CCNA – Subnetting 2

Note: If you are not sure about Subnetting, please read our Subnetting Tutorial – Subnetting Made Easy.

Question 1

Explanation

60 hosts < 64 = 2⁶ -> we need a subnet mask of at least 6 bit 0s -> “/26″. The question requires “wasting the fewest addresses” which means we have to allow only 62 hosts-per-subnet -> B is correct.

Note: Answer D is not correct because 192.168.1.56 is not a network address (/26 -> increment: 64)

Question 2

Explanation

All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24″-> /22 is the best choice -> D is correct.

Question 3

Explanation

We need to summarize 4 subnets so we have to move left 2 bits (2² = 4). In this question we can guess the initial subnet mask is /24 because 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0 belong to different networks. So “/24″ moves left 2 bits -> /22.

Question 4

Explanation

We need to remember the default subnet mask of class B is 255.255.0.0. Next, the company requires a minimum of 300 sub-networks so we have to use at least 512 sub-networks (because 512 is the minimum power of 2 and greater than 300). Therefore we need to get 9 bits for network mask (2⁹=512), leaving 7 bits for hosts which is 2⁷= 128 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.

We can increase the sub-networks to 1024 ( 1024 = 2¹⁰), leaving 6 bits for hosts that is 2⁶= 64 > 50 hosts. This scheme satisfies the requirement, too -> E is correct.

Notice: The question asks “The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet” but this is a typo, you should understand it as “”The company needs a minimum of 300 sub-networks and a minimum of 50 host addresses per subnet”.

Question 5

Explanation

To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (2⁴ = 16). Notice that we have to move the borrowed bits to the left (not right) because we are summarizing.

The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0, 172.16.3.0… are different networks).

Therefore “move 4 bits to the left” of “/24″ will give us “/20″ -> C is the correct answer.

Question 6

Explanation

First we should notice that different VLANs must use different sub-networks. In this case Host A (172.16.1.126) and Host B (172.16.1.129) are in different VLANs and must use different sub-networks. Therefore the subnet mask in use here should be 255.255.255.128. In particular, it is 172.16.1.0/25 with 2 sub-networks:

+ Sub-network 1: 172.16.1.0 -> 172.16.1.127 (assigned to VLAN 1)
+ Sub-network 2: 172.16.1.128 -> 172.16.1.255 (assigned to VLAN 2)

-> B is correct.

The IP address 172.16.1.25, which is in the same sub-network with host A so it can be assigned to VLAN 1 -> C is correct.

To make different VLANs communicate with each other we can configure sub-interfaces (with a different IP address on each interface) on the LAN interface of the router -> F is correct.

Question 7

Explanation

“The network administrator needs to address seven LANs” means we have 7 subnets < 8 = 2³, so we need to borrow 3 bits from the host part (to create 8 subnets). But the title said “subnet 0 is not being used”, we cannot use the first so in fact we only have 8 – 1 = 7 subnets. We are using class C address block which has 8 bits 0 (the default subnet mask of class C is 255.255.255.0), so the number of bit 0 left is 8 – 3 = 5. Therefore the hosts per subnet will be 2⁵ – 2 = 30 -> E is correct.

Note: There was some confusion here. The title only said “subnet 0 is not being used”, but it did not mention that the command “no ip subnet-zero” is used. Maybe that means we can still use the last subnet (called the All-Ones subnet). In other words, maybe the title implied that “the subnet 0 can be used but the network administrator ignored it for safe”. Thus the last subnet can still be used.

Question 8

Explanation

Network 172.1.4.0/25 and network 172.1.4.128/25 can be grouped to a single network 172.1.4.0/24

Network 172.1.4.0/24 + Network 172.1.5.0/24 + Network 172.1.6.0/24 + Network 172.1.7.0/24 can be grouped to a single network 172.1.4.0/22 because we have all 4 subnetworks so we can move left 2 bits (2²=4).

Question 9

Explanation

We have 4 routes learned by EIGRP:

D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1

These subnets are all /30 (as it says “192.168.25.0/30 is subnetted, 4 subnets”. We have 4 successive subnets = 2² so we can go back 2 bits -> the summarized subnet mask is 30 – 2 = 28 and the summarized network is 192.168.25.16.