An IPv4 host has the address 172.16.45.14 with a /30 subnet mask. To which subnetwork address does this host belong on the 172.16.45.0 network?

Introduction / Context:
In IPv4 subnetting, a very common task is to determine the network or subnetwork address to which a particular host belongs. This question uses a private Class B range with a Classless Inter Domain Routing (CIDR) mask of /30, which is frequently used for point to point links in Cisco certification exams and real networks. Understanding how to calculate block sizes and network boundaries for a /30 subnet is essential for routing, access control lists, and troubleshooting connectivity problems.

Given Data / Assumptions:

• Host IPv4 address: 172.16.45.14.
• Prefix length: /30, which corresponds to a subnet mask of 255.255.255.252.
• We assume standard classless addressing with no special summarisation tricks.
• We need the network (subnetwork) address that this host falls into on the 172.16.45.0 network.

Concept / Approach:
To find the subnetwork, we focus on the octet where subnetting occurs. A /30 mask means 30 bits are network bits and 2 bits are host bits. In dotted decimal, /30 is 255.255.255.252. The block size in the affected octet is calculated as 256 minus the mask value in that octet. For 252 this block size is 4. Each subnet therefore spans 4 addresses, with the first being the network address and the last being the broadcast address. The usable host addresses are the two addresses between them. We find the largest multiple of the block size that is less than or equal to the host value in that octet, which gives the network address range.

Step-by-Step Solution:
1. Write the subnet mask: /30 = 255.255.255.252. 2. Identify the varying octet. The first three octets 172.16.45 are fixed by the mask. The subnetting happens in the fourth octet. 3. Compute the block size in the fourth octet: 256 - 252 = 4. 4. List the possible network starts in the fourth octet: 0, 4, 8, 12, 16, 20, and so on. 5. Find where 14 fits. It lies between 12 and 16, so the subnet starts at 12. 6. Therefore the subnetwork address is 172.16.45.12. The usable host addresses in this subnet are 172.16.45.13 and 172.16.45.14, and the broadcast address is 172.16.45.15.

Verification / Alternative check:
We can verify by converting the last octet to binary and applying the mask. The last octet 14 in binary is 00001110. The mask 252 in binary is 11111100. A bitwise AND of these two values is 00001100 which is decimal 12. Therefore, the network portion for the host 172.16.45.14 with mask 255.255.255.252 is indeed 172.16.45.12, which confirms our earlier block size method. Both decimal block method and binary method agree, so the answer is consistent and reliable.

Why Other Options Are Wrong:
172.16.45.0: This is the first network of the /30 ranges on that third octet, but it covers hosts .1 and .2, not .14.
172.16.45.4: This subnet covers addresses 172.16.45.4 to 172.16.45.7. The host 14 is not inside this range.
172.16.45.8: This subnet covers addresses 172.16.45.8 to 172.16.45.11 only. Again, 14 is outside this range.
172.16.45.16: This would be the start of the next subnet after the one beginning at 12, and would contain host addresses 17 and 18, not 14.

Common Pitfalls:
Learners often forget to compute the block size correctly and may treat the subnet as if it uses a larger host range. Another frequent mistake is to misinterpret the broadcast address as the network address. Some candidates also mix up hexadecimal or binary conversions, or assume that a /30 subnet offers more than two usable hosts. Remember that /30 gives a block size of 4 addresses, so there are always only two usable host addresses. Keeping the block size rule clear in your mind prevents many such miscalculations during exams and troubleshooting.

Final Answer:
172.16.45.12