๐Ÿ“ก Networking Midterm โ€” Intensive Study Guide (33 Question Type Variants)

Each topic is presented with multiple variant phrasings โ€” different ways the same concept can be tested. Master the concept, not just the wording.

Table of Contents

Question Type 1: Switch MAC Address Learning (Homework Events)

How does a switch dynamically populate its MAC address table based on observed source addresses?
๐Ÿ“– Source: doc.txt (Homework) โ€” "When a frame arrives, the switch records the source MAC and ingress port. If the destination is unknown, the switch floods."
Variant A โ€” Scenario: Empty Table
A brand-new switch has just been powered on. Host A (Port 1) sends a frame destined for Host C (Port 3). Describe (a) what the switch writes to its table, and (b) what it does with the frame.
Reveal Answer
(a) It records Host A's MAC โ†’ Port 1. (b) It floods the frame out every port except Port 1.

The switch always learns from the SOURCE address. Since it has no entry for Host C yet, it must fail-open by flooding.

Variant B โ€” Multiple Choice: Known Destination
Host B sends a frame to Host A. The MAC table already maps Host A โ†’ Port 1. What does the switch do?
Reveal Answer
Correct Answer: B) Forward exclusively out Port 1

The switch has a valid table entry for Host A and can make a targeted forwarding decision, preserving bandwidth on all other segments.

Variant C โ€” True/False: Learning Direction
True or False: A switch learns the DESTINATION MAC address from an incoming frame and records it in the table.
Reveal Answer
False

Switches always learn from the SOURCE MAC field of frames they receive. The destination is looked up for forwarding, not for learning.

Variant D โ€” Short Answer: Table Aging
Why do switches have a MAC address aging timer (typically ~5 minutes)?
Reveal Answer
To accommodate physical topology changes.

If a laptop moves from Port 1 to Port 5, the old Port 1 entry must expire so the switch can relearn the new location. Otherwise, frames would be black-holed to the dead port.

Variant E โ€” Scenario: Multi-Event Sequence (from doc.txt Events iโ€“v)
Given three hosts (A on Port 1, B on Port 2, C on Port 3) and an initially empty table, trace these events: (i) Aโ†’C, (ii) Cโ†’A, (iii) Bโ†’A, (iv) Aโ†’B, (v) Bโ†’C. For each event, state what the switch learns and whether it floods or forwards.
Reveal Answer
(i) Learn Aโ†’P1, flood. (ii) Learn Cโ†’P3, forward to P1 (A is known). (iii) Learn Bโ†’P2, forward to P1. (iv) A already known โ€” no new learning; forward to P2 (B is known). (v) B already known โ€” no new learning; forward to P3 (C is known).

After events iโ€“iii, all three hosts are in the table. Events ivโ€“v can both be selectively forwarded with zero flooding.


Question Type 2: Ethernet Distance & Equipment Constraints

What are the physical limits of copper Ethernet, and how do hubs vs. switches differ in their handling of collision domains?
๐Ÿ“– Source: doc.txt (Homework) โ€” "The maximum segment length for UTP copper Ethernet is 100 meters. Exceeding this requires fiber optic cabling or repeaters subject to the 5-4-3 rule."
Variant A โ€” Scenario: The 2-Mile Cable
Could a company run a single copper Ethernet cable 2 miles between buildings?
Reveal Answer
No. Standard Ethernet over copper (100BASE-TX, 1000BASE-T) is limited to 100 meters per segment.

Signal attenuation over copper degrades beyond 100m. For 2 miles, single-mode fiber optic cable is required โ€” it uses light pulses immune to electrical attenuation limits.

Variant B โ€” Multiple Choice: 5-4-3 Rule
Your company daisy-chains 25 Ethernet hubs (repeaters) in a row to extend range. What fundamental rule does this violate?
Reveal Answer
Correct Answer: A) The 5-4-3 Rule

In classic Ethernet, total propagation delay across too many repeaters causes late collisions. A single collision domain cannot exceed a defined physical diameter.

Variant C โ€” True/False: Hub vs Switch Collision Domains
True or False: Replacing an 8-port Hub with an 8-port Switch divides 1 large collision domain into 8 separate micro-collision domains.
Reveal Answer
True

A Hub is Layer 1 โ€” it electrically repeats signals everywhere (1 collision domain). A Switch is Layer 2 โ€” it buffers frames per port, creating isolated collision domains. With full-duplex, collisions are eliminated entirely.

Variant D โ€” Fill in the Blank
The maximum cable length for standard copper UTP Ethernet (Cat5e/Cat6) is _____ meters per segment.
Reveal Answer
100 meters

This is a fundamental constant. Beyond 100m, signal integrity cannot be guaranteed for reliable collision detection or data recovery.

Variant E โ€” Matching: Device โ†’ Layer
Match each device to its OSI Layer: (a) Hub, (b) Switch, (c) Router
Device OSI Layer Collision Domain Behavior
Hub Layer 1 (Physical) Shares one collision domain across ALL ports
Switch Layer 2 (Data Link) Creates a separate collision domain per port
Router Layer 3 (Network) Creates a separate broadcast domain per interface
Reveal Answer
Hub = L1 (shared collision domain), Switch = L2 (per-port collision domains), Router = L3 (per-interface broadcast domains)

This hierarchy is fundamental: each higher-layer device adds more isolation and intelligence.


Question Type 3: Store-and-Forward vs Cut-Through Switching

How does a switch determine when to begin transmitting a frame out an egress port?
๐Ÿ“– Source: Module 2 Summary โ€” "Store-and-Forward buffers the entire frame and checks CRC before forwarding. Cut-Through begins forwarding after reading only the destination MAC."
Variant A โ€” Multiple Choice: Error Protection
Which switching method protects against forwarding corrupted frames?
Reveal Answer
Correct Answer: B) Store-and-Forward

Store-and-Forward buffers the entire frame into memory, then checks the Frame Check Sequence (FCS/CRC) at the end. If the CRC fails, the frame is silently dropped.

Variant B โ€” Scenario: High-Frequency Trading
Why might a trading firm prefer Cut-Through switching?
Reveal Answer
Significantly lower latency.

Cut-Through starts forwarding after reading only the first 6 bytes (Destination MAC), without waiting for the full payload. This shaves microseconds critical to financial trading.

Variant C โ€” Multiple Choice: Fragment-Free
What does the 'Fragment-Free' switching method specifically filter out?
Reveal Answer
Correct Answer: A) Runt frames from collisions

Fragment-Free buffers exactly 64 bytes (the minimum legal Ethernet frame size). Any frame shorter than 64 bytes is assumed to be a collision fragment and is dropped.

Variant D โ€” Ordering
Rank these switching methods from lowest latency to highest latency: Store-and-Forward, Cut-Through, Fragment-Free.
Reveal Answer
1) Cut-Through (lowest), 2) Fragment-Free, 3) Store-and-Forward (highest)

Cut-Through reads 6 bytes, Fragment-Free reads 64 bytes, Store-and-Forward reads the entire frame. More buffering = more delay.

Variant E โ€” True/False
True or False: Cut-Through switching can detect and drop frames with CRC errors before forwarding them.
Reveal Answer
False

Cut-Through starts forwarding before the CRC at the end of the frame has even arrived. It cannot detect corruption, which is the fundamental trade-off for its speed advantage.


Question Type 4: Bandwidth, Throughput & Delay Components

What determines end-to-end throughput, and what are the four sources of packet delay?
๐Ÿ“– Source: Module 1 Summary โ€” "End-to-end throughput is limited by the bottleneck link. Delay = Processing + Queuing + Transmission + Propagation."
Variant A โ€” Multiple Choice: Bottleneck
A file crosses three links: 1 Gbps, 100 Mbps, and 10 Gbps. What is the maximum end-to-end throughput?
Reveal Answer
Correct Answer: D) 100 Mbps โ€” the bottleneck link

Like water through a pipe, end-to-end throughput is limited by the narrowest section. The 100 Mbps link constrains the entire flow.

Variant B โ€” Fill in the Blank: Four Delay Types
The four primary sources of packet delay at a router are: ________, ________, ________, and ________.
Reveal Answer
Processing delay, Queuing delay, Transmission delay, Propagation delay

Processing = inspecting headers. Queuing = waiting in buffers. Transmission = pushing bits onto the wire (depends on bandwidth + packet size). Propagation = signal travel time (depends on distance + speed of light).

Variant C โ€” Calculation
A router receives a 1000-bit packet over a 10 Mbps link using store-and-forward. What is the transmission delay?
Reveal Answer
100 microseconds (0.1 ms)

Transmission Delay = Packet Size / Link Bandwidth = 1000 bits / 10,000,000 bits/sec = 0.0001 seconds = 100 ฮผs.

Variant D โ€” Conceptual
Explain the difference between Transmission Delay and Propagation Delay.
Reveal Answer
Transmission delay is the time to push all bits of a packet onto the wire. Propagation delay is the time for a single bit to physically travel across the wire.

Transmission depends on packet size and link speed. Propagation depends on physical distance and the speed of light/electricity in the medium. Doubling the link speed halves transmission delay but has zero effect on propagation delay.

Variant E โ€” True/False
True or False: Upgrading a link from 1 Gbps to 10 Gbps will reduce propagation delay.
Reveal Answer
False

Propagation delay depends only on physical distance and the speed of light in the medium. A faster link reduces transmission delay, not propagation delay.


Question Type 5: OSI 7-Layer Model vs TCP/IP 4-Layer Model

Compare the theoretical OSI stack against the practical TCP/IP model used on the real internet.
๐Ÿ“– Source: Module 1 Summary โ€” "TCP/IP collapses OSI Layers 5, 6, 7 into a single Application layer, and merges Layers 1โ€“2 into Network Access."
Variant A โ€” Multiple Choice: Collapsed Layers
The TCP/IP Application layer collapses which three OSI layers?
Reveal Answer
Correct Answer: A) Application, Presentation, Session (Layers 7, 6, 5)

TCP/IP treats encryption (Presentation) and connection management (Session) as the responsibility of the application itself, rather than separate protocol layers.

Variant B โ€” Matching: Protocol โ†’ Layer
Match each protocol to its TCP/IP layer: HTTP, TCP, IP, Ethernet.
Protocol TCP/IP Layer OSI Equivalent
HTTP Application Layer 7
TCP Transport Layer 4
IP Internet Layer 3
Ethernet Network Access Layers 1โ€“2
Reveal Answer
HTTP = Application, TCP = Transport, IP = Internet, Ethernet = Network Access

The Network Access layer in TCP/IP covers both the Physical and Data Link functions of the OSI model.

Variant C โ€” Fill in the Blank: PDU Names
At the Transport layer the PDU is called a ________. At the Network layer it is a ________. At the Data Link layer it is a ________.
Reveal Answer
Segment, Packet (or Datagram), Frame

Each layer wraps the previous PDU with its own header: Transport adds ports (Segment), Network adds IPs (Packet), Data Link adds MACs (Frame).

Variant D โ€” Scenario: Encapsulation Trace
When you type a URL into a browser, data flows down the stack. Describe what header is added at each layer as a web request goes from Application to Physical.
Reveal Answer
Application: HTTP headers โ†’ Transport: TCP header (src/dst port) โ†’ Internet: IP header (src/dst IP) โ†’ Network Access: Ethernet header (src/dst MAC) + trailer (FCS) โ†’ Physical: bits on the wire.

This encapsulation process wraps higher-layer data inside lower-layer envelopes. The receiver reverses the process (de-encapsulation) layer by layer.

Variant E โ€” True/False
True or False: The OSI model has more layers than TCP/IP because OSI was designed as a practical implementation standard.
Reveal Answer
False

OSI is the theoretical reference model (7 layers). TCP/IP is the practical implementation (4 layers) that actually runs the internet. OSI has more layers because it over-specifies boundaries that real protocols don't separate.


Question Type 6: Application Layer Protocols (DNS, HTTP, SMTP)

Identify the functions, port numbers, and transport protocols used by key application-layer services.
๐Ÿ“– Source: Module 1 Summary โ€” "DNS uses UDP/53 for queries. HTTP uses TCP/80. SMTP uses TCP/25 for mail relay."
Variant A โ€” Multiple Choice: DNS
DNS typically uses which transport protocol and port for standard queries?
Reveal Answer
Correct Answer: B) UDP Port 53

DNS needs fast, single-packet query-response exchanges. TCP Port 53 is reserved for bulk zone transfers between DNS servers.

Variant B โ€” Matching: Protocol โ†’ Port โ†’ Function
Match each protocol to its port and function.
Protocol Port Function
HTTP TCP/80 Web page retrieval
HTTPS TCP/443 Encrypted web traffic
DNS UDP/53 Name resolution
SMTP TCP/25 Outbound email relay
IMAP TCP/143 Inbound email retrieval
DHCP Server UDP/67 IP address leasing
DHCP Client UDP/68 IP address acquisition
FTP TCP/20-21 File transfer
Reveal Answer
See table above for complete mapping

Memorize these port numbers โ€” they appear frequently on networking exams.

Variant C โ€” Conceptual: Push vs Pull
Explain why SMTP is called a 'push' protocol and IMAP is called a 'pull' protocol.
Reveal Answer
SMTP pushes mail from sender โ†’ server โ†’ server. IMAP pulls mail from server โ†’ client on demand.

SMTP initiates transmission proactively. IMAP waits for the client to request messages. POP3 (Port 110) is a simpler pull protocol that downloads-and-deletes.

Variant D โ€” Scenario: Persistent Connections
Why do modern browsers use persistent HTTP/1.1 connections instead of non-persistent HTTP/1.0?
Reveal Answer
To eliminate the overhead of repeating the TCP 3-way handshake for every embedded object (image, script, stylesheet) on a page.

A persistent connection allows pipelining multiple requests over a single TCP socket, dramatically reducing page load time.

Variant E โ€” True/False
True or False: DNS always uses TCP because reliability is essential for name resolution.
Reveal Answer
False

Standard DNS queries use UDP for speed and simplicity. The response typically fits in a single packet. TCP is used only for zone transfers or oversized responses.


Question Type 7: CSMA/CD: Collision Detection on Wired Ethernet

Explain the Carrier Sense Multiple Access with Collision Detection algorithm step by step.
๐Ÿ“– Source: Module 2 Summary & Lecture 2.2 โ€” "1) Listen before talking. 2) If busy, wait. 3) If idle, transmit. 4) If collision detected, send JAM, then exponential backoff."
Variant A โ€” Multiple Choice: Carrier Sense
What does 'Carrier Sense' mean in CSMA/CD?
Reveal Answer
Correct Answer: B) Listening to the wire to check if it is idle

Before transmitting, a host checks the electrical voltage on the wire. If another signal is detected, it defers. If quiet, it attempts to transmit.

Variant B โ€” Short Answer: JAM Signal
When a collision is detected mid-transmission, what is the FIRST action taken before backing off?
Reveal Answer
Send a 32-bit JAM signal.

The JAM amplifies the collision to ensure every device on the segment recognizes that a collision occurred and discards the corrupted fragments.

Variant C โ€” Calculation: Exponential Backoff
After the 3rd consecutive collision on the same frame, what is the range of random slot values the host picks from?
Reveal Answer
Correct Answer: C) [0, 7]

After collision n, the host picks a random number from [0, 2^n โˆ’ 1]. For n=1: [0,1]. n=2: [0,3]. n=3: [0,7]. n=4: [0,15]. This Binary Exponential Backoff progressively spreads out retransmission attempts.

Variant D โ€” Fill in the Blank: Minimum Frame Size
CSMA/CD requires a minimum Ethernet frame size of _____ bytes to guarantee that collisions are detected before the sender finishes transmitting.
Reveal Answer
64 bytes

If a frame is shorter than 64 bytes, the sender could finish transmitting before the collision signal propagates back from the far end of the network. This is why Ethernet pads small payloads.

Variant E โ€” True/False: Modern Relevance
True or False: CSMA/CD is actively used on modern full-duplex Gigabit Ethernet links.
Reveal Answer
False

Full-duplex links use separate transmit and receive wire pairs, making simultaneous bidirectional communication collision-free. CSMA/CD is disabled entirely.


Question Type 8: Ethernet Frame Structure & EtherType

What metadata fields surround the payload in an IEEE 802.3 Ethernet frame?
๐Ÿ“– Source: Module 2 Summary & Lecture 2.2 โ€” "The Ethernet frame contains Preamble+SFD, Destination MAC, Source MAC, EtherType/Length, Payload, and FCS."
Variant A โ€” Multiple Choice: Demultiplexing
Which Ethernet header field tells the receiver whether to hand the payload to IPv4 or ARP?
Reveal Answer
Correct Answer: C) EtherType (Type/Length field)

EtherType 0x0800 = IPv4. 0x0806 = ARP. 0x86DD = IPv6. This field acts as the Layer 2 demultiplexer.

Variant B โ€” Short Answer: Preamble Purpose
What is the purpose of the 8-byte Preamble and SFD at the start of every Ethernet frame?
Reveal Answer
To synchronize the receiver's clock to the incoming bit stream before actual data begins.

The Preamble is 7 bytes of alternating 1-0 patterns. The Start Frame Delimiter (SFD) is 1 byte ending in '11', signaling that the next bit is the start of the actual frame.

Variant C โ€” Fill in the Blank: MAC Structure
A MAC address is _____ bytes long. The first 3 bytes are the _____ (vendor ID) and the last 3 bytes are the _____.
Reveal Answer
6 bytes. OUI (Organizationally Unique Identifier). Device-specific serial number.

This two-part structure (OUI + device ID) guarantees globally unique addresses. Example: AA:BB:CC:11:22:33 where AA:BB:CC identifies the manufacturer.

Variant D โ€” Calculation: Frame Sizes
What are the minimum and maximum valid Ethernet frame sizes (excluding Preamble/SFD)?
Reveal Answer
Minimum: 64 bytes. Maximum: 1518 bytes (or 1522 with 802.1Q VLAN tag).

The 64-byte minimum ensures collision detection works. The 1518-byte maximum was chosen to limit the time any single station can monopolize the shared medium.

Variant E โ€” True/False: FCS Location
True or False: The Frame Check Sequence (FCS) is located in the Ethernet header, before the payload data.
Reveal Answer
False

The FCS is a 4-byte CRC located in the TRAILER, after the payload. It must come last because it is a checksum calculated over the entire header + payload.


Question Type 9: Spanning Tree Protocol (STP) โ€” Root Election & Loop Prevention

How does STP prevent broadcast storms by electing a root switch and pruning redundant links?
๐Ÿ“– Source: Lecture 3.1 โ€” "Root Node: Switch with the smallest ID. Path cost: inversely proportional to bandwidth. BPDU: Hello messages sent on all interfaces."
Variant A โ€” Multiple Choice: Root Election
Five switches (IDs: 5, 12, 3, 8, 1) join a network. Which switch becomes the STP root?
Reveal Answer
Correct Answer: C) Switch 1 (lowest ID)

STP always elects the switch with the smallest Bridge ID as the root. Lower ID = higher priority. All other switches calculate their shortest path back to this root.

Variant B โ€” Short Answer: BPDU
What is a BPDU and what three fields does it contain?
Reveal Answer
Bridge Protocol Data Unit โ€” a multicast 'hello' message containing: (1) the sender's Switch ID, (2) the believed Root ID, and (3) the path cost to reach the root.

BPDUs are sent to multicast address 01:80:c2:00:00:00. They are processed by switches only, never forwarded to end devices. Switches use BPDUs to converge on a loop-free topology.

Variant C โ€” Scenario: Tie-Breaking
Switches A (ID=3) and B (ID=7) both have a cost of 4 to reach the root. A third switch C has two ports connecting to each. How does C decide which port to enable?
Reveal Answer
C enables the port connecting to Switch A (lower ID = 3 beats 7).

STP tie-breaking order: (1) shortest path cost, (2) lowest neighbor switch ID, (3) lowest port number. Since costs are equal, the lower switch ID wins.

Variant D โ€” Ordering: STP Interface Pruning Rules
Put these four STP pruning rules in their correct priority order.
Reveal Answer
Rule 1: B (root port), Rule 2: C (designated port for upstream), Rule 3: D (designated port for shared segment), Rule 4: A (host-only port)

The root port faces 'upward' toward the root. Designated ports face 'downward' providing service to other switches or hosts. Any port not selected by these rules is BLOCKED.

Variant E โ€” True/False: Partitioning
True or False: If a cable failure partitions the network into two disconnected pieces, STP will build a single shared spanning tree across both pieces.
Reveal Answer
False

If the network partitions, each disconnected piece independently elects its own root and builds its own separate spanning tree. There is no communication between the two halves.


Question Type 10: VLANs โ€” Virtual Network Segmentation

How do VLANs logically partition a physical switch into separate broadcast domains?
๐Ÿ“– Source: Lecture 3.2 โ€” "Each VLAN = subgroup of switch ports. Forwarding between VLANs is done via routing (like separate switches). Dynamic membership possible."
Variant A โ€” Multiple Choice: Broadcast Domain
A 24-port switch is configured with 3 VLANs (8 ports each). How many broadcast domains exist?
Reveal Answer
Correct Answer: B) 3 (one per VLAN)

Each VLAN creates its own isolated broadcast domain. A broadcast from a host in VLAN 10 is only seen by other hosts in VLAN 10. Traffic between VLANs requires a router.

Variant B โ€” Scenario: Security Use Case
A company has HR (handling salaries) and Engineering on the same floor. They share one physical switch. How can a network admin prevent Engineering from sniffing HR traffic?
Reveal Answer
Place HR on VLAN 10 and Engineering on VLAN 20. Traffic is isolated at Layer 2.

VLANs provide logical isolation without requiring separate physical switches. HR broadcast traffic never reaches Engineering ports, and vice versa.

Variant C โ€” Fill in the Blank: Inter-VLAN Communication
For hosts on VLAN 10 to communicate with hosts on VLAN 20, traffic must pass through a _______.
Reveal Answer
Router (Layer 3 device)

VLANs are separate broadcast domains. Just like hosts on different physical LANs, inter-VLAN communication requires routing โ€” a Layer 3 function.

Variant D โ€” Select All That Apply
Which of the following are benefits of VLANs? (Select all that apply)
Reveal Answer
Correct: A, B, C, E

VLANs do NOT eliminate IP addresses (D is wrong). They increase the count of broadcast domains while reducing the size of each, group users logically, and provide security isolation.

Variant E โ€” True/False: Dynamic Membership
True or False: Once a port is assigned to a VLAN, it can never be reassigned to a different VLAN without replacing the switch.
Reveal Answer
False

VLAN port assignments are purely software-configured and can be changed dynamically. This is one of the key advantages โ€” flexibility without physical rewiring.


Question Type 11: SDN & OpenFlow โ€” Separating Control and Data Planes

How does Software Defined Networking centralize routing decisions while switches handle packet forwarding?
๐Ÿ“– Source: Lecture 3.2 โ€” "SDN separates control plane (updates flow tables, decides routing) from data plane (receives/forwards packets based on flow table). Three layers: Application โ†’ Control โ†’ Infrastructure."
Variant A โ€” Multiple Choice: Plane Separation
In SDN, which component decides how packets should be routed?
Reveal Answer
Correct Answer: B) The SDN Controller (centralized)

SDN's core innovation is centralizing the control plane in a software controller ('the brain'). Switches in the infrastructure layer simply execute the forwarding rules they receive.

Variant B โ€” Matching: SDN Layers
Match each SDN layer to its function.
Layer Function Example
Application Layer Requests network resources/policies Firewall app, Load balancer app
Control Layer Centralized routing decisions, populates flow tables SDN Controller (e.g., ONOS, OpenDaylight)
Infrastructure Layer Physical/virtual switches that forward packets OpenFlow-enabled switches
Reveal Answer
Application = policy requests, Control = routing brain, Infrastructure = packet forwarding

This three-layer separation is what gives SDN its flexibility. Applications can dynamically change network behavior through the controller API.

Variant C โ€” Scenario: OpenFlow Learning (Multi-Table)
In the optimized OpenFlow multi-table approach, a packet with an unknown source AND unknown destination arrives. Walk through the process.
Reveal Answer
Table 0 checks destination โ†’ no match โ†’ flood the packet + send to Table 1. Table 1 checks source โ†’ no match โ†’ send packet info to SDN Controller. Controller installs new rules in both tables for future packets.

The multi-table approach is more efficient than single-table because it separates destination lookup (T0) from source learning (T1). Once rules are installed, subsequent packets are handled entirely by the switch without involving the controller.

Variant D โ€” Short Answer: OpenFlow Protocol
What is OpenFlow and what role does it play in SDN?
Reveal Answer
OpenFlow is the first standardized SDN protocol. It defines the communication interface between the SDN controller and OpenFlow-enabled switches/routers.

OpenFlow provides the 'secure channel' through which the controller installs, modifies, and deletes flow table entries on switches.

Variant E โ€” Select All That Apply: SDN Benefits
Which are legitimate benefits of SDN? (Select all that apply)
Reveal Answer
Correct: A, B, D, E

SDN still requires physical switches (C is wrong). The switches just become simpler 'dumb' forwarders while the intelligence is centralized in software.


Question Type 12: Wireless Fundamentals โ€” SNR, Channel Width & Shannon Capacity

How do signal-to-noise ratio and channel bandwidth determine maximum achievable data rate?
๐Ÿ“– Source: Lecture 4.1 โ€” "Max Data Rate = Channel Width ร— logโ‚‚(1 + SNR). SNR=127, Width=1 MHz โ†’ 7 Mbps."
Variant A โ€” Calculation: Shannon Capacity
A wireless channel has a width of 1 MHz and an SNR of 63. What is the theoretical maximum data rate?
Reveal Answer
6 Mbps

Max Rate = 1 MHz ร— logโ‚‚(1 + 63) = 1 MHz ร— logโ‚‚(64) = 1 MHz ร— 6 = 6 Mbps. The Shannon-Hartley theorem sets the absolute upper bound on error-free communication.

Variant B โ€” Multiple Choice: SNR Impact
If you double the SNR from 63 to 127 (keeping channel width at 1 MHz), how does the max data rate change?
Reveal Answer
Correct Answer: B) Increases from 6 to 7 Mbps

logโ‚‚(1+127) = logโ‚‚(128) = 7. The logarithmic relationship means doubling SNR yields diminishing returns โ€” just +1 Mbps in this case. Increasing channel width is more effective.

Variant C โ€” Short Answer: Collision Detection Problem
Why is collision detection impractical in wireless networks?
Reveal Answer
A transmitting radio cannot simultaneously listen because its own strong outgoing signal overwhelms any distant incoming signal.

In wired Ethernet, voltage levels are comparable and collisions are easily detected. In wireless, the transmitter's own signal is millions of times stronger than any remote signal, making detection impossible.

Variant D โ€” Scenario: Hidden Node
Hosts A and C are both in range of Host B, but A and C cannot hear each other. Both transmit to B simultaneously. What happens?
Reveal Answer
Their signals collide at B, corrupting both transmissions. Neither A nor C detects the collision because they cannot hear each other.

This is the Hidden Node Problem. It is solved by the RTS/CTS protocol: A sends RTS to B, B replies CTS heard by everyone in range (including C), and C defers.

Variant E โ€” Fill in the Blank: FM Radio
A standard FM radio channel has a width of _____ kHz (_____ MHz).
Reveal Answer
200 kHz (0.2 MHz)

FM stations are spaced 200 kHz apart in the frequency spectrum. This is why adjacent stations don't interfere with each other.


Question Type 13: Wi-Fi CSMA/CA & Collision Avoidance

How does Wi-Fi handle media access without the ability to detect collisions?
๐Ÿ“– Source: Lecture 4.2 โ€” "Wi-Fi cannot detect collisions in progress. Receiver sends ACK after successful receipt; no ACK = assumed collision โ†’ retransmit. Exponential backoff range starts 0โ€“31, max 1024."
Variant A โ€” Multiple Choice: CSMA/CA vs CSMA/CD
What is the KEY difference between CSMA/CA (Wi-Fi) and CSMA/CD (Ethernet)?
Reveal Answer
Correct Answer: B) CSMA/CA avoids collisions proactively; CSMA/CD detects them after they occur

Since wireless radios cannot detect collisions in progress, Wi-Fi uses Collision AVOIDANCE (careful timing, backoff, ACK confirmation) instead of Collision DETECTION.

Variant B โ€” Short Answer: ACK Mechanism
How does a Wi-Fi sender know its transmission was successful?
Reveal Answer
The receiver sends a link-layer ACK after SIFS time. If no ACK is received, the sender assumes a collision occurred and schedules a retransmission.

This is fundamentally different from wired Ethernet which detects collisions electrically. Wi-Fi must wait for positive confirmation of receipt.

Variant C โ€” Matching: Wi-Fi Timing Parameters
Match each timing parameter to its value.
Parameter Value Purpose
Slot Time 20 ยตs Base unit for backoff calculations
DIFS 50 ยตs Wait before initiating new transmission
SIFS 10 ยตs Short gap before ACK/CTS responses
Max Backoff 1024 slots Upper limit after repeated failures
Reveal Answer
Slot=20ยตs, DIFS=50ยตs, SIFS=10ยตs, Max=1024 slots

SIFS < DIFS guarantees that ACK/CTS responses always win priority over new transmissions. This prevents new senders from colliding with acknowledgments.

Variant D โ€” Calculation: Backoff Ranges
A Wi-Fi device has experienced 3 consecutive collisions. What is the current backoff slot range?
Reveal Answer
Correct Answer: D) 0โ€“255

Wi-Fi backoff starts at 0โ€“31 (first attempt). After collision 1: 0โ€“63. Collision 2: 0โ€“127. Collision 3: 0โ€“255. It keeps doubling up to a maximum of 1024.

Variant E โ€” True/False: Max Attempts
True or False: After 7 consecutive failed transmission attempts, a Wi-Fi device discards the packet and resets.
Reveal Answer
True

After 7 failures, the device gives up on that specific packet, resets its backoff counter, and starts fresh. The higher-layer protocol (e.g., TCP) would handle retransmission.


Question Type 14: RTS/CTS โ€” Solving the Hidden Node Problem

How does the RTS/CTS handshake reserve the wireless channel and prevent hidden-node collisions?
๐Ÿ“– Source: Lecture 4.2 โ€” "RTS includes receiver identity + data size โ†’ receiver replies CTS after SIFS โ†’ all nearby devices pause. Solves hidden-node problem."
Variant A โ€” Scenario: Hidden Nodes A-B-C
Host A wants to send a large file to AP B. Host C is in range of B but NOT of A. Without RTS/CTS, what goes wrong? With RTS/CTS, how is this solved?
Reveal Answer
Without: C transmits simultaneously (can't hear A), causing collision at B. With RTS/CTS: A sends RTS to B โ†’ B broadcasts CTS (heard by C) โ†’ C sees CTS and defers for the specified duration.

The CTS message contains a duration field telling ALL nearby nodes how long to wait, even if they never heard the original RTS.

Variant B โ€” Ordering: RTS/CTS Steps
Put the RTS/CTS exchange in the correct order: (a) All nearby nodes defer, (b) Sender transmits RTS, (c) Receiver replies CTS after SIFS, (d) Sender transmits data frame, (e) Receiver sends ACK.
Reveal Answer
b โ†’ c โ†’ a โ†’ d โ†’ e

The RTS/CTS exchange happens BEFORE the actual data. This is overhead, which is why RTS/CTS is often disabled for short packets and only used for large transmissions.

Variant C โ€” Multiple Choice: When to Use
RTS/CTS is most beneficial for:
Reveal Answer
Correct Answer: B) Large data frames where collision cost is high

RTS/CTS adds overhead (extra round-trip). For small packets, the overhead exceeds the cost of simply retransmitting after a collision. For large packets, avoiding a collision saves significant time.

Variant D โ€” True/False
True or False: RTS/CTS is enabled by default on most modern Wi-Fi access points.
Reveal Answer
False

Most deployments disable RTS/CTS by default because the overhead is not worth it for typical web traffic. It becomes valuable in environments with many hidden nodes.

Variant E โ€” Fill in the Blank
The CTS frame includes a ________ field that tells all nearby stations how long to remain silent.
Reveal Answer
Duration (or NAV โ€” Network Allocation Vector)

Nearby stations set their NAV timer based on the CTS duration. They will not attempt to transmit until the NAV timer expires, even if the channel sounds idle.


Question Type 15: Wi-Fi Rate Scaling & Automatic Rate Fallback (ARF)

How does Wi-Fi dynamically adjust its transmission speed based on channel conditions?
๐Ÿ“– Source: Lecture 4.2 โ€” "Rate decreases after 2 consecutive failures. Rate increases after 10 consecutive successes. 802.11g rates: 54, 48, 36, 24, 18, 12, 9, 6 Mbps."
Variant A โ€” Calculation: Effective Throughput
A device transmits at 36 Mbps with a 25% frame loss rate. What is the effective throughput? Is this better than 24 Mbps with 0% loss?
Reveal Answer
36 ร— 0.75 = 27 Mbps effective. Yes, 27 Mbps > 24 Mbps, so the higher rate with some loss is still better.

Rate scaling must consider this trade-off: a higher bit rate with moderate loss can deliver more effective throughput than a lower bit rate with no loss.

Variant B โ€” Multiple Choice: ARF Thresholds
Under Automatic Rate Fallback (ARF), when does the device step its rate DOWN?
Reveal Answer
Correct Answer: B) After 2 consecutive failures

ARF uses asymmetric thresholds: quick to downshift (2 failures) but slow to upshift (10 successes). This conservative approach prioritizes reliability over speed.

Variant C โ€” Short Answer: What Constitutes a 'Failure'?
In the context of ARF, what specifically counts as a transmission failure?
Reveal Answer
A missing link-layer ACK from the receiver.

If the sender doesn't receive an ACK within the expected time window after SIFS, the frame is considered lost โ€” either corrupted or collided.

Variant D โ€” Matching: Advanced Rate Algorithms
Match each rate adaptation algorithm to its approach.
Algorithm Approach
ARF Step down after 2 failures, up after 10 successes
RBAR Receiver measures SNR and selects the optimal rate
CARA Monitors channel busy after SIFS to distinguish collision from noise
Reveal Answer
ARF = failure counting, RBAR = SNR-based, CARA = collision-aware channel monitoring

RBAR requires receiver feedback (more overhead). CARA improves on ARF by distinguishing between collisions (don't change rate) and noise (do change rate).

Variant E โ€” True/False: Rate Ordering
True or False: 802.11g supports 10 discrete rate levels between 6 Mbps and 54 Mbps.
Reveal Answer
False

802.11g supports 8 rate levels: 6, 9, 12, 18, 24, 36, 48, 54 Mbps. Each uses different modulation and coding schemes.


Question Type 16: IP Subnetting โ€” CIDR Notation, Subnet Masks & Address Ranges

Given a CIDR block, calculate the network address, broadcast address, and number of usable hosts.
๐Ÿ“– Source: Module 5 Summary โ€” "A /24 prefix means 8 host bits โ†’ 256 addresses (254 usable). Subnet mask for /24 is 255.255.255.0."
Variant A โ€” Calculation: /24 Network
Given the address 192.168.10.50/24, what are the network address, broadcast address, and number of usable hosts?
Reveal Answer
Network: 192.168.10.0. Broadcast: 192.168.10.255. Usable hosts: 254 (2โธ โˆ’ 2).

/24 means the first 24 bits are the network prefix. The remaining 8 bits give 256 addresses. Subtract 2 (network address and broadcast) = 254 usable.

Variant B โ€” Calculation: /26 Subnet
A company is assigned 10.0.0.0/26. How many IP addresses are in this block? How many are usable for hosts?
Reveal Answer
Total: 64 addresses (2โถ). Usable hosts: 62 (2โถ โˆ’ 2).

/26 means 26 network bits, leaving 6 host bits. 2โถ = 64 total addresses. The first (10.0.0.0) is the network address and the last (10.0.0.63) is the broadcast.

Variant C โ€” Multiple Choice: Subnet Mask
What is the subnet mask for a /20 prefix?
Reveal Answer
Correct Answer: B) 255.255.240.0

/20 means 20 bits of 1s: 11111111.11111111.11110000.00000000 = 255.255.240.0. The third octet has only the top 4 bits set (128+64+32+16 = 240).

Variant D โ€” Fill in the Blank: Same Subnet Test
To determine if two hosts are on the same subnet, perform a bitwise ________ of each host's IP with the subnet mask and compare the results.
Reveal Answer
AND

IP AND Mask = Network Address. If both hosts produce the same network address, they are on the same subnet and can communicate directly at Layer 2.

Variant E โ€” True/False: Address Classes
True or False: Modern Internet routing still relies on classful addressing (Class A, B, C) rather than CIDR.
Reveal Answer
False

Classful addressing was replaced by CIDR (Classless Inter-Domain Routing) in 1993 to reduce routing table size and allow flexible, variable-length subnet masks.


Question Type 17: MIMO & Wi-Fi Antenna Techniques

How does MIMO use multiple antennas to increase wireless throughput over a single frequency?
๐Ÿ“– Source: Lecture 4.3 โ€” "Multiple antennas at both sender and receiver (N antennas for N streams). All on same frequency, different data streams. Leverages multipath interference."
Variant A โ€” Multiple Choice: MIMO Basics
MIMO uses N antennas to send N independent data streams. What frequency arrangement do these streams use?
Reveal Answer
Correct Answer: B) All streams on the SAME frequency channel

MIMO's key innovation is spatial multiplexing โ€” multiple data streams on the same frequency, differentiated by spatial paths (multipath reflections). This multiplies throughput without needing more spectrum.

Variant B โ€” Matching: Antenna Configurations
Match each antenna configuration to its description.
Configuration Description Key Feature
SISO Single-Input Single-Output One antenna each side โ€” basic Wi-Fi
SIMO Single-Input Multiple-Output Receiver picks the strongest signal from multiple antennas
MISO Multiple-Input Single-Output Transmitter selects best antenna (needs receiver feedback)
MIMO Multiple-Input Multiple-Output N antennas each side โ†’ N independent streams
Reveal Answer
SISO=1ร—1 basic, SIMO=1ร—N receiver diversity, MISO=Nร—1 transmit diversity, MIMO=Nร—N spatial multiplexing

SIMO and MISO provide diversity (reliability) but not multiplexing. Only full MIMO provides Nร— throughput gain.

Variant C โ€” True/False: Clear Space Performance
True or False: MIMO works BETTER in a clear, open field than in an indoor environment with walls and reflections.
Reveal Answer
False

MIMO leverages multipath reflections to distinguish spatial streams. In clear open space with only line-of-sight paths, the streams cannot be separated. Indoor environments with many reflecting surfaces are ideal.

Variant D โ€” Short Answer: Mathematical Separation
If antenna B1 receives S1+S2 and antenna B2 receives S1/2+S2, how are the original streams recovered?
Reveal Answer
Solve the system of equations: B1=S1+S2 and B2=S1/2+S2. Subtract: B1โˆ’B2 = S1/2, so S1 = 2(B1โˆ’B2). Then S2 = B1โˆ’S1.

The receiver uses linear algebra to separate the mixed streams, exploiting the different attenuation paths. More antennas = more equations = more streams that can be separated.

Variant E โ€” Fill in the Blank
In SIMO, the receiver has multiple antennas and picks the ________ signal.
Reveal Answer
stronger (or best/clearest)

SIMO provides receive diversity โ€” it doesn't multiply throughput, but it improves reliability by selecting the antenna with the best signal quality.


Question Type 18: Wi-Fi Modes: Infrastructure vs Ad Hoc & ESS Roaming

Compare the two fundamental Wi-Fi operating modes and explain how ESS enables seamless roaming.
๐Ÿ“– Source: Lecture 4.3 โ€” "Infrastructure Mode: devices communicate only through AP. Ad Hoc: devices communicate directly, no internet. ESS: Multiple APs sharing same SSID, enables roaming."
Variant A โ€” Matching: Mode Comparison
Match each Wi-Fi mode to its characteristics.
Mode AP Required? Internet Access? Use Case
Infrastructure Yes Yes (via AP to wired network) Normal home/office Wi-Fi
Ad Hoc No No (device-to-device only) Temporary file sharing, MANET
Reveal Answer
Infrastructure = AP-based with internet, Ad Hoc = peer-to-peer without internet

Infrastructure mode routes ALL traffic through the AP, even between two devices sitting next to each other. This creates a two-step forwarding inefficiency.

Variant B โ€” Scenario: Coffee Shop Roaming
A large coffee shop has 3 APs with the same SSID 'CoffeeWiFi'. When does a customer's laptop switch from AP1 to AP2?
Reveal Answer
The laptop stays associated with AP1 until AP1's signal drops below a threshold, then it reassociates with AP2 (which has a stronger signal).

In an ESS (Extended Service Set), the device is 'sticky' โ€” it holds onto the original AP as long as possible before switching to avoid unnecessary handoffs.

Variant C โ€” Multiple Choice: SSID
What is an SSID?
Reveal Answer
Correct Answer: B) A human-readable network name (Service Set Identifier)

The SSID is the network name you see when scanning for Wi-Fi. A single AP can even support multiple SSIDs with different security policies.

Variant D โ€” Short Answer: Probe Requests
What privacy concern arises from Wi-Fi probe requests?
Reveal Answer
Probe requests broadcast the device's MAC address and list of previously connected network names, enabling tracking across locations.

By monitoring probe requests, a tracker can identify a specific device and log its movement through space. MAC address randomization partially mitigates this.

Variant E โ€” True/False: Ad Hoc Internet
True or False: In Ad Hoc mode, devices can access the internet through multi-hop routing.
Reveal Answer
False (in basic Ad Hoc mode)

Standard Ad Hoc mode provides only direct peer-to-peer connectivity without internet. MANETs (Mobile Ad Hoc Networks) can theoretically relay traffic via multi-hop, but standard Ad Hoc does not.


Question Type 19: WPA2 Security & the Four-Way Handshake

How does WPA2 establish a secure session between a device and an access point?
๐Ÿ“– Source: Lecture 4.3 โ€” "WPA2: AES encryption with CCMP. Four-Way Handshake: AP sends nonce โ†’ Station sends nonce+MIC โ†’ PTK computed โ†’ GTK delivered encrypted."
Variant A โ€” Ordering: Four-Way Handshake
Put the WPA2 four-way handshake steps in order.
Reveal Answer
B โ†’ A โ†’ D โ†’ C

Step 1: AP provides its random nonce. Step 2: Station provides its nonce + proof of key knowledge (MIC). Step 3: Both derive PTK from shared master key + both nonces; AP delivers the group key (GTK). Step 4: Station confirms.

Variant B โ€” Multiple Choice: Encryption
WPA2 uses which encryption algorithm?
Reveal Answer
Correct Answer: B) AES with CCMP

WPA2 replaced the broken WEP/RC4 encryption with AES (Advanced Encryption Standard) using CCMP (Counter Mode with CBC-MAC Protocol). This provides both confidentiality and integrity.

Variant C โ€” Scenario: KRACK Attack
In the KRACK (Key Reinstallation Attack), what specific vulnerability does the attacker exploit?
Reveal Answer
The attacker blocks the station's acknowledgment (step 4), causing the AP to retransmit step 3. This forces the station to reinstall the same encryption key, resetting the nonce counter and reusing the keystream.

Reusing a keystream with different plaintext allows the attacker to XOR the two ciphertexts and recover the key. KRACK is a protocol-level flaw, not a flaw in AES itself.

Variant D โ€” Matching: WPA2-Personal vs Enterprise
Compare WPA2-Personal and WPA2-Enterprise.
Feature WPA2-Personal (PSK) WPA2-Enterprise
Key Type Pre-Shared Key (same for all) Unique key per device
Management Simple passphrase RADIUS server required
Revocation Change password for everyone Revoke individual device access
Best For Home networks Corporate environments
Reveal Answer
PSK = shared password (simple), Enterprise = per-device keys via RADIUS (scalable)

Enterprise mode is essential for organizations because you can revoke a single employee's access without changing the password for everyone else.

Variant E โ€” Fill in the Blank: Key Types
The ________ (PTK) encrypts unicast traffic, while the ________ (GTK) encrypts broadcast/multicast traffic.
Reveal Answer
Pairwise Transient Key (PTK); Group Temporal Key (GTK)

PTK is unique per device-AP pair (derived from both nonces + master key). GTK is shared among all connected devices for group communications.


Question Type 20: WiMAX vs LTE & Cellular Scheduling

How do WiMAX and LTE differ from Wi-Fi in their approach to spectrum access and scheduling?
๐Ÿ“– Source: Lecture 4.4 โ€” "RTT significant at distance (66ยตs at 10km). Contention-based access impractical โ†’ scheduling used instead. ITU 4G: 100 Mbps mobile, 1 Gbps stationary."
Variant A โ€” Matching: WiMAX vs LTE vs Wi-Fi
Compare the three wireless technologies.
Feature Wi-Fi (802.11) WiMAX (802.16) LTE
Range ~100 meters Up to tens of km 1โ€“10 km
Spectrum Unlicensed Licensed or unlicensed Mainly licensed
Access Method Contention (CSMA/CA) Scheduled slots Scheduled slots
Original Design WLAN Stationary broadband Mobile cellular
Reveal Answer
Wi-Fi = short-range/contention, WiMAX = long-range/scheduled/broadband, LTE = mobile/scheduled/cellular

The key insight is that contention-based access (CSMA/CA) breaks down over long distances because RTT becomes too large for collision avoidance timing to work.

Variant B โ€” Calculation: RTT at Distance
Why is CSMA/CA impractical for a base station 10 km away? What is the approximate RTT?
Reveal Answer
At 10 km, RTT โ‰ˆ 66 ยตs. This exceeds Wi-Fi's slot time (20 ยตs) and makes collision detection timing unreliable.

Wi-Fi's CSMA/CA timing was designed for ~100m ranges (RTT < 1 ยตs). At cellular distances, the propagation delay alone exceeds the protocol's timing parameters.

Variant C โ€” Multiple Choice: 4G Standard
The ITU IMT-Advanced 4G standard (2008) specifies what data rates?
Reveal Answer
Correct Answer: B) 100 Mbps mobile, 1 Gbps stationary

These are the official ITU benchmarks. Early LTE and WiMAX deployments marketed themselves as '4G' before meeting these targets.

Variant D โ€” Short Answer: Ranging
What is 'ranging' in WiMAX, and why is it necessary?
Reveal Answer
Ranging measures the RTT to each subscriber to determine their distance. The base station then tells farther subscribers to start transmitting earlier so all signals arrive at the correct time slot.

Without ranging, signals from distant subscribers would arrive late and overlap with the next time slot, causing interference.

Variant E โ€” True/False: QoS Approach
True or False: LTE emphasizes subscriber-to-base QoS while WiMAX emphasizes end-to-end QoS.
Reveal Answer
False (it's reversed)

WiMAX prioritizes subscriber-to-base QoS. LTE emphasizes end-to-end QoS. LTE's approach is more comprehensive, covering the entire path from device to destination.


Question Type 21: VPN, Carrier Ethernet & Token Ring

Compare alternative LAN technologies: tunneling (VPN), long-distance Ethernet (Carrier), and contention-free access (Token Ring).
๐Ÿ“– Source: Lecture 4.5 โ€” "VPN creates virtual links via Internet; tunnel between remote and office LAN. Carrier Ethernet: reserved lines, >10 Gbps. Token Ring: only token holder can transmit โ€” contention-free."
Variant A โ€” Scenario: Remote Worker VPN
A remote worker connects via VPN to her office network (10.1.1.0/24). Her home IP is 192.168.0.5. How does the VPN make her appear to be on the office network?
Reveal Answer
The VPN client creates a virtual interface (e.g., tun0) assigned an office IP like 10.1.1.50. All office-bound traffic is encrypted by the VPN client, tunneled through the internet, and decrypted by the VPN server on the office LAN.

From the office network's perspective, the remote worker's packets originate from 10.1.1.50 โ€” she appears to be a local device. The internet is simply a transport pipe.

Variant B โ€” Multiple Choice: Why Token Ring Lost
IBM Token Ring was technically elegant (zero collisions). Why did Ethernet win?
Reveal Answer
Correct Answer: B) Ethernet was cheaper, simpler, and easier to expand

Token Ring's complexity (token management, ring recovery) made it expensive. Ethernet's 'good enough' approach with simple collision handling won on cost and scalability.

Variant C โ€” Fill in the Blank: Token Ring Access
In Token Ring, only the device holding the ________ is permitted to transmit. After transmission, it passes the ________ to the next node.
Reveal Answer
token; token

The token is a special control frame circulating around the ring. This ensures contention-free access โ€” no collisions are ever possible โ€” but adds latency waiting for the token.

Variant D โ€” True/False: Carrier Ethernet
True or False: Carrier Ethernet uses contention-based access (CSMA/CD) over long-distance reserved lines.
Reveal Answer
False

Carrier Ethernet uses reserved communication lines โ€” there is no contention. It provides reliability, standardized QoS, and scalability beyond 10 Gbps over distances far exceeding traditional copper Ethernet's 100m limit.

Variant E โ€” Select All: VPN Uses
Which are valid uses of a VPN? (Select all that apply)
Reveal Answer
Correct: A, B, C, E

A VPN does NOT increase bandwidth (D). It adds encryption overhead that may slightly DECREASE throughput. Its value is in security, privacy, and virtual network access.


Question Type 22: Satellite Internet โ€” GEO vs LEO Orbits

How do geostationary and low-earth-orbit satellite systems differ in latency, coverage, and cost?
๐Ÿ“– Source: Lecture 4.4 โ€” "Geosynchronous orbit: 35,786 km. RTT ~1000 ms. LEO: 500โ€“1,500 km, RTTs closer to terrestrial. Starlink: ~7,000 satellites."
Variant A โ€” Matching: GEO vs LEO
Compare GEO and LEO satellite internet.
Feature GEO (Geostationary) LEO (Low Earth Orbit)
Altitude 35,786 km 500โ€“1,500 km
Typical RTT ~1,000 ms Close to terrestrial (~20-40 ms)
Satellites Needed 3 for global coverage Thousands (e.g., Starlink ~7,000)
Best For Rural broadband, broadcast Interactive apps, gaming, VPN
Reveal Answer
GEO = high altitude, high latency, few satellites. LEO = low altitude, low latency, many satellites.

GEO satellites stay fixed relative to Earth (one covers 1/3 of the globe). LEO satellites orbit rapidly and must be replaced frequently, but offer dramatically lower latency.

Variant B โ€” True/False: GEO Gaming
True or False: GEO satellite internet is ideal for real-time online gaming and VPN connections.
Reveal Answer
False

With ~1,000 ms RTT, GEO satellite internet makes real-time interactive applications (gaming, VoIP, VPN handshakes) nearly unusable. It works for bulk downloads using accelerator techniques.

Variant C โ€” Calculation
A GEO satellite orbits at 35,786 km. Light travels at ~300,000 km/s. What is the one-way propagation delay? The minimum RTT?
Reveal Answer
One-way: 35,786 / 300,000 โ‰ˆ 119 ms. RTT: โ‰ˆ 477 ms (signal goes up, down, up, down = 4ร— one-way).

The theoretical minimum is ~477 ms, but real-world processing adds overhead, pushing typical RTT to ~1,000 ms. The speed of light is the fundamental bottleneck.

Variant D โ€” Fill in the Blank
A residential satellite dish is typically ________ cm in diameter, and the transmitter uses ________ watts of power.
Reveal Answer
70โ€“100 cm; 1โ€“2 watts

The dish focuses the weak signal from space into the receiver. The small transmitter power is sufficient because the satellite has a large, sensitive antenna.

Variant E โ€” Short Answer: LEO Challenge
What is the primary engineering challenge of LEO satellite internet?
Reveal Answer
Maintaining 24-hour global coverage requires launching and maintaining thousands of satellites because each LEO satellite only covers a small area and orbits rapidly.

SpaceX's Starlink has ~7,000 satellites for this reason. GEO only needs 3 for global coverage, but at the cost of high latency.


Question Type 23: Virtual Circuits vs Datagram Forwarding

Contrast connection-oriented virtual circuit switching with connectionless datagram routing.
๐Ÿ“– Source: Lecture 4.5 โ€” "Virtual circuits: connection setup required, VCI addressing, routers maintain state, QoS guaranteed. Datagrams: no setup, destination address per packet, easy router crash recovery."
Variant A โ€” Matching: VC vs Datagram
Compare Virtual Circuits and Datagram forwarding.
Feature Virtual Circuits Datagram Forwarding
Connection Setup Required before data flows Not needed
Addressing Connection ID (VCI) per link Full destination address per packet
Router State Must maintain connection tables Minimal state (routing tables only)
QoS Can guarantee (reserved capacity) Best-effort, no guarantees
Router Crash All connections through that router are lost Easy recovery; packets reroute around failure
Packet Headers Small (just VCI) Larger (full destination address)
Reveal Answer
VCs = connection-oriented, stateful, guaranteed QoS, fragile. Datagrams = connectionless, stateless, best-effort, resilient.

The Internet (IP) uses datagrams. Telephone networks and ATM use virtual circuits. The trade-off is QoS guarantees vs resilience and simplicity.

Variant B โ€” Scenario: Router Crash
A router in the middle of the network crashes and reboots. How does this affect (a) virtual circuit connections and (b) datagram traffic?
Reveal Answer
(a) All virtual circuits through that router are destroyed because the connection state is lost. They must be re-established. (b) Datagram traffic automatically reroutes around the failed router once routing tables converge.

This is why the Internet chose datagrams โ€” resilience to failures was a core design requirement (ARPANET military origins).

Variant C โ€” Multiple Choice: VCI Scope
A Virtual Circuit Identifier (VCI) is:
Reveal Answer
Correct Answer: B) Locally unique per link

Each router remaps the VCI: incoming (VCI_in, port_in) โ†’ outgoing (VCI_out, port_out). This allows efficient locally-scoped numbering rather than requiring globally unique IDs.

Variant D โ€” Short Answer: ATM Cell Size
ATM (a virtual circuit technology) uses fixed-size cells of how many data bytes? Why so small?
Reveal Answer
48 bytes of data (53 bytes total with header). Small cells minimize queuing delay for time-sensitive traffic like voice.

Voice traffic is very sensitive to jitter. Small, fixed-size cells ensure predictable, low-latency forwarding โ€” a key QoS advantage of virtual circuits.

Variant E โ€” True/False
True or False: The modern Internet uses virtual circuits for all IP traffic.
Reveal Answer
False

The Internet uses datagram (connectionless) forwarding via IP. Each packet is independently routed. MPLS (Multi-Protocol Label Switching) adds some VC-like features within carrier networks, but end-to-end IP is datagram-based.


Question Type 24: IPv4 Header โ€” Fields, TTL, and Best-Effort Delivery

What are the key fields in an IPv4 header and what does 'best-effort delivery' mean?
๐Ÿ“– Source: Lecture 5.1 โ€” "IPv4 designed in 1981. Three core tasks: standard packet format, global addressing, path forwarding. TTL decremented at each hop, packet dropped at 0. Best effort = no delivery guarantee."
Variant A โ€” Matching: IPv4 Header Fields
Match each IPv4 header field to its purpose.
Field Purpose
Version Identifies IP version (4 for IPv4)
IHL Header length in 32-bit words
TTL Hop countdown โ€” packet dropped at 0 to prevent infinite loops
Protocol Identifies upper-layer protocol (TCP=6, UDP=17)
Checksum Validates header integrity
Source/Dest IP Global addressing for sender and receiver
ID/Flags/Offset Fragmentation management
DS/ECN Quality of Service and congestion signals
Reveal Answer
See table for complete field-purpose mapping

The Protocol field acts as a demultiplexer (like EtherType at Layer 2). The TTL field is critical for preventing routing loops from consuming bandwidth forever.

Variant B โ€” Multiple Choice: Best-Effort
What does 'best-effort delivery' mean for IP?
Reveal Answer
Correct Answer: B) Routers try to deliver but don't guarantee success

IP is stateless and connectionless. If a router is congested, it simply drops packets. Reliable delivery is the job of upper-layer protocols like TCP.

Variant C โ€” Scenario: TTL Protection
A routing table error creates a loop between Router A and Router B. Without TTL, what would happen? With TTL, what happens?
Reveal Answer
Without TTL: packets loop forever between A and B, consuming bandwidth and eventually congesting the link. With TTL: the packet's TTL decreases by 1 at each hop and is dropped when it reaches 0.

TTL is a safety valve. A typical initial TTL is 64 or 128. Even in a loop, the packet is destroyed after at most that many hops.

Variant D โ€” Short Answer: Smurf Attack
How does a Smurf Attack exploit IP source address spoofing?
Reveal Answer
The attacker sends ICMP echo requests (pings) to a broadcast address with the VICTIM's spoofed source IP. All hosts on the network reply to the victim, flooding it with traffic.

This works because IP does not verify source addresses. The local router accepts the spoofed packet and forwards it. This is a classic amplification DDoS attack.

Variant E โ€” Fill in the Blank: Three Core Tasks
The Network Layer's three core tasks are: (1) ________, (2) ________, and (3) ________.
Reveal Answer
(1) Standard packet format (datagrams), (2) Global addressing scheme (IP addresses), (3) Path forwarding scheme (routing/next-hop decisions)

These three tasks enable any device on Earth to send a packet to any other device, regardless of the underlying link-layer technology.


Question Type 25: IP Fragmentation & MTU

When and why does IP fragment packets, and how are fragments reassembled?
๐Ÿ“– Source: Lecture 5.2 โ€” "MTU = Maximum Transfer Unit. Fragmentation splits packets when next network's MTU is smaller. Example: 1500-byte packet โ†’ 996-byte + 524-byte fragments."
Variant A โ€” Scenario: MTU Mismatch
A 1500-byte IP packet (20-byte header + 1480-byte payload) must cross a link with MTU = 1000 bytes. How is it fragmented?
Reveal Answer
Fragment 1: 20-byte header + 976-byte payload = 996 bytes. Fragment 2: 20-byte header + 504-byte payload = 524 bytes. Total payload: 976 + 504 = 1480 bytes.

Each fragment gets its own IP header (20 bytes overhead). The payload is split so each fragment fits within the 1000-byte MTU. The More Fragments flag is set on all fragments except the last.

Variant B โ€” Multiple Choice: Where Reassembly Occurs
Where are IP fragments reassembled?
Reveal Answer
Correct Answer: C) Only at the final destination host

Intermediate routers do NOT reassemble fragments โ€” they may need to fragment further if they encounter an even smaller MTU. Only the destination has all fragments and performs reassembly using the ID, Flags, and Offset fields.

Variant C โ€” Fill in the Blank: Header Fields
Three IPv4 header fields used for fragmentation are: ________, ________, and ________.
Reveal Answer
Identification (ID), Flags (MF=More Fragments, DF=Don't Fragment), Fragment Offset

ID groups fragments from the same original packet. MF flag indicates more fragments follow. Fragment Offset tells the destination where each fragment's data fits in the original payload.

Variant D โ€” True/False: Fragmentation at Routers
True or False: Only the original sender can fragment an IP packet.
Reveal Answer
False

Any router along the path can fragment a packet if the next-hop link has a smaller MTU than the packet size. However, if the Don't Fragment (DF) flag is set, the router drops the packet and sends an ICMP error instead.

Variant E โ€” Short Answer: IP Options
Name two optional IP header fields and their purposes.
Reveal Answer
Record Route: intermediate routers store their IP addresses in the packet. Timestamp: routers insert timestamps. Source Routing: sender specifies the preferred route.

These options are rarely used in practice due to security concerns and processing overhead, but they exist in the specification (up to 10 option words available).


Question Type 26: Special IP Addresses & Multi-Homing

Identify reserved IP address ranges and explain why IP addresses belong to interfaces, not hosts.
๐Ÿ“– Source: Lecture 5.2 โ€” "IP address belongs to an interface, NOT directly to a host. 127.0.0.1 = loopback. Private ranges: 10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16."
Variant A โ€” Matching: Special Addresses
Match each address or range to its purpose.
Address/Range Purpose
127.0.0.1 Loopback (localhost) โ€” self-referencing
10.0.0.0/8 Private address range (Class A)
172.16.0.0/12 Private address range (Class B)
192.168.0.0/16 Private address range (Class C)
Host bits all 1s (e.g., x.x.x.255) Broadcast address for the subnet
Host bits all 0s (e.g., x.x.x.0) Network address (also broadcast in BSD Unix)
Reveal Answer
See table for mappings. These addresses are never routed on the public internet.

Private addresses are used inside organizations and translated to public IPs by NAT. The loopback address lets a host communicate with itself for testing.

Variant B โ€” Scenario: Multi-Homed Server
A server has two network cards: eth0 (10.1.1.5) connected to the office LAN, and eth1 (192.168.2.10) connected to the lab network. How many IP addresses does this server have?
Reveal Answer
Two โ€” one per network interface.

IP addresses are assigned to INTERFACES, not hosts. A multi-homed server has one IP per network card. A router typically has many interfaces, each with its own IP on a different subnet.

Variant C โ€” Multiple Choice: Loopback
What is the purpose of the loopback address 127.0.0.1?
Reveal Answer
Correct Answer: B) To allow a host to communicate with itself

The loopback interface is an abstract, software-only interface. Packets sent to 127.0.0.1 never leave the host โ€” they are immediately looped back to the receiving socket. Essential for local development and testing.

Variant D โ€” True/False: Private Addresses Online
True or False: A packet with source address 192.168.1.100 can be routed across the public internet.
Reveal Answer
False

Private addresses (10.x, 172.16-31.x, 192.168.x) are not routable on the public internet. A NAT device must translate them to a public IP before they leave the local network.

Variant E โ€” Fill in the Blank
A /24 network has 256 total addresses but only _____ usable host addresses because the _______ and _______ addresses are reserved.
Reveal Answer
254 usable; network address (all 0s in host part) and broadcast address (all 1s in host part)

The network address identifies the subnet itself. The broadcast address reaches all hosts on the subnet. Neither can be assigned to a device.


Question Type 27: IP Routing โ€” Longest Prefix Match & Forwarding Tables

How does a router use longest prefix matching to forward packets to the correct next hop?
๐Ÿ“– Source: Lecture 5.3 โ€” "Classless IP Forwarding: match destination address prefix to router interfaces. When multiple routes match, the most specific (longest prefix) wins."
Variant A โ€” Scenario: Multiple Matches
A router has these forwarding entries: Interface A = 10.3.0.0/16, Interface E = 10.3.32.0/19. A packet arrives for 10.3.37.103. Which interface is used?
Reveal Answer
Interface E (10.3.32.0/19) โ€” the longest prefix match.

Both A (/16) and E (/19) match the destination. /19 is more specific (19 bits match vs 16), so E wins. This is the Longest Prefix Match rule โ€” always choose the most specific route.

Variant B โ€” Multiple Choice: Forwarding Algorithm
When a router receives a packet with destination D, what is the first thing it does?
Reveal Answer
Correct Answer: B) Compares D's prefix against interface addresses

The router first checks if D is on a directly-connected subnet (local delivery). If no interface matches, it consults the forwarding table for next-hop entries and uses longest prefix match.

Variant C โ€” Calculation: Prefix Matching
Router has interface B = 10.4.0.0/22. Does destination 10.4.1.101 match this interface?
Reveal Answer
Yes. /22 means the first 22 bits must match. 10.4.0.0 = 00001010.00000100.000000|00.00000000. 10.4.1.101 = 00001010.00000100.000001|01.01100101. First 22 bits: both are 00001010.00000100.0000 โ€” they match.

To verify a match, convert both addresses to binary and compare the first k bits (where k is the prefix length).

Variant D โ€” Short Answer: Default Route
What happens when NO forwarding table entry matches a packet's destination?
Reveal Answer
The packet is sent to the default route (0.0.0.0/0) if one exists. If no default route is configured, the packet is dropped and an ICMP Destination Unreachable message is sent back to the sender.

The default route (prefix length 0) matches everything โ€” it is the 'catch-all'. It is always the shortest prefix, so any more specific match will win over it.

Variant E โ€” True/False: Longest Prefix
True or False: If two forwarding table entries both match a destination, the router picks the one with the SHORTEST prefix.
Reveal Answer
False

Longest Prefix Match โ€” the most specific (longest) matching prefix is always chosen. A /24 match wins over a /16 match because it is more specific.


Question Type 28: Subnetting โ€” Dividing Address Space

How does an organization divide its allocated IP block into smaller subnets?
๐Ÿ“– Source: Lecture 5.3 โ€” "Subnet: smaller IPv4 network within a larger address space. Example: 147.126.0.0/16 divided into /24 and /20 subnets. Hierarchical routing: external gateway โ†’ subnet router โ†’ host."
Variant A โ€” Calculation: Subnetting a /16
An organization owns 147.126.0.0/16. They create four /24 subnets. List them and calculate hosts per subnet.
Reveal Answer
147.126.0.0/24 (254 hosts), 147.126.1.0/24 (254 hosts), 147.126.2.0/24 (254 hosts), 147.126.3.0/24 (254 hosts).

Each /24 has 8 host bits = 256 addresses โˆ’ 2 reserved = 254 usable hosts. The /16 block contains 256 possible /24 subnets (from 147.126.0.0/24 through 147.126.255.0/24).

Variant B โ€” Scenario: Hierarchical Routing
An external host sends a packet to 147.126.2.50. Describe the routing path.
Reveal Answer
The packet first reaches the organization's external gateway (which handles 147.126.0.0/16). The gateway forwards based on the /24 subnet match to the subnet router for 147.126.2.0/24. That router delivers directly to host .50.

This hierarchical approach means external routers only need ONE entry for the entire /16 block. Internal subnet structure is invisible to the outside world.

Variant C โ€” Multiple Choice: Variable Subnets
Can a /16 block contain both /24 and /20 subnets simultaneously?
Reveal Answer
Correct Answer: B) Yes โ€” VLSM

VLSM allows subnets of different sizes within the same address space. A /20 provides 4094 hosts (for a large department) while /24s provide 254 hosts each (for smaller groups). The real lecture example includes both /24 and /20 subnets under 147.126.0.0/16.

Variant D โ€” Fill in the Blank: Subnet Benefit
Subnetting reduces the size of ________ domains and allows hierarchical routing that hides internal topology from ________ routers.
Reveal Answer
broadcast; external (internet)

Each subnet is its own broadcast domain. External routers only see the aggregate /16 prefix, not the internal subnet structure. This reduces routing table size across the internet.

Variant E โ€” True/False
True or False: Every device on the internet can see the internal subnet structure of any organization's network.
Reveal Answer
False

Hierarchical routing hides internal subnets. External routers only know how to reach the organization's aggregate prefix (e.g., /16). The internal /24 or /20 divisions are invisible outside.


Question Type 29: DHCP โ€” Dynamic Address Assignment (DORA)

How does a host automatically obtain an IP address when joining a network?
๐Ÿ“– Source: Lecture 5.4 โ€” "DHCP 4-Step Process: Discover โ†’ Offer โ†’ Request โ†’ ACK. DHCP server typically collocated with AP/router."
Variant A โ€” Ordering: DORA Process
Put the DHCP steps in correct order: ACK, Offer, Request, Discover.
Reveal Answer
1) Discover (broadcast from client), 2) Offer (server proposes IP), 3) Request (client accepts), 4) ACK (server confirms lease).

Remember 'DORA': Discover โ†’ Offer โ†’ Request โ†’ ACK. Discover and Offer are technically optional (client can skip directly to Request if it knows the server), but the full sequence is the standard.

Variant B โ€” Scenario: No DHCP Server
A laptop connects to a new Wi-Fi network but no DHCP server is available. What happens?
Reveal Answer
The laptop broadcasts a DHCP Discover message but receives no Offer. After timeout, it either assigns itself a link-local address (169.254.x.x) or reports 'No network access'.

APIPA (Automatic Private IP Addressing) assigns 169.254.x.x as a fallback. These addresses allow local subnet communication only โ€” no internet routing is possible.

Variant C โ€” Multiple Choice: Why Broadcast?
Why must the initial DHCP Discover message be broadcast?
Reveal Answer
Correct Answer: B) The client has no IP yet and doesn't know the server

A new device has no IP address, no subnet mask, and no knowledge of the DHCP server's location. Broadcasting (to 255.255.255.255) ensures the Discover reaches any DHCP server on the local subnet.

Variant D โ€” Fill in the Blank: Address Hierarchy
The global IP address hierarchy is: ________ allocates to 5 regional registries โ†’ registries allocate to ________ โ†’ ISPs break addresses into ________ โ†’ hosts get IPs via static assignment or ________.
Reveal Answer
ICANN/IANA; ISPs; subnets; DHCP

ICANN (via IANA) manages the global pool. The 5 RIRs (ARIN, APNIC, RIPE, LACNIC, AFRINIC) distribute to ISPs in their regions.

Variant E โ€” Matching: Regional Registries
Match each Regional Internet Registry to its region.
Registry Region
ARIN North America
APNIC Asia Pacific
RIPE NCC Europe
LACNIC Latin America
AFRINIC Africa
Reveal Answer
See table. As of 2017, 4 of 5 RIRs have exhausted their free IPv4 space.

IPv4 exhaustion is the primary driver for IPv6 adoption and the widespread use of NAT. Only AFRINIC had remaining free IPv4 space as of 2017.


Question Type 30: NAT โ€” Network Address Translation

How does NAT allow multiple internal hosts to share a single public IP address?
๐Ÿ“– Source: Lecture 5.5 โ€” "NAT router rewrites source address from private IP to router's public IP. Port remapping resolves conflicts. External sites cannot scan internal hosts."
Variant A โ€” Scenario: Two Internal Hosts
Host A (192.168.0.6, port 3000) and Host B (192.168.0.7, port 3000) both request the same web server (213.25.63.25:80). The NAT router's public IP is 200.1.2.37. How does NAT handle this?
Reveal Answer
A's packets: source rewritten to 200.1.2.37:3000. B's packets: source rewritten to 200.1.2.37:3001 (port remapped to avoid conflict). The NAT table maps 3000โ†’A and 3001โ†’B for incoming replies.

When two internal hosts use the same source port, NAT must remap one to a different external port. The forwarding table tracks which internal host corresponds to each external port.

Variant B โ€” Multiple Choice: Reply Routing
When an external server sends a reply to the NAT router's public IP, how does the router know which internal host to deliver to?
Reveal Answer
Correct Answer: B) It looks up the destination port in its NAT forwarding table

The NAT table maps each pair back to the original . This is why NAT uses port remapping as the key multiplexing mechanism.

Variant C โ€” Short Answer: NAT Security Benefit
How does NAT provide a form of security for internal hosts?
Reveal Answer
Internal hosts have private IPs that are invisible from the internet. External attackers cannot directly initiate connections to internal hosts because there is no public IP to target.

NAT acts as a natural firewall. Unsolicited inbound connections are dropped because they don't match any entry in the NAT forwarding table. Only responses to internally-initiated connections are forwarded.

Variant D โ€” Matching: NAT Forwarding Table
Complete the NAT forwarding table.
Internal Connection External Representation External Destination
(remapped)
Reveal Answer
See table. Host B's port was remapped from 3000โ†’3001 to avoid conflict with Host A's connection.

The NAT table must maintain a unique external for every active internal connection. Port remapping is only needed when internal connections would create duplicates.

Variant E โ€” True/False: NAT and IPv4 Exhaustion
True or False: NAT is one of the key technologies that has delayed the urgency of IPv6 adoption.
Reveal Answer
True

By allowing entire networks to share a single public IPv4 address, NAT dramatically reduced the consumption rate of public IPv4 addresses. Without NAT, IPv4 would have been exhausted much earlier.


Question Type 31: ARP โ€” Address Resolution Protocol

How does a host discover the MAC address of another host on the same subnet?
๐Ÿ“– Source: Module 2 Summary โ€” "ARP resolves IP addresses to MAC addresses within a local subnet. ARP Request is broadcast; ARP Reply is unicast."
Variant A โ€” Scenario: First Packet
Host A (10.0.0.1) wants to send a packet to Host B (10.0.0.2) on the same subnet. A knows B's IP but NOT B's MAC address. What happens?
Reveal Answer
A broadcasts an ARP Request: 'Who has 10.0.0.2? Tell 10.0.0.1.' B receives the broadcast and sends a unicast ARP Reply: 'I am 10.0.0.2, my MAC is XX:XX:XX:XX:XX:XX.' A caches this mapping and sends the data frame.

ARP bridges the gap between Layer 3 (IP) and Layer 2 (MAC). Without ARP, hosts would have no way to create Ethernet frames for IP packets destined for local hosts.

Variant B โ€” Multiple Choice: ARP Scope
ARP operates within what scope?
Reveal Answer
Correct Answer: B) Only within the same broadcast domain

ARP uses Layer 2 broadcast, which does not cross router boundaries. For hosts on different subnets, the sender must ARP for its default gateway's MAC address instead.

Variant C โ€” Fill in the Blank: Different Subnet
If Host A wants to reach Host C on a DIFFERENT subnet, A's ARP request resolves the MAC address of the ________, not Host C.
Reveal Answer
default gateway (router)

A sends the packet to the router's MAC address, but with Host C's IP as the destination. The router then re-ARPs on the destination subnet to find Host C's MAC.

Variant D โ€” True/False: ARP Caching
True or False: ARP results are cached permanently and never expire.
Reveal Answer
False

ARP cache entries have a timeout (typically 1-20 minutes depending on the OS). Entries must expire so that IP-to-MAC mappings can update when devices move or change.

Variant E โ€” Short Answer: ARP vs DNS
How do ARP and DNS differ in what they resolve?
Reveal Answer
DNS resolves human-readable domain names (e.g., google.com) to IP addresses. ARP resolves IP addresses to MAC addresses on the local subnet.

DNS is Layer 7 (Application), global scope. ARP is Layer 2/3 boundary, local subnet scope only. DNS uses UDP/TCP; ARP uses raw Ethernet frames.


Question Type 32: Network Topology & Physical Design

Compare common network topologies and their failure modes.
๐Ÿ“– Source: Module 1 Summary & Lecture 2.1 โ€” "Common topologies: Bus, Star, Ring, Mesh. Star is most common in modern LANs. Mesh provides redundancy but is expensive."
Variant A โ€” Matching: Topology Comparison
Match each topology to its key characteristics.
Topology Structure Single Point of Failure Cost
Bus All devices share one cable The cable itself Low
Star All devices connect to a central switch/hub The central switch Moderate
Ring Each device connects to exactly 2 neighbors Any single link (unless dual-ring) Moderate
Full Mesh Every device connected to every other None (maximum redundancy) Very high
Reveal Answer
Bus = shared cable, Star = central device, Ring = circular chain, Mesh = full interconnect

Modern LANs almost universally use Star topology with Ethernet switches at the center. Mesh is reserved for WANs and critical backbone connections.

Variant B โ€” Scenario: Star Failure
A company uses a Star topology with a single central switch. The switch fails. What happens to the network?
Reveal Answer
Complete network outage โ€” all devices lose connectivity because every communication must pass through the central switch.

The central device is the single point of failure in Star topology. This is mitigated by using redundant switches, stacking, or a partial mesh at the core.

Variant C โ€” Multiple Choice: Modern LANs
Which topology is most commonly used in modern Ethernet LANs?
Reveal Answer
Correct Answer: C) Star

Star topology with an Ethernet switch at the center provides easy management, simple troubleshooting (each host has its own cable), and supports full-duplex per-port connections.

Variant D โ€” Short Answer: Mesh Formula
How many direct links are needed for a full mesh of 10 devices?
Reveal Answer
45 links. Formula: n(nโˆ’1)/2 = 10(9)/2 = 45.

Full mesh grows quadratically. For 100 devices, you'd need 4,950 links โ€” impractical for LANs but valuable for small groups of critical routers.

Variant E โ€” True/False: Bus Topology
True or False: Bus topology is still the dominant choice for new network installations.
Reveal Answer
False

Bus topology (shared coaxial cable) is obsolete. A break anywhere in the cable disrupts the entire network. Star topology replaced it because faults are isolated to individual links.


Question Type 33: Comprehensive Synthesis โ€” End-to-End Packet Journey

Trace a packet's complete journey from web browser to remote server across all network layers.
๐Ÿ“– Source: All Modules โ€” Integration of Layers 1โ€“5 concepts
Variant A โ€” Scenario: Full HTTP Request Trace
You type 'http://example.com/page' in your browser. The page is one hop away through your home router. Trace the complete packet journey from keystroke to page display, naming every protocol and layer involved.
Reveal Answer
1) DNS (Application/UDP) resolves example.com โ†’ IP address. 2) Browser creates HTTP GET request. 3) TCP (Transport) performs 3-way handshake (SYN, SYN-ACK, ACK) to port 80. 4) HTTP request is encapsulated: TCP segment โ†’ IP packet (src=your IP, dst=server IP) โ†’ Ethernet frame. 5) Host ARPs for router MAC. 6) Frame sent to router via Wi-Fi (CSMA/CA) or Ethernet. 7) Router performs NAT (rewrites source to public IP). 8) Router forwards IP packet toward server (next-hop routing, longest prefix match). 9) Server receives, de-encapsulates, processes HTTP request. 10) Response travels back (NAT reverses source/destination). 11) Browser renders HTML.

This question integrates DNS, TCP, HTTP, IP, ARP, NAT, CSMA/CA or CSMA/CD, MAC learning, and routing โ€” virtually every midterm topic.

Variant B โ€” Short Answer: Layer-by-Layer Headers
List every header/trailer added as an HTTP request travels from Application layer down to the Physical layer.
Reveal Answer
Application: HTTP headers (GET, Host, etc.) โ†’ Transport: TCP header (src port, dst port, seq#, ack#, flags) โ†’ Network: IP header (src IP, dst IP, TTL, Protocol) โ†’ Data Link: Ethernet header (src MAC, dst MAC, EtherType) + FCS trailer โ†’ Physical: Preamble/SFD + electrical/radio signals.

Each layer encapsulates the previous layer's PDU inside its own header. The receiver reverses this process, stripping each header to extract the next layer's data.

Variant C โ€” Scenario: Troubleshooting Chain
A user reports: 'I can ping 8.8.8.8 but cannot load google.com.' What is the most likely problem?
Reveal Answer
DNS resolution is failing. The host can reach external IPs (network layer works) but cannot resolve domain names to IP addresses.

This is a classic Layer 7 (Application/DNS) problem. The ability to ping an IP proves Layers 1-3 are functional. The failure of name resolution isolates the issue to DNS.

Variant D โ€” Scenario: Can Reach Gateway But Not Internet
A user can ping the default gateway (192.168.1.1) but cannot reach any external IP like 8.8.8.8. What are the most likely causes?
Reveal Answer
The router's WAN link is down, the router's NAT is misconfigured, or the ISP connection is severed. Layers 1-2 to the gateway work, but the router cannot forward beyond its LAN interface.

Pinging the gateway confirms local network connectivity. The failure to reach external IPs points to a problem at or beyond the gateway โ€” WAN link, ISP, or routing configuration.

Variant E โ€” Multiple Choice: Layer Isolation
A host cannot ping ANY device, including its own default gateway. At which layer should you begin troubleshooting?
Reveal Answer
Correct Answer: D) Physical/Data Link

If even the local gateway is unreachable, the problem is at the lowest layers. Check: cable connected? Wi-Fi associated? NIC enabled? Link light on? This bottom-up troubleshooting approach is the standard methodology.