Subnetting Notes

Class A 0-126 Class B 128-191 Class C 192-223 2^7 = 128 2^6 = 64 2^5 = 32 2^4 = 16 2^3 = 8 2^2 = 4 2^1 = 2 2^0 = 1 There are only 8 subnets in any octet. 128 = 128 192 = 128+64 224 = 128+64+32 240 = 128+64+32+16 248 = 128+64+32+16+8 252 = 128+64+32+16+8+4 254 = 128+64+32+16+8+4+2 255 = 128+64+32+16+8+4+2+1

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Given: 192.168.1.0 255.255.255.0 Create 3 subnets that contains at least 1x50 ip and 2x14 usable ip addresses, no waste address space. Subnet#1 192.168.1.0 1) How many bits to create 50 hosts network? 128 64

| 32 16 8 4 2 1

Network | max hosts = 63 Next usable subnet? Subnet#2 192.168.1.64(0+64) What is the mask for subnet#1? 128+64 = 192 Subnet#1 192.168.1.0/26(CIDR) 255.255.255.192 What is the broadcast address for subnet#1? 192.168.1.63(64-1) What actual hosts: 192.168.1.1 – 192.168.1.62 2) How many bits to create 14 hosts network? 128 64 32 16 | 8 4 2 1 Network | max hosts = 15 Next usable subnet? Subnet#3 192.168.1.80(64+16) What is the broadcast address for subnet#2? 192.168.1.79(80-1) What is the mask for subnet#2? 128+64+32+16 = 240 Subnet#2 192.168.1.64/28(CIDR) 255.255.255.240 What actual hosts: 192.168.1.65 – 192.168.1.78 3) How many bits to create 14 hosts network? 128 64 32 16 | 8 4 2 1 Network | max hosts = 15 Next usable subnet? Subnet#4 192.168.1.96(80+16) What is the broadcast address for subnet#3? Page 2 of 4

192.168.1.95(96-1) What is the mask for subnet#2? 128+64+32+16 = 240 Subnet#2 192.168.1.64/28(CIDR) 255.255.255.240 What actual hosts: 192.168.1.81 – 192.168.1.95 Next available for future use: 192.168.1.96 – 192.168.1.255 Exercise: 192.168.156.0/22 begin 192.168.157.0 128 hosts /24 192.168.157.0 255.255.255.0 192.168.157.1 0.0.0.255 90 hosts /25 192.168.158.0 255.255.255.128 192.168.158.1 0.0.0.127 60 hosts /26 192.168.159.0 255.255.255.192 192.168.159.1 0.0.0.63 60 hosts /26 192.168.158.128 255.255.255.192 192.168.158.129 0.0.0.63 60 hosts /26 192.168.158.192 255.255.255.192 192.168.158.193 0.0.0.63 30 hosts /30 192.168.159.64 255.255.255.224 192.168.159.65 0.0.0.31 2 2 2 2

hosts hosts hosts hosts

/30 /30 /30 /30

192.168.159.96 255.255.255.252 192.168.159.97 0.0.0.3 192.168.159.100 255.255.255.252 192.168.159.101 0.0.0.3 192.168.159.96 255.255.255.252 192.168.159.98 0.0.0.3 192.168.159.100 255.255.255.252 192.168.159.102 0.0.0.3

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6 Steps to subnet 1) How many sub-networks do you need? 128 for 2 networks from 1 bit: 0 and 128 192 for 4 networks from 2 bit: 0, 64, 128, and 192. 2) How many bits did you have to use? 1 bits for 2 networks. 2 bits for 4 networks. 3) What is your subnet mask? 128 = 128 for 2 networks 192+64 = 192 for 4 networks. 4) What is your block-size? 128 for 2 networks. 64 for 4 networks. 5) What are your subnets? From network section. 0+64 = 64 64+64 =128 128+64 = 192 192+64 = 256 6) What are the numbers of hosts and IP ranges? From host section. 2^7-2 = 126 for 2 networks 2^6-1 = 62 for 4 networks Example: Basic classfull 1) 2) 3) 4) 5) 6)

44 networks 6 bits 252 4 0,4,8,12…252 2

VLSM classless Start with /26 = 192 1) 4 more subnets 2) 2 bits for 4 more subnet 3) 240 = 192+32+16 4) 16 5) 192,208,224.240 6) 193-206,209-222,225-238,241-254

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