Explanation/Reference:
Section: Addressing and Routing Protocols in an Existing Network Explanation Explanation:
Of the available choices, you are least likely to allocate a /25 subnet. The largest broadcast domain in this scenario contains 50 hosts. A /25 subnet can contain up to 126 assignable hosts. In this scenario, allocating a /25 subnet would reserve half the 192.168.51.0/24 network for a single virtual LAN (VLAN).
The total number of hosts for which you need addresses in this scenario is 86. Therefore, you would only need to use half the /24 subnet if all 86 hosts were residing in the same VLAN.
You should begin allocating address ranges starting with the largest group of hosts to ensure that the entire group has a large, contiguous address range available. Subnetting a contiguous address range in structured, hierarchical fashion enables routers to maintain smaller routing tables and eases administrative burden when troubleshooting.
You are likely to use a /26 subnet. In this scenario, the largest VLAN contains 50 hosts. If you were to divide the 192.168.51.0/25 subnet into two /26 subnets, the result would be two new subnets capable of supporting up to 62 assignable hosts: the 192.168.51.0/26 subnet and the 192.168.51.64/26 subnet.
Therefore, you should start subnetting with a /26 network. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.64/26 subnet to SwitchD's VLAN.
You are likely to use a /27 subnet. The nextlargest broadcast domain in this scenario is the SwitchC subnet, which contains 20 hosts. If you were to divide the 192.168.51.0/26 subnet into two /27 subnets, the result would be two new subnets capable of supporting up to 30 assignable hosts: the 192.168.51.0/27 subnet and the 192.168.51.32/27 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.32/27 subnet to SwitchC's VLAN.
You are likely to use a /28 subnet. The nextlargest broadcast domain in this scenario is the SwitchB subnet, which contains 10 hosts. If you were to divide the 192.168.51.0/27 subnet into two /28 subnets, the result would be two new subnets capable of supporting up to 14 assignable hosts: the 192.168.51.0/28 subnet and the 192.168.51.16/28 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.16/28 subnet to SwitchB's VLAN.
You are likely to use a /29 subnet. The nextlargest broadcast domain in this scenario is the SwitchA subnet, which contains four hosts. If you were to divide the 192.168.51.0/28 subnet into two /29 subnets, the result would be two new subnets capable of supporting up to six assignable hosts: the 192.168.51.0/29 subnet and the 192.168.51.8 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.8/29 subnet to SwitchA's VLAN.
You are likely to use a /30 subnet. The final subnet in this scenario is the link between RouterA and RouterB, which contains two hosts. If you were to divide the 192.168.51.0/29 subnet into two /30 subnets, the result would be two new subnets capable of supporting two assignable hosts each: the
192.168.51.0/30 subnet and the 192.168.51.4/30 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.4/30 subnet to the link between RouterA and RouterB. This would leave the 192.168.51.0/30 subnet unallocated. However, you could further divide the 192.168.51.0/30 subnet into single /32 host addresses that could then be used for loopback IP addressing on the routers.
Reference:
CCDA 200-310 Official Cert Guide, Chapter 8, IPv4 Address Subnets, pp. 302-310 CCDA 200-310 Official Cert Guide, Chapter 8, Plan for a Hierarchical IP Address Network, pp. 311-312 Cisco: IP Addressing and Subnetting for New Users