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NEW QUESTION: 1
View the Exhibit.

You administer the network shown above. You want to summarize the networks connected to RouterA so that a single route is inserted into RouterB's routing table.
Which of the following is the smallest summarization for the three networks?
A. 172.16.1.0/16
B. 172.16.1.0/23E.
C. 172.16.1.0/22
D. 172.16.1.0/25
E. 172.16.1.0/18
Answer: C
Explanation:
Explanation/Reference:
Section: Addressing and Routing Protocols in an Existing Network Explanation Explanation:
The smallest summarization for the three networks connected to RouterA is 172.16.1.0/22, which is equivalent to a network address of 172.16.1.0 and a subnet mask of 255.255.252.0. In this scenario, the Class B 172.16.0.0/16 network has been divided into 256 /24 subnets. Three of the first four subnets in the Class B range have been assigned to network interfaces on RouterA: 172.16.0.0/24, 172.16.1.0/24, and
172.16.3.0/24. Absent from the network assignments is the 172.16.2.0/24 subnet. However, there is no way to summarize the address range without including the 172.16.2.0/24 subnet. Therefore, the smallest summarization you can create would summarize four subnets into a single /22 subnet.
A /22 subnet creates 64 subnetworks capable of supporting 1,022 assignable host IP addresses each. The assignable address range of the 172.16.0.0/22 subnet begins with 172.16.0.1 and ends with 172.16.3.255.
This range includes all possible assignable IP addresses in the /24 subnets that are directly connected to RouterA. It also includes all possible assignable IP addresses in the 172.16.2.0/24 subnet.
Subnetting a contiguous address range in structured, hierarchical fashion enables routers to maintain smaller routing tables and eases administrative burden when troubleshooting. Conversely, a discontiguous IP version 4 (IPv4) addressing scheme can cause routing tables to bloat because the subnets cannot be summarized. Summarization minimizes the size of routing tables and advertisements and reduces a router's processor and memory requirements.
Summarizing the three /24 networks with a /16 subnet would create too large of a summarization, because the /16 subnet contains the entire Class B range of 172.16.0.0 IP addresses. The first assignable IP address in the 172.16.0.0/16 range is 172.16.0.1. The last assignable IP address is 172.16.255.255. The range would therefore summarize 256 /24 subnets, not four.
Summarizing the three /24 networks with a /18 subnet would create too large of a summarization. A /18 subnet creates four possible subnets containing 16,382 assignable host IP addresses each. The first assignable IP address in the 172.16.0.0/18 range is 172.16.0.1. The last assignable IP address is
172.16.63.255. The range would therefore summarize 64 /24 subnets, not four.
Summarizing the three /24 networks with a /23 subnet would create too small of a summarization. A /23 subnet creates 128 possible subnets containing 510 assignable host IP addresses each. The first assignable IP address in the 172.16.0.0/23 range is 172.16.0.1. The last assignable IP address is
172.16.1.255. This range would therefore exclude the 172.16.3.0/24 subnet connected to RouterA.
Summarizing the three /24 networks with a /25 subnet would not work, because a /25 subnet divides the
172.16.0.0/24 subnet instead of summarizing. A /25 subnet creates 512 possible subnets containing 126 assignable host IP addresses each. The first assignable IP address in the 172.16.0.0/25 range is
172.16.0.1. The last assignable IP address is 172.16.0.127. This subnet would therefore contain only half of one the subnets that is directly connected to RouterA.
Reference:
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

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