Question

Answer the following questions.

Show all work leading to your answer.

1. On a network with an IP address of 140.133.28.72 and a subnet mask of 255.248.0.0, what is the network ID?
2. You have decided to create 254 subnets on your Class B network. What subnet mask will you use to accomplish this? What is the maximum number of hosts you can assign to any single subnet?
3. Your workstation's IP address is 10.35.88.12, and your supervisor's workstation's IP address is 10.35.91.4. When you send data from your workstation to your supervisor's workstation, what is the most likely IP address of the first default gateway that will accept and interpret your transmission?

Answer 1

1. 140.133.28.72 - host ip address

255.248.0.0 - subnet mask, which in binary:

11111111.11111000.00000000.00000000

Thus the number of network bits is 13

So the most significant 13 bits in the given above ip address is the network id.

The first octet(8 bits) 140 will be the first octet in network id as well. In the next octet left most 5 bits are considered for network id.

The second octet 133 in binary is 1000101. The leftmost 5 bits 10000. The remaining 3 bits in that octet will be zero. The next two octets will be zero.

Therefore the second octet will be 128.

140.128.0.0 is the network id.

2. In class B the default mask is /16.

To create 254 subnets, 8 bits are required (2^8=256). This 8 bits are borrowed from 16 bits of host part, thus making the number of bits for host part to 8(16-8=8).

With 24 bits (16+8=24 bits) for network part, the subnet mask is /24, which is 255.255.255.0 in decimal dotted notation.

And 8 bits for host part, the number of hosts available in each subnet is 2^8-2= 256-2=254(2 ip addresses one for network ip address and other for broadcast ip address).

3. Work station ip address: 10.35.88.12

Supervisor work station ip address: 10.35.91.4

The ip address that starts with 10 belongs to class A network. The subnet mask is 255.0.0.0

Thus the first default gateway to receive is: 10.0.0.0