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ccna salary per month.To determine that the address of its type (192.168.1.132) lies between two subnet numbers (192.168.1.128 and 192.168.1.192 ), this means that steps 5 and 6 need to be performed for both subnets (and possibly the one before them). However, this is ignored here, and steps 5 and 6 are performed for all subnets, as this provides good practice. Don't forget that the directed broadcast address is 1 smaller than the next subnet number, and all addresses between the subnet number and the directed broadcast address are host addresses. Table 6.10 shows the results after performing steps 5 and 6, from which it can be seen that 192.168.1.132 is a host address whose subnet number and directed broadcast address are 192.168.1.128 and 192.168.1.191o, respectivelyTable 6.10 Network Addresses, Directed Broadcast Addresses, and Host Addresses in 192.168.1.0/26 Network Addresses Host Addresses Fixed Ho Broadcast Addresses192.168.1.0 192.168.1.1.192.168.1.62 192.168.1.63192.168.1.64 192.168.1.65.192.168.1.126 192.168.1.127192.168.1.128 192.168.1.129.192.168.1.190 192.168.1.191192.168.1.192 192.168.1.193.192.168.1.253 192.168.1.2546.4.4 Example of determining the type of IP address 3The previous two examples were both very simple because the given address belongs to a Class C network. This example uses a Class B network to make the problem a little more complicated.In step 1, the address 172.16.5.0255.255.254.0 is given, which can also be expressed as 172.16.5.0/23, and most candidates were unable to correctly determine its type in the exam. First, guess whether the address is a network address, a directed broadcast address, or a host address, and then perform the previous steps to find out the answer. However, if you remember the trick described earlier, you can determine that the address is not a directed broadcast address based on the fact that the host portion of the address is even, reducing the likelihood to two.In step 2, you need to find the relevant byte-the byte where the network portion and host portion boundaries are located. Here, the boundary is located in byte 3 of the subnet mask (254) It is important to note that the entire byte 4 belongs to the host portion. In step 3, you need to identify the adjacent sub