Subnetting Part-01

Question-01

Given the network diagram below, and the results of the sh ip route command, on which interface will the destination IP address 17.102.36.153 be found? For your answer simply enter the corresponding letter (i.e. A).

A) Fa0/1

B) Fa0/2

C) Fa0/3

router# sh ip route

Codes: L - local, C - connected

[output cut]

C	17.102.36.128/25 is directly connected, FastEthernet0/1
L	17.102.36.129/32 is directly connected, FastEthernet0/1
C	17.102.37.0/26 is directly connected, FastEthernet0/2
L	17.102.37.1/32 is directly connected, FastEthernet0/2
C	17.102.36.0/25 is directly connected, FastEthernet0/3
L	17.102.36.1/32 is directly connected, FastEthernet0/3        

The correct answer is: A

There are 2^h-2 usable hosts in a network, where h is the number of host bits. Find the usable hosts in each block until you find block which holds the given IP

Question-02

Which of the following contains more network bits? For your answer simply write 'A' or 'B'.

The correct answer is: B

Convert the CIDR and the network mask into binary, and from there see which has more 1's.

Question-03

What is the maximum number of valid hosts one will have from the network 192.168.66.0/25

The correct answer is: 126

There are 2^h-2 usable hosts in a network, where h is the number of hot bits.

Question-04

The following shows a 'router on a stick' scenario. Given the partial?show run?command information from the router:

interface FastEthernet1/1.2
encapsulation dot1Q 2
ip address 192.168.0.1 255.255.255.192
interface FastEthernet1/1.3
encapsulation dot1Q 3
ip address 192.168.0.65 255.255.255.192
interface FastEthernet1/1.4
encapsulation dot1Q 4
ip address 192.168.0.129 255.255.255.192        

You plan to add two new VLANS: VLAN 5 and VLAN 6. VLAN 5 should support 14 hosts, while VLAN 6 should support 9 hosts. Which of the following answers makes sense. For your answer simply give the letter of the correct choice (e.g. 'A').

A) 192.168.0.209 255.255.255.240 on Fa1/1.6

B) 192.168.0.46 255.255.255.240 on Fa1/1.5

C) 192.168.0.43 255.255.255.240 on Fa1/1.5

D) 192.168.0.131 255.255.255.240 on Fa1/1.5

The correct answer is: A

Make sure the subnet mask provides a large enough block for the needed hosts. Once you find the needed netmask, make sure that the possible IP address is not within any of the previously existing networks.

Question-05

What is the broadcast address of the network 10.39.246.248/15

The correct answer is: 10.39.255.255

The broadcast address is one less than the next network address. Count in block sizes. The block size is either 256 minus the netmask, or 2^32-slash.

Question-06

The following shows a network with two subnets. A device on the network is misconfigured. Select the letter of the device which has a an invalid IP address. Your answer should simply be the letter of the deivce (i.e. A).

The correct answer is: A

Find the block size and make sure that all the IPs fall within that block size. Then check to make sure no broadcast and network addresses are being assigned to hosts. Remember, the usable hosts are those between the broadcast and the network addresses.

Question-07

Which letter VLAN would the host 192.168.33.1 be placed? For your answer simply select the letter of the Vlan (i.e. 'A')

The correct answer is: A

Count the powers of two?until you find the first power of two which is above the needed amount of hosts for the larger network. This is the amount of host bits you will need.

Question-08

What is the maximum number of valid hosts one will have from the network 192.168.209.0 ?? 255.255.255.248

The correct answer is: 6

There are 2^h-2 usable hosts in a network, where h is the number of hot bits.

Question-09

Enter the wildcard mask for the given subnets (i.e. 0.0.0.255).

The correct answer is: 0.0.0.63

A wildcard is an inverted subnet mask. For example, a subnet mask of 255.255.255.0 has a wildcard mask of 0.0.0.255. The binary bits are inverted. This is useful for things such as OSPF routing.

Question-10

You have a subnetwork, 192.168.186.0/24. It is divided into subnet A and subnet B. Your boss wants to add a third subnet, C, with 1 hosts. Is this possible? If yes, provide the network address of the new subnet you will be adding (i.e. 192.168.2.14). If it is not possible, type 'no' as your answer.

The correct answer is: 192.168.186.192

Calculate how much space subnet A and B take up. To calculate how many ip addresses a subnet takes up, look at the network mask. In this case we are working in the last octet. Subtract the netmask value of the last octet from 256 to find how many IPs each subnet mask uses. For example, subnet A uses 256-128=128 IP addresses. Do the same for subnet B, to get 64 IPs.

Knowing how many IPs we have used up, we need to find how many we need. We need 1 hosts. We need to find a block size minus two which will fit our needed hosts. That block size is how many IPs we need. In this case we need 4 IP addresses.

The network address for subnet C would be the first IP of the next block for the netmask of subnet B, In this case 192.168.186.192.

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