Chapter 8
Advance Internetworking (old course)
TU BIM past exam questions
Past questions and answers
17 questions set from this chapter. Most repeated first.
- 2016 AD Model · 1 mark
List the function of Network Layer.
Answer
- Logical addressing (IP addresses of source and destination)
- Routing: finding the best path from source to destination
- Packet forwarding from input to output interface
- Fragmentation and reassembly of packets
- Host-to-host (source to destination) packet delivery
- Congestion control and error reporting (e.g. ICMP)
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- 2016 AD Model · 1 mark
List the issues of IPv4.
Answer
- Address exhaustion: the 32-bit space (about addresses) is too small
- Large routing tables because of weak hierarchical addressing
- No built-in security (IPsec is optional)
- No quality-of-service support for real-time traffic (audio, video)
- Header of variable length with options, and checksum recomputed at every router, which slows processing
- Needs manual or DHCP configuration and relies on NAT as a workaround
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- 2016 AD Model · 1 mark
Define Routing.
Answer
Routing is the process of selecting the best path for packets to travel from a source to a destination across one or more networks. Routers do this using routing tables built by static configuration or routing protocols (RIP, OSPF, BGP).
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- 2016 AD Model · 1 mark
Define multicast.
Answer
Multicast is one-to-many (or many-to-many) communication in which a packet is sent once to a group of interested receivers identified by a group address (IPv4 class D, 224.0.0.0 to 239.255.255.255), rather than separately to each receiver.
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- 2016 AD Model · 1 mark
Differentiate between Integrated and differentiate service.
Answer
| Integrated Services (IntServ) | Differentiated Services (DiffServ) |
|---|---|
| Per-flow quality of service, with resources reserved for each flow | Per-class QoS; packets are marked and aggregated into classes |
| Uses RSVP signalling to reserve resources end to end | No signalling; the DS field (DSCP) in the IP header is used |
| Routers must keep per-flow state, so it scales poorly | Stateless core, scales well |
| Gives hard guarantees | Gives relative (statistical) priority |
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- 2016 AD Model · 1 mark
Define playback buffer.
Answer
A playback buffer is a buffer at the receiver of streaming audio or video that stores data before it is played. It absorbs delay variation (jitter) in packet arrival so that playback is smooth, at the cost of a small start-up delay.
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- 2016 AD Model · 1 mark
What is virtual circuit forwarding table?
Answer
In virtual-circuit switching, each switch keeps a table whose entry maps an incoming interface and incoming VCI to an outgoing interface and outgoing VCI: (In-interface, In-VCI) to (Out-interface, Out-VCI). A packet carries only the VCI, and the switch uses this table to forward it and rewrite the VCI. Entries are created when the circuit is set up and removed when it is torn down.
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- 2016 AD Model · 1 mark
Differentiate between client server and peer to peer architecture.
Answer
| Client-server | Peer-to-peer (P2P) |
|---|---|
| Dedicated server provides services; clients request them | All nodes (peers) act as both client and server |
| Centralized control and management | Decentralized |
| Server is a single point of failure and bottleneck | More robust; no single point of failure |
| Scaling needs more server capacity | Capacity grows as peers join |
| Easier to secure and back up | Harder to secure and control |
| Examples: web, email | Examples: BitTorrent, file sharing |
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- 2016 AD Model · 1 mark
What is multi homing.
Answer
Multi-homing is connecting a host or network to more than one network or ISP at the same time, to provide redundancy (backup if one link fails), better performance and load sharing.
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- 2016 AD Model · 1 mark
What is IP aggregation?
Answer
IP (route) aggregation, also called route summarization or supernetting, combines several contiguous address blocks into one larger prefix, e.g. 192.168.0.0/24 to 192.168.3.0/24 become 192.168.0.0/22. It reduces the size of routing tables and the routing updates exchanged.
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- 2016 AD Model · 4 marks
How Packet loss in RTP protocol can be recovered?
Answer
RTP (Real-time Transport Protocol) itself gives no retransmission or guaranteed delivery. It numbers packets with sequence numbers so the receiver can detect loss, and loss is hidden or recovered by application-level techniques:
- Forward Error Correction (FEC): the sender adds redundant packets, e.g. an XOR of every packets, so one lost packet among the group can be rebuilt without retransmission.
- Interleaving: units of audio/video are rearranged before sending so a lost packet causes small scattered gaps rather than one large gap, which is easier to conceal.
- Redundant (piggybacked) encoding: a lower-quality copy of the previous packet is carried in the next packet, so if one is lost the next packet supplies a replacement.
- Receiver-based concealment: the receiver repeats the previous packet, interpolates or substitutes silence/noise, using sequence numbers and timestamps to detect the gap.
- Selective retransmission: if delay allows, the receiver reports loss through RTCP (receiver reports / NACK) and the sender resends the lost packet or adapts its rate.
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- 2016 AD Model · 4 marks
From the following information, map IP multicast address with ethernet address.
Physical Address: 68-94-23-40-E9-9C
IP multicast Address :229.122.54.99
Answer
Rule: an IP multicast address is mapped to an Ethernet multicast address by placing the low-order 23 bits of the IP address into the low-order 23 bits of the Ethernet address 01-00-5E-00-00-00. (The physical address 68-94-23-40-E9-9C is a unicast address of the host; it plays no part in the mapping.)
IP multicast address: 229.122.54.99
The low 23 bits come from the last three octets 122.54.99, with the top bit of the second octet cleared (122 is less than 128, so it is unchanged):
- 122 = 7A
- 54 = 36
- 99 = 63
Ethernet multicast address:
Since 5 bits of the IP group address (the upper 5 of the 28-bit group ID) are ignored, 32 different multicast IP addresses map to the same Ethernet address.
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- 2016 AD Model · 4 marks
If you are assigned an IP address 92.16.1.0/24 and plans to deploy CIDR. Here are some requirement which you have to fulfill for Subnet A = 120 Host, Subnet B = 60 Host, Subnet C = 30 Host, Subnet D = 10 Host Subnet E=5. You are also required to calculate subnetmask, range, netid, broadcast id for each subnet.
Answer
Given block 92.16.1.0/24 (256 addresses). Allocate the largest subnet first (VLSM). Host requirement needs hosts.
| Subnet | Hosts | Block size | Mask |
|---|---|---|---|
| A | 120 | 128 (, 126 usable) | /25 = 255.255.255.128 |
| B | 60 | 64 (, 62 usable) | /26 = 255.255.255.192 |
| C | 30 | 32 (, 30 usable) | /27 = 255.255.255.224 |
| D | 10 | 16 (, 14 usable) | /28 = 255.255.255.240 |
| E | 5 | 8 (, 6 usable) | /29 = 255.255.255.248 |
| Subnet | Network ID | Host range | Broadcast ID |
|---|---|---|---|
| A | 92.16.1.0/25 | 92.16.1.1 to 92.16.1.126 | 92.16.1.127 |
| B | 92.16.1.128/26 | 92.16.1.129 to 92.16.1.190 | 92.16.1.191 |
| C | 92.16.1.192/27 | 92.16.1.193 to 92.16.1.222 | 92.16.1.223 |
| D | 92.16.1.224/28 | 92.16.1.225 to 92.16.1.238 | 92.16.1.239 |
| E | 92.16.1.240/29 | 92.16.1.241 to 92.16.1.246 | 92.16.1.247 |
Addresses 92.16.1.248 to 92.16.1.255 remain unused for future growth.
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- 2016 AD Model · 4 marks
Identify the first six /42 address blocks out of ABCD:EFAB::/32
Answer
The block is ABCD:EFAB::/32. A /42 block uses 10 more bits (bits 33 to 42) which fall in the top 10 bits of the third 16-bit group. Consecutive /42 blocks therefore differ by in the third group.
The first six /42 blocks:
ABCD:EFAB:0000::/42ABCD:EFAB:0040::/42ABCD:EFAB:0080::/42ABCD:EFAB:00C0::/42ABCD:EFAB:0100::/42ABCD:EFAB:0140::/42
Each block has addresses, and there are /42 blocks in the /32.
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- 2016 AD Model · 4 marks
From the following figure find the shortest path from node S to all other nodes.
Answer
The shortest paths from source S to all other nodes are found with Dijkstra's algorithm (link-state shortest path), which is used in OSPF.
Algorithm
- Set and for every other node; the permanent set .
- Initialise for the neighbours of S.
- Repeat: choose the node with the smallest , add it to , and for each neighbour of not in update , recording as the predecessor when the value improves.
- Stop when all nodes are in . Trace the predecessors back to S to get each path.
Worked illustration. Take the link costs S-A = 2, S-B = 5, A-B = 1, A-C = 4, B-C = 2, B-D = 3, C-D = 1.
| Step | Node added | D(A) | D(B) | D(C) | D(D) |
|---|---|---|---|---|---|
| 0 | S | 2 | 5 | inf | inf |
| 1 | A | 2 | 3 (via A) | 6 (via A) | inf |
| 2 | B | 2 | 3 | 5 (via B) | 6 (via B) |
| 3 | C | 2 | 3 | 5 | 6 (via B or C) |
| 4 | D | 2 | 3 | 5 | 6 |
Shortest paths from S: S-A (cost 2); S-A-B (cost 3); S-A-B-C (cost 5); S-A-B-D (cost 6).
For the figure in the question, the same table is built with its link costs; the node with the minimum label is made permanent at each step until all nodes are reached.
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- 2016 AD Model · 5 marks
Explain DCCP Connection Initiation and Connection Termination process. Explain how congestion is controlled with CCID 2.
Answer
DCCP (Datagram Congestion Control Protocol) is a connection-oriented, unreliable transport protocol that provides congestion control without guaranteed delivery, suited to streaming media and telephony.
Connection initiation (three-way handshake plus data)
- The client sends DCCP-Request (with a service code) and moves from CLOSED to REQUEST.
- The server, in LISTEN state, replies with DCCP-Response (acknowledging the request) and moves to RESPOND.
- The client sends DCCP-Ack (or a DCCP-DataAck carrying data); both ends move to OPEN. During the handshake the endpoints can negotiate features such as the congestion control ID (CCID).
Connection termination
- The end that wants to close (usually the server) sends DCCP-CloseReq, asking the other end to close. The other end replies with DCCP-Close, and the first sends DCCP-Reset to complete the termination.
- Alternatively, either side sends DCCP-Close directly; the peer answers with DCCP-Reset (reason "Closed") and the connection is released, the closing side waiting in a TIMEWAIT state (about 2 maximum segment lifetimes) so old packets expire.
- DCCP-Reset is also sent to abort a connection abnormally.
Congestion control with CCID 2 (TCP-like)
CCID 2 uses an AIMD window scheme similar to TCP:
- The sender keeps a congestion window
cwndand a slow-start thresholdssthresh. - Receivers send Ack Vectors (in DCCP-Ack) that report which packets arrived, ECN-marked, or were lost.
- Slow start:
cwnddoubles each round trip untilssthresh; then congestion avoidance increasescwndby one packet per RTT (additive increase). - On loss or an ECN mark (detected through the Ack Vector)
cwndandssthreshare halved (multiplicative decrease), at most once per window. - A timeout reduces
cwndto 1. - Lost packets are not retransmitted, but the sending rate is controlled, giving a bursty, rapidly changing rate, which suits applications that want to use available bandwidth quickly.
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- 2016 AD Model · 5 marks
List the difference between peer to peer network and client server network. Explain various peer discovering methods in P2P network.
Answer
Differences between peer-to-peer (P2P) and client-server networks
| Basis | Peer-to-peer | Client-server |
|---|---|---|
| Roles | Every node is both client and server | Separate dedicated servers and clients |
| Control | Decentralised, no central authority | Centralised control of resources and users |
| Cost | Cheap, no special server hardware or NOS | Expensive, needs powerful server and NOS |
| Scalability | Performance falls as the network grows; suits small networks | Scales well to large networks |
| Security | Weak, each peer manages its own security | Strong, central security and user accounts |
| Backup | Done on each machine | Done centrally |
| Reliability | No single point of failure, but data availability depends on peers | Server failure stops the service |
| Examples | BitTorrent, Gnutella, small office LAN | Web, email, banking systems |
Peer discovery methods in a P2P network
Peer discovery is how a new node finds other peers to join the network.
- Centralised directory (index server): A central server keeps a list of peers and the resources they share. A new peer registers with it and queries it to find others, as in Napster. It is simple and fast, but the server is a single point of failure.
- Flooding / broadcast query: A peer sends its query to all its neighbours, which forward it to their neighbours until a TTL (time to live) expires, as in early Gnutella. It needs no central server but creates heavy traffic.
- Bootstrap nodes / well-known peers: A new peer connects to a pre-configured list of known peers (hard-coded addresses, web caches or DNS seeds). It then learns more peers from them. Gnutella and Bitcoin use this.
- Distributed Hash Table (DHT): Peers and resources get IDs in a common key space, and each peer keeps a routing table. A lookup is routed to the responsible peer in about hops, as in Chord, Kademlia and Pastry.
- Peer exchange (gossip): Connected peers exchange lists of other peers they know, so the neighbour list grows without a central server (e.g. PEX in BitTorrent).
- Tracker-based discovery: In BitTorrent, a tracker returns the list of peers sharing a given file.
- Local broadcast / multicast: On a LAN, a peer announces itself using broadcast or multicast messages (e.g. mDNS, UPnP/SSDP).
Answer checked by Claude Sonnet · Oct 21, 2026
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Questions from TU BIM Fifth Semester IT 228 Artificial Intelligence Model Question 2024 (Faculty of Management model question 2024, BIM fifth semester, via EducateNepal), TU BIM Fifth Semester IT 242 Software Design and Development Model Question 2024 (Faculty of Management model question 2024, BIM fifth semester, via EducateNepal), TU BIM Fifth Semester IT 243 Programming with Python Model Question 2024 (Faculty of Management model question 2024, BIM fifth semester, via EducateNepal), TU BIM Fifth Semester IT 244 Information Security Model Question 2024 (Faculty of Management model question 2024, BIM fifth semester, via EducateNepal), TU BIM Fifth Semester MKT 201 Fundamentals of Marketing Model Question 2024 (Faculty of Management model question 2024, BIM fifth semester, via EducateNepal), TU BIM Fifth Semester IT 221 Computer Graphics 2016 (model question) (Faculty of Management model question 2016 (old course), via EducateNepal), TU BIM Fifth Semester IT 222 Java Programming II 2016 (model question) (Faculty of Management model question 2016 (old course), via EducateNepal) and TU BIM Fifth Semester IT 223 Advance Internetworking 2016 (model question) (Faculty of Management model question 2016 (old course), via EducateNepal). Answers are written for this site; check them against your class notes.