Planning and designing

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Questions derived from the 640-801 - Cisco Certified Network Associate (single-exam option) Cisco Self-Test Software Practice Test.

Objective: Planning and designing
SubObjective: Design an IP addressing scheme to meet design requirements

Item Number: 640-801.1.2.3
Multiple Answer, Multiple Choice

You have a Class C network using the subnet mask 255.255.255.224. All the available subnets that were created have been used except for one. You need to take the remaining subnet (205.13.137.64/27) and use VLSM to create four smaller subnets to use on point-to-point links within your network.

Which three of the following addresses are valid host addresses that can be created on any of these new subnets? (Choose three.)

1. 205.13.137.69/30
2. 205.13.137.13/30
3. 205.13.137.88/30
4. 205.13.137.74/30
5. 205.13.137.85/30

Answer:
A. 205.13.137.69/30
D. 205.13.137.74/30
E. 205.13.137.85/30

Tutorial:
The correct answers are 205.13.137.69/30, 205.13.137.74/30 and 205.13.137.85/30. Here is how we solve the problem.

Take the one remaining subnet (205.13.137.64) and split it into multiple, smaller subnets. Each of these smaller subnets needs two IP addresses. The current mask that we are using is 255.255.255.224 (/27). This means all the binary digits in the first three octets and the first three binary digits in the fourth octet belong to the network portion of the address. The remaining five binary digits of the fourth octet belong to the host portion of the address. Because the dividing line between the network portion of the address and the host portion of the address is in the fourth octet, this is where we will place our focus. Below is a binary representation of the fourth octet.

There is only one subnet with 30 hosts remaining. You need more subnets with two hosts on each. To accomplish this, use the 64 subnet and take three more binary digits of the fourth octet to use for the network portion of the address. Now there are a total of six binary digits for the network portion and two binary digits for the host portion. The first three binary digits of the fourth octet are always set at 010 because you are using the 64 subnet.

The fourth through the sixth bits can be used to create different subnets with two bits remaining for creating two hosts on each subnet. The valid subnets are:

• 68
• 72
• 76
• 80
• 84
• 88
• 92

Each subnet has two hosts. For example, the 88 subnet has hosts 89 and 90 (205.13.137.89/30 and 205.13.137.90/30). Also, each subnet has a network address and a host address. These are not valid host addresses. For example, the 88 subnet has a network address of 88 and a broadcast address of 91(205.13.137.88/30 and 205.13.137.91/30). Because all bits of the IP addresses except for the last two are used for the network portion of the address, the new mask is /30 (255.255.255.252).

205.13.137.69/30 is on the 68 subnet, 205.13.137.74/30 is on the 72 subnet and 205.13.137.85/30 is on the 84 subnet. None of the other options is a valid host address on any of the newly created subnets.

For more information about subnetting, see IP Addressing and Subnetting for New Users at
http://www.cisco.com/warp/public/701/3.html

Reference:
1. INTRO Student Guide v1.0a - Volume 2 - IP Subnetting
- Network Address Planning