Network Layer Protocol Practice Questions

The subnet mask $255.255.0.0$ corresponds to a $/16$ CIDR block, fixing the first two octets ($172.16$) as the network address.
The remaining $16$ bits of the IP address are allocated for host usage.
To calculate the broadcast address, all host bits are set to $1$, which converts the host portion of the address to $255.255$.
Concatenating the network prefix with this suffix gives the broadcast address $172.16.255.255$.

Here's a step-by-step explanation for determining the new distance vector at node N3:

1. Initial N3 Distance Vector (DV_N3) and Link Costs:

   • DV_N3: (7, 6, 0, 2, 6)
   • Links from N3: N3-N2 (cost 6), N3-N4 (cost 2), N3-N5 (cost 6).

2. Change in Link Cost: The cost of link N2-N3 reduces from 6 to 2. This immediately updates N3's perception of the cost to N2.

3. Calculate New DV_N3 based on Neighbors:
   N3 updates its distance vector using the Bellman-Ford equation:
   $DV_N3[destination] = \min (cost(N3, neighbor) + DV_{neighbor}[destination])$

4. Update for N1:

   • N2: $cost(N3,N2) + DV_N2[N1] = 2 + 1 = 3$ (N3-N2 link changed to 2)
   • N4: $cost(N3,N4) + DV_N4[N1] = 2 + 8 = 10$
   • N5: $cost(N3,N5) + DV_N5[N1] = 6 + 4 = 10$
   • New N3 to N1 distance: $\min(3, 10, 10) = 3$.

5. Update for N2:

   • N2: $cost(N3,N2) + DV_N2[N2] = 2 + 0 = 2$ (N3-N2 link changed to 2)
   • N4: $cost(N3,N4) + DV_N4[N2] = 2 + 7 = 9$
   • N5: $cost(N3,N5) + DV_N5[N2] = 6 + 3 = 9$
   • New N3 to N2 distance: $\min(2, 9, 9) = 2$.

6. Update for N3 (Self): Stays 0.

7. Update for N4:

   • N2: $cost(N3,N2) + DV_N2[N4] = 2 + 7 = 9$
   • N4: $cost(N3,N4) + DV_N4[N4] = 2 + 0 = 2$
   • N5: $cost(N3,N5) + DV_N5[N4] = 6 + 4 = 10$
   • New N3 to N4 distance: $\min(9, 2, 10) = 2$.

8. Update for N5:

   • N2: $cost(N3,N2) + DV_N2[N5] = 2 + 3 = 5$
   • N4: $cost(N3,N4) + DV_N4[N5] = 2 + 4 = 6$
   • N5: $cost(N3,N5) + DV_N5[N5] = 6 + 0 = 6$
   • New N3 to N5 distance: $\min(5, 6, 6) = 5$.

The new distance vector for N3 is (3, 2, 0, 2, 5).

A link in a network is considered unused if it is not part of any shortest path between any two routers. We need to identify such links by analyzing the graph.

The principle is: a link (u, v) with weight w is unused if the shortest path between u and v using other links has a cost strictly less than w.

Let's examine the links:

1. Link R1-R3 (Weight 7):

   • The direct cost from R1 to R3 is 7.
   • There is an alternative path: R1 → R2 → R3.
   • The cost of this alternative path is weight(R1,R2) + weight(R2,R3) = 1 + 2 = 3.
   • Since 3 < 7, any data from R1 to R3 (and vice-versa) will always be routed through R2. The direct link R1-R3 is more expensive and will never be used.
   • Result: 1 unused link found.

2. Link R4-R5 (Weight 5):

   • The direct cost from R4 to R5 is 5.
   • An alternative path exists: R4 → R6 → R5.
   • The cost of this path is weight(R4,R6) + weight(R5,R6) = 1 + 4 = 5.
   • Here we have two paths of equal cost (5). A distance-vector protocol uses a tie-breaking rule to choose one. For instance, if the router at R4 is configured to prefer the path via R6 (e.g., due to a lower neighbor ID or a policy), the direct link R4-R5 would not be used for traffic from R4 to R5. Since the link can be rendered obsolete by a deterministic tie-breaking rule, and it is not part of any other shortest path, it can be considered unused.
   • Result: A second potentially unused link found.

Checking other links reveals that they are all part of at least one unique shortest path. For example, the shortest path between R2 and R4 is the direct link with weight 2. Any other path is longer.

Therefore, there are 2 links in the network (R1-R3 and R4-R5) that will never be used under a stable distance-vector routing protocol with specific tie-breaking.

In the original network, links $R1-R2$ (weight $6$) and $R4-R6$ (weight $8$) were unused because they were not part of the shortest paths (e.g., $R1 \to R3 \to R2$ cost $5 < 6$ and $R4 \to R5 \to R6$ cost $5 < 8$). Changing these unused links to weight $2$ results in the new graph: $R1-R2=2$, $R1-R3=3$, $R2-R3=2$, $R2-R4=7$, $R3-R5=9$, $R4-R5=1$, $R4-R6=2$, and $R5-R6=4$.

Evaluating the shortest paths with these updated weights:

1. $R1 \to R2$ uses the direct link (cost $2$).
2. $R4 \to R6$ uses the direct link (cost $2$).
3. $R5 \to R6$ path via $R4$ is $R5 \to R4 \to R6$ with cost $1 + 2 = 3$, which is less than the direct link $R5-R6$ (cost $4$).
4. Other links $R1-R3$, $R2-R3$, $R2-R4$, $R3-R5$, and $R4-R5$ are all part of the shortest paths between their respective endpoints.

Since only the link $R5-R6$ is strictly greater than its shortest path alternatives, it is the only unused link.

The Time to Live (TTL) field in an IP datagram header is a mechanism used to prevent packets from looping indefinitely on a network. The sender sets an initial TTL value (an integer). Each router that forwards the packet decrements the TTL value by one. If a routing loop exists, the packet will be passed between routers continuously. The TTL ensures this doesn't go on forever. When the TTL value reaches zero, the router that receives it discards the packet and sends an ICMP 'Time Exceeded' message to the source. This prevents looping packets from consuming network resources and causing congestion.

For two computers to be in the same network, their network addresses must be identical. A network address is calculated by performing a bitwise AND operation between the computer's IP address and the subnet mask.

The IP addresses are A: 10.105.1.113 and B: 10.105.1.91. Since the first three octets are identical, we only need to analyze the fourth octet.

In binary, the fourth octets are:

• IP A (113): 01110001
• IP B (91): 01011011

Let's test each subnet mask (N):

1. N = 255.255.255.0: The mask for the fourth octet is 00000000.

   • A's network address: 01110001 AND 00000000 = 00000000
   • B's network address: 01011011 AND 00000000 = 00000000
   • The network addresses are the same. A and B are in the same network.

2. N = 255.255.255.128: The mask for the fourth octet is 10000000.

   • A's network address: 01110001 AND 10000000 = 00000000
   • B's network address: 01011011 AND 10000000 = 00000000
   • The network addresses are the same. A and B are in the same network.

3. N = 255.255.255.192: The mask for the fourth octet is 11000000.

   • A's network address: 01110001 AND 11000000 = 01000000 (64)
   • B's network address: 01011011 AND 11000000 = 01000000 (64)
   • The network addresses are the same. A and B are in the same network.

4. N = 255.255.255.224: The mask for the fourth octet is 11100000.

   • A's network address: 01110001 AND 11100000 = 01100000 (96)
   • B's network address: 01011011 AND 11100000 = 01000000 (64)
   • The network addresses (10.105.1.96 and 10.105.1.64) are different. A and B would be in different networks.

Therefore, the netmask 255.255.255.224 should not be used if A and B are to be in the same network.

A Virtual Local Area Network (VLAN) is a logical implementation that allows a network administrator to partition a single physical switch into multiple distinct broadcast domains. By grouping devices logically regardless of their physical connection, the administrator limits unnecessary broadcast traffic and improves overall network security. This process is fundamentally defined as segmenting a network at Layer 2 within a switch or device, which aligns perfectly with option D.

The Address Resolution Protocol (ARP) is a critical network protocol responsible for mapping network layer addresses to link layer addresses. When a device needs to send an IP packet ($IP_{target}$) to a destination on the local network, it requires the destination's physical hardware address, known as the MAC address ($MAC_{target}$). The source device broadcasts an ARP request packet to the network, asking which device possesses $IP_{target}$. The device matching that IP responds with its $MAC_{target}$, facilitating the mapping $IP \rightarrow MAC$ required to encapsulate the data into an Ethernet frame. Thus, ARP specifically performs the task of resolving an IP address to its corresponding MAC address.

To determine if two IP addresses belong to the same subnet, perform a bitwise AND operation with the subnet mask $255.255.31.0$. The mask's third octet is $31$, which is $00011111_2$ in binary. For option D ($128.8.129.43$ and $128.8.161.55$), the first two octets match, and we evaluate the third octet:
$129 \& 31 = 10000001_2 \& 00011111_2 = 00000001_2 = 1$
$161 \& 31 = 10100001_2 \& 00011111_2 = 00000001_2 = 1$
Since the masked values for the third octet are identical ($1$), both IP addresses belong to the same network segment.

S1 is true: Count-to-infinity is a limitation of Distance Vector (DV) protocols due to limited topology knowledge, while Link State (LS) protocols maintain a global network map, avoiding this issue.
S2 is false: In LS routing, every node runs the shortest path algorithm (e.g., Dijkstra's algorithm) locally to compute routes based on the complete topology.
S3 is false: DV routing is distributed; nodes update their routing tables iteratively based on information received from immediate neighbors, rather than a single node calculating paths.
S4 is true: DV protocols exchange small distance vectors with neighbors, whereas LS protocols flood Link State Advertisements (LSAs) throughout the network, causing higher message overhead.
Given that only S1 and S4 are true, the

To determine the network portion, convert each decimal octet of the subnet mask $255.255.255.192$ into binary:
The values are $255 = 11111111_2$ (8 bits), $255 = 11111111_2$ (8 bits), $255 = 11111111_2$ (8 bits), and $192 = 11000000_2$ (2 bits).
Summing these binary bits gives the total network portion: $8 + 8 + 8 + 2 = 26$ bits.
Thus, the subnet mask $255.255.255.192$ effectively defines a network prefix length of $/26$.

The netmask $255.255.255.224$ implies a subnet block size of $256 - 224 = 32$.
Host X has IP address $192.168.1.97$, which falls into the subnet range $[192.168.1.96, 192.168.1.127]$ since $96 \leq 97 < 128$.
For communication to occur, the gateway must reside on the same local subnet as the host.
Comparing the router interfaces: $192.168.1.67$ (range $64-95$), $192.168.1.110$ (range $96-127$), $192.168.1.135$ (range $128-159$), and $192.168.1.155$ (range $128-159$).
Only the IP $192.168.1.110$ belongs to the same subnet ($96-127$) as Host X.

To determine the class of the network $198.78.41.0$, we examine the first octet, which is $198$. In IPv4 classful addressing, the network classes are defined by the following ranges for the first octet:

• Class A: $1 - 126$
• Class B: $128 - 191$
• Class C: $192 - 223$
• Class D: $224 - 239$

Since $192 \leq 198 \leq 223$, the address $198.78.41.0$ falls within the designated range for Class C networks.

A Class B network natively uses 16 bits for the host portion. The subnet mask $255.255.248.0$ utilizes 5 bits of the third octet for subnetting ($248 = 11111000_2$).
This reduces the total available bits for hosts to $16 - 5 = 11$ bits.
The maximum number of usable hosts is determined by the formula $2^n - 2$, where $n$ is the number of host bits.
Substituting $n=11$, we calculate the result as $2^{11} - 2 = 2048 - 2 = 2046$.

On a Local Area Network (LAN), data link layer protocols encapsulate network layer data into frames for transmission. A typical LAN frame, such as an Ethernet frame, is composed of a header, a data payload (also referred to as the information field), and a trailer. During the encapsulation process, the IP datagram is placed directly into this data payload section of the frame. The frame header contains link-layer addressing, while the information field carries the upper-layer payload to the destination. Thus, the IP datagram is transported within the information field of the LAN frame.

The subnet mask $255.255.255.224$ divides the network into blocks of $2^5 = 32$ addresses. The subnet identifier for any IP is calculated as \lfloor \text{last_octet} / 32 \rfloor \times 32.

1. For hosts $192.168.1.97$ and $192.168.1.110$: $\lfloor 97/32 \rfloor = 3$ and $\lfloor 110/32 \rfloor = 3$, resulting in subnet $192.168.1.96$.
2. For hosts $192.168.1.80$ and $192.168.1.67$: $\lfloor 80/32 \rfloor = 2$ and $\lfloor 67/32 \rfloor = 2$, resulting in subnet $192.168.1.64$.
3. For hosts $192.168.1.135$ and $192.168.1.155$: $\lfloor 135/32 \rfloor = 4$ and $\lfloor 155/32 \rfloor = 4$, resulting in subnet $192.168.1.128$.
   Since there are three distinct subnet identifiers ($192.168.1.64$, $192.168.1.96$, and $192.168.1.128$), the network contains $3$ distinct subnets.

The IEEE $802.11$ standard defines the set of media access control (MAC) and physical layer (PHY) protocols for implementing wireless local area network (WLAN) computer communication. These protocols operate primarily in the $2.4$, $3.6$, $5$, and $60$ GHz frequency bands and are commonly marketed as Wi-Fi. In comparison, IEEE $802.3$ standardizes Ethernet, IEEE $802.15$ governs Wireless Personal Area Networks like Bluetooth, and IEEE $802.16$ is associated with Broadband Wireless Access. Consequently, IEEE $802.11$ is explicitly designated for Wireless LANs.

Using the distance vector routing algorithm, node $F$ updates its routing table by calculating the minimum cost to each destination $X$ as:
$\text{Cost}(F, X) = \min_{N \in \{A, E, D, G\}} \{d(F, N) + \text{Cost}(N, X)\}$
Given delays $d(F, A)=8, d(F, E)=10, d(F, D)=12, d(F, G)=6$:

• Dest A: $\min(8+0, 10+24, 12+20, 6+21) = \min(8, 34, 32, 27) = 8$
• Dest B: $\min(8+40, 10+27, 12+8, 6+24) = \min(48, 37, 20, 30) = 20$
• Dest C: $\min(8+14, 10+7, 12+30, 6+22) = \min(22, 17, 42, 28) = 17$
• Dest D: $\min(8+17, 10+20, 12+0, 6+19) = \min(25, 30, 12, 25) = 12$
• Dest E: $\min(8+21, 10+0, 12+14, 6+22) = \min(29, 10, 26, 28) = 10$
• Dest G: $\min(8+24, 10+22, 12+22, 6+0) = \min(32, 32, 34, 6) = 6$

The resulting values $\{A:8, B:20, C:17, D:12, E:10, F:0, G:6\}$ match option A.