IP Addressing & Subnetting / Sumiet23
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The document discusses IP address classes and subnetting. It provides details on class A, B, C and D IP addresses including their public and private IP ranges and subnet masks. It also defines key subnetting concepts like CIDR (Classless Inter-Domain Routing) notation, the number of network and host bits for each class, how to calculate the total number of subnets and maximum number of hosts per subnet. There are examples provided on how to determine these values given an IP address and subnet mask.
IP Addressing & Subnetting / Sumiet23
- 1. Public IP Private IP Subnet Mask Class A 1.0.0.0 – 126.0.0.0 10.0.0.0 – 10.255.255.255 255.0.0.0 Class B 128.0.0.0 – 191.0.0.0 172.16.0.0 – 172.31.255.255 255.255.0.0 Class C 192.0.0.0 – 223.0.0.0 192.168.0.0 – 192.168.255.255 255.255.255.0 Class D 224.0.0.0 – 239.0.0.0 x 255.255.255.255 Class E 240.0.0.0 – 255.0.0.0 x 255.255.255.255 First 8 Bits = Network ID A First 8 Bits = Network ID First 16 Bits = Network ID B First 12 Bits = Network ID First 24 Bits = Network ID C First 16 Bits = Network ID = Bits 8 + 8+ 8 + 8 (11111111.11111111.11111111.00000000) = 32 bits = 8 bits (11111111) = Mask Bit (ON) 0 = Unmask Bit (OFF)
- 2. CIDR : Total ON bits of whole Subnet Mask. X : Total ON bits in octets of Class wise Host ID. Y : Total OFF bits in octets of Class wise Host ID. Overall Subnets : 2X Possible Hosts per Subnet : 2Y – 2 Valid Subnets : 256 – calculated Subnet. Possible Hosts : IPs in range of Valid Subnets. Click here for details in IP Addressing & Subnetting.
- 3. Solved Examples Q : Give the number of Total Subnets, Hosts per Subnet, Valid Subnets and Possible Hosts of the IP - 10.0.0.0 with a subnet mask of 255.255.240.0 Answer : X : 8+4+0=12 Y : 0+4+8=12 CIDR : 8+8+4+0=20 Overall Subnets : 2x = 64(6) x 64(6) = 4096(12) Hosts per Subnet : 2y-2 = 4096(12)-2 = 4094 Valid Subnets : 256-Calculated Subnet = 256-240 = 16 Possible Hosts : 10.0.0.0 - 10.0.15.255 10.0.16.0 - 10.0.255.255 10.0.32.0 - 10.0.47.255
- 4. 7 6 5 4 3 2 1 0 2 2 2 2 2 2 2 2 128 64 32 16 8 4 2 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 Decimal 56… then Binary…? Let’s count the on & off bits in the value 56.. So, as counting in descending order.. The ON bits in value – 54 are – 32, 8, 4, 128 & 64 can’t be subtracted from 56. 2, & 1. Hence.. 56 – 32 = 14 Now what can be subtracted from 14..? Therefore the Binary value of 56 is Exactly.. 14 – 8 = 6 00101110 Therefore.. 6 – 4 = 2 2 – 2= 0
- 5. Decimal Hexadecimal 0 0 First divide the DEC number by 16. 1 1 Now write down the remainder . The remainder is converted to HEX. 2 2 Now divide Quotient by 16. 3 3 Again, the remainder is converted to HEX. 4 4 And it keeps on like this.. 5 5 Finally, we put the numbers together. The remainder from our first division goes into the unit place. 6 6 Our remainder from our second division goes into the tens place. And it keeps on like this.. 7 7 So now our answer is ready…!! 8 8 9 9 DEC 56.. Then HEX..? 10 A So the answer is 56 / 16 = 3.8 8 38 3 / 16 = 0.3 3 11 B 12 C 13 D 4096 256 16 1 14 E 3 8 15 F Hex 38 = (8 x 1) + (3 x 16) = 56