# IP Subnetting, Variable Length Subnet Masks (VLSMs), & Troubleshooting IP

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Introduction to Cisco Networking Technologies Study Guide & CCNA INTRO: Introduction to Cisco Networking Technologies Study Guide (Exam 640-821): Chapter 3

### Why subnet?

Reduced network traffic

### Why subnet?

Optimized network performance

### Why subnet?

Simplified management

### Why subnet?

Facilitated spanning of large geographical distances

### IP Subnet-Zero

allows you to use the first and last subnet in your network design; turned this command on by default

### Classful Routing

all nodes in the network use the same subnet mask

### Variable Length Subnet Masks (VLSMs)

each network segment can use a different subnet mask

32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address

/25 (with ip subnet-zero), /26, /27, /28, /29, /30

10000000 = 128

11000000 = 192

11100000 = 224

11110000 = 240

11111000 = 248

11111100 = 252

### How many subnets?

2^x where x is the number of masked bits (or 1's)

### How many valid hosts per subnet?

2y - 2 where y is the number of unmasked bits (or 0&#039;s)

### What are the valid subnets?

The number just before the next subnet; the broadcast of the last subnet is always 255

### What are the valid hosts in each subnet?

The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

255.255.255.128

2

126

0 & 128

127 & 255

### Given 192.168.10.0/28, What are the valid hosts?

.1 - .126 & .129 - .254

4

16,382

### Given 172.16.0.0/18, What are the valid subnets?

0.0, 64.0, 128.0, and 192.0

63.255, 127.255, 191.255, 255.255

### Given 172.16.0.0/18, What are the valid hosts?

.0.1 - .63.254, .64.1 - .127.254, .128.1 - .191.254, .192.1 - .255.254

the 10.32 subnet; The broadcast is 10.63

The subnet is 172.16.64.0. The broadcast must be 172.16.127.255

The subnet is 172.16.32.0, and the broadcast must be 172.16.63.25

This subnet address must be in the 172.16.32.0 subnet, and the broadcast must be 172.16.47.255

The subnet is 172.16.45.12, with a broadcast of 172.16.45.15

### Variable Length Subnet Masks (VLSMs)

Create many networks using subnet masks of different lengths from one network

### Classful Routing

all interfaces within the classful address space have the same subnet mask

RIPv1 and IGRP

### Classless Routing Protocols

RIPv2, EIGRP, or OSPF

### Variable Length Subnet Masks (VLSMs)

Use different size masks on each router interface

### 4 Troubleshooting Steps

Ping 127.0.0.1 (loopback), Ping the local host, Ping Default Gateway (router), Ping remote destination

Traceroute, Arp -a, Ipconfig /all

Tracert, Show ip arp

### Unable to Ping Loopback

IP stack failure; Reinstall TCP/IP

### Unable to Ping Local Host

Problem with the NIC; Replace the NIC

### Unable to Ping Default Gateway

Local physical network problem between NIC and router

### Unable to Ping Remote Destination

Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

### Able to ping but still unable to communicate

Possible DNS problem

### 192.168.100.25/30

/30 is 255.255.255.252. The valid subnet is 192.168.100.24, broadcast is 192.168.100.27, and valid hosts are 192.168.100.25 and 26

### 192.168.100.37/28

/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts 33-46

### 192.168.100.66/27

/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64. The host is in the 32 subnet, broadcast address of 63. Valid host range of 33-62.

### 192.168.100.17/29

/29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid host 17-22

### 192.168.100.99/26

/26 is 255.255.255.192. The fourth octet has a block size of 64. 0, 64, 128. The host is in the 64 subnet, broadcast of 127. Valid host 65-126

### 192.168.100.99/25

/25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid host 1-126

### You have a Class B network and need 29 subnets. What is your mask?

This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21

/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0, 8, 16. The host is in the 8 subnet, broadcast is 15

### How many hosts are available with a Class C /29 mask?

/29 is 255.255.255.248, which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

### What is the subnet for host ID 10.16.3.65/23?

/23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet, the broadcast address is 16.3.255

### What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?

A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B or C network address? Not at all. The amount of hosts bits would never change.

### You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?

A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).

### What is the subnetwork address for a host with the IP address 200.10.5.68/28?

This is a pretty simple question. A /28 is 255.255.255.240, which means that our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the 64 subnet.

### The network address of 172.16.0.0/19 provides how many subnets and hosts?

A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits but provides 13 host bits, or 8 subnets, each with 8,190 hosts.

### Which two statements describe the IP address 10.16.3.65/23?

The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.

### If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?

A /30, regardless of the class of address, has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0, 4, 8, 12, 16, etc. Address 14 is obviously in the 12 subnet.

### On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?

A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.

### What is the subnetwork number of a host with an IP address of 172.16.66.0/21?

A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.

### You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to router interface?

A /29 (255.255.255.248), regardless of the class of address, has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN, including the router interface.

### You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?

A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.

### You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?

A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.

### You need to subnet a network that has five subnets, each with at least 16 hosts. Which classful subnet mask would you use?

You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts—this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.

### A network administrator is connecting hosts A and B directly thorough their Ethernet interfaces as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts?

First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easily solution is just to set both masks to 255.255.255.0 (/24).

### If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?

A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on—which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.

### Using the following illustration, what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered valid for this question

A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet, not starting at subnet-zero. 16, 32, 48, 64, 80, 96, 112, 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address, which is 143).The valid host range for the eight subnet (not using subnet zero) is 129 through 142. The last available IP address available in the eighth subnet is 142.

### Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again, the zero subnet should not be considered valid for this question

A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.

### Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?

A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets, each with 30 hosts. However, if the command ip subnet-zero is not used, then only 6 subnets would be available for use.

### You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?

A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.

### Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host ID's on the LAN interface attached to the router?

The routers IP address on the E0 interface is 172.16.2.1/23, which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet, the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.

### To test the IP stack on your local host, which IP address would you ping?

To test the local stack on your host, ping the loopback interface of 127.0.0.1

Example: