ccna rs 200-301

ccna rs 200-301.For example, if you have a Class C network, and subnetting is done, you need 5 subnet bits and 4 host bits, or a total of 9 bits. Unfortunately, however, Class C networks only have 8 host bits by default, so this scenario is not feasible. In this case, one Class B network or two Class C networks are required. As another example, if you have a Class C network, and subnetting requires 3 subnet bits and 4 host bits, or 7 bits in total, the Class C network has 8 host bits by default, and you only need 7 bits, which provides some flexibility in that you can use the one with more outgoing to create more subnets, or you can have the subnet contain more hosts.Back to the original example: Class C network 192.168.1.0 is assigned, and there are 14 subnets in the network, each of which contains no more than 14 hosts.Each subnet contains no more than 14 hosts. . 2S>= 14 (subnets): S = 4, which will create 16 subnets. . 2H-2>= 14 (hosts): H = 4, the subnet can contain up to 14 hosts.. S + H v= 8 (default number of host bits for a class C network): 4 + 4 <= 8O Here is a detailed description of these steps. First, you need to find a curtain of 2 that is greater than or equal to the number of subnets needed. In this example, it needs to be the 4th curtain of 2: 2 4> =14, which satisfies the subnetting requirement since only 14 subnets are needed (only 14 network segments). If subnet 0 is not available, 5 instead of 4 subnet bits will be needed, since the first and last 4^ nets are not available. However, it is assumed here that subnet 0 is available.Next, the number of host bits needed (i.e., where H) needs to be calculated using the formula 2 2>=14. In this example, 24.2 = 14, so 4 host bits are needed to support 14 hosts. Don't forget to subtract 2, since the first address in the subnet is the network address and the last address is the directed broadcast address.Finally, since a Class C network is assigned, there are only 8 host bits by default. You need to make sure that the number of bits used to divide the subnet and the number of host bits does not exceed this default value. In this example, 4 + 4 = 8, so there is no problem. If the total number of bits exceeds 8, you need two Class C networks or one Class B network. If the total number of bits is less than 8, the extra bits can be used to create more subnets and/or hosts. If there are extra bits, it is necessary to look deeper into the network and thus decide whether more subnets should be created or whether each subnet should be able to support more hosts depending on future expansion.6.3.3 Step 3: Determining the Subnet MaskThe toughest task is complete and the remaining 4 steps are easy. You now know the number of subnet bits needed, but when subnetting, the network portion of the subnet mask will contain both network bits and subnet bits. As mentioned earlier, the default number of network bits for Class A addresses is 8,Class B is 16, and Class C is 24.Just add these bits to the number of subnet bits to get the total number of bits in the network part. In this example, 24 + 4 = 28. To simplify