IP Address Class, Network ID, Subnet ID, and Host ID: 25-Comp-B5 Computer Communications, December 2016, Q6

Question 6: IP Address Class, Network ID, Subnet ID, and Host ID (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
IP address	198.53.147.45
Subnet mask	255.255.255.224

Find. The address class, the (classful) network ID, the subnet ID, and the host ID, with the reasoning behind each.

Approach. Classify the address from its leading octet against the classful ranges; find the classful network ID by zeroing the host portion of that class's default mask; then find the subnet ID by bit-wise ANDing the full address with the GIVEN subnet mask, and the host ID as the bits left over once the subnet mask's 1-bits are removed.

Classify the address. The first octet is $198$, which falls in the Class C range $192\text{-}223$: $$\boxed{\text{Class C}}$$
Classful network ID. A default Class C mask is $255.255.255.0$ (/24): the first 3 octets are the network, the last is host. Zeroing the host octet of the given address: $$\text{Network ID} = \boxed{198.53.147.0}$$
Interpret the given subnet mask. $255.255.255.224$ borrows bits from the classful host octet: $224 = 11100000_2$, i.e. the top 3 bits of the last octet are now subnet bits (/27 overall $= 24+3$), leaving 5 host bits. This carves the class C's 256 host addresses into $2^{3}=8$ subnets of $2^{5}=32$ addresses each (30 usable per subnet, after excluding the subnet's own network and broadcast addresses).
Subnet ID. AND the full address with the subnet mask, octet by octet (only the last octet changes): $45 = 00101101_2$, $224=11100000_2$, so $45\ \&\ 224 = 00100000_2 = 32$. $$\text{Subnet ID} = \boxed{198.53.147.32}\ \ (\text{i.e. } 198.53.147.32/27)$$ This is the third of the 8 subnets ($0,32,64,96,128,160,192,224,\ldots$), covering host addresses $32$–$63$.
Host ID. The bits NOT covered by the mask (the low 5 bits of the last octet) identify the host within its subnet: $45\ \&\ (\sim 224\ \&\ 255) = 45\ \&\ 31 = 00001101_2 = 13$. $$\text{Host ID} = \boxed{13}\ \ (\text{host } 13\ \text{within subnet } 198.53.147.32/27)$$

Quantity	Result
Class	C
Network ID (classful)	198.53.147.0
Subnet ID	198.53.147.32 (/27)
Host ID	13
Subnets available / hosts per subnet	8 subnets × 30 usable hosts

Topic: Classful IP addressing and subnetting.

Key relations: Class C first-octet range $192\text{-}223$, default mask $/24$; subnet bits $s$ borrowed from the host portion give $2^{s}$ subnets of $2^{(32-\text{prefix})}$ addresses each; $\text{Subnet ID} = \text{address}\ \&\ \text{mask}$; $\text{Host ID} = \text{address}\ \&\ \overline{\text{mask}}$.

Why this works: a subnet mask's 1-bits mark exactly which bits of the address are treated as "network/subnet" versus "host"; bit-wise ANDing the address with the mask zeroes out every host bit, leaving the network+subnet prefix, while ANDing with the mask's complement isolates only the host bits — this is the mechanical basis of how a router decides whether a destination is on the local subnet.

Common pitfall: reporting the classful network ID (198.53.147.0) when the question actually wants the SUBNET ID (198.53.147.32) — the two are different once subnetting borrows bits from the host octet; also forgetting that the subnet's own network address (32) and broadcast address (63) are reserved and not assignable to a host, leaving only 30 (not 32) usable host addresses per subnet.

Source: Kurose & Ross, Computer Networking: A Top-Down Approach, 8th ed., Ch.4 §4.3 (The Internet Protocol: IPv4 Addressing); Stallings, Data and Computer Communications, 10th ed., Ch.18 §18.2 (Internet Protocol, subnetting).

Question 6 of 7: IP Address Class, Network ID, Subnet ID, and Host ID

Question 6: IP Address Class, Network ID, Subnet ID, and Host ID (20 marks)