
Let’s take an IP address : 192.168.1.0
For 100 hosts ,
2h-2 >= required hosts
27-2 >= 100
128-2 >=100
The subnet mask is 255.255.255.128
Number of IP addresses 27 = 128.
So the first subnet will be 192.168.1.0 – 192.168.1.127
The CIDR value is /25
For 40 hosts ,
2h-2 >= required hosts
26-2 >= 40
64-2 >=40
The subnet mask is 255.255.255.192
Number of IP addresses 26 = 64.
So the Second subnet will be 192.168.1.128 – 192.168.1.191
The CIDR value is /26
For 20 hosts ,
2h-2 >= required hosts
25-2 >= 20
32-2 >=20
The subnet mask is 255.255.255.224
Number of IP addresses 25 = 32.
So the Third subnet will be 192.168.192. – 192.168.1.223
The CIDR value is /27
For 10 hosts ,
The subnet mask is ?
Number of IP addresses ?
The Fourth subnet Ip ranges ?
The CIDR value ?
The table representation for the above example is,
Block Network Network ID Range of Broadcast ID Subnet Mask CIDR
Size hosts Value
128 1 192.168.1.0 1.1 – 1.126 192.168.1.127 255.255.255.128 /25
64 2 192.168.1.128 1.129 – 1.190 192.168.1.191 255.255.255.192 /26
32 3 192.168.1.192 1.193 - 1.222 192.168.1.223 255.255.255.224 /27
16 4 192.168.1.224 1.225 – 1.238 192.168.1.239 255.255.255.240 /28
