Q2GATE 2026MSQ

An ISP having an address block $202.16.0.0/15$ assigns a block of $6000$ IP addresses to a client, using the classless internet domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?

🚀 Solve in Practice Mode

📖 Explanation

$\textbf{Answer: (A), (B), (C)}$ $\textbf{Solution:}$ $\text{ISP Address Block =} \textbf{ 202.16.0.0/15}$ $\text{Client needs =} \textbf{ 6000 IP addresses}$ $\textbf{Step 1: Find Number of Host Bits Required}$ $2^n > 6000$ $2^{12} = 4096 < 6000$ $2^{13} = 8192 > 6000$ $\text{Minimum host bits required} = 13$ $\textbf{ Step 2: Check Base Network Details}$ $\text{202.16.0.0/15, Network bits = 15, Host bits available = 32 - 15 = 17}$ $\text{Since client needs 13 host bits, allocation is possible.}$ $\textbf{ Valid Subnets Inside 202.16.0.0/15}$ $\begin{array}{|l|l|} \hline \textbf{Possible Subnet IDs (/19)} & \textbf{Binary Representation} \\ \hline \color{green}202.16.0.0/19 & \color{green}11001010 . 00010000 . 000xxxxx . xxxxxxxx \\ \hline \color{green}202.16.32.0/19 & \color{green}11001010 . 00010000 . 001xxxxx . xxxxxxxx \\ \hline 202.16.64.0/19 & 11001010 . 00010000 . 010xxxxx . xxxxxxxx \\ \hline 202.16.96.0/19 & 11001010 . 00010000 . 011xxxxx . xxxxxxxx \\ \hline 202.16.128.0/19 & 11001010 . 00010000 . 100xxxxx . xxxxxxxx \\ \hline 202.16.160.0/19 & 11001010 . 00010000 . 101xxxxx . xxxxxxxx \\ \hline 202.16.192.0/19 & 11001010 . 00010000 . 110xxxxx . xxxxxxxx \\ \hline 202.16.224.0/19 & 11001010 . 00010000 . 111xxxxx . xxxxxxxx \\ \hline 202.17.0.0/19 & 11001010 . 00010001 . 000xxxxx . xxxxxxxx \\ \hline 202.17.32.0/19 & 11001010 . 00010001 . 001xxxxx . xxxxxxxx \\ \hline \color{green}202.17.64.0/19 & \color{green}11001010 . 00010001 . 010xxxxx . xxxxxxxx \\ \hline 202.17.96.0/19 & 11001010 . 00010001 . 011xxxxx . xxxxxxxx \\ \hline 202.17.128.0/19 & 11001010 . 00010001 . 100xxxxx . xxxxxxxx \\ \hline 202.17.160.0/19 & 11001010 . 00010001 . 101xxxxx . xxxxxxxx \\ \hline 202.17.192.0/19 & 11001010 . 00010001 . 110xxxxx . xxxxxxxx \\ \hline 202.17.224.0/19 & 11001010 . 00010001 . 111xxxxx . xxxxxxxx \\ \hline \end{array}$ $\textbf{Option Check}$ $\text{(A) 202.16.0.0/19 Valid}$ $\text{(B) 202.17.64.0/19 Valid}$ $\text{(C) 202.16.32.0/19 Valid}$ $\text{(D) 202.17.24.0/19 Invalid (Not aligned with subnet boundary)}$ $\textbf{Here Super-netting actually refers to: }$ $\textbf{Allocated block + remaining blocks should reconstruct the original ISP network without address loss.}$ $\textbf{Quick Trick:}$ For a /19 subnet inside the given /15 block, the subnet ID is formed using: $\text{The last bit of the second octet & The first 3 bits of the third octet}$ These together determine the subnet portion beyond the original /15 network. For a valid subnet ID: All host bits must be 0 Only the network bits (as defined by the prefix length) can vary. $\text{In option (D) 202.17.24.0/19, the value 24 \in the third octet corresponds to binary 00011000.}$ Here, the 4th and 5th bits of the third octet are set to 1, but these bits fall \in the host portion for a /19 network. Since host bits must be zero \in a valid network address, this violates CIDR alignment rules. Hence, option (D) is invalid. $\boxed{\because \text{All other options correctly align with /19 boundaries and follow the CIDR rule. Option (A), (B) & (C) are valid.}}$

Q3GATE 2025NAT

Suppose a message of size 15000 bytes is transmitted from a source to a destination using IPv4 protocol via two routers as shown in the figure. Each router has a defined maximum transmission unit (MTU) as shown in the figure, including IP header. The number of fragments that will be delivered to the destination is _________. (Answer in integer)

🚀 Solve in Practice Mode

📖 Explanation

The problem asks us to determine the number of fragments when a 15000-byte message is sent through two routers with specified Maximum Transmission Units (MTUs). The IP header size is 20 bytes, and this is included in the MTU.

Original Message Size: The total message size is 15000 bytes. This includes the data and the initial IP header. So, the data size is $15000 - 20 = 14980$ bytes.
Router-1 Processing (MTU = 5000 bytes):
- The maximum data an IP packet can carry through Router-1 is $5000 - 20 = 4980$ bytes.
- To find the number of fragments from Router-1, divide the total data by the maximum data per fragment: $\lceil 14980 / 4980 \rceil = \lceil 3.008 \rceil = 4$ fragments.
- The sizes of these fragments will be: $4976, 4976, 4976, 52$ bytes of data (ensuring data is a multiple of 8, the fragmentation unit). Each will have a 20-byte IP header, making total sizes $4996, 4996, 4996, 72$ bytes.
Router-2 Processing (MTU = 3000 bytes):
- The maximum data an IP packet can carry through Router-2 is $3000 - 20 = 2980$ bytes.
- Now, each of the 4 fragments from Router-1 must pass through Router-2.
- For the first three fragments (4976 bytes of data each): $\lceil 4976 / 2976 \rceil = \lceil 1.67 \rceil = 2$ fragments each. This gives $3 \times 2 = 6$ fragments.
- For the last fragment (52 bytes of data): $\lceil 52 / 2976 \rceil = 1$ fragment.
- Total fragments at the destination: $6 + 1 = 7$ .

The number of fragments delivered to the destination is 7.

[CORRECT_OPTION: 7]

Q7GATE 2025MCQ

Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes. Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?

🚀 Solve in Practice Mode

📖 Explanation

The problem describes a UDP segment carrying 7488 bytes of payload. This segment needs to be transmitted over an Ethernet network with IPv4.
The maximum segment size (MSS) for Ethernet data field is 1500 bytes.
The IPv4 header is 20 bytes (since no options field is used).
The UDP header is 8 bytes.
Therefore, the total size of the UDP segment (including its header and payload) is $7488 + 8 = 7496$ bytes.
The maximum payload an IPv4 datagram can carry for Ethernet is $1500 - 20 (\text{IPv4 header}) = 1480$ bytes.
Now, we need to fragment the UDP segment of 7496 bytes into multiple IPv4 datagrams, each carrying at most 1480 bytes of UDP data.

Number of fragments = $\lceil \frac{\text{Total UDP segment size}}{\text{Maximum UDP data per fragment}} \rceil = \lceil \frac{7496}{1480} \rceil = \lceil 5.064 \rceil = 6$ .
So, there will be 6 fragments.

Let's calculate the size of each fragment:
The first 5 fragments will each carry the maximum 1480 bytes of UDP data.
Total data in first 5 fragments = $5 \times 1480 = 7400$ bytes.
Remaining UDP data = $7496 - 7400 = 96$ bytes.
The last fragment will carry the remaining 96 bytes of UDP data.
The size of the last fragment, including the IPv4 header, is $96 (\text{UDP data}) + 20 (\text{IPv4 header}) = 116$ bytes.

Therefore, the total number of fragments is 6, and the size of the last fragment (including IPv4 header) is 116 bytes.
This corresponds to option (D).

Q12GATE 2024NAT

Consider the entries shown below in the forwarding table of an $\mathbb{P}$ router. Each entry consists of an IP prefix and the corresponding next hop router for packets whose destination IP address matches the prefix. The notation " $\mathrm{/N}$ " in a prefix indicates a subnet mask with the most significant $\mathrm{N}$ bits set to 1.

\begin{array}{|l|c|} \hline Prefix & Next\; hop \; router \\ \hline 10.1 .1 .0 / 24 & R1 \\ \hline 10.1 .1 .128 / 25 & R2 \\ \hline 10.1 .1 .64 / 26 & R3 \\ \hline 10.1 .1 .192 / 26 & R4 \\ \hline \end{array}

This router forwards 20 packets each to 5 hosts. The IP addresses of the hosts are 10.1.1.16, 10.1.1.72, 10.1.1.132, 10.1.1.191, and 10.1.1.205. The number of packets forwarded via the next hop router $R 2$ is

🚀 Solve in Practice Mode

📖 Explanation

We need to determine which next hop router (R1, R2, R3, R4) handles packets for the given IP addresses, considering the longest prefix match rule.
The IP addresses are: 10.1.1.16, 10.1.1.72, 10.1.1.132, 10.1.1.191, and 10.1.1.205.
Prefixes in the forwarding table:
R1: 10.1.1.0/24 (00000000)
R2: 10.1.1.128/25 (10000000)
R3: 10.1.1.64/26 (01000000)
R4: 10.1.1.192/26 (11000000)

Let's convert the last octet of each IP address to binary and compare with prefixes for 25 and 26 bits:

10.1.1.16:
$16 = 00010000_2$ .
Matches 10.1.1.0/24. (Prefix 00000000) $\rightarrow$ R1.
10.1.1.72:
$72 = 01001000_2$ .
Matches 10.1.1.0/24 (Prefix 00000000).
Matches 10.1.1.64/26 (Prefix 01000000). Longest prefix match is /26. $\rightarrow$ R3.
10.1.1.132:
$132 = 10000100_2$ .
Matches 10.1.1.0/24 (Prefix 00000000).
Matches 10.1.1.128/25 (Prefix 10000000). Longest prefix match is /25. $\rightarrow$ R2.
10.1.1.191:
$191 = 10111111_2$ .
Matches 10.1.1.0/24 (Prefix 00000000).
Matches 10.1.1.128/25 (Prefix 10000000). Longest prefix match is /25. $\rightarrow$ R2.
10.1.1.205:
$205 = 11001101_2$ .
Matches 10.1.1.0/24 (Prefix 00000000).
Matches 10.1.1.128/25 (Prefix 10000000).
Matches 10.1.1.192/26 (Prefix 11000000). Longest prefix match is /26. $\rightarrow$ R4.

Summary of next hop routers for each IP address:
10.1.1.16 $\rightarrow$ R1
10.1.1.72 $\rightarrow$ R3
10.1.1.132 $\rightarrow$ R2
10.1.1.191 $\rightarrow$ R2
10.1.1.205 $\rightarrow$ R4

Packets forwarded via next hop router R2 are for 10.1.1.132 and 10.1.1.191.
There are 2 IP addresses that use R2.
The router forwards 20 packets to each of the 5 hosts.
So, for each IP address, 20 packets are forwarded.
Number of packets forwarded via R2 = 2 IP addresses $\times$ 20 packets/IP address = 40 packets.

Q13GATE 2024NAT

Consider sending an IP datagram of size 1420 bytes (including 20 bytes of IP header) from a sender to a receiver over a path of two links with a router between them. The first link (sender to router) has an MTU (Maximum Transmission Unit) size of 542 bytes, while the second link (router to receiver) has an MTU size of 360 bytes. The number of fragments that would be delivered at the receiver is _____

🚀 Solve in Practice Mode

📖 Explanation

Given:
IP datagram size = 1420 bytes (includes 20 bytes IP header)
First link MTU = 542 bytes
Second link MTU = 360 bytes

The MTU (Maximum Transmission Unit) is the largest size of a packet that a layer can pass. For IP datagrams, this means the maximum total size, including the IP header.
The payload size for a fragment is MTU - IP header size. Here IP header size is 20 bytes.

Step 1: Fragmentation at the first link (MTU = 542 bytes)
Maximum payload size for first link = $542 - 20 = 522$ bytes.
Original IP datagram payload size = $1420 - 20 = 1400$ bytes.

Number of fragments from first link:
Fragment 1 payload: 522 bytes. Offset = 0.
Fragment 2 payload: 522 bytes. Offset = 522.
Fragment 3 payload: Remaining payload = $1400 - 522 - 522 = 1400 - 1044 = 356$ bytes. Offset = $522+522=1044$ .
Total fragments = 3.

Step 2: Fragmentation at the second link (MTU = 360 bytes)
Maximum payload size for second link = $360 - 20 = 340$ bytes.
Each of the 3 fragments from Step 1 must pass through the second link.

Fragment 1 (from step 1): Total size = 542 bytes (20B header + 522B payload).
Payload size = 522 bytes.
Sub-fragment 1.1 payload: 340 bytes. Offset = 0.
Sub-fragment 1.2 payload: $522 - 340 = 182$ bytes. Offset = 340.
Fragment 1 from step 1 splits into 2 sub-fragments.

Fragment 2 (from step 1): Total size = 542 bytes (20B header + 522B payload).
Payload size = 522 bytes.
Sub-fragment 2.1 payload: 340 bytes. Offset = 0.
Sub-fragment 2.2 payload: $522 - 340 = 182$ bytes. Offset = 340.
Fragment 2 from step 1 splits into 2 sub-fragments.

Fragment 3 (from step 1): Total size = 376 bytes (20B header + 356B payload).
Payload size = 356 bytes.
Sub-fragment 3.1 payload: 340 bytes. Offset = 0.
Sub-fragment 3.2 payload: $356 - 340 = 16$ bytes. Offset = 340.
Fragment 3 from step 1 splits into 2 sub-fragments.

Total number of fragments delivered at the receiver = $2 + 2 + 2 = 6$ .
Note: The offset in IP fragmentation is in units of 8 bytes. The payload sizes (522, 356, 340, 182, 16) are all multiples of 8 or adjusted to be so by padding at lower layers. However, the calculation of fragment counts relies on the maximum payload size, not necessarily aligning to 8-byte boundaries at each step of fragmentation, but rather for the actual data transfer itself. The question just asks for the number of fragments, which depends on total size vs max size per fragment.
All payload sizes (522, 356) and maximum payload (340) for the next hop are not necessarily multiples of 8. For actual fragmentation, payloads must be multiples of 8, except for the last fragment. However, for determining the number of fragments, it's typically $ceil(\text{total payload} / \text{max payload per fragment})$ .
$ceil(522/340) = ceil(1.53) = 2$ .
$ceil(356/340) = ceil(1.04) = 2$ .
So, calculation for number of fragments is correct.

Q15GATE 2023NAT

The forwarding table of a router is shown below.

\begin{array}{|c|c|c|c|} \hline \\ Subnet \; Number&Subnet \;Mask&Interface ID \\ \hline 200.150.0.0 &255.255.0.0& 1\\ \hline 200.150.64.0 &255.255.224.0 &2\\ \hline 200.150.68.0 &255.255.255.0 &3\\ \hline 200.150.68.64 &255.255.255.224 &4 \\ \hline Default & &0\\ \hline \end{array}

A packet addressed to a destination address 200.150.68.118 arrives at the router. It will be forwarded to the interface with ID _____.

🚀 Solve in Practice Mode

Q16GATE 2022MSQ

Consider routing table of an organization's router shown below:

\begin{array}{|l|l|l|} Subnet Number & Subnet Mask & Next Hop \\ 12.20.164.0 & 255.255.252.0 & R1 \\ 12.20.170.0 & 255.255.254.0 & R2 \\ 12.20.168.0 & 255.255.254.0 & Interface 0 \\ 12.20.166.0 & 255.255.254.0 & Interface 1 \\ default & ~ & R3 \\ \hline \end{array}

Which of the following prefixes in CIDR notation can be collectively used to correctly aggregate all of the subnets in the routing table?

🚀 Solve in Practice Mode

📖 Explanation

We need to find a CIDR prefix that aggregates all the given subnets.
The subnets are:

$12.20.164.0/22$ (255.255.252.0)
$12.20.170.0/23$ (255.255.254.0)
$12.20.168.0/23$ (255.255.254.0)
$12.20.166.0/23$ (255.255.254.0)

Let's convert the third octet of the IP addresses to binary to find the common prefix. The first two octets ( $12.20$ ) are common.

$164 = 1010\,0100_2$
$170 = 1010\,1010_2$
$168 = 1010\,1000_2$
$166 = 1010\,0110_2$

Let's list them:
$164.0/22$ : $1010\,0100\,0000\,0000$ (network bits for /22 are first 22 bits). The 3rd octet is bits 17-24. So $1010\,01_2$ is network.
$166.0/23$ : $1010\,0110\,0000\,0000$ (network bits for /23 are first 23 bits). So $1010\,011_2$ is network.
$168.0/23$ : $1010\,1000\,0000\,0000$ (network bits for /23 are first 23 bits). So $1010\,100_2$ is network.
$170.0/23$ : $1010\,1010\,0000\,0000$ (network bits for /23 are first 23 bits). So $1010\,101_2$ is network.

Let's find the longest common prefix for these binary representations of the third octet.
164: $1010\,0100$
166: $1010\,0110$
168: $1010\,1000$
170: $1010\,1010$
The common prefix for the third octet is $1010\_$ . The first 4 bits are common.
This means bits 17-20 are common. The first 16 bits (12.20) are common.
So, the common prefix is $12.20.$ plus the first 4 bits of the 3rd octet.
The common prefix length is $16 + 4 = 20$ bits.
The common prefix is $12.20.1010\,0000_2.0 = 12.20.160.0$ .
So the aggregate CIDR prefix is $12.20.160.0/20$ .

Now let's check the options given.
(A) $12.20.164.0/20$ . The actual IP address $12.20.164.0$ is within the range of $12.20.160.0/20$ .
$12.20.160.0/20$ means the first 20 bits are fixed.
$12.20.1010\,0000_2.\text{xxxx xxxx}_2$
The range of addresses covered by $12.20.160.0/20$ is from $12.20.160.0$ to $12.20.175.255$ .
160 is $1010\,0000_2$ . 175 is $1010\,1111_2$ .
All given subnets are within this range:

$12.20.164.0$ (10100100) is in 1010xxxx (160-175). Yes.
$12.20.170.0$ (10101010) is in 1010xxxx (160-175). Yes.
$12.20.168.0$ (10101000) is in 1010xxxx (160-175). Yes.
$12.20.166.0$ (10100110) is in 1010xxxx (160-175). Yes.
So, $12.20.160.0/20$ aggregates all of them. The given option is $12.20.164.0/20$ . While the actual network prefix is $12.20.160.0/20$ , an option can use any address within that aggregate. So $12.20.164.0/20$ is a valid way to specify the aggregate prefix.

Let's look at other options to make sure:
(B) $12.20.164.0/22$ . This is a specific subnet, not an aggregation of all. Its prefix is $12.20.10100100_2 /22$ . This is only the first subnet.
(C) $12.20.164.0/21$ . Prefix $12.20.1010010_2 /21$ . Range $12.20.160.0$ to $12.20.167.255$ . This does not include $170.0$ and $168.0$ .
(D) $12.20.168.0/22$ . Prefix $12.20.10101000_2 /22$ . Range $12.20.168.0$ to $12.20.171.255$ . This does not include $164.0$ and $166.0$ .

Therefore, option (A) correctly identifies the aggregate prefix.

Q17GATE 2022NAT

Consider a network with three routers P, Q, R shown in the figure below. All the links have cost of unity. The routers exchange distance vector routing information and have converged on the routing tables, after which the link Q-R fails. Assume that P and Q send out routing updates at random times, each at the same average rate. The probability of a routing loop formation (rounded off to one decimal place) between P and Q, leading to count-to-infinity problem, is ____

🚀 Solve in Practice Mode

📖 Explanation

The problem describes a network with three routers P, Q, R, all links having a cost of unity. Initially, the routers have converged on routing tables. Then the link Q-R fails. We need to find the probability of a routing loop formation between P and Q, leading to a count-to-infinity problem.

Initial state (all links up, cost 1):

P's routing table (destination, next-hop, cost):
- To Q: Q, 1
- To R: Q, 2 (P->Q->R)
Q's routing table:
- To P: P, 1
- To R: R, 1
R's routing table:
- To P: Q, 2 (R->Q->P)
- To Q: Q, 1

Now, the link Q-R fails.

R realizes its direct link to Q is down. R's route to Q becomes infinite.
Q realizes its direct link to R is down. Q's route to R becomes infinite.

This failure information needs to be propagated. Routers P and Q send updates at random times.
Consider the situation from Q's perspective for destination R:
Q detects R is unreachable directly. It sets $Cost(Q,R) = \infty$ .
If Q sends an update to P, P knows $Cost(Q,R)=\infty$ .
If P sends an update to Q, P's route to R is via Q ( $Cost(P,R) = Cost(P,Q) + Cost(Q,R)$ ).

The "count-to-infinity" problem arises in Distance Vector Routing when a router incorrectly believes a path to a destination through a neighbor is still valid, even though that neighbor has lost its direct path to the destination.
Scenario for a loop between P and Q for destination R:

Q's link to R fails. Q sets $Cost(Q,R) = \infty$ .
Before Q can send its updated (infinite) cost for R to P, P sends its routing update to Q.
P's routing table still says $Cost(P,R) = Cost(P,Q) + Cost(Q,R) = 1+1 = 2$ (via Q).
Q receives P's update. Q now sees a "new" path to R: via P with cost $Cost(Q,P) + Cost(P,R) = 1+2 = 3$ .
Q updates its table to $Cost(Q,R) = 3$ (via P), believing it can reach R through P. This is incorrect because P's route to R is actually through Q. This forms a loop: Q -> P -> Q -> R.
P then receives an update from Q indicating $Cost(Q,R)=3$ . P will then update its cost to R via Q as $1+3=4$ . This count-to-infinity continues.

A loop occurs if P sends an update to Q before Q sends its update (infinity) to P.
There are two routers (P and Q) involved in exchanging updates.
Each router sends updates at random times. We can consider two events:

Event A: P sends update to Q.
Event B: Q sends update to P.

The routing loop for destination R forms if P sends its update to Q first.
If Q sends its update to P first, P learns that R is unreachable via Q (cost $\infty$ ). P will then update its own route to R as $\infty$ . No loop.
If P sends its update to Q first, Q learns a path to R via P (cost 2+1=3), while its direct path is down. Q updates its table to use P. Then P receives Q's update, sees a path to R via Q with cost 3+1=4, and so on. A loop forms.

Assuming random times for updates, there is a 50% probability that P sends its update to Q before Q sends its update to P.
So, the probability of a routing loop formation is 0.5.
Rounded off to one decimal place, this is 0.5.

Q18GATE 2022MCQ

Consider an enterprise network with two Ethernet segments, a web server and a firewall, connected via three routers as shown below. What is the number of subnets inside the enterprise network?

🚀 Solve in Practice Mode

📖 Explanation

A subnet is a logical subdivision of an IP network. Each separate broadcast domain or LAN segment typically corresponds to a subnet. Routers are used to connect different subnets.
Let's identify the distinct network segments in the diagram:

The segment connecting "To Internet" to the first "Firewall" is one network segment. This is external.
The segment between the "Firewall" and "Router" (labeled "Router" above "Web server") is one segment.
The segment connecting the first "Router" to the "Web server" is another segment.
The segment connecting the first "Router" to the second "Router" (labeled "Router" below "Web server") is a third segment.
The segment connecting the second "Router" to the "Ethernet" (left) is a fourth segment.
The segment connecting the second "Router" to the "Ethernet" (right) is a fifth segment.
Each interface of a router (or firewall) that connects to a distinct network segment defines a boundary for a subnet. The given enterprise network contains three routers and one firewall.
Counting the internal network segments:

Firewall-Router segment (1)
Router-Web server segment (2)
Router-Router segment (3)
Router-Ethernet segment (left) (4)
Router-Ethernet segment (right) (5)
The link between the Firewall and "To Internet" is generally not considered internal to the enterprise network, but a boundary.
The solution states "First firewall is dividing is two subnets and second router further divided is to two subnets are is via web server and the other is via other router '3 subnets'". This is not clear.

Let's identify distinct broadcast domains (subnets).

Firewall to Router 1.
Router 1 to Web Server.
Router 1 to Router 2.
Router 2 to Ethernet segment 1.
Router 2 to Ethernet segment 2.
This implies 5 subnets.
However, the solution gives "3 subnets". This is a much smaller number.
Let's reconsider the network structure. A router separates broadcast domains. Each interface on a router belongs to a different subnet.

The firewall has at least two interfaces: one to "To Internet" and one to the first router. The segment to the first router is a subnet.
The first router has interfaces to: Firewall, Web server, and the second router. These are three distinct subnets.
The second router has interfaces to: the first router, Ethernet segment 1, Ethernet segment 2. These are three distinct subnets.
Summing these up:

Segment between Firewall and Router 1.
Segment between Router 1 and Web Server.
Segment between Router 1 and Router 2.
Segment between Router 2 and Ethernet segment 1.
Segment between Router 2 and Ethernet segment 2.
This still totals to 5 distinct subnets within the enterprise.
If the solution's "3 subnets" is correct, there must be a different interpretation.
Could it be that the "Ethernet" segments are the same subnet? Unlikely, as they are shown connected to separate ports of a router.
The description "First firewall is dividing is two subnets and second router further divided is to two subnets are is via web server and the other is via other router '3 subnets'" is grammatically difficult to parse.
If the question is asking for subnets beyond what the firewall creates, or if there's a specific grouping.
Let's assume there are two Ethernets on the other side of the second router.
Subnets are separated by routers.
The LAN containing the web server. (Connected to Router 1).
The link connecting Router 1 and Router 2.
The LANs connected to Router 2 (two Ethernets). If these two Ethernets are distinct segments, then they are 2 subnets.

If we count each distinct physical/logical network segment that requires its own IP network address:

Between firewall and first router.
Between first router and web server.
Between first router and second router.
Between second router and first Ethernet segment.
Between second router and second Ethernet segment.
This totals to 5.

Let's review the solution's exact wording: "First firewall is dividing is two subnets and second router further divided is to two subnets are is via web server and the other is via other router '3 subnets'."
This phrase indicates the total count is 3. This interpretation is extremely difficult given standard network topology. A firewall also separates subnets.
If we consider only the internal subnets formed by the routers, excluding the segment to the firewall.
Router 1 is connected to:

Web server segment (1)
Router 2 (link segment) (2)
Router 2 is connected to:
Router 1 (link segment, already counted)
Ethernet (left) (3)
Ethernet (right) (4)
This totals 4.

The phrase "First firewall is dividing is two subnets" could mean the firewall separates the internet from the enterprise, creating 1 internal subnet.
The "second router further divided is to two subnets" could mean 2 more subnets.
Total = $1+2 = 3$ . This is a possible interpretation but requires making assumptions about how "dividing" is meant.

Subnet for the web server (behind Router 1).
Subnet for the left Ethernet (behind Router 2).
Subnet for the right Ethernet (behind Router 2).
This interpretation excludes the interconnecting links between firewall/routers from being counted as distinct subnets, which is highly unusual. Usually, each point-to-point link between routers is its own subnet.
If the question asks for user-facing subnets (LANs where hosts like the web server or client machines are), then there are three: the Web server's segment, and the two Ethernet segments. The links between routers and the firewall are then considered infrastructure rather than distinct user subnets. This is the only way to arrive at 3.
Given the ambiguity, the most common interpretation in network design would be 5 subnets. However, to match the answer '3', we must assume the question implicitly refers to the three "endpoints" of the internal network reachable by hosts (the web server and the two Ethernet segments).

Q19GATE 2021MSQ

Consider a computer network using the distance vector routing algorithm in its network layer. The partial topology of the network is shown below. The objective is to find the shortest-cost path from the router R to routers P and Q. Assume that R does not initially know the shortest routes to P and Q. Assume that R has three neighboring routers denoted as X, Y and Z. During one iteration, R measures its distance to its neighbors X, Y, and Z as 3, 2 and 5, respectively. Router R gets routing vectors from its neighbors that indicate that the distance to router P from routers X, Y and Z are 7, 6 and 5, respectively. The routing vector also indicates that the distance to router Q from routers X, Y and Z are 4, 6 and 8 respectively. Which of the following statement(s) is/are correct with respect to the new routing table o R, after updation during this iteration?

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📖 Explanation

The distance vector routing algorithm calculates the shortest path from R to P and Q.
Distances from R to neighbors: $d(R,X)=3$ , $d(R,Y)=2$ , $d(R,Z)=5$ .
Routing vectors received by R:
From X: $d(X,P)=7$ , $d(X,Q)=4$ .
From Y: $d(Y,P)=6$ , $d(Y,Q)=6$ .
From Z: $d(Z,P)=5$ , $d(Z,Q)=8$ .

To P:
$d(R,P)$ via X = $d(R,X) + d(X,P) = 3 + 7 = 10$ .
$d(R,P)$ via Y = $d(R,Y) + d(Y,P) = 2 + 6 = 8$ .
$d(R,P)$ via Z = $d(R,Z) + d(Z,P) = 5 + 5 = 10$ .
Minimum $d(R,P) = 8$ via Y. So, statement C is correct.

To Q:
$d(R,Q)$ via X = $d(R,X) + d(X,Q) = 3 + 4 = 7$ .
$d(R,Q)$ via Y = $d(R,Y) + d(Y,Q) = 2 + 6 = 8$ .
$d(R,Q)$ via Z = $d(R,Z) + d(Z,Q) = 5 + 8 = 13$ .
Minimum $d(R,Q) = 7$ via X. So, statement B is correct.

Therefore, statements B and C are correct.

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