1What is the size of the standard IPv4 header without any options?
IPv4 Header
Easy
A.20 bytes
B.40 bytes
C.64 bytes
D.32 bytes
Correct Answer: 20 bytes
Explanation:
The base IPv4 header is 20 bytes long. If options are included, it can be up to 60 bytes.
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2Which field in the IPv4 header is used to prevent packets from endlessly circulating in a network?
IPv4 Header
Easy
A.Header Checksum
B.Fragment Offset
C.Protocol
D.Time to Live (TTL)
Correct Answer: Time to Live (TTL)
Explanation:
The Time to Live (TTL) field is decremented by each router. When it reaches zero, the packet is discarded.
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3What is the fixed length of the base IPv6 header?
IPv6 Header
Easy
A.60 bytes
B.20 bytes
C.32 bytes
D.40 bytes
Correct Answer: 40 bytes
Explanation:
Unlike IPv4, the base IPv6 header has a fixed length of 40 bytes to simplify processing.
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4Which IPv4 header field is equivalent to the 'Hop Limit' field in the IPv6 header?
IPv6 Header
Easy
A.Type of Service
B.Time to Live (TTL)
C.Header Length
D.Protocol
Correct Answer: Time to Live (TTL)
Explanation:
The 'Hop Limit' field in IPv6 serves the same purpose as the 'Time to Live' (TTL) field in IPv4.
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5How many bits are there in an IPv6 address?
IPv6 Addressing
Easy
A.32 bits
B.128 bits
C.256 bits
D.64 bits
Correct Answer: 128 bits
Explanation:
An IPv6 address is 128 bits long, vastly increasing the number of available addresses compared to IPv4.
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6Which of the following is the loopback address in IPv6?
IPv6 Addressing
Easy
A.::1
B.FE80::1
C.::0
D.127.0.0.1
Correct Answer: ::1
Explanation:
In IPv6, the loopback address is represented as ::1, equivalent to 127.0.0.1 in IPv4.
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7What is the primary purpose of Network Address Translation (NAT)?
Network Address Translation (NAT)
Easy
A.To translate private IP addresses to public IP addresses
B.To calculate the shortest routing path
C.To assign MAC addresses to devices
D.To encrypt network traffic
Correct Answer: To translate private IP addresses to public IP addresses
Explanation:
NAT is used to map multiple private IP addresses to a single public IP address, helping to conserve IPv4 addresses.
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8Which table does a NAT router maintain to keep track of translated addresses?
Network Address Translation (NAT)
Easy
A.NAT Translation Table
B.Routing Table
C.ARP Table
D.MAC Table
Correct Answer: NAT Translation Table
Explanation:
A NAT router uses a translation table to map internal private IP and port numbers to external public IP and port numbers.
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9Which device operates primarily at the Network Layer to forward packets between different networks?
NETWORK LAYER: Routing
Easy
A.Repeater
B.Router
C.Hub
D.Switch
Correct Answer: Router
Explanation:
Routers operate at the Network Layer (Layer 3) and forward data packets between different computer networks.
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10What is the main function of the routing table?
NETWORK LAYER: Routing
Easy
A.To determine where to forward a packet based on its destination IP address
B.To assign IP addresses to hosts dynamically
C.To store MAC addresses of local devices
D.To translate domain names to IP addresses
Correct Answer: To determine where to forward a packet based on its destination IP address
Explanation:
A routing table contains rules (routes) that dictate where data packets should be directed to reach their destination.
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11Which of the following is an example of an Interior Gateway Protocol (IGP)?
Unicast routing protocols
Easy
A.HTTP
B.BGP
C.SMTP
D.OSPF
Correct Answer: OSPF
Explanation:
OSPF (Open Shortest Path First) is an Interior Gateway Protocol used for routing within an autonomous system.
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12What does BGP stand for in the context of routing protocols?
Unicast routing protocols
Easy
A.Binary Gateway Protocol
B.Broadband Gateway Protocol
C.Border Gateway Protocol
D.Basic Gateway Protocol
Correct Answer: Border Gateway Protocol
Explanation:
BGP stands for Border Gateway Protocol, which is the primary routing protocol used to route traffic across the internet.
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13What is the primary goal of a routing algorithm?
Routing algorithms
Easy
A.To encrypt data packets
B.To find the best path from a source to a destination
C.To assign IP addresses to devices
D.To detect transmission errors
Correct Answer: To find the best path from a source to a destination
Explanation:
Routing algorithms are designed to determine the most efficient path for a packet to travel from its source to its destination.
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14Which category of routing algorithm updates its routing decisions continuously based on network topology changes?
Routing algorithms
Easy
A.Dynamic Routing
B.Static Routing
C.Fixed Routing
D.Default Routing
Correct Answer: Dynamic Routing
Explanation:
Dynamic routing algorithms adjust automatically to changes in network topology and traffic.
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15Which famous algorithm is commonly used to find the shortest path in network routing?
Routing Algorithm- Shortest path algorithm
Easy
A.RSA Algorithm
B.Dijkstra's Algorithm
C.Bubble Sort
D.Binary Search
Correct Answer: Dijkstra's Algorithm
Explanation:
Dijkstra's algorithm is a well-known shortest path algorithm used extensively in link-state routing protocols like OSPF.
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16In Distance Vector Routing, what does a router share with its immediate neighbors?
Distance vector Routing
Easy
A.Its entire routing table
B.Only the state of its links
C.Its MAC address
D.The entire network topology
Correct Answer: Its entire routing table
Explanation:
In Distance Vector routing protocols (like RIP), routers periodically share their entire routing table with their direct neighbors.
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17What is the 'Count-to-Infinity' problem primarily associated with?
Distance vector Routing
Easy
A.Distance Vector Routing
B.Link State Routing
C.Shortest Path Routing
D.Static Routing
Correct Answer: Distance Vector Routing
Explanation:
The 'Count-to-Infinity' problem is a known routing loop issue in Distance Vector routing protocols.
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18In Link State Routing, what information does each router broadcast to all other routers in the network?
Link State routing
Easy
A.The state of its directly connected links
B.Its entire routing table
C.The MAC addresses of its neighbors
D.Its IP address only
Correct Answer: The state of its directly connected links
Explanation:
In Link State routing, every router floods information about its direct links to all other routers, allowing each to build a complete map of the network.
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19Which routing protocol is a classic example of Link State Routing?
Link State routing
Easy
A.OSPF
B.BGP
C.RIP
D.EIGRP
Correct Answer: OSPF
Explanation:
OSPF (Open Shortest Path First) is a widely used routing protocol based on the Link State algorithm.
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20What is the metric typically used by shortest path algorithms to determine the 'best' path?
Routing Algorithm- Shortest path algorithm
Easy
A.Manufacturer name
B.Cost (e.g., delay, distance, or bandwidth)
C.Device color
D.Operating system version
Correct Answer: Cost (e.g., delay, distance, or bandwidth)
Explanation:
Shortest path algorithms calculate the path with the lowest overall cost, which can be based on metrics like distance, delay, or bandwidth.
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21An IPv4 datagram is fragmented into three pieces. The total length of the original datagram payload was 4000 bytes. If the first two fragments each carry 1480 bytes of payload, what will be the value of the fragment offset field in the third fragment's IPv4 header?
IPv4 Header
Medium
A.1480
B.370
C.185
D.2960
Correct Answer: 370
Explanation:
The fragment offset specifies the starting byte of the payload in the fragment relative to the original payload, measured in units of 8 bytes. The first fragment carries 1480 bytes, the second carries another 1480 bytes. The third fragment starts at byte 2960. The offset value is .
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22If the Header Length (HLEN) field in an IPv4 header contains the binary value 0111, what is the total size of the IPv4 header in bytes, and how many bytes of options are included?
IPv4 Header
Medium
A.28 bytes total, 0 bytes of options
B.32 bytes total, 12 bytes of options
C.28 bytes total, 8 bytes of options
D.7 bytes total, 0 bytes of options
Correct Answer: 28 bytes total, 8 bytes of options
Explanation:
The HLEN field measures the header length in 4-byte words. A value of 0111 (7 in decimal) means the header is bytes. Since the standard base header is 20 bytes, there are bytes of options.
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23In the IPv6 base header, which field replaces the IPv4 'Time to Live' (TTL) field to prevent packets from endlessly circulating in the network?
IPv6 Header
Medium
A.Traffic Class
B.Next Header
C.Flow Label
D.Hop Limit
Correct Answer: Hop Limit
Explanation:
The 'Hop Limit' field in IPv6 serves the exact same purpose as the TTL field in IPv4. It is decremented by 1 by each router that forwards the packet, and the packet is discarded when the limit reaches zero.
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24Which of the following correctly describes how IPv6 handles packet fragmentation?
IPv6 Header
Medium
A.Fragmentation is handled entirely by intermediate routers using the standard IPv6 base header.
B.Fragmentation is performed only by the source node, using a Fragment Extension Header.
C.Fragmentation is performed by the destination node during the reassembly process.
D.Fragmentation is not allowed in IPv6; packets exceeding the MTU are always silently dropped.
Correct Answer: Fragmentation is performed only by the source node, using a Fragment Extension Header.
Explanation:
In IPv6, intermediate routers do not fragment packets to increase processing speed. If fragmentation is necessary, the source node performs it and indicates it using a Fragment Extension Header.
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25Which of the following is the correct, most compressed representation of the IPv6 address 2001:0db8:0000:0000:0001:0000:0000:0001?
IPv6 Addressing
Medium
A.2001:db8::1::1
B.2001:db8::1:0:0:1
C.2001:db8:0:0:1::1
D.2001:db8::1:0:0:1 and 2001:db8:0:0:1::1 are both correct, but 2001:db8::1:0:0:1 is not the most compressed
Correct Answer: 2001:db8::1:0:0:1
Explanation:
The zero compression rule (::) can only be used once in an IPv6 address to avoid ambiguity. Thus, 2001:db8::1:0:0:1 compresses the first block of zeros. Compressing the second block instead (2001:db8:0:0:1::1) is also valid, but the first is standard when block sizes are equal.
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26An IPv6 Anycast address is syntactically indistinguishable from a Unicast address. How does a router know how to route a packet destined for an Anycast address?
IPv6 Addressing
Medium
A.The router broadcasts the packet to all interfaces, and the nearest one accepts it.
B.The router checks the 'Anycast Flag' in the IPv6 base header.
C.The router routes it to the nearest interface sharing that address based on routing protocol metrics.
D.The router queries a DNS server to resolve the Anycast address to a specific MAC address.
Correct Answer: The router routes it to the nearest interface sharing that address based on routing protocol metrics.
Explanation:
Anycast addresses are assigned to multiple interfaces (usually on different nodes). The routing infrastructure routes the packet to the 'nearest' interface possessing that address, according to the routing protocol's measure of distance.
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27In a scenario where multiple internal hosts use a single public IP address to access the Internet simultaneously, which specific mechanism does NAT utilize to distinguish between the return traffic for different internal hosts?
Network Address Translation (NAT)
Medium
A.Modifying the IPv4 Protocol field
B.Static NAT mapping
C.Port Address Translation (PAT) using Layer 4 port numbers
D.Using different MAC addresses for each internal host
Correct Answer: Port Address Translation (PAT) using Layer 4 port numbers
Explanation:
Port Address Translation (PAT), also known as NAT overload, translates both the IP address and the Layer 4 (TCP/UDP) port numbers. This unique combination of public IP and mapped port number allows the router to multiplex multiple private IP streams over a single public IP.
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28A NAT router modifies a packet originating from a private network. Which of the following fields in the IPv4 header MUST the NAT router recalculate before forwarding the packet?
Network Address Translation (NAT)
Medium
A.Fragment Offset
B.Type of Service (ToS)
C.Identification
D.Header Checksum
Correct Answer: Header Checksum
Explanation:
Because NAT changes the Source IP address in the IPv4 header, the router must recalculate the IPv4 Header Checksum to reflect this change; otherwise, the packet will be dropped as corrupt by the next hop.
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29Which of the following best differentiates a 'forwarding table' from a 'routing table' in modern routers?
NETWORK LAYER: Routing
Medium
A.A routing table is used exclusively for static routes, while a forwarding table is used for dynamic protocols like OSPF.
B.A routing table is built by control plane protocols, whereas a forwarding table is used by the data plane to actually switch packets.
C.A routing table contains only MAC addresses, while a forwarding table contains IP addresses.
D.There is no difference; they are exactly the same data structure.
Correct Answer: A routing table is built by control plane protocols, whereas a forwarding table is used by the data plane to actually switch packets.
Explanation:
The routing table (RIB) is created by routing algorithms (control plane). The router extracts the best paths from the routing table to populate the forwarding table (FIB), which is optimized for high-speed packet lookup and switching (data plane).
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30A network administrator needs a routing protocol to exchange routing information between two different Autonomous Systems (AS). Which protocol is designed specifically for this purpose?
Unicast routing protocols
Medium
A.OSPF (Open Shortest Path First)
B.RIP (Routing Information Protocol)
C.IS-IS (Intermediate System to Intermediate System)
D.BGP (Border Gateway Protocol)
Correct Answer: BGP (Border Gateway Protocol)
Explanation:
BGP is an Exterior Gateway Protocol (EGP) designed specifically to exchange routing and reachability information between different Autonomous Systems (AS) on the Internet. OSPF, RIP, and IS-IS are Interior Gateway Protocols (IGPs).
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31In the Routing Information Protocol (RIP), the maximum valid hop count is 15. What happens when a router running RIP receives a routing update for a network with a metric of 15?
Unicast routing protocols
Medium
A.The router accepts the update, but if it forwards traffic to that network, the metric becomes 16, representing infinity/unreachable.
B.The router discards the update as the network is considered unreachable.
C.The router generates an ICMP Time Exceeded message.
D.The router resets the hop count to 0 and forwards it to the next neighbor.
Correct Answer: The router accepts the update, but if it forwards traffic to that network, the metric becomes 16, representing infinity/unreachable.
Explanation:
In RIP, a metric of 16 is defined as infinity (unreachable). If a router receives an update with a metric of 15, the network is 15 hops away from the sender. For the receiving router, it becomes hops away, making it unreachable.
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32Which of the following is a key characteristic that distinguishes dynamic routing algorithms from static routing?
Routing algorithms
Medium
A.Dynamic routing algorithms only work for single-path topologies.
B.Dynamic algorithms do not use routing tables.
C.Static routing causes significantly higher network overhead due to periodic updates.
D.Dynamic algorithms automatically adapt to topology changes and link failures.
Correct Answer: Dynamic algorithms automatically adapt to topology changes and link failures.
Explanation:
Dynamic routing algorithms continuously exchange messages to learn the network topology, allowing them to automatically reroute traffic if a link fails or a new path is added. Static routes must be manually updated by an administrator.
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33Which of the following solutions is NOT used to mitigate the 'Count-to-Infinity' problem commonly found in basic routing algorithms?
Routing algorithms
Medium
A.Split Horizon
B.Poison Reverse
C.Defining a maximum metric (infinity)
D.Link State Advertisements (LSA)
Correct Answer: Link State Advertisements (LSA)
Explanation:
Count-to-infinity is a problem specific to Distance Vector routing. Split Horizon, Poison Reverse, and defining a maximum metric (like 16 in RIP) are solutions for it. LSAs are a feature of Link State routing, which does not suffer from the count-to-infinity problem because routers have a complete topology map.
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34In Dijkstra's shortest path algorithm, what does the set represent during the execution of the algorithm?
Routing Algorithm- Shortest path algorithm
Medium
A.The set of directly connected neighbors of the source node.
B.The set of nodes whose shortest path from the source is definitively known.
C.The set of nodes that are currently unreachable.
D.The set of all nodes in the network graph.
Correct Answer: The set of nodes whose shortest path from the source is definitively known.
Explanation:
In Dijkstra's algorithm, keeps track of the set of nodes for which the minimum cost path from the source node has already been finalized.
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35Consider a network graph where link costs can be negative. Why might Dijkstra's algorithm fail to find the correct shortest path in this scenario?
Routing Algorithm- Shortest path algorithm
Medium
A.Dijkstra's algorithm is a decentralized algorithm and cannot see the entire graph at once.
B.Dijkstra's algorithm assumes that adding an edge to a path can never decrease the total path cost.
C.The algorithm enters an infinite loop when encountering negative link costs.
D.Dijkstra's algorithm only supports hop-count metrics, not arbitrary cost values.
Correct Answer: Dijkstra's algorithm assumes that adding an edge to a path can never decrease the total path cost.
Explanation:
Dijkstra's algorithm uses a greedy approach, definitively closing nodes (adding them to ) under the assumption that path costs only increase. Negative weights break this assumption, making the Bellman-Ford algorithm a better choice.
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36According to the Bellman-Ford equation used in Distance Vector Routing, the cost of the least-cost path from node to node is computed as:
Distance vector Routing
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Bellman-Ford equation states that the minimum cost from to is the minimum over all neighbors of the cost to reach , , plus the minimum cost from to , .
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37How does the 'Split Horizon' rule improve the performance of Distance Vector routing?
Distance vector Routing
Medium
A.It splits the routing table into smaller segments to speed up lookup times.
B.It forces routers to exchange routing tables only with routers in different Autonomous Systems.
C.It prevents a router from advertising a route back out the same interface from which the route was learned, preventing two-node routing loops.
D.It ensures that updates are sent only when a topology change occurs, rather than periodically.
Correct Answer: It prevents a router from advertising a route back out the same interface from which the route was learned, preventing two-node routing loops.
Explanation:
Split Horizon dictates that if node A learns about a route to network X through node B, node A will not advertise that route back to node B. This prevents immediate routing loops between two adjacent routers.
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38In Link State routing, which mechanism is used to ensure that all routers in the domain eventually receive the same routing information?
Link State routing
Medium
A.Reliable Flooding of Link State Advertisements (LSAs)
B.Reverse Path Forwarding (RPF)
C.Spanning Tree Protocol (STP)
D.Distance Vector exchanging
Correct Answer: Reliable Flooding of Link State Advertisements (LSAs)
Explanation:
Link State routing protocols (like OSPF) rely on routers generating Link State Advertisements (LSAs) and reliably flooding them to all other routers in the area, ensuring everyone builds identical topological maps.
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39What is the primary computational advantage of Link State (LS) routing over Distance Vector (DV) routing?
Link State routing
Medium
A.LS routing requires significantly less memory on each router compared to DV routing.
B.LS routing protocols do not require the use of any shortest-path algorithms.
C.LS routing is less susceptible to routing loops because each router computes the shortest path using a complete topology map.
D.LS routers only need to communicate with their immediate neighbors, drastically reducing network traffic.
Correct Answer: LS routing is less susceptible to routing loops because each router computes the shortest path using a complete topology map.
Explanation:
In LS routing, every router possesses the entire network graph and computes paths independently (e.g., via Dijkstra's). This complete view prevents issues like the count-to-infinity problem and transient routing loops common in DV routing, where routers only have partial views.
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40When a router receives a packet destined for an IP address that does not match any specific subnet entry in its routing table, what action does it take?
NETWORK LAYER: Routing
Medium
A.It encapsulates the packet into a DNS request to find the proper route.
B.It forwards the packet to the Default Route (e.g., 0.0.0.0/0), if configured.
C.It broadcasts the packet to all active interfaces.
D.It randomly selects a next-hop router to load balance unknown traffic.
Correct Answer: It forwards the packet to the Default Route (e.g., 0.0.0.0/0), if configured.
Explanation:
The default route (represented as 0.0.0.0/0 in IPv4) acts as a catch-all gateway. If no specific match is found in the routing table (longest prefix match), the router forwards the packet to the default gateway.
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41An IPv4 datagram of size 4000 bytes (including a standard 20-byte header) is to be forwarded over a link with an MTU of 1500 bytes. Which of the following correctly describes the fragmentation offset and the More Fragments (MF) flag of the third fragment?
IPv4 Header
Hard
A.Offset = 370, MF = 0
B.Offset = 185, MF = 0
C.Offset = 370, MF = 1
D.Offset = 2960, MF = 0
Correct Answer: Offset = 370, MF = 0
Explanation:
The payload size is 3980 bytes. The link MTU is 1500 bytes, so max payload per fragment is 1480 bytes (must be a multiple of 8). Fragment 1: 1480 bytes, Offset = 0, MF = 1. Fragment 2: 1480 bytes, Offset = 1480/8 = 185, MF = 1. Fragment 3: 1020 bytes, Offset = 2960/8 = 370. Since it's the last fragment, MF = 0.
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42When a router receives an IPv4 packet and decrements the Time to Live (TTL) field by 1, how is the IPv4 header checksum most efficiently updated without recomputing the entire checksum?
IPv4 Header
Hard
A.By adding $256$ (0x0100) to the existing checksum using one's complement arithmetic.
B.By subtracting $1$ from the checksum using two's complement arithmetic.
C.By adding $1$ to the existing checksum using one's complement arithmetic.
D.By applying an XOR operation between the old TTL and the new TTL.
Correct Answer: By adding $256$ (0x0100) to the existing checksum using one's complement arithmetic.
Explanation:
According to RFC 1141, since the TTL is the high byte of the 16-bit word (TTL, Protocol), decrementing the TTL by 1 subtracts 0x0100 from the header. To maintain the checksum, 0x0100 must be added to the old checksum using one's complement arithmetic.
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43In IPv6, if a packet requires Hop-by-Hop Options, Routing, and Fragment extension headers, what is the mandatory order in which these headers must appear after the main IPv6 header?
IPv6 Header
Hard
A.Routing, Hop-by-Hop Options, Fragment
B.Hop-by-Hop Options, Routing, Fragment
C.Fragment, Routing, Hop-by-Hop Options
D.Hop-by-Hop Options, Fragment, Routing
Correct Answer: Hop-by-Hop Options, Routing, Fragment
Explanation:
RFC 8200 specifies the recommended order of extension headers: Hop-by-Hop Options (which must be first if present), Destination Options (for routing), Routing, Fragment, Authentication, Encapsulating Security Payload, Destination Options, and finally upper-layer headers.
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44What is the primary implication of a router altering the 20-bit Flow Label field in an IPv6 header during transit?
IPv6 Header
Hard
A.It is permissible only if the router is performing NAT64.
C.It triggers an ICMPv6 Parameter Problem message.
D.It violates RFC 6437, as the Flow Label must be delivered unchanged to maintain flow state.
Correct Answer: It violates RFC 6437, as the Flow Label must be delivered unchanged to maintain flow state.
Explanation:
According to RFC 6437, the IPv6 Flow Label is an end-to-end field. Forwarding nodes must not change the Flow Label value if it is non-zero, to ensure consistent flow identification for QoS and load balancing (like ECMP).
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45What is the Solicited-Node Multicast Address for an IPv6 interface assigned the global unicast address 2001:db8:85a3::8a2e:370:7334?
IPv6 Addressing
Hard
A.ff02::1:ff37:7334
B.ff02::1:ff70:7334
C.ff02::1:ff00:7334
D.ff02::1:ff8a:2e37
Correct Answer: ff02::1:ff70:7334
Explanation:
The Solicited-Node Multicast Address is formed by appending the lower 24 bits (6 hex digits) of the IPv6 address to the prefix ff02::1:ff00:0/104. The lower 24 bits of the given address are 70:7334. Therefore, the address is ff02::1:ff70:7334.
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46A host generates an IPv6 interface identifier using the modified EUI-64 format from its MAC address 00:1A:2B:3C:4D:5E. What will be the resulting 64-bit interface identifier?
IPv6 Addressing
Hard
A.001a:2bff:fe3c:4d5e
B.021a:2bfe:ff3c:4d5e
C.021a:2bff:fe3c:4d5e
D.021a:2b3c:4d5e:ffff
Correct Answer: 021a:2bff:fe3c:4d5e
Explanation:
To form a modified EUI-64 ID, ff:fe is inserted in the middle of the MAC address (00:1A:2B:FF:FE:3C:4D:5E), and the Universal/Local (U/L) bit (the 7th bit of the first byte) is inverted. The first byte 00 (binary 00000000) becomes 02 (binary 00000010). The result is 021a:2bff:fe3c:4d5e.
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47In the context of STUN and WebRTC, which type of NAT creates a unique external port mapping not just for the internal IP/port, but specifically for each distinct external destination IP and port combo, thus breaking standard UDP hole punching?
Network Address Translation (NAT)
Hard
A.Restricted Cone NAT
B.Full Cone NAT
C.Symmetric NAT
D.Port-Restricted Cone NAT
Correct Answer: Symmetric NAT
Explanation:
Symmetric NAT assigns a new external port for every new destination IP and port that the internal host contacts. Because the external port used to contact a STUN server will differ from the one used to contact a peer, standard UDP hole punching fails.
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48A network utilizes 'Hairpinning' (NAT loopback). An internal host tries to reach internal host using host 's external (public) IP. Which of the following best describes the packet modification at the NAT router?
Network Address Translation (NAT)
Hard
A.The router alters both the source IP (to its own internal IP) and the destination IP (to host B's internal IP).
B.The router alters only the source IP to its public IP.
C.The router alters only the destination IP to host B's internal IP.
D.The router drops the packet because private IP addresses cannot route to public IPs on the same interface.
Correct Answer: The router alters both the source IP (to its own internal IP) and the destination IP (to host B's internal IP).
Explanation:
For hairpinning to work correctly without asynchronous routing (which breaks TCP), the NAT router must translate the destination IP to B's private IP, AND translate the source IP to the router's internal IP (SNAT). If it didn't SNAT, B would reply directly to A, but A expects a reply from the public IP, dropping the packet.
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49A router's forwarding table contains the following entries: 192.168.0.0/16 via Interface 1, 192.168.128.0/17 via Interface 2, and 192.168.192.0/18 via Interface 3. A packet arrives destined for 192.168.200.5. Which interface will it use, and why?
NETWORK LAYER: Routing
Hard
A.Interface 1, because /16 encompasses the entire class C range.
B.Interface 3, due to Longest Prefix Match (LPM).
C.Interface 2, because 128.0/17 covers 200.5 and has a higher priority than /18.
D.It will be load-balanced between Interface 2 and 3.
Correct Answer: Interface 3, due to Longest Prefix Match (LPM).
Explanation:
The destination 192.168.200.5 matches all three prefixes. 192.168.192.0/18 covers from 192.168.192.0 to 192.168.255.255. Since /18 is the most specific match (longest prefix length), the router forwards the packet out Interface 3.
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50Consider a router configured with policy-based routing (PBR). If a packet matches both a destination-based route in the routing table (RIB) and a PBR route-map specifying a next-hop based on the source IP, what is the default behavior in most commercial routers (e.g., Cisco)?
NETWORK LAYER: Routing
Hard
A.The RIB route takes precedence because destination routing is strictly prioritized at the network layer.
B.The packet is dropped due to routing ambiguity.
C.The router applies ECMP, load-balancing between the PBR next-hop and the RIB next-hop.
D.The PBR route-map takes precedence, overriding the destination-based RIB entry.
Correct Answer: The PBR route-map takes precedence, overriding the destination-based RIB entry.
Explanation:
Policy-Based Routing (PBR) intercepts packets before they are routed using the standard routing table (RIB). If a packet matches a PBR policy, the policy dictates the next hop, overriding the destination-based routing lookup.
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51In the Border Gateway Protocol (BGP), what is the primary purpose of the 'iBGP split-horizon' rule, and what architectural solution is typically implemented to bypass its scaling limitations?
Unicast routing protocols
Hard
A.It prevents loops between different ASes; solved by using the AS-PATH attribute.
B.It prevents routing loops within a single AS by dropping routes learned from one iBGP peer if sent to another; solved by Route Reflectors or BGP Confederations.
C.It ensures symmetric routing; solved by Multi-Exit Discriminator (MED).
D.It prevents routes learned from eBGP from being advertised to iBGP; solved by Next-Hop-Self.
Correct Answer: It prevents routing loops within a single AS by dropping routes learned from one iBGP peer if sent to another; solved by Route Reflectors or BGP Confederations.
Explanation:
The iBGP split-horizon rule states that a router cannot advertise a route learned from an iBGP neighbor to another iBGP neighbor. This prevents loops but requires a full mesh of iBGP peers. Route Reflectors and Confederations are used to break this full-mesh requirement for scalability.
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52In OSPF, a Not-So-Stubby Area (NSSA) allows the injection of external routes while retaining stub area characteristics. Which LSA type is uniquely generated by an ASBR within an NSSA to advertise these external routes, and what happens to it at the Area Border Router (ABR)?
Unicast routing protocols
Hard
A.Type 7 LSA; the ABR translates it into a Type 5 LSA before flooding it into the backbone.
B.Type 7 LSA; the ABR translates it into a Type 3 LSA to hide external topology.
C.Type 5 LSA; the ABR floods it unchanged into the backbone (Area 0).
D.Type 4 LSA; the ABR translates it into a Type 3 LSA for inter-area routing.
Correct Answer: Type 7 LSA; the ABR translates it into a Type 5 LSA before flooding it into the backbone.
Explanation:
NSSAs use Type 7 LSAs to carry external route information. Because Type 5 LSAs are not allowed in stub/NSSA areas, the ASBR generates a Type 7 LSA. When this LSA reaches the ABR, the ABR translates it into a standard Type 5 external LSA to be injected into Area 0.
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53Which of the following scenarios describes a topology where the 'Split Horizon with Poisoned Reverse' mechanism fails to prevent the Count-to-Infinity problem?
Routing algorithms
Hard
A.A linear network (A-B-C-D) where the connection between A and B fails.
B.A two-node network where the link between the nodes fails.
C.A three-node network with a routing loop involving all three nodes (a ring topology).
D.Split Horizon with Poisoned Reverse is mathematically proven to prevent Count-to-Infinity in all network topologies.
Correct Answer: A three-node network with a routing loop involving all three nodes (a ring topology).
Explanation:
Split Horizon with Poisoned Reverse only breaks two-node routing loops (e.g., A routes through B, and B routes through A). If a loop involves three or more nodes (A points to B, B points to C, C points to A), the poison reverse messages do not propagate across the entire loop simultaneously, and count-to-infinity can still occur.
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54Consider a Distance Vector routing network. Node X routes traffic to Destination D via Node Y. Node X's current distance to D is . If X receives an update from Node Z indicating a path to D with cost , under which specific condition MUST X unconditionally accept Z's update, regardless of whether ?
Routing algorithms
Hard
A.X must accept it if the sequence number from Z is strictly less than X's current sequence number.
B.X must accept it if Z has a lower router ID than Y.
C.X must accept it if Z is the current next-hop for destination D (i.e., Y = Z).
D.X must never accept an update that increases its cost.
Correct Answer: X must accept it if Z is the current next-hop for destination D (i.e., Y = Z).
Explanation:
In Distance Vector algorithms, if the router providing the update is the CURRENT next-hop to the destination, the receiving router must accept the update even if the metric increases. This indicates that the path through that next-hop has degraded, and the router must update its table to reflect reality.
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55In an implementation of Link-State Routing, Dijkstra's algorithm is run using a Fibonacci Heap priority queue. For a network with routers and links, what is the asymptotic time complexity of computing the shortest path tree?
Routing Algorithm- Shortest path algorithm
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Using a Fibonacci heap, the extract-min operation takes amortized time, and the decrease-key operation takes amortized time. Since extract-min is called times and decrease-key is called at most times, the overall time complexity is .
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56Equal-Cost Multi-Path (ECMP) routing requires routers to identify multiple shortest paths of identical cost. How must Dijkstra's algorithm be modified to support ECMP?
Routing Algorithm- Shortest path algorithm
Hard
A.Instead of keeping a single predecessor for each node, a set/list of predecessors must be maintained for nodes when discovering an equal-cost path.
B.The algorithm must run a second time in reverse (Destination to Source) to verify symmetric paths.
C.The distance array must be changed to allow negative weights.
D.The priority queue must pop all nodes with the same cost simultaneously.
Correct Answer: Instead of keeping a single predecessor for each node, a set/list of predecessors must be maintained for nodes when discovering an equal-cost path.
Explanation:
Standard Dijkstra's tracks a single previous_node to construct the shortest path tree. To find all shortest paths (for ECMP), when the algorithm encounters a path to a node that exactly equals the current known shortest distance, it adds the new preceding node to a list of predecessors rather than overwriting the single predecessor.
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57In the Distributed Bellman-Ford algorithm used by Distance Vector routing, a link cost increases significantly. Why does this trigger the 'Bad News Travels Slowly' phenomenon?
Distance vector Routing
Hard
A.Negative edge weights cause the algorithm to stall until the network administrator clears the route.
B.Nodes iteratively rely on each other's outdated information, incrementing metrics gradually until infinity is reached.
C.Routers must wait for hold-down timers to expire before processing any updates.
D.The algorithm requires a full topology sync, consuming massive bandwidth.
Correct Answer: Nodes iteratively rely on each other's outdated information, incrementing metrics gradually until infinity is reached.
Explanation:
When a link metric increases (or fails), nodes adjacent to the link may falsely believe they can route through their neighbors, completely unaware that those neighbors were relying on them for the path. They bounce increasing metrics back and forth (count-to-infinity) until the metric reaches the defined maximum ('infinity').
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58What happens if a negative weight cycle exists in a routing domain utilizing the Bellman-Ford algorithm, and how is this handled in practical Distance Vector protocols like RIP?
Distance vector Routing
Hard
A.The algorithm fails to converge as path costs decrease infinitely. RIP avoids this inherently because network links cannot have negative latencies or costs.
B.The algorithm flips the negative weights to positive. RIP uses absolute values of link metrics.
C.The algorithm terminates early. RIP handles this using split horizon.
D.The algorithm computes a path of zero cost. RIP handles this by setting a floor value of 0.
Correct Answer: The algorithm fails to converge as path costs decrease infinitely. RIP avoids this inherently because network links cannot have negative latencies or costs.
Explanation:
In graph theory, Bellman-Ford cannot find shortest paths if a negative weight cycle exists because walking the cycle endlessly decreases the cost. In practical routing protocols like RIP, link metrics represent physical realities (hops, delay) which are strictly positive, making negative weight cycles impossible.
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59In Link State routing, Sequence Numbers are used to distinguish new LSAs from old ones. To prevent the sequence number wrap-around problem (where an LSA with a small sequence number is incorrectly deemed newer after the space wraps), which specific arithmetic space is typically employed by modern protocols like OSPF?
Link State routing
Hard
A.Linear sequence space starting from 0x80000001 to 0x7FFFFFFF, using signed 32-bit integers with a MaxAge timer.
B.Lollipop sequence space, starting with negative numbers and transitioning to a circular space.
C.Unsigned 64-bit sequence space to ensure wrap-around never practically happens.
D.Linear sequence space with an absolute hard stop at , requiring router reboot.
Correct Answer: Linear sequence space starting from 0x80000001 to 0x7FFFFFFF, using signed 32-bit integers with a MaxAge timer.
Explanation:
OSPF uses a linear sequence space utilizing signed 32-bit numbers, starting at 0x80000001 (Initial) and ending at 0x7FFFFFFF (Max). It relies on the MaxAge timer to age out old LSAs before sequence wrap-around can cause conflicts, rather than using the older lollipop method (used in early ARPANET).
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60During a massive network instability event (a 'broadcast storm' of topology changes), how do Link State (LS) and Distance Vector (DV) protocols fundamentally differ in their computational degradation at the router level?
Link State routing
Hard
A.Both protocols suffer identically because they both use Bellman-Ford algorithms under the hood.
B.LS limits updates to immediate neighbors, saving CPU; DV floods the entire network, saving memory.
C.LS suffers severe CPU spikes due to repeated Dijkstra executions; DV suffers network congestion due to full routing table exchanges.
D.LS degrades primarily in memory due to table expansions; DV degrades in CPU due to complex matrix multiplications.
Correct Answer: LS suffers severe CPU spikes due to repeated Dijkstra executions; DV suffers network congestion due to full routing table exchanges.
Explanation:
In a highly unstable network, an LS protocol receives many tiny topology changes (LSAs) and must repeatedly re-run the CPU-intensive Dijkstra's algorithm. A DV protocol reacts by frequently broadcasting its entire (or partial) routing table to neighbors, which eats up link bandwidth and causes network congestion (Bad news travels slowly/routing loops).