Refer to the exhibit. Packets are flowing from 192.168.10.1 to the destination at IP address 192.168.20.75. Which next hop will the router select for the packet?
B is correct because
/24 starts from 192.168.20.0 - 192.168.20.254
/25 startes from 192.168.20.0- 192.168.20.127 so the question is destined to 192.168.20.75 which is within this range.
The answer is B
The longest prefix match is used when we have several routes leading to the destination address. The longest prefix match must include the destination IP address. Both 10.10.10.2 and 10.10.10.11 serve as a valid next hop to reach 192.168.20.75 but we will use 10.10.10.1, since the route associated with it has a longer prefix length (more specific)
10.10.10.1 for the route 192.168.20.0/26 (192.168.20.1-192.168.20.62) INCORRECT
10.10.10.11 for route 192.168.20.0/25 (192.168.20.1-192.168.20.126) CORRECT
10.10.10.12 for route 192.168.20.0/27 (192.168.20.1-192.168.20.30) INCORRECT
10.10.10.14 for route 0.0.0.0/0 INCORRECT (Default route is only used if there is no match for the destination's IP address in the routing table)
The router will select 10.10.10.12 as the next hop for the packet.
The routing table shows that there are four routes to the destination subnet 192.168.20.0/24. The first three routes are for more specific subnets of 192.168.20.0/26, 192.168.20.0/27, and 192.168.20.0/25. When there are multiple routes to the same destination subnet, the router will select the route with the longest prefix match. In this case, the route with the longest prefix match is 192.168.20.0/27, which has a prefix length of 27. The next hop for this route is 10.10.10.12.
The fourth route in the routing table is for the default route, which is used when the destination IP address does not match any of the other routes in the table. The default route has a prefix length of 0, which means that it matches any destination IP address. The next hop for the default route is 10.10.10.14.
Since the destination IP address in this case matches the route with the longest prefix match (192.168.20.0/27), the router will select the next hop for that route, which is 10.10.10.12.
The 192.168.20.0/27 match indicates that the subnet is subnet 20.0 not subnet 20.64, ergo an address of 192.168.20.75 would not be in the subnet for that prefix.
To determine the next hop for the packet from 192.168.10.1 to the destination IP address 192.168.20.75, we need to find the most specific matching route in the routing table. Let's analyze the given routing table:
192.168.20.0/26 (90/24513456) via 10.10.10.1
192.168.20.0/24 (120/5) via 10.10.10.2
192.168.20.0/27 (90/4123710) via 10.10.10.12
192.168.20.0/25 (90/14464211) via 10.10.10.11
0.0.0.0/0 (1/0) via 10.10.10.14
The destination IP address 192.168.20.75 falls within the following subnets:
192.168.20.0/26
192.168.20.0/24
192.168.20.0/27
192.168.20.0/25
Among these, the most specific match (with the longest prefix) is the /27 subnet (3rd entry). Thus, the router will select the next hop 10.10.10.12 for the packet. The correct answer is:
C. 10.10.10.12
Hamish88 wrote: ".... The highest IP address that can fall in the 192.168.20.0/27 range is 192.168.20.30."
What do you mean by hamish88 didn't justify the answer? He literally explained that the highest useable IP address range in the /27 block is 192.168.20.30.
Can you even read? read again.
/27 means 5 bits in the host portion. 2^5= 32 (-2) = 30 assignable IP addresses so the range will be 192.168.20.1 to 192.168.20.30. The question says : to the destination at IP address 192.168.20.75 therefore it can't be C
Bro 192.168.20.75 is not in the range of /27.
In a /27 subnet, there are 32 IP addresses in total. However, the first address (192.168.20.0) is the network address, and the last address (192.168.20.31) is the broadcast address. Therefore, these two addresses are not usable for host assignment.
B is correct
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