a) How many bits of an IPv4 address are required to define 16 subnets? How many

ID: 3772101 • Letter: A

Question

a) How many bits of an IPv4 address are required to define 16 subnets? How many bits of the IPv4 address are required to define 12 subnets?

b) Assume that you have been assigned the 204.135.1.0/24 address block and were asked to subnet this block into subnets so that a subnet can support no less than 20 hosts. What is the maximum number of subnets that you can create? Enumerate the subnets that you had created.

c) Aggregate the following set of four /24 network addresses to the highest degree possible.

d) Aggregate the following set of four /24 network addresses to the highest degree possible. 196.48.132.0/24 196.48.133.0/24 196.48.134.0/24 196.48.135.0/24 210.58.146.0/24 210.58.147.0/24 210.58.148.0/24 210.58.149.0/24 e) If you succeed in aggregating four /16 network addresses, what will be the prefix length of the aggregated block? The prefix length is expressed in / notation.

Explanation / Answer

Answer:

To calculate the number of subnets that exists in the given network is defined by the mask bits. Depending on those, the number of subnets is calculated.

If the mask bits are not provided, then need to choose the host bits in the network.

To get the 16 subnets, the formula to calculate through the host bits is given as,

So, the number of host bits is obtained as,

So, among the 32-bit version of IPv4, 4-bits are used for host addresses, the remaining bits specify the network mass.

So, the number bits that are used to specify the 16-subnets is.

Therefore, 28 mask bits are required to define the subnets, or 4 host bits are required to define the subnets.

To get the 12 subnets, there are no definitely defined bits. For simple calculations, of the subnets or host addresses the value should be power of 2.

Since, 12-subnets are present in-between 8 and 16 subnets of the predefined subnets with respect to the existing bits.

If 8 – subnet is chosen then other 4 subnets will be missing.

If 16 – subnet is chosen then 4 subnets will be wasted.

For, simple understanding, consider 16 – subnet value.

So, from bit (a), 28 mask bits are required to define the subnets, or 4 host bits are required to define the subnets.

The maximum number of subnets is given for the provided address is as follows:

The number of mask bits its 24.

So, the remaining host bits are:

Since, given network address is a Class C.

So, the number of subnets that can be formed is,

The number of hosts each subnet is obtained as,

Therefore, in each subnet there will be 254 hosts which are greater than the 20 hosts.

The subnets would be:

Subnet number

Network address

Broadcast address

Host addresses range in each subnet

204.135.1.1

204.135.1.254

204.135.1.254 - 204.135.1.254

204.135.2.1

204.135.2.254

204.135.2.254 - 204.135.2.254

255

204.135.255.1

204.135.255.254

204.135.255.254 - 204.135.255.254

Given, addresses are 196.48.132.0/24 196.48.133.0/24 196.48.134.0/24 196.48.135.0/24

Convert the given IP addresses into equivalent binary format.

196.48.132.0

11000100.0011000.10000100.000000

196.48.133.0

11000100.0011000.10000101.000000

196.48.134.0

11000100.0011000.10000110.000000

196.48.135.0

11000100.0011000.10000111.000000

Observe the bits in the third octet, the first 6 bits are common, the rest of the two bits are being changing.

So, the common prefix is 196.48.132.0 - 11000100.0011000.100001 00.000000

The CIDR aggregation is 196.48.132.0/24

d, e)

Given, addresses are 210.58.146.0/24 210.58.147.0/24 210.58.148.0/24 210.58.149.0/24

Convert the given IP addresses into equivalent binary format.

210.58.146.0

11010010. 00111010.10010010.000000

210.58.147.0

11010010. 00111010.10010011.000000

210.58.148.0

11010010. 00111010.10010100.000000

210.58.149.0

11010010. 00111010.10010101.000000

Here, the given set can-not summarized to single /23.

210.58.146.0 - 11010010. 00111010.1001001 0.000000

210.58.148.0 - 11010010. 00111010.10010100.000000

The CIDR aggregation is 210.58.146.0 /23 and 210.58.148.0 /23