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5 Base position Decimal value Convert decimal to binary To convert a decimal number in binary we would use addition till number method. In this method we start adding from left to get target value. If after adding right position value, sum is lower than target number, keep adding, or if sum is greater than target number skip the position value. Only the value of on bit [1] will be added in sum. Off bit [0] has zero value. For example, convert decimal number 117 in binary. Target decimal number 117 Move direction From Left ===========================> to Right Base position Decimal value Bit status Decimal value in addition Binary value of 117 is Decimal calculation Bit in binary 128 is greater than 117 off the bit 0+64 = 64 is less than 117 on the bit = 96 is less than 117 on the bit = 112 is less than 117 on the bit = 120 is greater than 117 off the bit = 116 is less than 117 0n the bit = 118 is greater than 117 off the bit = 117 is equivalent to 117 on the bit Convert binary in decimal

6 To convert a binary in decimal we will follow above method in reverse mode. We will find the decimal value of on binary bit position and add them. For example convert binary number in decimal. Target binary number Move direction From Left ===========================> to Right Base position Decimal value Bit status Decimal value in addition Decimal value of is 170 [ ]. Binary bit Decimal value 1 On bit Off bit 0 1 On bit 64 0 Off bit 0 1 On bit 32 0 Off bit 0 1 On bit 8 0 Off bit 0 1 On bit 2 0 Off bit 0 Review above method and do as much practice of following as you can. Pick any number from and convert it in binary. Pick any combination from and convert it in decimal. Better you are with this conversion, the better you will do on the subnetting. Combination provided by binary position Now you know the process of converting binary in decimal and decimal in binary. Our next step is to figure out the combination provided by a binary position. First bit provide two combination 0 or 1. If we take two bits than we have four combinations 00,01,10,11. IP address has 32 bits, so you should be able to find the number of combinations provided by each bit position until position number 32.

11 Total subnets ( 2 N ) :- 2 1 = 2. Block size (256 - subnet mask) : = 128. Valid subnets ( Count blocks from 0) :- 0,128 Total hosts (2 H ) :- 2 7 = 128 Valid hosts per subnet ( Total host - 2 ) : = 126 Subnets Subnet 1 Subnet 2 Network ID First host Last host Broadcast ID /26 CIDR /26 has subnet mask and 192 is in binary. We used two host bits in network address. H Default subnet mask of class C ====== N N = 2 H = 6 Total subnets ( 2 N ) :- 2 2 = 4. Block size (256 - subnet mask) : = 64. Valid subnets ( Count blocks from 0) :- 0,64,128,192 Total hosts (2 H ) :- 2 6 = 64 Valid hosts per subnet ( Total host - 2 ) : = 62 Subnets Subnet 1 Subnet 2 Subnet 3 Subnet 4 Network ID First host Last host Broadcast ID

12 /27 CIDR /27 has subnet mask and 224 is in binary. We used three host bits in network address. H Default subnet mask of class C ===== N N = 3 H = 5 Total subnets ( 2 N ) :- 2 3 = 8. Block size (256 - subnet mask) : = 32. Valid subnets ( Count blocks from 0) :- 0,32,64,96,128,160,192,224 Total hosts (2 H ) :- 2 5 = 32 Valid hosts per subnet ( Total host - 2 ) : = 30 Subnets Subnet1 Subnet2 Subnet3 Subnet4 Subnet5 Subnet6 Subnet7 Subnet8 Network ID First host Last host Broadcast ID / CIDR /28 has subnet mask and 240 is in binary. We used four host bits in network address. H Default subnet mask of class C ==== N N = 4 H = 4 Total subnets ( 2 N ) :- 2 4 = 16. Block size (256 - subnet mask) : = 16. Valid subnets ( Count blocks from 0) :-0,16,32,48,64,80,96,112,128,144,160,176,192,208,224,240 Total hosts (2 H ) :- 2 4 = 16 Valid hosts per subnet ( Total host - 2 ) : = 14

13 "I hope you have understand the pattern of making the subnet chart with above information's. Let's join me in this practice. For this example and next two examples I will fill only two subnets [first and last one], it's your turn to fill the middle subnets." Subnets Subnet 1 Network ID 0 First host 1 Last host 14 Broadcast ID 15 Subnet 2 To Subnet 15 (Filled by you)... Subnet /29 CIDR /29 has subnet mask and 248 is in binary. We used five host bits in network address. H Default subnet mask of class C === N N = 5 H = 3 Total subnets ( 2 N ) :- 2 5 = 32. Block size (256 - subnet mask) : = 8. Valid subnets ( Count blocks from 0) :- 0,8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,152,160,168,176,184,192,200,208, 216,224,232,240,248 Total hosts (2 H ) :- 2 3 = 8 Valid hosts per subnet ( Total host - 2 ) :- 8-2 = 6 Subnets Subnet 1 Subnet 2 To Subnet Subnet 32

14 Network ID 0 First host 1 Last host 6 Broadcast ID 7 31 (Filled by you) /30 CIDR /30 has subnet mask and 252 is in binary. We used six host bits in network address. H Default subnet mask of class C == N N = 6 H = 2 Total subnets ( 2 N ) :- 2 6 = 64. Block size (256 - subnet mask) : = 4. Valid subnets ( Count blocks from 0) :- 0,4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100,104,108,112,116,1 20,124,128,132,136,140,144,148,152,156,160,164,168,172,176,180,184,188,192,196,200,204,208,2 12,216,220,224,228,232,236,240,244,248,252 Total hosts (2 H ) :- 2 2 = 4 Valid hosts per subnet ( Total host - 2 ) :- 4-2 = 2 Subnets Subnet 1 Subnet 2 To Subnet 63 Subnet 64 (Filled by you) Network ID First host Last host 2 254

15 Broadcast ID At the end of this long article I have a small word for you, that is practice. Believe me in exam you are going to face many questions related to subnetting. More practice will make you more comfortable in exam. For more article visit our site ComputerNetworkingNotes.com

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