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Addresses in a network withand without subnetting

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Default mask and subnet mask

Example 1

What is the subnetwork address if thedestination address is 200.45.34.56 and thesubnet mask is 255.255.240.0?

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Solution

11001000 00101101 00100010 00111000

11111111 11111111 11110000 00000000

11001000 00101101 00100000 00000000

The subnetwork address is 200.45.32.0.

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Example 2

A company is granted the site address201.70.64.0 (class C). The company needssix subnets. Design the subnets.

Solution

The number of 1s in the defaultmask is 24 (class C).

/205

Solution (Continued)

The company needs six subnets. This number6 is not a power of 2. The next number that isa power of 2 is 8 (23). We need 3 more 1s inthe subnet mask. The total number of 1s inthe subnet mask is 27 (24 + 3).

The total number of 0s is 5 (32 − 27). Themask is

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Solution (Continued)

11111111 11111111 11111111 11100000or

255.255.255.224

The number of subnets is 8.The number of addresses in each subnetis 25 (5 is the number of 0s) or 32.

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Example 2

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Example 3

A company is granted the site address181.56.0.0 (class B). The company needs1000 subnets. Design the subnets.

Solution

The number of 1s in the default mask is 16(class B).

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Solution (Continued)

The company needs 1000 subnets. Thisnumber is not a power of 2. The next numberthat is a power of 2 is 1024 (210). We need 10more 1s in the subnet mask.The total number of 1s in the subnet mask is26 (16 + 10).The total number of 0s is 6 (32 − 26).

/2010

Solution (Continued)

The mask is11111111 11111111 11111111 11000000

or255.255.255.192.

The number of subnets is 1024.The number of addresses in each subnet is 26

(6 is the number of 0s) or 64.

/2011

Example 3

/2012

If we don’t subnet and use 255.255.255.0 as our subnet mask then we use all of our IP addresses on one network.

This is not an efficient use of our Class C network.

For each network you loose two IP address, one for the network and one for the broadcast.

11111111.11111111.11111111.00000000 = 255.255.255.0 = class C mask

11111111.11111111.11111111.11111111 = 255.255.255.255 = broadcast

Class C Subnet Effective Effective#Bits Mask Subnets Hosts

0 255.255.255.0 1 2541 255.255.255.128 2 1262 255.255.255.192 4 623 255.255.255.224 8 304 255.255.255.240 16 145 255.255.255.248 32 66 255.255.255.252 64 2

/2013

192.168.100.96 Call this 96 network255.255.255.240 subnet mask

192.168.100.97 thru 192.168.100.111 are valid IP address for the 96 network255.255.255.240 = 11111111.11111111.11111111.11110000

11110000 – 96 network111100011111001011110011111101001111010111110110111101111111100011111001111110101111101111111100111111011111111011111111 - broadcast /2014

We’ll use the 192.168.100.0 network /2015

192.168.100.0 is our network and we have 254 total IP addresses

Building C – 40 users – 62 hosts – 255.255.255.192 – subnet 1Building B – 20 users – 30 hosts – 255.255.255.224 – subnet 2Building A – 08 users – 14 hosts – 255.255.255.240 – subnet 3Firewall – 02 users – 04 hosts – 255.255.255.252 – subnet 4

subnet 1 192.168.100.0 0 is the network255.255.255.192 63 is the broadcast

subnet 2 192.168.100.64 64 is the network255.255.255.192 95 is the broadcast

subnet 3 192.168.100.96 96 is the network255.255.255.240 111 is the broadcast

subnet 4 192.168.100.112 112 is the network255.255.255.252 115 is the broadcast

192.168.100.116 is the next usable network/2016

/2017

Slash notation is also called CIDR

notation.

/2018

Example 4

A small organization is given a block with the beginningaddress and the prefix length 205.16.37.24/29 (in slashnotation). What is the range of the block?

Solution

The beginning address is 205.16.37.24. To find thelast address we keep the first 29 bits and change thelast 3 bits to 1s.

Beginning:11001111 00010000 00100101 00011000Ending :11001111 00010000 00100101 00011111

There are only 8 addresses in this block.

/2019

Example 5

We can find the range of addresses in Example 4 byanother method. We can argue that the length of the suffixis 32 − 29 or 3. So there are 23 = 8 addresses in this block.If the first address is 205.16.37.24, the last address is205.16.37.31 (24 + 7 = 31).

/2020

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