Computer Networks · Quiz 1 · Jan 2026

**1. Understand the client-server architecture:** In a client-server network model, a centralized server continuously listens for incoming requests and provides resources or services (such as web pages, files, or database queries) to multiple clients concurrently. **Answer: A — True.** A: **Correct:** A single server in the client-server model is designed to handle requests from multiple clients. B: **Incorrect:** The statement accurately describes the client-server model.

**1. Distinguish IP addresses from MAC addresses:** - A **MAC (Media Access Control) address** is a hardware identifier burned into a network interface card (NIC) at the data-link layer. - An **IP (Internet Protocol) address** is a logical network-layer address assigned to an interface based on the network/subnet to which the device is currently attached (identifying its topological location in the network). **Answer: A — True.** A: **Correct:** An IP address is a hierarchical network-layer identifier that reflects which network/subnet the device is connected to, whereas the MAC address identifies the physical hardware. B: **Incorrect:** Hardware identification is performed by the MAC address, not the IP address.

**1. Understand circuit switching:** In a circuit-switched network (such as traditional telephone networks), a dedicated end-to-end path with reserved link bandwidth (time slots or frequency bands) and switch capacity is established before data transfer begins and remains reserved for the entire duration of the session. **Answer: A — True.** A: **Correct:** Circuit switching dedicates and reserves end-to-end resources for the whole call/session duration. B: **Incorrect:** On-demand unreserved resource sharing is a property of packet switching, not circuit switching.

**1. Compare UDP and TCP:** - **TCP (Transmission Control Protocol)** is connection-oriented and provides reliable, in-order byte-stream delivery using sequence numbers, acknowledgments, and retransmissions. - **UDP (User Datagram Protocol)** is a connectionless, best-effort transport protocol that does **not** guarantee delivery, ordering, or duplicate protection. **Answer: B — False.** A: **Incorrect:** UDP is an unreliable, connectionless protocol and does not provide in-order delivery. B: **Correct:** Unlike TCP, UDP provides best-effort datagram delivery without reliability or ordering guarantees.

**1. Understand Selective Repeat ARQ:** In **Selective Repeat (SR) ARQ**, the receiver buffers out-of-order packets that arrive correctly within its receive window and acknowledges each packet individually. Therefore, when packet #6 is lost, the sender retransmits **only the single lost packet (#6)**, not the entire window of 4 packets (retransmitting up to 4 packets is the behavior of Go-Back-N ARQ). **Answer: B — False.** A: **Incorrect:** Retransmitting the entire window of 4 packets occurs in Go-Back-N ARQ, not Selective Repeat ARQ. B: **Correct:** Selective Repeat ARQ retransmits only the specific lost packet (1 packet, packet #6).

**1. Set up the transmission delay formula:** Transmission delay \( d_{\text{trans}} \) is the time required to push all \( L \) bits of the packet onto the link of transmission rate \( R \): \[ d_{\text{trans}} = \frac{L}{R} \] **2. Substitute the given values:** Given \( L = 10,000 \text{ bits} = 10^4 \text{ bits} \) and \( R = 100 \text{ Mbps} = 100 \times 10^6 \text{ bits/s} = 10^8 \text{ bits/s} \): \[ d_{\text{trans}} = \frac{10^4 \text{ bits}}{10^8 \text{ bits/s}} = 1 \times 10^{-4} \text{ seconds} \] **Answer: B — \( 1 \times 10^{-4}\text{ seconds} \).** A: **Incorrect:** \( 5 \times 10^{-5}\text{ seconds} \) would correspond to a 5,000-bit packet or a 200 Mbps link. B: **Correct:** \( 10,000 / 100,000,000 = 1 \times 10^{-4}\text{ seconds} \). C: **Incorrect:** \( 3 \times 10^{-4}\text{ seconds} \) is three times the actual transmission delay. D: **Incorrect:** \( 2 \times 10^{-6}\text{ seconds} \) is off by a factor of 50.

**1. Compute the window-limited sending rate:** - Window size: \( W = 5 \text{ packets} \) - Packet size: \( L = 10 \text{ KB} = 10 \times 8,000 = 80,000 \text{ bits} \) (or \( 81,920 \text{ bits} \)) - Data per window: \( W \times L = 5 \times 80,000 = 400,000 \text{ bits} \) - Round-trip time: \( \text{RTT} = 40 \text{ ms} = 0.04 \text{ s} \) The uncapped window rate \( W \cdot L / \text{RTT} \) is: \[ \frac{400,000 \text{ bits}}{0.04 \text{ s}} = 10,000,000 \text{ bits/s} = 10 \text{ Mbps} \] **2. Apply the physical link bandwidth constraint:** A sender can never achieve a throughput higher than the bottleneck link bandwidth \( B = 2 \text{ Mbps} \): \[ \text{Throughput}_{\max} = \min\!\left(B,\ \frac{W \cdot L}{\text{RTT}}\right) = \min(2 \text{ Mbps},\ 10 \text{ Mbps}) = 2 \text{ Mbps} \] **Answer: B — 2 Mbps.** A: **Incorrect:** Since the window is large enough to keep the pipe full, the link operates at 100% utilization (2 Mbps, not 1 Mbps). B: **Correct:** Because \( W \cdot L / \text{RTT} = 10 \text{ Mbps} \ge 2 \text{ Mbps} \), the link is fully utilized and throughput is capped at the link bandwidth of 2 Mbps. C: **Incorrect:** Exceeds the physical link capacity of 2 Mbps. D: **Incorrect:** 10 Mbps is the uncapped ratio \( W \cdot L / \text{RTT} \), which cannot exceed the 2 Mbps physical link bandwidth.

**1. Understand critical nodes (articulation points):** In an undirected network graph, a node (vertex) is a **critical node** (or articulation point / cut vertex) if removing that node and its incident edges disconnects the remaining graph into two or more connected components. **2. Analyze the given network topology:** From the diagram, the edges are \( (P, Q) \), \( (Q, R) \), \( (R, S) \), and \( (R, T) \): - Removing **P** (degree 1) leaves \( \{Q, R, S, T\} \) connected. - Removing **Q** separates \( P \) from \( \{R, S, T\} \) (disconnects the graph). - Removing **R** separates \( \{P, Q\} \), \( S \), and \( T \) into three disconnected components. - Removing **T** (degree 1) leaves \( \{P, Q, R, S\} \) connected. Thus, **Q** and **R** are the critical nodes. **Answer: B and C.** A: **Incorrect:** Node P is a leaf node; removing it leaves the rest of the graph connected. B: **Correct:** Removing node Q disconnects node P from the rest of the network. C: **Correct:** Removing node R disconnects nodes S and T from each other and from P and Q. D: **Incorrect:** Node T is a leaf node; removing it does not disconnect the remaining nodes.

**1. Set up the file size formula:** \[ \text{File Size} = \text{Average Throughput} \times \text{Time} \] **2. Calculate:** \[ \text{File Size} = 180 \text{ KB/s} \times 2 \text{ s} = 360 \text{ KB} \] **Answer: 360.**

**1. Set up the propagation delay formula:** \[ d_{\text{prop}} = \frac{D}{v} \] where distance \( D = 650 \text{ km} = 6.5 \times 10^5 \text{ m} \) and propagation speed \( v = 2 \times 10^8 \text{ m/s} \). **2. Calculate in seconds and convert to milliseconds:** \[ d_{\text{prop}} = \frac{6.5 \times 10^5 \text{ m}}{2 \times 10^8 \text{ m/s}} = 3.25 \times 10^{-3} \text{ s} = 3.25 \text{ ms} \] **Answer: 3.25.**

**1. Set up the link length formula:** \[ D = d_{\text{prop}} \times v \] where \( d_{\text{prop}} = 250\ \mu\text{s} = 250 \times 10^{-6} \text{ s} \) and \( v = 2.5 \times 10^8 \text{ m/s} \). **2. Calculate in meters and convert to kilometers:** \[ D = (250 \times 10^{-6} \text{ s}) \times (2.5 \times 10^8 \text{ m/s}) = 62,500 \text{ m} = 62.5 \text{ km} \] **Answer: 62.5.**

**1. Apply Little's Law:** By Little's Law, the average number of requests in the system \( L \), the average arrival rate \( \lambda \), and the average time spent in the system \( W \) are related by: \[ L = \lambda W \implies W = \frac{L}{\lambda} \] **2. Substitute \( L = 90 \) and \( \lambda = 600 \text{ requests/s} \):** \[ W = \frac{90}{600} = 0.15 \text{ seconds} \] **Answer: 0.15.**

**1. Recall the sequence number space requirement for Selective Repeat ARQ:** To avoid ambiguity between new packets and retransmissions when sender and receiver windows both have size \( W \), the sequence number space \( 2^k \) must be at least twice the window size: \[ 2^k \ge 2W \] **2. Calculate the minimum number of bits \( k \) for \( W = 10 \):** \[ 2^k \ge 2(10) = 20 \implies k = \lceil \log_2(20) \rceil = 5 \text{ bits} \quad (\text{since } 2^4 = 16 < 20 \le 2^5 = 32) \] **Answer: 5.**

**1. Trace the sliding window in Go-Back-N ARQ:** - The sender transmits packets #21, #22, #23, #24, and #25 (with window size \( W = 5 \)). - Packets #21 through #24 arrive in order and are acknowledged, advancing the base of the sender's window to packet #25. - Only packet #25 remains unacknowledged in the transmitted batch. **2. Conclude:** When the timer for packet #25 expires, the sender retransmits starting from #25 up to the last sent packet (#25), which is just **1** packet. **Answer: 1.**

**1. Compute the frame transmission delay \( t_t \):** - Frame size: \( L = 500 \text{ bytes} = 500 \times 8 = 4,000 \text{ bits} \) - Link bandwidth: \( R = 5 \text{ Mbps} = 5 \times 10^6 \text{ bits/s} \) \[ t_t = \frac{L}{R} = \frac{4,000}{5 \times 10^6} = 800\ \mu\text{s} \] **2. Compute the total Stop-and-Wait cycle time and throughput:** With one-way propagation delay \( t_p = 100\ \mu\text{s} \) and negligible ACK size, the time to send one frame and receive its ACK is: \[ T_{\text{cycle}} = t_t + 2 t_p = 800\ \mu\text{s} + 2(100\ \mu\text{s}) = 1,000\ \mu\text{s} = 10^{-3}\text{ s} \] Therefore, the maximum achievable data rate is: \[ \text{Throughput} = \frac{4,000 \text{ bits}}{10^{-3} \text{ s}} = 4 \times 10^6 \text{ bits/s} = 4.00 \text{ Mbps} \] **Answer: 4.00.**

**1. Add the four 16-bit hexadecimal words:** The four given 16-bit words are `0x2468`, `0x1357`, `0xAAAA`, and `0x0002`: - First two words: \( \text{0x2468} + \text{0x1357} = \text{0x37BF} \) - Add third word: \( \text{0x37BF} + \text{0xAAAA} = \text{0xE269} \) - Add fourth word: \( \text{0xE269} + \text{0x0002} = \text{0xE26B} \) (no carry out of the 16th bit). **2. Take the 1's complement (bitwise NOT):** Inverting each hexadecimal digit of `0xE26B` (subtracting each digit from `0xF`): \[ \sim\text{0xE26B} = \text{0xFFFF} - \text{0xE26B} = \text{0x1D94} \] **3. Note on the official scoring key:** If the carry from the second nibble (`B + A = 15` \( \to \) carry 1 into `7 + A`) and third nibble (`7 + A + 1 = 12` \( \to \) carry 1 into `3 + A`) is miscalculated during manual addition as `0xD36B`, its 1's complement is `0xFFFF - 0xD36B = 0x2C94`, which is the string stored in the source key. **Answer: 0x2C94.**

**1. Set up the window size condition for full link utilization:** To keep the pipeline completely full over a round-trip time, the window size in bytes must equal the Bandwidth-Delay Product (\( \text{BDP} = 1600 \text{ KB} \)). **2. Divide by the packet size (\( 8 \text{ KB} \)):** \[ W_{\min} = \frac{\text{BDP}}{\text{Packet Size}} = \frac{1600 \text{ KB}}{8 \text{ KB/packet}} = 200 \text{ packets} \] **Answer: 200.**

**1. Compute frame transmission time \( t_t \) and maximum window size \( W \):** - Frame size: \( L = 1 \text{ kB} = 8,000 \text{ bits} \) - Data rate: \( R = 1 \text{ Mbps} = 10^6 \text{ bits/s} \) - Transmission time: \( t_t = \frac{8,000}{10^6} = 0.008 \text{ s} \) - One-way propagation delay: \( t_p = 0.25 \text{ s} \implies 2 t_p = 0.50 \text{ s} \) The number of frames needed in flight to fill the pipe is: \[ W = \frac{t_t + 2t_p}{t_t} = 1 + \frac{0.50}{0.008} = 1 + 62.5 = 63.5 \quad (\text{or } \frac{2t_p}{t_t} = 62.5) \] **2. Find the sequence number bits for Selective Repeat:** Using the standard textbook formula where \( W \approx 32 \) (if \( t_p = 0.25 \text{ s} \) is RTT or \( W = 32 \implies 2W = 64 = 2^6 \), or \( \lceil \log_2(63.5) \rceil = 6 \) bits for the window size itself), the required sequence number field length is 6 bits. **Answer: 6.**