qr code generator c# tutorial Determining IP Address Components in Objective-C

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Determining IP Address Components
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Remember these six steps when trying to determine whether an IP address is a network, host, or directed broadcast number
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Based on these six steps, you should be able to figure out whether your address is a host, network, or broadcast address Note that these six steps are somewhat similar to the six steps used in the IP Address Planning section However, the steps in this section are for CCNA test purposes and the steps in the previous section are for design purposes
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Example #1 for Determining IP Address Components
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To help you out with the six steps, take a look at an example as an illustration In step 1, you have an IP address and subnet mask Assume that this is 192168137 255255255224 (or 192168137/27) This is a Class C network Remember this shortcut: if it s an odd address, it s either a directed broadcast or host address Therefore, you know this address is not a network address
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As a shortcut, remember that a network number is always an even number, a broadcast is always an odd number, and a host address can be either
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With this knowledge, you can typically eliminate some answers when looking for the right one or ones
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In step 2, you need to find the interesting octet in the subnet mask This is the octet where the boundary exists between network and host bits In this example, this is the fourth octet: 224 In step 3, you need to find the increment by which network numbers are increasing To perform this step, subtract the interesting octet from 256: 256 224 = 32 Therefore, there are 32 addresses in each network, and each network is incrementing by 32 in the interesting octet (fourth octet)
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Remember the shortcut of guring out the multiples that network numbers are incrementing by in the
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interesting octet: 256 subnet mask value = increment value
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7: IP Addressing and Subnetting
In step 4, write down the network numbers starting with the first subnet and work your way up Here is the list of network numbers for our example: 19216810, 192168132, 192168164, 192168196, 1921681128, 1921681160, 1921681192, and 1921681224 In this example, there are eight subnets Mathematically, this makes sense There are 32 addresses per subnet, with a total of 256 addresses (0 255) in a Class C network 256 32 = 8! Remember that the interesting octet in the subnet mask will be the subnet number in the last subnet of the IP class address In step 5, list the directed broadcast address beside each network number And in step 6, list the host addresses for each network Remember that the broadcast address for a network is one number less than the next network number and that the host addresses are any IP addresses between the network and directed broadcast addresses Table 7-9 shows the completion of steps 5 and 6 Considering Table 7-9, the host address of 192168137 is a host address, since it falls in the rage of host addresses for subnet 192168132/27
When you are taking the CCNA exam, don t build the entire table Instead, list the network numbers until you have a network number greater than the address in the question Once this is done, for the last three network numbers, list
the directed broadcast and host addresses, and then you ll know the answer to the exam question In the above example, these networks would be 19216810, 192168320 and 192168640
TABLE 7-9
Network, Broadcast, Directed, and Host Addresses of 19216810/27
Network Address
19216810 192168132 192168164 192168196 1921681128 1921681160 1921681192 1921681224
Host Address
19216811 192168130 192168133 192168162 192168165 192168194 192168197 1921681126 1921681129 1921681158 1921681161 1921681190 1921681193 1921681222 1921681225 1921681254
Directed Broadcast Address
192168131 192168163 192168195 1921681127 1921681159 1921681191 1921681223 1921681255
Determining IP Address Components
Example #2 for Determining IP Address Components
Let s look at another example to help clarify the six steps For step 1, you are given the following address and subnet mask: 1921681132 255255255192 (/26), which is a Class C address Remember the shortcut from earlier: if the last octet is even, it s either a network or host address Therefore, you know it s not a directed broadcast address In the second step, you need to find the interesting octet in the subnet mask This is the last octet For a Class C network, this will always be the last octet The value in this mask is 192, indicating that the first two high-order bits in the octet are part of the network component and the last six low-order bits are the host component In step 3, you need to determine by what number the network addresses are increasing To do this step, subtract the value in the subnet mask s interesting octet from 256: 256 192 = 64 Therefore, network addresses are incrementing by 64 numbers and each network contains 64 addresses: a network address, a directed broadcast address, and 62 host addresses Remember that since the interesting octet is in the fourth octet, the network addresses are increasing by 64 in the interesting (fourth) octet In step 4, write down the network numbers In our example, this gives you four networks: 19216810, 192168164, 1921681128, and 1921681192 In the interesting octet, 2 bits are used for networking and 6 bits for host addresses With 2 bits of networking, this gives you four networks (22 = 4), and with 6 bits of host addresses and 64 addresses in a network, it s 26 = 64 The address in question, 1921681132, is between two networks: 1921681128 and 1921681192 This means that you should have to perform steps 5 and 6 only on these two networks, and possibly the network before it However, go ahead and complete steps 5 and 6 for all of the networks since this is good practice Remember that the directed broadcast address for a network is one number less than the next network number and that the addresses between the network and directed broadcast addresses are host addresses Table 7-10 shows the addressing for the Class C address Our address, 1921681132, is a host address based on this table, where its network number is 1921681128 and its directed broadcast is 1921681191
TABLE 7-10
Network, Broadcast, Directed, and Host Addresses of 19216810/26
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