1) If a packet is to be delivered to exactly one host in a network, what kind of address should be used to specify the destination?

a. Unicast address
b. Broadcast address
c. Anycast address
d. None of these

Answer: c

Solution: Unicast address is used if a packet is to be delivered to a specific host. Broadcast address is used if a packet has to be delivered to all the hosts within a network or subnetwork. Anycast address is used if a packet has to be delivered to exactly one of the hosts in a network or subnetwork. Hence, the correct option is (c).

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2) What is the purpose of HELLO packets in the OSPF protocol?

a. Keep the links between neighbor routers alive.
b. Update routing table by exchanging information between neighbors.
c. For broadcasting link state advertisement information.
d. All of these.

Answer: a

Solution: A HELLO packet received from a neighboring router indicates that the corresponding communication link is up and running. In the OSPF protocol, if the HELLO packet is not received for 40 seconds, it indicates failure of the neighbor or the communication link. Hence the correct option is (a).

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3) When an entire IPv6 packet is included as payload inside an IPv4 packet, it is called

a. Encapsulation
b. Tunneling
c. Decapsulation
d. None of these.

Answer: b

Solution: When entire IPv6 packets are encapsulated within IPv4 packets, it is called tunneling. The IPv6 packet gets transmitted as data over an IPv4 network. Hence, the correct option is (b).

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4) Consider the following routing table in a router. On which interface will an IP packet with destination address 144.25.64.120 be forwarded?
Destination Subnet Mask Interface
144.25.0.0 255.255.0.0 Eth0
144.25.64.0 255.255.224.0 Eth1
144.25.68.0 255.255.255.0 Eth2
144.25.68.64 255.255.255.224 Eth3
default 0.0.0.0 Eth2

a. Eth0
b. Eth1
c. Eth2
d. Eth3

Answer: b

Solution:
Row 1: 144.25.64.120 AND 255.255.0.0 = 144.25.0.0  Matches with destination address
Row 2: 144.25.64.120 AND 255.255.224.0 = 144.25.64.0  Matches with destination address
Row 3: 144.25.64.120 AND 255.255.255.0 = 144.25.64.0  No match
Row 4: 144.25.64.120 AND 255.255.255.224 = 144.25.64.112  No match
Row 2 provides the longest prefix match; hence the packet will be forwarded to Eth1.

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5) An entry in the routing table has 135.46.56.0 as the destination and /22 as the subnet mask. What will be the network address?

a. 135.46.56.0
b. 135.46.0.0
c. 135.46.48.0
d. None of these.

Answer: a

Solution: In binary notation, 135.46.56.0 = 10000111 00101110 00111000 00000000
If we use /22 as the subnet mask, this means that the first 22 bits of the address must be used to get the network address. If we do this, we get 10000111 00101110 00111000 00000000 = 135.46.56.0 Hence, correct option is (a).