Subnetting IPv4 Addresses

Please read the following text carefully BEFORE answering the questions below. You will be expected to have completed these questions as they form part of the marks for this part of your log book submission.

Step 1 - IP Address Basics

If your organization has a /8 (class "A") IP network address, the first octet (8 bits) is assigned by InterNIC and your organization can use the remaining 24 bits to define up to 16,777,214 hosts on your network. This is a lot of hosts! It is not possible to put all of these hosts on one physical network without separating them with routers and subnets. A workstation may be on one network or subnet and a server may be on another network or subnet. When the workstation needs to retrieve a file from the server it will need to use its subnet mask to determine the network or subnet that the server is on. The purpose of a subnet mask is to help hosts and routers determine the network location where a destination host can be found. Refer to the following table to review IP address classes, default subnet masks and the number of networks and hosts that can be created with each class of network address.

Cls

1st Octet Decimal Range

1st Octet High Order Bits

Network / Host ID (N=Network, H=Host)

Default Subnet Mask

Number of Networks

Hosts per Network (usable addresses)

A

1 - 126*

0

N.H.H.H

255.0.0.0

126 (27 - 2)

16,777,214 (2 24 - 2)

B

128 - 191

1 0 

N.N.H.H

255.255.0.0

16,384 (214)

65,534 (2 16 - 2)

C

192 - 223

1 1 0

N.N.N.H

255.255.255.0

2,097,152 (221)

254 (2 8 - 2)

D

224 - 239

1 1 1 0

Reserved for Multicasting

E

240 - 254 

1 1 1 1 0

Experimental, used for research

Step 2 - The "ANDing" process.

Hosts and routers use the "ANDing" process to determine if a destination host is on the same network or not. The ANDing process is done each time a host wants to send a packet to another host on an IP network. If you want to connect to a server, you may know the IP address of the server you want to connect to or you may just enter the host name (e.g. engweb.info) and a Domain Name Server (DNS) will convert the host name to an IP address.

First the source host will compare (AND) its own IP address to its own subnet mask. The result of this ANDing is to identify the network where the source host resides.

Secondly it will compare the destination IP address to its own subnet mask. The result of the 2nd ANDing will be the network that the destination host is on.

If the source network address and the destination network address are the same they can communicate directly.

If the results are different then they are on different networks or subnets and will need to communicate through routers or may not be able to communicate at all.

ANDing depends on the subnet mask. The subnet mask for a /24 (Class C) network is 255.255.255.0 or 11111111.111111111.111111111.00000000. This is compared to the source IP address bit for bit. The first bit of the IP address is compared to the first bit of the subnet mask and the second bit to the second etc. If the two bits are both ones, then the ANDing result is a ONE. If the two bits are a zero and a one or two zeros then the ANDing result is a ZERO. Basically this means that a combination of 2 ones results in a ONE, anything else is a zero. The result of the ANDing process is the network or subnet number that the source or destination address is on.

Step 3 - Two /24 (Class C) networks using the default subnet mask.

This example will show how a /24 subnet mask can be used to determine which network a host is on. A  subnet mask does not break an address into subnets. If the given subnet mask is used then the network is not being "subnetted". 

Host X (source) on network 200.1.1.0 has an IP address of 200.1.1.5 and wants to send a packet to host Z (destination) on network 200.1.2.0 and has an IP address of 200.1.2.8. All hosts on each network are connected to hubs or switches and then to a router. 

Remember that with a /24 (Class C) network address your Regional Internet Registry (RIR) assigns the first 3 octets (24 bits) as the network address so these are two different class C networks. This leaves one octet (8 bits) for hosts so each class C network could have up to 254 hosts (28  = 256 - 2 = 254).

The ANDing process will help the packet get from host 200.1.1.5 on network 200.1.1.0 to host 200.1.2.8 on network 200.1.2.0 using the following steps.

3a)   Host X compares its own IP address to its own subnet mask using the ANDing process

Host X IP address 200.1.1.5

11001000.00000001.00000001.00000101

Subnet Mask 255.255.255.0

11111111.11111111.11111111.00000000

ANDing Result (200.1.1.0)

11001000.00000001.00000001.00000000

NOTE:

The result of step 3a of the ANDing process is the network address of host X which is 200.1.1.0

3b    Next host X compares the IP address of the Host Z destination to its own subnet mask using the ANDing process.

Host Z IP address 200.1.2.8

11001000.00000001.00000010.00001000

Subnet Mask 255.255.255.0

11111111.11111111.11111111.00000000

ANDing Result (200.1.2.0)

11001000.00000001.00000010.00000000

NOTE:

The result of step 3b ANDing process is the network address of host Z which is 200.1.2.0.

3C     Host X compares the ANDing results from step A and the ANDing result from step B and they are different. Host X now knows that host Z is not in its Local Area Network (LAN) and it must send the packet to its "Default Gateway" which is the IP address of the router interface of 200.1.1.1 on network 200.1.1.0. The router will then repeat the ANDing process to determine which router interface to send the packet out.

Step 4 - One  /24 (Class C) network using a Custom subnet mask.

This example uses a single /24 network address (200.1.1.0) and will show how a /24 subnet mask can be used to determine which subnetwork (or subnet) a host is on and to route packets from one subnetwork to another. Remember that with a /24 network address your RIR assigns the first 3 octets (24 bits) as the network address. This leaves 8 bits (one octet) for hosts so each class C network could have up to 254 hosts (28  = 256 - 2 = 254).

Perhaps you want less than 254 host (workstations and servers) all on one network and you want to create 2 sub-networks and separate them with a router for security reason or to reduce traffic. This will create smaller independent broadcast domains and can improve network performance and increase security since these subnetworks will be separated by a router. Assume you will need at least 2 subnetworks and at least 50 hosts per subnetwork. Since you only have one Class C network address you have only 8 bits in the fourth octet available for a total of 254 possible hosts, you must create a Custom Subnet mask. You will use the custom subnet mask to "BORROW" bits from the host portion of the address. The following steps will help accomplish this:

The first step to "subnetting" is to determine how many subnets are needed. In this case you will need 2 subnetworks. To see how many bits you should borrow from the host portion of the network address, add the bit values from right to left until the total (decimal value) is greater than the number of subnets you will need. Since we need 2 subnets, add the one bit and the two bit which equals three. This is over the number of subnets we need, so we will need to borrow at least two bits from the host address starting from the left side of the octet that contains the host address.

Network address: 200.1.1.0

 

 

 

 

 

 

 

 

4th octet Host address bits:

1

1

1

1

1

1

1

1

Host address bit values (from right)

128

64

32

16

8

4

2

1

(Add bits starting from the right side (the 1 and the 2) until you get more than the number of subnets needed)
Once we know how many bits to borrow we take them from the left side of the first octet of the host address. Every bit we borrow from the host leaves fewer bits for the hosts. Even though we increase the number of subnets, we decrease the number of hosts per subnet. Since we need to borrow 2 bits from the left side, we must show that new value in our subnet mask. Our existing default subnet mask was 255.255.255.0 and our new "Custom" subnet mask is 255.255.255.192. The 192 comes from the value of the first two bits from the left (128 + 64 = 192). These bits now become 1s and are part of the overall subnet mask. This leaves 6 bits for host IP addresses or 26 = 64 hosts per subnet.

4th Octet borrowed bits for subnet:

1

1

1

1

1

1

1

1

Subnet bit values: (from left side)

128

64

32

16

8

4

2

1

     With this information you can build the following table. The first two bits are the Subnet binary value. The last 6 bits are the host bits. By borrowing 2 bits from the 8 bits of the host address you can create 4 subnets with 64 hosts each. The 4 networks created are the "0" net, the "64" net, the "128" net and the "192" net. 

Subnet No.

Subnet bits borrowed Binary value

Subnet bits Decimal Value

Host bits possible binary values (range) (6 bits)

Subnet / Host Decimal range

Useable?

Subnet #0

00

0

000000 - 111111

0 - 63

YES

Subnet #1

01  

 

64 000000 - 111111

64 - 127

YES

Subnet #2

10   

 

128 000000 - 111111

128 - 191

YES

Subnet #3

11 

 

192 000000 - 111111 

192 - 254

YES
Notice that the first subnet always starts at 0 and, in this case, increases by 64 which is the number of hosts on each subnet. One way to determine the number of hosts on each subnet or the start of each subnet is to take the remaining host bits to the power of 2. Since we borrowed two of the 8 bits for subnets and have six bits left, the number of hosts per subnet is 26 or 64. Another way to figure the number of host per subnet or the "increment" from one subnet to the next is to subtract the subnet mask value in decimal (192 in the fourth octet) from 256 (which is maximum number of possible combinations of 8 bits) which equals 64. This means you start at 0 for the first network and add 64 for each additional subnetwork. If we take the second subnet (the 64 net) as an example the IP address of 200.1.1.64 cannot be used for a host ID because it is the "network ID" or "wire address" of the "64" subnet (host portion is all zeros) and the IP address of 200.1.1.127 cannot be used because it is the broadcast address for the "64" net (host portion is all ones).

Step 5 - One /24 (Class C) network using a Custom Subnet Mask.

Task: Use the following information and the previous examples to answer the following subnet related questions.
Explanation: Your company has applied for and received a /24 (Class C) network address of 197.15.22.0. You want to subdivide your physical network into 8 subnets, which will be interconnected by routers. You will need at least 25 hosts per subnet. You will need to use a custom subnet mask and will have a router between the subnets to route packets from one subnet to another.
Determine the number of bits you will need to borrow from the host portion of the network address and then the number of bits left for host addresses. 

Fill in the table below and answer the following questions:

 

Subnet No.

Subnet bits borrowed (Binary value)

Subnet bits (Decimal value)

Host bits possible binary values range

Subnet / Host Decimal range

Broadcast Address

Subnet #0

 

 

 

 

 

Subnet #1

 

 

 

 

 

Subnet #2

 

 

 

 

 

Subnet #3

 

 

 

 

 

Subnet #4

 

 

 

 

 

Subnet #5

 

 

 

 

 

Subnet #6

 

 

 

 

 

Subnet #7

 

 

 

 

 

Notes:

 

QUESTIONS: Use the table you just developed above to help answer the following questions:

  1. Which octet(s) represent the network portion of a /24 (Class C) IP address?

  1. Which octet(s) represent the host portion of a /24 (Class C) IP address?

  1. What is the binary equivalent of the /24 (Class C) network address in the scenario (197.15.22.0)?
    Decimal Network address: __________ . __________ . __________ . __________
    Binary Network address:    __________ . __________ . __________ . __________

  2. How many high-order bits were borrowed from the host bits in the fourth octet?

  1. What subnet mask must you use (show the subnet mask in decimal and binary)?
    Decimal Subnet mask: __________ . __________ . __________ . __________
    Binary subnet mask:    __________ . __________ . __________ . __________

 

  1. What is the maximum number of subnets that can be created with this subnet mask?

  1. What is the maximum number of useable subnets that can be created with this mask?

  1. How many bits were left in the 4th octet for host IDs ?

  1. How many hosts per subnet can be defined with this subnet mask?

  1. What is the maximum number of hosts that can be defined for all subnets with this scenario (assuming you cannot use the lowest and highest subnet numbers and cannot use the lowest and highest host ID on each subnet)?

  1. Is 197.15.22.63 a valid host IP address with this scenario?

  1. Why or why not ?

  1. Is 197.15.22.160 a valid host IP address with this scenario?

  1. Why or why not?

  1. Host "A" has an IP address of 197.15.22.126. Host "B" has an IP address of 197.15.22.129. Are these hosts on the same subnet?

  1. Is a router required to communicate between Host "A"  and Host "B" ?