Network Layer Protocol Practice Questions

To find the potential candidate IP address ranges, we first determine the required subnet size for 1500 computers.
Since $2^{10} = 1024$ and $2^{11} = 2048$, an organization with 1500 computers needs at least 11 host bits. This implies a subnet mask of $32 - 11 = 21$ bits, or $/21$.

The ISP's available address space is $202.61.0.0/17$. This means the first 17 bits are fixed ($202.61.00000000.00000000$).
The network address for a $/21$ subnet must fall within this $/17$ range. To minimize routing entries, the ISP should aggregate the allocated $/21$ range within its $/17$ block.
A $/17$ block covers IP addresses from $202.61.0.0$ to $202.61.127.255$.
In binary, the last octet for a $/17$ is $00000000$ to $01111111$.
The third octet for $202.61.X.X/21$ ranges from $202.61.0.X$ to $202.61.255.X$.
For a $/21$ subnet, the block size for the third octet is $2^{(24-21)} = 2^3 = 8$.
Possible starting third octets within the $/17$ range are multiples of 8: $0, 8, 16, \dots, 120$.
Examining the options:
I. $202.61.84.0/21$: $84$ is not a multiple of 8. This is not a valid starting address for a $/21$ block.
II. $202.61.104.0/21$: $104 = 13 \times 8$. This is a valid starting address within the $/17$ range.
III. $202.61.64.0/21$: $64 = 8 \times 8$. This is a valid starting address within the $/17$ range.
IV. $202.61.144.0/21$: $144 > 127$. This address falls outside the ISP's $/17$ block.

Therefore, only options II and III are potential candidates.

To determine if the machines are on the same subnet, we perform a bitwise AND between each IP address and the subnet mask (255.255.255.252). The result is the subnet ID.

• M (100.10.5.2) AND (255.255.255.252) = 100.10.5.0
• N (100.10.5.5) AND (255.255.255.252) = 100.10.5.4
• P (100.10.5.6) AND (255.255.255.252) = 100.10.5.4
  Machines N and P have the same subnet ID (100.10.5.4), while M has a different one. Therefore, only N and P belong to the same subnet.

The total length of the IP packet is 4500 bytes, with a 20-byte header. The total data to be fragmented is $4500 - 20 = 4480$ bytes. The Maximum Transmission Unit (MTU) is 600 bytes. Each fragment will have its own 20-byte IP header.

The maximum data payload in each fragment is $600 - 20 = 580$ bytes. The data length must be a multiple of 8. The largest multiple of 8 less than or equal to 580 is $576$ ($72 \times 8$).

The fragments are created as follows:

• Fragment 1: Carries data bytes 0 to 575. The offset is $\frac{0}{8} = 0$.
• Fragment 2: Carries data bytes 576 to 1151. The offset is $\frac{576}{8} = 72$.
• Fragment 3: Carries data bytes 1152 to 1727. The offset is $\frac{1152}{8} = 144$.

The fragmentation offset value for the third fragment is 144.

In IPv4 addressing, Class B networks use the first two octets for the network ID and the last two octets for the host ID.
This structure corresponds to a CIDR notation of $/16$, where the first $16$ bits are set to $1$ and the remaining $16$ bits are set to $0$.
The binary representation of this subnet mask is $11111111.11111111.00000000.00000000$.
Converting these binary octets to decimal notation results in $255.255.0.0$.
Thus, $255.255.0.0$ is the standard default subnet mask for a Class B network.

The IPv4 header has a maximum size of 60 bytes. With a fixed base header of 20 bytes, this leaves a maximum of 40 bytes for options.
The Record Route (RR) option uses 3 bytes for its own metadata: 1 byte for option type, 1 byte for length, and 1 byte for a pointer. This leaves 40 - 3 = 37 bytes available for storing the IP addresses.
Since each IPv4 address requires 4 bytes, the maximum number of addresses that can be recorded is $\lfloor \frac{37}{4} \rfloor = 9$.

(a) Multicast group membership is managed by the $IGMP$ protocol ($ii$).
(b) An Interior gateway protocol operates within an autonomous system, such as $OSPF$ ($iii$).
(c) An Exterior gateway protocol handles routing between autonomous systems, like $BGP$ ($iv$).
(d) $RIP$ is a routing protocol based on the Distance Vector routing algorithm ($i$).
Therefore, the matching order is a-$ii$, b-$iii$, c-$iv$, and d-$i$.

Let's evaluate each statement:
I. RIP is a distance-vector routing protocol, which is correct.
II. RIP uses UDP for sending its routing updates. This is also correct.
III. OSPF does not use TCP or UDP. It operates directly on top of the IP protocol. Thus, this statement is incorrect.
IV. OSPF is a link-state routing (LSR) protocol, which is correct.
Therefore, statements I, II, and IV are the correct ones.

Dynamic routing protocols automate network maintenance by performing neighbor discovery and managing routing tables, which satisfies Option A. They facilitate the exchange of routing information with adjacent devices to propagate changes throughout the network, satisfying Option B. By utilizing consensus algorithms to calculate optimal paths, these protocols ensure that routers converge and reach a consistent agreement regarding the overall network topology, satisfying Option C. Since these protocols collectively perform all these functions to maintain efficient network operations, Option D is correct.

The question asks which protocol does not resolve one form of address to another. ARP resolves an IP address to a MAC address. RARP does the reverse, resolving a MAC address to an IP address. DNS resolves a domain name to an IP address. All three perform a mapping or resolution. DHCP, however, is a protocol used to dynamically assign an IP address and other network configuration parameters to a host. It does not resolve an existing address into another form.

For a Class B address, the first 16 bits are for the network ID.
Given a 6-bit subnet number, the number of subnets is $2^6 - 2 = 64 - 2 = 62$ (excluding all-zeros and all-ones subnets).
The total host bits become $32 \text{ (total bits)} - 16 \text{ (network bits)} - 6 \text{ (subnet bits)} = 10$ host bits.
The maximum number of hosts per subnet is $2^{10} - 2 = 1024 - 2 = 1022$ (excluding the network and broadcast addresses).

The total data payload to be transmitted is the datagram size minus the header size, which is $1000 - 20 = 980$ bytes. The maximum transmission unit (MTU) is 100 bytes, so the maximum data payload per fragment is the MTU minus the header size, or $100 - 20 = 80$ bytes. To find the number of fragments, we divide the total data payload by the maximum payload per fragment: $\lceil \frac{980}{80} \rceil = \lceil 12.25 \rceil = 13$. Therefore, 13 fragments are required.

The Dynamic Host Configuration Protocol (DHCP) is a network management protocol designed to automate device configuration on networks.
DHCP enables a client host to dynamically request and receive a unique $IP$ address and associated parameters during each bootup process.
While BOOTP was a predecessor, it lacks the advanced lease management features that characterize modern DHCP functionality.
Conversely, RARP maps hardware addresses to $IP$ addresses, and ARP maps $IP$ addresses to physical MAC addresses, neither handling dynamic configuration.
Thus, DHCP is the primary protocol for dynamic $IP$ allocation.

The given network is 200.10.11.144/27. The /27 suffix indicates that the first 27 bits are the network portion, leaving $32 - 27 = 5$ bits for the host portion.

The last octet, 144, is 10010000 in binary. The network mask for /27 in the last octet is 11100000 (224). The network address is 144 AND 224, which is 128. So, the network is 200.10.11.128/27.

The broadcast address is found by setting all 5 host bits to 1. The network part is 100. So, the last octet is 10011111, which is 159. The last usable host address is one less than the broadcast address, which is 158.

To find the network address, perform a bitwise AND operation between the host IP address and the subnet mask.
The 3rd octet of the IP ($112$) in binary is $01110000_2$, and the 3rd octet of the mask ($224$) is $11100000_2$.
Calculating the bitwise AND: $01110000_2 \land 11100000_2 = 01100000_2$.
Converting the binary result $01100000_2$ back to decimal yields $96$.
Combining the network bits, the resulting network address is $125.134.96.0$.

The OSI model is categorized into lower media-dependent layers (Physical, Data Link, and Network) and upper host-dependent layers (Transport, Session, Presentation, and Application). The lower layers (1-3) focus on the physical transmission and routing of data packets across the network path between nodes. In contrast, the upper layers (4-7) reside within the end-systems and are responsible for end-to-end data processing, session management, and formatting. Because these layers manage communication directly between the source and destination hosts, they are defined as the host-to-host layers.

The question refers to an IPv4 address, which is the foundational addressing standard for traditional TCP/IP networks. An IPv4 address is composed of $32$ bits, organized into four separate $8$-bit octets (calculated as $4 \times 8 = 32$). This bit length creates a unique address space of $2^{32}$ possible combinations, allowing each host to be uniquely identified for network communication. Consequently, the standard internet address assigned to a host is 32 bits.

The router uses the principle of longest prefix matching to determine the next hop for a given IP address. This means it finds all matching routes in its table and selects the one with the most specific (longest) netmask.

1. 128.96.171.92: Matches 128.96.170.0/23. Next hop is Interface 0 (a).
2. 128.96.167.151: Matches both 128.96.166.0/23 and 128.96.164.0/22. The /23 mask is longer, so the next hop is R2 (c).
3. 128.96.163.151: Does not match any specific route. It will use the default route 0.0.0.0, so the next hop is R4 (e).
4. 128.96.165.121: Matches 128.96.164.0/22. Next hop is R3 (d).

The correct mapping is i-a, ii-c, iii-e, iv-d.

To determine the correct interface, perform a bitwise AND operation between the destination address $144.16.68.117$ and each subnet mask:

1. eth0 ($/16$): $144.16.68.117 \land 255.255.0.0 = 144.16.0.0$ (Matches prefix).
2. eth1 ($/19$): $144.16.68.117 \land 255.255.224.0 = 144.16.64.0$ (Matches prefix).
3. eth2 ($/24$): $144.16.68.117 \land 255.255.255.0 = 144.16.68.0$ (Matches prefix).
4. eth3 ($/27$): $144.16.68.117 \land 255.255.255.224 = 144.16.68.96 \neq 144.16.68.64$ (No match).
   Routers use the longest prefix matching rule. Comparing the matching masks ($/16$, $/19$, and $/24$), the longest is $/24$, which corresponds to interface eth2.

When a router forwards an IP packet, it modifies certain header fields. The Time to Live (TTL) is decremented at each hop to prevent indefinite packet looping. If the TTL changes, the header checksum must be recalculated. The length field might also change if the packet is fragmented. However, the Source IP Address, which identifies the original sender, must remain unchanged throughout the packet's journey to the destination.