Understanding Subnetting in Software

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Understanding Subnetting
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The lowest-order bit will be the bit on the far right side The work area is shown in the following table, and you can see that with the two networks you have a rst valid address for each network of 106401 and 1012801
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First Octet Second Octet Third Octet Fourth Octet (Decimal) (Binary) (Binary) (Binary) Calculation Original IP
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10 10 10 10 10 0 00000000 01000000 10000000 11000000 0 00000000 00000000 00000000 00000000 0 00000000 00000001 00000001 00000000 106400 1012801
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Now that you have calculated the rst valid address for each of the two networks, you will need to calculate the broadcast address The broadcast address is the address that any system will send data in order to ensure that each system on the network reads the data To calculate the broadcast address, you will enable all of the host bits and get the outcome in the following table:
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First Octet Second Octet Third Octet Fourth Octet (Decimal) (Binary) (Binary) (Binary) Original IP
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10 10 10 10 10 0 00000000 01111111 10111111 11000000 0 00000000 11111111 11111111 00000000 0 00000000 11111111 11111111 00000000
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Calculation
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10127255255 10191255255
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As you can see, with all the host bits enabled, if you convert that to decimal, you get 10127255255 and 10191255255 for the broadcast addresses of your two networks Notice that the rst two bits from the left in the second octet have not been changed in this entire process, but they are used in the conversion of that octet from binary to decimal
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5:
Subnetting and Routing
Remember that it is illegal to assign a host an address that has all host bits set to 0 or all host bits set to 1 These are illegal, because all host bits set to 0 is reserved for the network ID and all host bits set to 1 is reserved for the broadcast address Now that you have calculated the new subnet mask, the network ID, the rst valid address, and the broadcast address for your two new subnets, the only additional information you need is the last valid address that may be assigned to hosts on each subnet To calculate the last valid host address of each subnet, simply subtract one from the broadcast address by disabling the low-order bit (the far rightmost host bit) To view what the binary and decimal representation looks like for our last valid address of each network take a look at the table below
First Octet Second Octet (Decimal) (Binary) Original IP
10 10 10 10 10 0 00000000 01111111 10111111 11000000
Third Octet Fourth Octet (Binary) (Binary)
0 00000000 11111111 11111111 00000000 0 00000000 11111110 11111110 00000000
Calculation
10127255254 10191255254
You have now calculated all of the information required to con gure the two physical network segments that you have created The following table summarizes the con guration for each of the two network segments, and Figure 5-2 displays how these two network segments will be con gured
Network First Valid Last Valid ID Address Address Subnet 1 Subnet 2
106400 1012800 106401 1012801
Broadcast Address
Subnet Mask
25519200 25519200
10127255254 10127255255 10191255254 10191255255
Understanding Subnetting
FIGURE 5-2 106403
Subnetting a Class A network into two segments
106401 Computer A
1012801 Computer B
106400 Network
Router
1012800 Network
Computer C 1012803 Server 106402
EXERCISE 5-1
Subnetting a Class A Address
In this exercise you will determine the ve pieces of information needed for a class A network that is being divided into four network segments
Question No 1
The network ID of the class A address is 120000 Take a few pieces of paper and calculate the new subnet mask, network ID, rst valid address, last valid address, and the broadcast address of the four subnets Once you have calculated your answer, ll in the following table Refer to the subnetting example in this chapter when you need to as a guide to help you calculate the answers for this exercise
5:
Subnetting and Routing
Network First Valid Last Valid ID Address Address Subnet 1 Subnet 2 Subnet 3 Subnet 4
Broadcast Address
Subnet Mask
Once you have your answers, check your work against the answer table under Answer No 1 on page 305
The Work
Given that we want to have four subnets, we will need to take 3 bits; to verify that answer we enable the bits in our temp work area to get the number 4 and then count the bits from right to left that are used to get to that last enabled bit 00000100 gives us the number 4, and we have used 4 bits from right to left to get that value We can also use our formula of 2masked bits 2 number of networks to verify that the number is correct, which means we get a formula of 23 2 6 We get six networks and we require only four networks This just means that we have 2 extra networks that will not be used If we were to use only 2 bits, we would have only two networks The next thing to calculate is the new subnet mask used by these four new networks The new subnet mask is determined by masking additional bits in the original subnet mask The original subnet mask was 1111111100000000000000000 0000000n and by masking 3 additional bits we get the following:
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