Question

Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3. Suppose all of the interfaces in each of these three subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support up to 63 interfaces, Subnet 2 is to support up to 95 interfaces, and Subnet 3 is to support up to 16 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints.

Answer 1

Answer:

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We have given three subnets Subnet 1 , Subnet 2 and Subnet 3

Prefix = 223.1.17/24

/24 --------> means First 3 octets are for the network and the last octet is for the host ( means for subnetting)

We know that in IP address there are four octets ( octets means each represents with 8 bits)

There is a total of 32 bits ( 4 octets * each octet contains 8 bits = 32 bits). Here we have given /24 means 8 *3 ( first 3 octets are the network part and the last octet for the host part and in the last octet subnetting will be done).

Now Subnet 1:

Required to support up to 63 interfaces

63 = 2^6

6 bits are needed here so First IP address = 223.1.17. 0 /26

Last IP address = 223.1.17.62/26

So the IP addresses of Subnet1 = 223.1.17.0/26 to 223.1.17.62/26

Subnet 2:

Required 95 interfaces

95 is close to 128 so 128 = 2^7

So Here 7 bits are required

Therefore First IP address = 223.1.17.63/25

Last IP address = 223.1.17.157/25

So the IP addresses of Subnet2 = 223.1.17.63/25 to 223.1.17.157/25

Subnet3:

16 interfaces

16 = 2^4, hence 4 bits are required

First IP address of Subnet 3 = 223.1.17.158/28 Last IP address = 223.1.17.178/28