Subnetting Exercise =================== INSTRUCTIONS: Subnet the group of IP add
ID: 3730934 • Letter: S
Question
Subnetting Exercise
===================
INSTRUCTIONS: Subnet the group of IP address in the class C 214.186.30.0 network into six subnets. Each subnet will be used by a maximum of 12 twelve hosts. Determine the subnet addresses, range of host addresses, and subnet mask for your subnets. You may use a subnet mask that provides for more than six subnets. Answer the following questions about the six subnets.
A. What subnet mask did you choose?
B. What are the six network addresses for your subnets?
1.
2.
3.
4.
5.
6.
C. What are the ranges of host addresses for each subnet?
1.
2.
3.
4.
5.
6.
D. What is the broadcast address for each subnet?
1.
2.
3.
4.
5.
6.
E. Using the subnet mask you selected above, demonstrate the “anding” process used by a computer to determine which of your subnetted networks a remote host with the IP address of 214.186.30.167 is on.
Explanation / Answer
Given IP address is 214.186.30.0 which a class C address with default mask 255.255.255.0. In class C, first 3 bits are reserved, following 21bits are for network address & last 8 bits are for the host.
Subnetting is the process of designating some high-order bits from the host part as part of the network prefix and adjusting the subnet mask appropriately.
In this example we need to create 6 subnets and thus would require atleast 3 bits(2^3 subnets) from host part to be moved to network part. Now from last 8 bits, 3bits are for subnet and 5bits are for hosts. Thus bits 25,26,27 will be now for network address.
For adjusting subnet mask set these three bits to 1, thus subnet mask would become 255.255.255.224. Now network prefix can be calculated by ANDing IP address and the subnet mask.
Binary form Dot-decimal notation
IP address 11010110.10111010.00011110.00000000 214.186.30.0
Subnet mask 11111111.11111111.11111111.11100000 255.255.255.224
Network prefix 11010110.10111010.00011110.00000000 214.186.30.0
Host part 00000000.00000000.00000000.00000010 0.0.0.0
A) Subnet Mask: 255.255.255.224
B)
Since we reserved 3 bits, there are 8 subnets possible:
11010110.10111010.00011110.00000000 214.186.30.0
11010110.10111010.00011110.00100000 214.186.30.32
11010110.10111010.00011110.01000000 214.186.30.64
11010110.10111010.00011110.01100000 214.186.30.96
11010110.10111010.00011110.10000000 214.186.30.128
11010110.10111010.00011110.10100000 214.186.30.160
11010110.10111010.00011110.11000000 214.186.30.192
11010110.10111010.00011110.11100000 214.186.30.224
C)
Range of host addresses:
214.186.30.0 214.186.30.0 - 214.186.30.31
214.186.30.32 214.186.30.32 - 214.186.30.63
214.186.30.64 214.186.30.64 - 214.186.30.95
214.186.30.96 214.186.30.96 - 214.186.30.127
214.186.30.128 214.186.30.128 - 214.186.30.159
214.186.30.160 214.186.30.160 - 214.186.30.191
214.186.30.192 214.186.30.192 - 214.186.30.223
214.186.30.224 214.186.30.224 - 255.255.255.255
D)
IPv4 uses the last address within a network, for broadcast transmission.Thus for these subnets broadcast addresses would be:
214.186.30.0 214.186.30.31
214.186.30.32 214.186.30.63
214.186.30.64 214.186.30.95
214.186.30.96 214.186.30.127
214.186.30.128 214.186.30.159
214.186.30.160 214.186.30.191
214.186.30.192 214.186.30.223
214.186.30.224 214.186.30.255
E)
IP Address : 214.186.30.167 => 11010110.10111010.00011110.10100111
Subnet Mask: 255.255.255.224=> 11111111.11111111.11111111.11100000
------------------------------------------------------
11010110.10111010.00011110.10100000 => 214.186.30.160 => Subnet 6 => Which is correct as per list of host addresses in problem C
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