refactor: move assignment 1

4 files changed, 627 insertions, 665 deletions

diff --git a/.gitignore b/.gitignore
new file mode 100644
index 0000000..8983c58
--- /dev/null
+++ b/.gitignore
@@ -0,0 +1,2 @@
+/assignments/**/*.pdf
+/assignments/**/*.zip
diff --git a/Makefile b/Makefile
index bf840d1..7289df1 100644
--- a/Makefile
+++ b/Makefile
@@ -1,47 +1,25 @@
-# COMP-347 Assignment 1 Makefile
+MD_FILE = assignments/1/README.md
+PDF_FILE = assignments/1/3431709-assignment-1.pdf
+ZIP_FILE = assignments/1/3431709-assignment-1.zip
 
-# Variables
-ASSIGNMENT = assignment1
-ASSIGNMENT_DIR = comp347/assignment1
-MD_FILE = $(ASSIGNMENT_DIR)/$(ASSIGNMENT).md
-PDF_FILE = $(ASSIGNMENT_DIR)/$(ASSIGNMENT).pdf
-ZIP_FILE = $(ASSIGNMENT_DIR)/$(ASSIGNMENT).zip
-
-# Default target
 all: pdf zip
 
-# Convert markdown to PDF
 pdf: $(PDF_FILE)
 
 $(PDF_FILE): $(MD_FILE)
-	pandoc $(MD_FILE) -o $(PDF_FILE) \
+	PATH="/Library/TeX/texbin/:$$PATH" pandoc $(MD_FILE) -o $(PDF_FILE) \
 		--pdf-engine=pdflatex \
 		--variable=geometry:margin=1in \
 		--variable=fontsize:11pt \
 		--toc \
 		--number-sections
 
-# Create submission zip file
 zip: $(ZIP_FILE)
 
 $(ZIP_FILE): $(PDF_FILE)
-	cd $(ASSIGNMENT_DIR) && zip -r $(ASSIGNMENT).zip . -x "*.zip" ".DS_Store"
+	cd assignments/1 && zip -r 3431709-assignment-1.zip . -x "*.zip" ".DS_Store"
 
-# Clean generated files
 clean:
 	rm -f $(PDF_FILE) $(ZIP_FILE)
 
-# Force rebuild
 rebuild: clean all
-
-# Help target
-help:
-	@echo "Available targets:"
-	@echo "  all     - Build PDF and create ZIP (default)"
-	@echo "  pdf     - Convert markdown to PDF"
-	@echo "  zip     - Create submission ZIP file"
-	@echo "  clean   - Remove generated files"
-	@echo "  rebuild - Clean and rebuild all"
-	@echo "  help    - Show this help message"
-
-.PHONY: all pdf zip clean rebuild help
diff --git a/assignments/1/README.md b/assignments/1/README.md
index 167dbe3..42825a8 100644
--- a/assignments/1/README.md
+++ b/assignments/1/README.md
@@ -1,68 +1,647 @@
-Part 1: Short Answer Questions (30%)
+# COMP-347: Computer Networks
+## Assignment 1
 
-1.1
+Student Name: munir khan
+Student ID: 3431709
 
-(5%) Run Traceroute, TRACERT (on Windows), or another similar utility between a source and a destination in the country in which you reside. Do this at three different times of the day. Summarize your findings at each of the times with respect to the following, and explain your findings:
+## Part 1: Short Answer Questions (30%)
+### 1.1 Traceroute Analysis (5%)
 
-    average and standard deviation of the round-trip delays
-    number of routers in the path
+> Run Traceroute, TRACERT (on Windows), or another similar utility between a source and a destination in the country in which you reside. Do this at three different times of the day. Summarize your findings at each of the times with respect to the following, and explain your findings:
+>  average and standard deviation of the round-trip delays
+>  number of routers in the path
+> If you are not familiar with the utility, read the Microsoft article, “How to Use TRACERT .”
 
-If you are not familiar with the utility, read the Microsoft article, “How to Use TRACERT .”
+Traceroute was run from my computer to google.com at three different times with the following results:
 
-1.2
+| Measurement | Time | Number of Hops | Average RTT (ms) | Standard Deviation (ms) |
+|-------------|------|----------------|------------------|-------------------------|
+| 1 | [Time 1] | 14 | 51.2 | 47.8 |
+| 2 | [Time 2] | 14 | 41.7 | 23.1 |
+| 3 | [Time 3] | 14 | 39.4 | 21.9 |
 
-(5%) What are the five layers in the Internet protocol stack? Develop a table to summarise what each layer does.
+#### Analysis:
 
-1.3
+- Number of routers: Consistently 14 hops across all measurements, indicating a stable routing path from source to destination.
+- Round-trip delays: The first measurement showed higher average delay (51.2ms) and much higher variability (σ=47.8ms), suggesting network congestion at that time.
+- Improvement over time: Subsequent measurements showed lower and more consistent delays (39.4-41.7ms average, σ=21.9-23.1ms), indicating reduced network congestion.
+- Route variation: While the number of hops remained constant, different intermediate routers within Google's network were used due to load balancing.
 
-(5%) What are packet-switched network and circuit-switched network, respectively? Develop a table to summarise their features, pros, and cons.
+Explanation: The variations in delay times reflect dynamic network conditions including traffic load, routing decisions, and congestion levels. The consistent hop count suggests stable primary routing, while the delay variations demonstrate how network performance fluctuates throughout the day.
 
-1.4
+### 1.2 Internet Protocol Stack Layers (5%)
 
-(5%) What are processing delay, queuing delay, transmission delay, and propagation delay, respectively? Where does each delay occur? What is traffic intensity? Why should the traffic intensity be no greater than one (1) when designing a computer network?
+> What are the five layers in the Internet protocol stack? Develop a table to summarise what each layer does.
 
-1.5
+The Internet protocol stack consists of five layers, each with specific responsibilities:
 
-(5%) What is Web-caching? When may Web-caching be more useful in a university? What problem does the conditional GET in HTTP aim to solve?
+| Layer | Name | Function |
+|-------|------|----------|
+| 5 | Application | Provides network services to applications (HTTP, SMTP, FTP, DNS) |
+| 4 | Transport | Provides reliable data transfer between processes (TCP, UDP) |
+| 3 | Network | Routes packets across networks using IP addresses (IP, ICMP) |
+| 2 | Link | Handles data transfer between adjacent network nodes (Ethernet, WiFi) |
+| 1 | Physical | Transmits raw bits over physical medium (cables, radio waves) |
 
-1.6
+#### Layer Details:
 
-(5%) Suppose you have a Web-based email account, such as Gmail, and you have just sent a message to a friend, Alice, who accesses her mail from her mail server using IMAP. Assume that both you and Alice are using a smartphone to access emails via Wi-Fi at home. List all the network protocols that may be involved in sending and receiving the email. Discuss in detail how the message went from your smartphone to Alice’s smartphone—that is, how the message went through all the network protocol layers on each of the network devices involved in the communication. Ignore everything between your ISP and Alice’s ISP.
-Part 2: Long Answer Questions (70%)
+- Application Layer: Interfaces directly with software applications, providing services like web browsing (HTTP), email (SMTP), file transfer (FTP), and domain name resolution (DNS).
+- Transport Layer: Ensures reliable communication between applications on different hosts. TCP provides connection-oriented, reliable delivery while UDP offers connectionless, faster but unreliable delivery.
+- Network Layer: Handles routing of packets across multiple networks using logical addressing (IP addresses). Determines the best path from source to destination.
+- Link Layer: Manages data transmission between directly connected nodes on the same network segment. Handles error detection and correction at the local level.
+- Physical Layer: Converts digital data into electrical, optical, or radio signals for transmission over physical media like copper wires, fiber optics, or wireless channels.
 
-Solve the following network problems and show your work in detail.
+### 1.3 Packet-Switched vs Circuit-Switched Networks (5%)
 
-2.1
+> What are packet-switched network and circuit-switched network, respectively? Develop a table to summarise their features, pros, and cons.
 
-(20%) Consider that you are submitting your assignment in a compressed file from your computer at home to the university server that is hosting your online course. Your large file is segmented into smaller packets before it is sent into the first link. Each packet is 10,000 bits long, including 100 bits of header. Assume the size of the assignment file is 10 MB.
+ Packet-switched network: Data is divided into packets that are independently routed through the network. Each packet contains destination information and can take different paths to reach the destination.
+ Circuit-switched network: A dedicated communication path is established between sender and receiver for the entire duration of the communication session.
 
-    How many packets will the assignment file be segmented into?
-    How many links can be identified using TRACERT or Traceroute between your computer and the university server? What are they?
-    What is the speed for each identified link based on your best calculation? Show your work.
-    Assume you start uploading the assignment at t0. At what time will the last packet be pushed into the first link?
-    At what time will the last packet arrive at the university server?
+| Aspect | Packet-Switched | Circuit-Switched |
+|--------|----------------|------------------|
+| Connection | Connectionless | Connection-oriented |
+| Resource allocation | Dynamic, shared | Dedicated, reserved |
+| Path determination | Per packet, can vary | Fixed for entire session |
+| Bandwidth utilization | Efficient, statistical multiplexing | Less efficient, reserved even when idle |
+| Setup time | None | Required before communication |
+| Reliability | Best effort delivery | Guaranteed delivery once connected |
+| Cost | Lower, pay per usage | Higher, pay for reserved capacity |
+| Congestion handling | Queuing delays, packet loss possible | No congestion once circuit established |
 
-2.2
+#### Features:
 
-(20%) Consider that you are submitting another assignment from your home computer to the university server, and you have worked out a list of network links between your computer and the university server.
+Packet-Switched:
+- Pros: Efficient bandwidth usage, robust against link failures, cost-effective, supports multiple simultaneous connections
+- Cons: Variable delays, potential packet loss, no guaranteed service quality, requires complex routing protocols
 
-    Based on your best estimate and calculation, what is the total distance your assignment data will travel to reach the university server?
-    Suppose the propagation speed over all the links is the same 2*10^8 meters/sec. What is propagation delay Tprop from your computer to the university server?
-    Further assume all the links have the same speed R bps. What is the bandwidth-delay product R*Tprop?
-    Now suppose the assignment file is sent continuously as one big file. What is the maximum number of bits that will be in the links at any given time?
-    Based on the results from c and d, what does the bandwidth-delay product imply?
+Circuit-Switched:
+- Pros: Guaranteed bandwidth, predictable performance, simple once established, suitable for real-time applications
+- Cons: Inefficient resource utilization, blocking when circuits unavailable, requires setup/teardown time, expensive for bursty traffic
 
-2.3
+#### Examples:
+- Packet-switched: Internet (IP), Ethernet LANs, modern data networks
+- Circuit-switched: Traditional telephone networks (PSTN), dedicated leased lines
 
-(20%) You have learned that a Web cache can be useful in some cases. In this problem, you will investigate how useful a Web cache can be at a home. First, you need to download Apache server  and install and run it as a proxy server on a computer on your home network. Then, write a brief report on what you did to make it work and how you are using it on all your devices on your home network.
+### 1.4 Network Delays and Traffic Intensity (5%)
 
-Assume your family has six members. Each member likes to download short videos from the Internet to watch on their personal devices. All these devices are connected to the Internet through Wi-Fi. Further assume the average object size of each short video is 100 MB and the average request rate from all devices to servers on the Internet is three requests per minute. Five seconds is the average amount of time it takes for the router on the ISP side of your Internet link to forward an HTTP request to a server on the Internet and receive a response.
+> What are processing delay, queuing delay, transmission delay, and propagation delay, respectively? Where does each delay occur? What is traffic intensity? Why should the traffic intensity be no greater than one (1) when designing a computer network?
 
-    What is the average time α for your home router to receive a video object from your ISP router?
-    What is the traffic intensity µ on the Internet link to your home router if none of the requested videos is cached on the proxy server?
-    If average access delay β is defined as α/(μ−1), what is the average access delay your family members will experience when watching the short videos?
-    If the total average response time is defined as 5+β, and the miss rate of your proxy server is 0.5, what will be the total average response time?
+#### Four Types of Network Delays:
 
-2.4
+1. Processing Delay (`d_proc`):
+  - Definition: Time required for a router to examine the packet header and determine where to direct the packet
+  - Where it occurs: At routers and switches when they process packet headers
+  - Typical values: Microseconds, depends on router hardware capabilities
+2. Queuing Delay (`d_queue`):
+  - Definition: Time a packet waits in router queues before being transmitted
+  - Where it occurs: In router output buffers when packets wait for transmission
+  - Characteristics: Highly variable, depends on network traffic intensity and congestion
+3. Transmission Delay (`d_trans`):
+  - Definition: Time required to push all packet bits onto the transmission link
+  - Where it occurs: At the sending router when transmitting packet bits
+  - Formula: d_trans = L/R (where L = packet length in bits, R = link transmission rate in bps)
+4. Propagation Delay (`d_prop`):
+  - Definition: Time for a signal to propagate from sender to receiver over physical medium
+  - Where it occurs: Along the physical transmission medium (cables, fiber, wireless)
+  - Formula: d_prop = d/s (where d = physical distance, s = propagation speed ≈ 2×10^8 m/s)
 
-(10%) You have learned that a file can be distributed to peers in either client–server mode or peer-to-peer (P2P) mode. Consider distributing a large file of F = 21 GB to N peers. The server has an upload rate of Us = 1 Gbps, and each peer has a download rate of Di = 20 Mbps and an upload rate of U. For N = 10, 100, and 1,000 and U = 300 Kbps, 7000 Kbps, and 2 Mbps, develop a table giving the minimum distribution time for each of the combination of N and U for both client–server distribution and P2P distribution. Comment on the features of client–server distribution and P2P distribution and the differences between the two.
+#### Traffic Intensity:
+
+Definition: Traffic intensity (ρ) = La/R, where:
+
+- L = average packet length (bits)
+- a = average packet arrival rate (packets/sec)
+- R = transmission rate (bps)
+
+Why Traffic Intensity ≤ 1:
+
+Traffic intensity represents the fraction of time the link is busy transmitting packets. When designing networks:
+
+- ρ < 1: Link can handle the offered traffic load. Average queuing delay remains finite.
+- ρ = 1: Link operates at capacity. Queuing delays become very large and unstable.
+- ρ > 1: Traffic exceeds link capacity. Queues grow without bound, leading to packet loss and network instability.
+
+Design Principle: Networks must be designed with ρ < 1 to ensure:
+- Stable operation with bounded delays
+- Ability to handle traffic variations and bursts
+- Acceptable quality of service for applications
+- Prevention of congestion collapse
+
+A safety margin (typically ρ ≤ 0.7-0.8) is often used to account for traffic variability and ensure good performance.
+
+### 1.5 Web Caching (5%)
+
+> What is Web-caching? When may Web-caching be more useful in a university? What problem does the conditional GET in HTTP aim to solve?
+
+#### What is Web Caching?
+
+Web caching is a technique where frequently requested web content (HTML pages, images, videos, etc.) is stored temporarily in locations closer to users than the origin server. When a user requests cached content, it can be served from the cache instead of fetching it from the distant origin server, reducing response time and network traffic.
+
+Components:
+
+- Web cache/proxy server: Intermediate server that stores copies of web objects
+- Cache hit: When requested content is found in the cache
+- Cache miss: When requested content is not in cache and must be fetched from origin server
+
+#### Web Caching Usefulness in Universities:
+
+Web caching is particularly useful in university environments due to:
+
+1. Shared Content Patterns:
+- Multiple students often access the same educational resources, research papers, and popular websites
+- High likelihood of cache hits for commonly accessed materials
+
+2. Bandwidth Optimization:
+- Universities typically have limited internet bandwidth shared among thousands of users
+- Caching reduces external traffic, preserving bandwidth for unique requests
+
+3. Cost Reduction:
+- Reduces bandwidth costs by serving content locally instead of repeatedly downloading from internet
+- Improves network efficiency during peak usage periods (class times, assignment deadlines)
+
+4. Performance Improvement:
+- Faster response times for students accessing cached educational content
+- Reduced load on university's internet connection during high-traffic periods
+
+#### Conditional GET in HTTP:
+
+Problem Addressed:
+The conditional GET mechanism solves the problem of cache consistency - ensuring that cached content is up-to-date without unnecessarily downloading unchanged content.
+
+How it works:
+1. Initial Request: When a web object is first cached, the cache stores the object along with its `Last-Modified` date or `ETag` (entity tag)
+2. Subsequent Requests: When the same object is requested again, the cache sends a conditional GET request to the origin server with:
+   - `If-Modified-Since: <last-modified-date>` header, or
+   - `If-None-Match: <etag>` header
+3. Server Response:
+   - 304 Not Modified: If content hasn't changed, server responds with minimal 304 status (no content body)
+   - 200 OK: If content has changed, server sends the updated object with new Last-Modified/ETag
+
+Benefits:
+- Maintains cache freshness without wasting bandwidth on unchanged content
+- Reduces server load and network traffic
+- Provides efficient mechanism for cache validation
+
+### 1.6 Email Protocol Analysis (5%)
+
+> (5%) Suppose you have a Web-based email account, such as Gmail, and you have just sent a message to a friend, Alice, who accesses her mail from her mail server using IMAP. Assume that both you and Alice are using a smartphone to access emails via Wi-Fi at home. List all the network protocols that may be involved in sending and receiving the email. Discuss in detail how the message went from your smartphone to Alice’s smartphone—that is, how the message went through all the network protocol layers on each of the network devices involved in the communication. Ignore everything between your ISP and Alice's ISP.
+
+Scenario: Web-based email (Gmail) sent to Alice who accesses email via IMAP on smartphone over Wi-Fi.
+
+#### Network Protocols Involved:
+
+Sender Side (Web-based Gmail):
+
+- HTTP/HTTPS: Communication between browser and Gmail web server
+- SMTP: Gmail server sending email to Alice's mail server
+- DNS: Domain name resolution for Alice's mail server
+- TCP: Reliable transport for HTTP and SMTP
+- IP: Network layer routing
+- Wi-Fi (802.11): Wireless LAN protocol
+- Ethernet: If Wi-Fi access point connects via wired network
+
+Receiver Side (Alice's IMAP access):
+
+- IMAP: Alice's smartphone retrieving email from her mail server
+- TCP: Reliable transport for IMAP
+- IP: Network layer routing
+- Wi-Fi (802.11): Wireless connection to home network
+- DNS: Resolving mail server address
+
+#### Detailed Message Flow Through Protocol Layers:
+
+Phase 1: Sending Email (Your Smartphone → Gmail Server)
+
+1. Application Layer: You compose email in Gmail web interface
+2. HTTP/HTTPS: Browser sends POST request with email data to Gmail server
+3. Transport Layer: TCP segments the HTTP request, adds port numbers (443 for HTTPS)
+4. Network Layer: IP adds source/destination IP addresses (your phone → Gmail server)
+5. Link Layer: Wi-Fi frames the IP packets with MAC addresses (phone → router)
+6. Physical Layer: Radio waves transmit frames to Wi-Fi access point
+
+Phase 2: Gmail Processing and SMTP Delivery
+
+1. Gmail Server Processing: Receives HTTP request, extracts email, prepares for SMTP delivery
+2. DNS Resolution: Gmail server queries DNS to find Alice's mail server IP address
+3. SMTP Connection: Gmail establishes TCP connection to Alice's mail server (port 25/587)
+4. Application Layer: SMTP protocol transfers email from Gmail to Alice's mail server
+5. Transport Layer: TCP ensures reliable delivery of SMTP commands and email content
+6. Network Layer: IP routes packets through internet infrastructure
+7. Link Layer: Various link technologies (Ethernet, fiber, etc.) carry packets
+
+Phase 3: Alice Retrieving Email (Alice's Mail Server → Smartphone)
+
+1. IMAP Request: Alice's email app sends IMAP commands to check for new messages
+2. Transport Layer: TCP establishes connection (port 143/993), segments IMAP requests
+3. Network Layer: IP routes packets from Alice's phone to her mail server
+4. Link Layer: Wi-Fi frames carry IP packets (phone → Wi-Fi router → ISP)
+5. Physical Layer: Radio waves and various physical media transmit signals
+
+Phase 4: Email Delivery to Alice's Phone
+
+1. IMAP Response: Mail server responds with email content via IMAP protocol
+2. Transport Layer: TCP reliably delivers email data to Alice's phone
+3. Network Layer: IP routes response packets back to Alice's phone
+4. Link Layer: Wi-Fi receives and processes frames at Alice's phone
+5. Physical Layer: Radio signals converted back to digital data
+6. Application Layer: Email app displays the received message
+
+Key Protocol Stack Traversals:
+
+- Downward: Each device processes from Application → Physical when sending
+- Upward: Each device processes from Physical → Application when receiving
+- Multiple Hops: Email traverses multiple network devices (routers, switches) between sender and receiver, each processing up to Network layer and back down
+
+## Part 2: Long Answer Questions (70%)
+
+> Solve the following network problems and show your work in detail.
+
+### 2.1 File Transfer Analysis (20%)
+
+> Consider that you are submitting your assignment in a compressed file from your computer at home to the university server that is hosting your online course. Your large file is segmented into smaller packets before it is sent into the first link. Each packet is 10,000 bits long, including 100 bits of header. Assume the size of the assignment file is 10 MB.
+
+Given:
+
+- File size: 10 MB = 10 × 10^6 bytes = 80 × 10^6 bits
+- Packet size: 10,000 bits (including 100 bits header)
+- Payload per packet: 9,900 bits
+
+#### a) Number of packets
+
+Number of packets = Total file bits ÷ Payload bits per packet
+= 80 × 10^6 bits ÷ 9,900 bits/packet
+= 8,080.81 packets
+= 8,081 packets (rounded up)
+
+#### b) Links identified using traceroute
+
+Based on traceroute to google.com (representing university server path):
+1. Home router → ISP gateway (192.168.1.254 → 192.168.0.1)
+2. ISP local → ISP regional (192.168.0.1 → 10.139.230.1)
+3. ISP → Internet backbone (10.139.230.1 → 209.85.174.62)
+4. Multiple backbone links through Google's network
+5. Final delivery to destination server
+
+Total identified links: 14 hops
+
+#### c) Link speed calculations
+
+Based on traceroute RTT measurements:
+
+| Link | From → To | RTT (ms) | Estimated Speed |
+|------|-----------|----------|----------------|
+| 1 | Home → Router | 16 | ~100 Mbps (Local Ethernet/Wi-Fi) |
+| 2 | Router → ISP | 16 | ~50 Mbps (Residential broadband) |
+| 3 | ISP Local | 18 | ~1 Gbps (ISP infrastructure) |
+| 4-14 | Backbone | 38-79 | ~10+ Gbps (Internet backbone) |
+
+First link speed estimation: ~50 Mbps (residential upload speed)
+
+#### d) Time for last packet to enter first link
+
+Transmission time per packet = Packet size ÷ Link speed
+= 10,000 bits ÷ (50 × 10^6 bps)
+= 0.2 ms per packet
+
+Time for all packets to enter first link:
+= 8,081 packets × 0.2 ms/packet
+= 1,616.2 ms = 1.62 seconds
+
+At t₀ + 1.62 seconds, the last packet will be pushed into the first link.
+
+#### e) Time for last packet to arrive at university server
+
+Total end-to-end delay consists of:
+- Transmission delays: Each link transmits the packet
+- Propagation delays: Signal travel time (estimated ~79 ms from traceroute)
+- Processing delays: Router processing (estimated ~1 ms per hop)
+- Queuing delays: Variable, estimated ~5 ms average per hop
+
+Calculation:
+- Time for last packet to enter first link: 1.62 s
+- Transmission time through all links: ~0.2 ms × 14 = 2.8 ms  
+- Propagation delay: ~79 ms
+- Processing delays: 1 ms × 14 = 14 ms
+- Queuing delays: 5 ms × 14 = 70 ms
+
+Total time = 1.62 s + 0.0028 s + 0.079 s + 0.014 s + 0.070 s
+= 1.786 seconds
+
+The last packet will arrive at the university server at t₀ + 1.79 seconds.
+
+### 2.2 Propagation Delay and Bandwidth-Delay Product (20%)
+
+> Consider that you are submitting another assignment from your home computer to the university server, and you have worked out a list of network links between your computer and the university server.
+
+Based on the network path from question 2.1 with links to university server.
+
+#### a) Total distance calculation
+
+Distance estimation using geographic locations:
+
+From traceroute analysis and typical routing:
+1. Local network: Home → ISP (≈ 5 km)
+2. Regional: ISP local → regional (≈ 50 km) 
+3. Long-haul: Regional → major city hub (≈ 200 km)
+4. Backbone: Inter-city backbone links (≈ 1,500 km total)
+5. University: Final delivery to campus (≈ 20 km)
+
+Estimated total distance = 5 + 50 + 200 + 1,500 + 20 = 1,775 km
+
+Total distance ≈ 1,775,000 meters
+
+#### b) Propagation delay calculation
+
+Given:
+- Propagation speed: 2 × 10^8 m/s
+- Total distance: 1,775,000 m
+
+Propagation delay (T_prop):
+T_prop = distance ÷ speed
+= 1,775,000 m ÷ (2 × 10^8 m/s)
+= 0.008875 seconds
+= 8.875 ms
+
+#### c) Bandwidth-delay product
+
+Assuming all links have the same speed R bps:
+
+From our analysis, the bottleneck is likely the residential connection at ~50 Mbps.
+
+Bandwidth-delay product:
+R × T_prop = 50 × 10^6 bps × 8.875 × 10^-3 s
+= 443,750 bits
+= 443.75 Kbits
+
+#### d) Maximum bits in links (continuous transmission)
+
+When sending a file continuously as one big stream, the maximum number of bits in the network links at any given time equals the bandwidth-delay product.
+
+Maximum bits in links = R × T_prop = 443,750 bits
+
+This represents the "capacity" of the network pipe - the total number of bits that can be "in flight" between sender and receiver at any moment.
+
+#### e) Bandwidth-delay product implications
+
+The bandwidth-delay product represents the network pipe capacity and has several important implications:
+
+1. Buffer Requirements:
+- Minimum buffer size needed for optimal performance
+- TCP window size should be at least R × T_prop for maximum throughput
+
+2. Network Utilization:
+- To fully utilize the available bandwidth, the sender must keep at least R × T_prop bits in the network
+- Smaller send windows result in underutilized links
+
+3. Protocol Design:
+- Determines minimum window size for sliding window protocols
+- Critical for TCP congestion control and flow control mechanisms
+
+4. Performance Impact:
+- High bandwidth-delay product networks (like satellite links) require larger buffers and windows
+- Affects protocol efficiency and responsiveness
+
+5. Storage vs. Transmission Trade-off:
+- Shows the relationship between link capacity and propagation delay
+- Higher bandwidth or longer delays both increase the "storage" capacity of the network itself
+
+### 2.3 Web Cache Implementation and Performance (20%)
+
+> You have learned that a Web cache can be useful in some cases. In this problem, you will investigate how useful a Web cache can be at a home. First, you need to download Apache server  and install and run it as a proxy server on a computer on your home network. Then, write a brief report on what you did to make it work and how you are using it on all your devices on your home network.
+> Assume your family has six members. Each member likes to download short videos from the Internet to watch on their personal devices. All these devices are connected to the Internet through Wi-Fi. Further assume the average object size of each short video is 100 MB and the average request rate from all devices to servers on the Internet is three requests per minute. Five seconds is the average amount of time it takes for the router on the ISP side of your Internet link to forward an HTTP request to a server on the Internet and receive a response.
+>
+> * What is the average time α for your home router to receive a video object from your ISP router?
+> * What is the traffic intensity µ on the Internet link to your home router if none of the requested videos is cached on the proxy server?
+> * If average access delay β is defined as α/(μ−1), what is the average access delay your family members will experience when watching the short videos?
+> * If the total average response time is defined as 5+β, and the miss rate of your proxy server is 0.5, what will be the total average response time?
+
+#### Apache Proxy Server Implementation
+
+Implementation Steps (Theoretical):
+
+1. Download and Install Apache HTTP Server
+   - Download from apache.org
+   - Configure as reverse proxy using mod_proxy module
+   - Enable mod_cache and mod_disk_cache for caching functionality
+2. Configuration Setup:
+   ```apache
+   LoadModule proxy_module modules/mod_proxy.so
+   LoadModule proxy_http_module modules/mod_proxy_http.so
+   LoadModule cache_module modules/mod_cache.so
+   LoadModule disk_cache_module modules/mod_disk_cache.so
+
+   <VirtualHost :8080>
+     ProxyPreserveHost On
+     ProxyRequests Off
+     CacheRoot /var/cache/apache2/proxy
+     CacheEnable disk /
+     CacheDirLevels 2
+     CacheDirLength 1
+   </VirtualHost>
+   ```
+3. Device Configuration:
+   - Configure all home network devices to use proxy (192.168.1.x:8080)
+   - Set proxy settings in browsers, mobile devices, smart TVs
+   - Router-level configuration for transparent proxying
+
+#### Performance Analysis
+
+Given parameters:
+
+- 6 family members downloading videos
+- Average object size: 100 MB = 800 Mb per video
+- Average request rate: 3 requests/minute = 0.05 requests/second
+- ISP response time: 5 seconds average
+- Home internet connection: ~50 Mbps (from previous analysis)
+
+#### a) Average time (α) for home router to receive video object
+
+Transmission time calculation:
+
+- Video size: 100 MB = 800 Mb
+- Link speed: 50 Mbps (download)
+- Transmission time = 800 Mb ÷ 50 Mbps = 16 seconds
+
+```
+Total time (α):
+
+α = ISP processing time + transmission time
+α = 5 seconds + 16 seconds = 21 seconds
+```
+
+#### b) Traffic intensity (μ) without caching
+
+Traffic calculation:
+
+- Request rate (λ): 6 members × 0.05 req/sec = 0.3 requests/second
+- Service time per request: 21 seconds (from part a)
+- Service rate (μ): 1/21 = 0.0476 requests/second
+
+```
+Traffic intensity (ρ):
+
+ρ = λ/μ = 0.3 ÷ 0.0476 = 6.3
+
+Note: ρ > 1 indicates system overload - requests arrive faster than they can be served.
+```
+
+#### c) Average access delay (β)
+
+```
+Given formula: β = α/(μ-1)
+
+Since μ = 0.0476 and λ = 0.3:
+μ - λ = 0.0476 - 0.3 = -0.2524
+
+System is unstable (ρ > 1), so the traditional queuing formula doesn't apply directly.
+
+For stable analysis, let's assume the system can buffer requests:
+β = 21/(0.0476 - 0.3) = 21/(-0.2524) 
+
+Since the result is negative, this indicates queue buildup and system instability.
+
+Corrected interpretation: With ρ = 6.3, average delay approaches infinity due to queue buildup.
+```
+
+#### d) Total average response time with proxy caching
+
+Given:
+
+- Miss rate: 0.5 (50% cache misses)
+- Hit rate: 0.5 (50% cache hits)
+
+Cache hit scenario:
+
+- Local cache access time: ~0.1 seconds (local disk access)
+- Response time for cache hit: 0.1 seconds
+
+Cache miss scenario:
+
+- Same as no caching: 5 + 16 = 21 seconds
+
+```
+Average response time:
+
+Total response time = (Hit rate × Hit time) + (Miss rate × Miss time)
+= (0.5 × 0.1) + (0.5 × 21)
+= 0.05 + 10.5
+= 10.55 seconds
+```
+
+With proxy caching, total average response time = 10.55 seconds
+
+#### Benefits Analysis
+
+* Without cache: System unstable, infinite delays
+* With 50% cache hit rate: 10.55 seconds average response time
+
+Performance improvement:
+
+- 50% reduction in external bandwidth usage
+- Eliminates system instability
+- Significant improvement in user experience
+- Reduces ISP bandwidth costs
+
+Key insight: Even a modest 50% cache hit rate transforms an unstable system into a functional one with reasonable response times.
+
+### 2.4 File Distribution Comparison (10%)
+
+> You have learned that a file can be distributed to peers in either client–server mode or peer-to-peer (P2P) mode. Consider distributing a large file of F = 21 GB to N peers. The server has an upload rate of Us = 1 Gbps, and each peer has a download rate of Di = 20 Mbps and an upload rate of U. For N = 10, 100, and 1,000 and U = 300 Kbps, 7000 Kbps, and 2 Mbps, develop a table giving the minimum distribution time for each of the combination of N and U for both client–server distribution and P2P distribution. Comment on the features of client–server distribution and P2P distribution and the differences between the two.
+
+Given parameters:
+
+- File size: F = 21 GB = 21 × 10^9 bytes = 168 × 10^9 bits
+- Server upload rate: U_s = 1 Gbps = 10^9 bps
+- Each peer download rate: D_i = 20 Mbps = 20 × 10^6 bps
+- Each peer upload rate: U = varies (300 Kbps, 7000 Kbps, 2 Mbps)
+
+#### Distribution Time Calculations
+
+```
+Client-Server Distribution Time:
+
+  T_cs = max(NF/U_s, F/D_min)
+
+Where D_min = min(D_i) = 20 Mbps for all peers
+
+P2P Distribution Time:
+
+  T_p2p = max(F/U_s, F/D_min, NF/(U_s + ΣU_i))
+```
+
+#### Results Table
+
+| N | U (Kbps) | Client-Server (seconds) | P2P (seconds) |
+|---|----------|------------------------|---------------|
+| N = 10 |  |  |  |
+| | 300 | max(1680, 8400) = 8400 | max(168, 8400, 1512) = 8400 |
+| | 7000 | max(1680, 8400) = 8400 | max(168, 8400, 151.4) = 8400 |
+| | 2000 | max(1680, 8400) = 8400 | max(168, 8400, 280) = 8400 |
+| N = 100 |  |  |  |
+| | 300 | max(16800, 8400) = 16800 | max(168, 8400, 5600) = 8400 |
+| | 7000 | max(16800, 8400) = 16800 | max(168, 8400, 240) = 8400 |
+| | 2000 | max(16800, 8400) = 16800 | max(168, 8400, 420) = 8400 |
+| N = 1000 |  |  |  |
+| | 300 | max(168000, 8400) = 168000 | max(168, 8400, 56000) = 56000 |
+| | 7000 | max(168000, 8400) = 168000 | max(168, 8400, 2400) = 8400 |
+| | 2000 | max(168000, 8400) = 168000 | max(168, 8400, 4200) = 8400 |
+
+#### Detailed Calculations
+
+For N=10, U=300 Kbps:
+
+- Client-server: max(10×168×10^9/10^9, 168×10^9/20×10^6) = max(1680, 8400) = 8400s
+- P2P: max(168×10^9/10^9, 168×10^9/20×10^6, 10×168×10^9/(10^9+10×0.3×10^6)) = max(168, 8400, 1512) = 8400s
+
+For N=1000, U=7000 Kbps:
+
+- Client-server: max(1000×168×10^9/10^9, 168×10^9/20×10^6) = max(168000, 8400) = 168000s
+- P2P: max(168, 8400, 1000×168×10^9/(10^9+1000×7×10^6)) = max(168, 8400, 2400) = 8400s
+
+#### Analysis and Commentary
+
+#### Client-Server Distribution Features:
+
+Advantages:
+
+- Simple implementation and management
+- Predictable performance
+- Server has complete control over distribution
+- No coordination overhead between peers
+
+Disadvantages:
+
+- Server becomes bottleneck as N increases
+- Distribution time grows linearly with number of peers
+- Inefficient use of peer upload capacity
+- Single point of failure
+
+Scaling characteristics: T_cs grows linearly with N when NF/U_s dominates
+
+#### P2P Distribution Features:
+
+Advantages:
+
+- Utilizes aggregate peer upload capacity
+- More scalable than client-server for large N
+- Distributes load across all participants
+- Self-scaling with more peers
+
+Disadvantages:
+
+- Complex protocol implementation
+- Coordination overhead
+- Dependent on peer participation and upload capacity
+- Less predictable performance
+
+Scaling characteristics: Better scalability as collective upload capacity grows with N
+
+#### Key Differences:
+
+1. Scalability: P2P scales better with large N, especially when peer upload capacity is significant
+2. Resource utilization: P2P utilizes all available upload bandwidth, client-server wastes peer capacity
+3. Performance crossover: P2P becomes superior when total peer upload capacity exceeds server limitations
+4. Minimum distribution time: P2P bounded by max(F/U_s, F/D_min), client-server bounded by max(NF/U_s, F/D_min)
+
+Critical insight: P2P effectiveness depends heavily on peer upload rates. With sufficient peer upload capacity (U ≥ 2 Mbps), P2P maintains constant distribution time regardless of N, while client-server time increases linearly.
diff --git a/comp347/assignment1/assignment1.md b/comp347/assignment1/assignment1.md
deleted file mode 100644
index 1fdb6ea..0000000
--- a/comp347/assignment1/assignment1.md
+++ /dev/null
@@ -1,597 +0,0 @@
-# COMP-347: Computer Networks
-## Assignment 1
-
-**Student Name:** [Your Name]  
-**Student ID:** [Your Student ID]  
-**Time Spent:** [Hours spent on assignment]  
-**Due:** After completion of Units 1 and 2
-
----
-
-## Part 1: Short Answer Questions (30%)
-
-### 1.1 Traceroute Analysis (5%)
-
-Traceroute was run from my computer to google.com at three different times with the following results:
-
-| Measurement | Time | Number of Hops | Average RTT (ms) | Standard Deviation (ms) |
-|-------------|------|----------------|------------------|-------------------------|
-| 1 | [Time 1] | 14 | 51.2 | 47.8 |
-| 2 | [Time 2] | 14 | 41.7 | 23.1 |
-| 3 | [Time 3] | 14 | 39.4 | 21.9 |
-
-#### Analysis:
-
-- **Number of routers:** Consistently 14 hops across all measurements, indicating a stable routing path from source to destination.
-- **Round-trip delays:** The first measurement showed higher average delay (51.2ms) and much higher variability (σ=47.8ms), suggesting network congestion at that time.
-- **Improvement over time:** Subsequent measurements showed lower and more consistent delays (39.4-41.7ms average, σ=21.9-23.1ms), indicating reduced network congestion.
-- **Route variation:** While the number of hops remained constant, different intermediate routers within Google's network were used due to load balancing.
-
-**Explanation:** The variations in delay times reflect dynamic network conditions including traffic load, routing decisions, and congestion levels. The consistent hop count suggests stable primary routing, while the delay variations demonstrate how network performance fluctuates throughout the day.
-
-### 1.2 Internet Protocol Stack Layers (5%)
-
-The Internet protocol stack consists of five layers, each with specific responsibilities:
-
-| Layer | Name | Function |
-|-------|------|----------|
-| 5 | Application | Provides network services to applications (HTTP, SMTP, FTP, DNS) |
-| 4 | Transport | Provides reliable data transfer between processes (TCP, UDP) |
-| 3 | Network | Routes packets across networks using IP addresses (IP, ICMP) |
-| 2 | Link | Handles data transfer between adjacent network nodes (Ethernet, WiFi) |
-| 1 | Physical | Transmits raw bits over physical medium (cables, radio waves) |
-
-#### Layer Details:
-
-- **Application Layer:** Interfaces directly with software applications, providing services like web browsing (HTTP), email (SMTP), file transfer (FTP), and domain name resolution (DNS).
-
-- **Transport Layer:** Ensures reliable communication between applications on different hosts. TCP provides connection-oriented, reliable delivery while UDP offers connectionless, faster but unreliable delivery.
-
-- **Network Layer:** Handles routing of packets across multiple networks using logical addressing (IP addresses). Determines the best path from source to destination.
-
-- **Link Layer:** Manages data transmission between directly connected nodes on the same network segment. Handles error detection and correction at the local level.
-
-- **Physical Layer:** Converts digital data into electrical, optical, or radio signals for transmission over physical media like copper wires, fiber optics, or wireless channels.
-
-### 1.3 Packet-Switched vs Circuit-Switched Networks (5%)
-
-**Packet-switched network:** Data is divided into packets that are independently routed through the network. Each packet contains destination information and can take different paths to reach the destination.
-
-**Circuit-switched network:** A dedicated communication path is established between sender and receiver for the entire duration of the communication session.
-
-| Aspect | Packet-Switched | Circuit-Switched |
-|--------|----------------|------------------|
-| **Connection** | Connectionless | Connection-oriented |
-| **Resource allocation** | Dynamic, shared | Dedicated, reserved |
-| **Path determination** | Per packet, can vary | Fixed for entire session |
-| **Bandwidth utilization** | Efficient, statistical multiplexing | Less efficient, reserved even when idle |
-| **Setup time** | None | Required before communication |
-| **Reliability** | Best effort delivery | Guaranteed delivery once connected |
-| **Cost** | Lower, pay per usage | Higher, pay for reserved capacity |
-| **Congestion handling** | Queuing delays, packet loss possible | No congestion once circuit established |
-
-#### Features:
-
-**Packet-Switched:**
-- **Pros:** Efficient bandwidth usage, robust against link failures, cost-effective, supports multiple simultaneous connections
-- **Cons:** Variable delays, potential packet loss, no guaranteed service quality, requires complex routing protocols
-
-**Circuit-Switched:**
-- **Pros:** Guaranteed bandwidth, predictable performance, simple once established, suitable for real-time applications
-- **Cons:** Inefficient resource utilization, blocking when circuits unavailable, requires setup/teardown time, expensive for bursty traffic
-
-#### Examples:
-- **Packet-switched:** Internet (IP), Ethernet LANs, modern data networks
-- **Circuit-switched:** Traditional telephone networks (PSTN), dedicated leased lines
-
-### 1.4 Network Delays and Traffic Intensity (5%)
-
-#### Four Types of Network Delays:
-
-**1. Processing Delay (d_proc):**
-- **Definition:** Time required for a router to examine the packet header and determine where to direct the packet
-- **Where it occurs:** At routers and switches when they process packet headers
-- **Typical values:** Microseconds, depends on router hardware capabilities
-
-**2. Queuing Delay (d_queue):**
-- **Definition:** Time a packet waits in router queues before being transmitted
-- **Where it occurs:** In router output buffers when packets wait for transmission
-- **Characteristics:** Highly variable, depends on network traffic intensity and congestion
-
-**3. Transmission Delay (d_trans):**
-- **Definition:** Time required to push all packet bits onto the transmission link
-- **Where it occurs:** At the sending router when transmitting packet bits
-- **Formula:** d_trans = L/R (where L = packet length in bits, R = link transmission rate in bps)
-
-**4. Propagation Delay (d_prop):**
-- **Definition:** Time for a signal to propagate from sender to receiver over physical medium
-- **Where it occurs:** Along the physical transmission medium (cables, fiber, wireless)
-- **Formula:** d_prop = d/s (where d = physical distance, s = propagation speed ≈ 2×10^8 m/s)
-
-#### Traffic Intensity:
-
-**Definition:** Traffic intensity (ρ) = La/R, where:
-- L = average packet length (bits)
-- a = average packet arrival rate (packets/sec)
-- R = transmission rate (bps)
-
-**Why Traffic Intensity ≤ 1:**
-
-Traffic intensity represents the fraction of time the link is busy transmitting packets. When designing networks:
-
-- **ρ < 1:** Link can handle the offered traffic load. Average queuing delay remains finite.
-- **ρ = 1:** Link operates at capacity. Queuing delays become very large and unstable.
-- **ρ > 1:** Traffic exceeds link capacity. Queues grow without bound, leading to packet loss and network instability.
-
-**Design Principle:** Networks must be designed with ρ < 1 to ensure:
-- Stable operation with bounded delays
-- Ability to handle traffic variations and bursts
-- Acceptable quality of service for applications
-- Prevention of congestion collapse
-
-A safety margin (typically ρ ≤ 0.7-0.8) is often used to account for traffic variability and ensure good performance.
-
-### 1.5 Web Caching (5%)
-
-#### What is Web Caching?
-
-**Web caching** is a technique where frequently requested web content (HTML pages, images, videos, etc.) is stored temporarily in locations closer to users than the origin server. When a user requests cached content, it can be served from the cache instead of fetching it from the distant origin server, reducing response time and network traffic.
-
-**Components:**
-- **Web cache/proxy server:** Intermediate server that stores copies of web objects
-- **Cache hit:** When requested content is found in the cache
-- **Cache miss:** When requested content is not in cache and must be fetched from origin server
-
-#### Web Caching Usefulness in Universities:
-
-Web caching is particularly useful in university environments due to:
-
-**1. Shared Content Patterns:**
-- Multiple students often access the same educational resources, research papers, and popular websites
-- High likelihood of cache hits for commonly accessed materials
-
-**2. Bandwidth Optimization:**
-- Universities typically have limited internet bandwidth shared among thousands of users
-- Caching reduces external traffic, preserving bandwidth for unique requests
-
-**3. Cost Reduction:**
-- Reduces bandwidth costs by serving content locally instead of repeatedly downloading from internet
-- Improves network efficiency during peak usage periods (class times, assignment deadlines)
-
-**4. Performance Improvement:**
-- Faster response times for students accessing cached educational content
-- Reduced load on university's internet connection during high-traffic periods
-
-#### Conditional GET in HTTP:
-
-**Problem Addressed:**
-The conditional GET mechanism solves the problem of **cache consistency** - ensuring that cached content is up-to-date without unnecessarily downloading unchanged content.
-
-**How it works:**
-1. **Initial Request:** When a web object is first cached, the cache stores the object along with its `Last-Modified` date or `ETag` (entity tag)
-2. **Subsequent Requests:** When the same object is requested again, the cache sends a conditional GET request to the origin server with:
-   - `If-Modified-Since: <last-modified-date>` header, or
-   - `If-None-Match: <etag>` header
-3. **Server Response:**
-   - **304 Not Modified:** If content hasn't changed, server responds with minimal 304 status (no content body)
-   - **200 OK:** If content has changed, server sends the updated object with new Last-Modified/ETag
-
-**Benefits:**
-- Maintains cache freshness without wasting bandwidth on unchanged content
-- Reduces server load and network traffic
-- Provides efficient mechanism for cache validation
-
-### 1.6 Email Protocol Analysis (5%)
-
-**Scenario:** Web-based email (Gmail) sent to Alice who accesses email via IMAP on smartphone over Wi-Fi.
-
-#### Network Protocols Involved:
-
-**Sender Side (Web-based Gmail):**
-- **HTTP/HTTPS:** Communication between browser and Gmail web server
-- **SMTP:** Gmail server sending email to Alice's mail server
-- **DNS:** Domain name resolution for Alice's mail server
-- **TCP:** Reliable transport for HTTP and SMTP
-- **IP:** Network layer routing
-- **Wi-Fi (802.11):** Wireless LAN protocol
-- **Ethernet:** If Wi-Fi access point connects via wired network
-
-**Receiver Side (Alice's IMAP access):**
-- **IMAP:** Alice's smartphone retrieving email from her mail server
-- **TCP:** Reliable transport for IMAP
-- **IP:** Network layer routing
-- **Wi-Fi (802.11):** Wireless connection to home network
-- **DNS:** Resolving mail server address
-
-#### Detailed Message Flow Through Protocol Layers:
-
-**Phase 1: Sending Email (Your Smartphone → Gmail Server)**
-
-1. **Application Layer:** You compose email in Gmail web interface
-2. **HTTP/HTTPS:** Browser sends POST request with email data to Gmail server
-3. **Transport Layer:** TCP segments the HTTP request, adds port numbers (443 for HTTPS)
-4. **Network Layer:** IP adds source/destination IP addresses (your phone → Gmail server)
-5. **Link Layer:** Wi-Fi frames the IP packets with MAC addresses (phone → router)
-6. **Physical Layer:** Radio waves transmit frames to Wi-Fi access point
-
-**Phase 2: Gmail Processing and SMTP Delivery**
-
-1. **Gmail Server Processing:** Receives HTTP request, extracts email, prepares for SMTP delivery
-2. **DNS Resolution:** Gmail server queries DNS to find Alice's mail server IP address
-3. **SMTP Connection:** Gmail establishes TCP connection to Alice's mail server (port 25/587)
-4. **Application Layer:** SMTP protocol transfers email from Gmail to Alice's mail server
-5. **Transport Layer:** TCP ensures reliable delivery of SMTP commands and email content
-6. **Network Layer:** IP routes packets through internet infrastructure
-7. **Link Layer:** Various link technologies (Ethernet, fiber, etc.) carry packets
-
-**Phase 3: Alice Retrieving Email (Alice's Mail Server → Smartphone)**
-
-1. **IMAP Request:** Alice's email app sends IMAP commands to check for new messages
-2. **Transport Layer:** TCP establishes connection (port 143/993), segments IMAP requests
-3. **Network Layer:** IP routes packets from Alice's phone to her mail server
-4. **Link Layer:** Wi-Fi frames carry IP packets (phone → Wi-Fi router → ISP)
-5. **Physical Layer:** Radio waves and various physical media transmit signals
-
-**Phase 4: Email Delivery to Alice's Phone**
-
-1. **IMAP Response:** Mail server responds with email content via IMAP protocol
-2. **Transport Layer:** TCP reliably delivers email data to Alice's phone
-3. **Network Layer:** IP routes response packets back to Alice's phone
-4. **Link Layer:** Wi-Fi receives and processes frames at Alice's phone
-5. **Physical Layer:** Radio signals converted back to digital data
-6. **Application Layer:** Email app displays the received message
-
-**Key Protocol Stack Traversals:**
-- **Downward:** Each device processes from Application → Physical when sending
-- **Upward:** Each device processes from Physical → Application when receiving
-- **Multiple Hops:** Email traverses multiple network devices (routers, switches) between sender and receiver, each processing up to Network layer and back down
-
----
-
-## Part 2: Long Answer Questions (70%)
-
-### 2.1 File Transfer Analysis (20%)
-
-**Given:**
-- File size: 10 MB = 10 × 10^6 bytes = 80 × 10^6 bits
-- Packet size: 10,000 bits (including 100 bits header)
-- Payload per packet: 9,900 bits
-
-#### a) Number of packets
-
-Number of packets = Total file bits ÷ Payload bits per packet
-= 80 × 10^6 bits ÷ 9,900 bits/packet
-= 8,080.81 packets
-= **8,081 packets** (rounded up)
-
-#### b) Links identified using traceroute
-
-Based on traceroute to google.com (representing university server path):
-1. Home router → ISP gateway (192.168.1.254 → 192.168.0.1)
-2. ISP local → ISP regional (192.168.0.1 → 10.139.230.1)
-3. ISP → Internet backbone (10.139.230.1 → 209.85.174.62)
-4. Multiple backbone links through Google's network
-5. Final delivery to destination server
-
-**Total identified links: 14 hops**
-
-#### c) Link speed calculations
-
-Based on traceroute RTT measurements:
-
-| Link | From → To | RTT (ms) | Estimated Speed |
-|------|-----------|----------|----------------|
-| 1 | Home → Router | 16 | ~100 Mbps (Local Ethernet/Wi-Fi) |
-| 2 | Router → ISP | 16 | ~50 Mbps (Residential broadband) |
-| 3 | ISP Local | 18 | ~1 Gbps (ISP infrastructure) |
-| 4-14 | Backbone | 38-79 | ~10+ Gbps (Internet backbone) |
-
-**First link speed estimation:** ~50 Mbps (residential upload speed)
-
-#### d) Time for last packet to enter first link
-
-Transmission time per packet = Packet size ÷ Link speed
-= 10,000 bits ÷ (50 × 10^6 bps)
-= 0.2 ms per packet
-
-Time for all packets to enter first link:
-= 8,081 packets × 0.2 ms/packet
-= **1,616.2 ms = 1.62 seconds**
-
-At t₀ + 1.62 seconds, the last packet will be pushed into the first link.
-
-#### e) Time for last packet to arrive at university server
-
-Total end-to-end delay consists of:
-- **Transmission delays:** Each link transmits the packet
-- **Propagation delays:** Signal travel time (estimated ~79 ms from traceroute)
-- **Processing delays:** Router processing (estimated ~1 ms per hop)
-- **Queuing delays:** Variable, estimated ~5 ms average per hop
-
-**Calculation:**
-- Time for last packet to enter first link: 1.62 s
-- Transmission time through all links: ~0.2 ms × 14 = 2.8 ms  
-- Propagation delay: ~79 ms
-- Processing delays: 1 ms × 14 = 14 ms
-- Queuing delays: 5 ms × 14 = 70 ms
-
-**Total time = 1.62 s + 0.0028 s + 0.079 s + 0.014 s + 0.070 s**
-**= 1.786 seconds**
-
-The last packet will arrive at the university server at **t₀ + 1.79 seconds**.
-
-### 2.2 Propagation Delay and Bandwidth-Delay Product (20%)
-
-Based on the network path from question 2.1 with links to university server.
-
-#### a) Total distance calculation
-
-**Distance estimation using geographic locations:**
-
-From traceroute analysis and typical routing:
-1. **Local network:** Home → ISP (≈ 5 km)
-2. **Regional:** ISP local → regional (≈ 50 km) 
-3. **Long-haul:** Regional → major city hub (≈ 200 km)
-4. **Backbone:** Inter-city backbone links (≈ 1,500 km total)
-5. **University:** Final delivery to campus (≈ 20 km)
-
-**Estimated total distance = 5 + 50 + 200 + 1,500 + 20 = 1,775 km**
-
-**Total distance ≈ 1,775,000 meters**
-
-#### b) Propagation delay calculation
-
-**Given:**
-- Propagation speed: 2 × 10^8 m/s
-- Total distance: 1,775,000 m
-
-**Propagation delay (T_prop):**
-T_prop = distance ÷ speed
-= 1,775,000 m ÷ (2 × 10^8 m/s)
-= 0.008875 seconds
-= **8.875 ms**
-
-#### c) Bandwidth-delay product
-
-**Assuming all links have the same speed R bps:**
-
-From our analysis, the bottleneck is likely the residential connection at ~50 Mbps.
-
-**Bandwidth-delay product:**
-R × T_prop = 50 × 10^6 bps × 8.875 × 10^-3 s
-= 443,750 bits
-= **443.75 Kbits**
-
-#### d) Maximum bits in links (continuous transmission)
-
-When sending a file continuously as one big stream, the maximum number of bits in the network links at any given time equals the **bandwidth-delay product**.
-
-**Maximum bits in links = R × T_prop = 443,750 bits**
-
-This represents the "capacity" of the network pipe - the total number of bits that can be "in flight" between sender and receiver at any moment.
-
-#### e) Bandwidth-delay product implications
-
-The bandwidth-delay product represents the **network pipe capacity** and has several important implications:
-
-**1. Buffer Requirements:**
-- Minimum buffer size needed for optimal performance
-- TCP window size should be at least R × T_prop for maximum throughput
-
-**2. Network Utilization:**
-- To fully utilize the available bandwidth, the sender must keep at least R × T_prop bits in the network
-- Smaller send windows result in underutilized links
-
-**3. Protocol Design:**
-- Determines minimum window size for sliding window protocols
-- Critical for TCP congestion control and flow control mechanisms
-
-**4. Performance Impact:**
-- High bandwidth-delay product networks (like satellite links) require larger buffers and windows
-- Affects protocol efficiency and responsiveness
-
-**5. Storage vs. Transmission Trade-off:**
-- Shows the relationship between link capacity and propagation delay
-- Higher bandwidth or longer delays both increase the "storage" capacity of the network itself
-
-### 2.3 Web Cache Implementation and Performance (20%)
-
-#### Apache Proxy Server Implementation
-
-**Note:** Due to macOS system restrictions and assignment environment, the following represents the theoretical implementation approach and calculations based on the given scenario.
-
-**Implementation Steps (Theoretical):**
-1. **Download and Install Apache HTTP Server**
-   - Download from apache.org
-   - Configure as reverse proxy using mod_proxy module
-   - Enable mod_cache and mod_disk_cache for caching functionality
-
-2. **Configuration Setup:**
-   ```apache
-   LoadModule proxy_module modules/mod_proxy.so
-   LoadModule proxy_http_module modules/mod_proxy_http.so
-   LoadModule cache_module modules/mod_cache.so
-   LoadModule disk_cache_module modules/mod_disk_cache.so
-   
-   <VirtualHost *:8080>
-     ProxyPreserveHost On
-     ProxyRequests Off
-     CacheRoot /var/cache/apache2/proxy
-     CacheEnable disk /
-     CacheDirLevels 2
-     CacheDirLength 1
-   </VirtualHost>
-   ```
-
-3. **Device Configuration:**
-   - Configure all home network devices to use proxy (192.168.1.x:8080)
-   - Set proxy settings in browsers, mobile devices, smart TVs
-   - Router-level configuration for transparent proxying
-
-#### Performance Analysis
-
-**Given parameters:**
-- 6 family members downloading videos
-- Average object size: 100 MB = 800 Mb per video
-- Average request rate: 3 requests/minute = 0.05 requests/second
-- ISP response time: 5 seconds average
-- Home internet connection: ~50 Mbps (from previous analysis)
-
-#### a) Average time (α) for home router to receive video object
-
-**Transmission time calculation:**
-- Video size: 100 MB = 800 Mb
-- Link speed: 50 Mbps (download)
-- Transmission time = 800 Mb ÷ 50 Mbps = 16 seconds
-
-**Total time (α):**
-α = ISP processing time + transmission time
-α = 5 seconds + 16 seconds = **21 seconds**
-
-#### b) Traffic intensity (μ) without caching
-
-**Traffic calculation:**
-- Request rate (λ): 6 members × 0.05 req/sec = 0.3 requests/second
-- Service time per request: 21 seconds (from part a)
-- Service rate (μ): 1/21 = 0.0476 requests/second
-
-**Traffic intensity (ρ):**
-ρ = λ/μ = 0.3 ÷ 0.0476 = **6.3**
-
-**Note:** ρ > 1 indicates system overload - requests arrive faster than they can be served.
-
-#### c) Average access delay (β)
-
-**Given formula:** β = α/(μ-1)
-
-Since μ = 0.0476 and λ = 0.3:
-μ - λ = 0.0476 - 0.3 = -0.2524
-
-**System is unstable** (ρ > 1), so the traditional queuing formula doesn't apply directly.
-
-For stable analysis, let's assume the system can buffer requests:
-β = 21/(0.0476 - 0.3) = 21/(-0.2524) 
-
-Since the result is negative, this indicates **queue buildup and system instability**.
-
-**Corrected interpretation:** With ρ = 6.3, average delay approaches infinity due to queue buildup.
-
-#### d) Total average response time with proxy caching
-
-**Given:**
-- Miss rate: 0.5 (50% cache misses)
-- Hit rate: 0.5 (50% cache hits)
-
-**Cache hit scenario:**
-- Local cache access time: ~0.1 seconds (local disk access)
-- Response time for cache hit: 0.1 seconds
-
-**Cache miss scenario:**
-- Same as no caching: 5 + 16 = 21 seconds
-
-**Average response time:**
-Total response time = (Hit rate × Hit time) + (Miss rate × Miss time)
-= (0.5 × 0.1) + (0.5 × 21)
-= 0.05 + 10.5
-= **10.55 seconds**
-
-**With proxy caching, total average response time = 10.55 seconds**
-
-#### Benefits Analysis
-
-**Without cache:** System unstable, infinite delays
-**With 50% cache hit rate:** 10.55 seconds average response time
-
-**Performance improvement:**
-- 50% reduction in external bandwidth usage
-- Eliminates system instability
-- Significant improvement in user experience
-- Reduces ISP bandwidth costs
-
-**Key insight:** Even a modest 50% cache hit rate transforms an unstable system into a functional one with reasonable response times.
-
-### 2.4 File Distribution Comparison (10%)
-
-**Given parameters:**
-- File size: F = 21 GB = 21 × 10^9 bytes = 168 × 10^9 bits
-- Server upload rate: U_s = 1 Gbps = 10^9 bps
-- Each peer download rate: D_i = 20 Mbps = 20 × 10^6 bps
-- Each peer upload rate: U = varies (300 Kbps, 7000 Kbps, 2 Mbps)
-
-#### Distribution Time Calculations
-
-**Client-Server Distribution Time:**
-T_cs = max(NF/U_s, F/D_min)
-
-Where D_min = min(D_i) = 20 Mbps for all peers
-
-**P2P Distribution Time:**
-T_p2p = max(F/U_s, F/D_min, NF/(U_s + ΣU_i))
-
-#### Results Table
-
-| N | U (Kbps) | Client-Server (seconds) | P2P (seconds) |
-|---|----------|------------------------|---------------|
-| **N = 10** |  |  |  |
-| | 300 | max(1680, 8400) = **8400** | max(168, 8400, 1512) = **8400** |
-| | 7000 | max(1680, 8400) = **8400** | max(168, 8400, 151.4) = **8400** |
-| | 2000 | max(1680, 8400) = **8400** | max(168, 8400, 280) = **8400** |
-| **N = 100** |  |  |  |
-| | 300 | max(16800, 8400) = **16800** | max(168, 8400, 5600) = **8400** |
-| | 7000 | max(16800, 8400) = **16800** | max(168, 8400, 240) = **8400** |
-| | 2000 | max(16800, 8400) = **16800** | max(168, 8400, 420) = **8400** |
-| **N = 1000** |  |  |  |
-| | 300 | max(168000, 8400) = **168000** | max(168, 8400, 56000) = **56000** |
-| | 7000 | max(168000, 8400) = **168000** | max(168, 8400, 2400) = **8400** |
-| | 2000 | max(168000, 8400) = **168000** | max(168, 8400, 4200) = **8400** |
-
-#### Detailed Calculations
-
-**For N=10, U=300 Kbps:**
-- Client-server: max(10×168×10^9/10^9, 168×10^9/20×10^6) = max(1680, 8400) = 8400s
-- P2P: max(168×10^9/10^9, 168×10^9/20×10^6, 10×168×10^9/(10^9+10×0.3×10^6)) = max(168, 8400, 1512) = 8400s
-
-**For N=1000, U=7000 Kbps:**
-- Client-server: max(1000×168×10^9/10^9, 168×10^9/20×10^6) = max(168000, 8400) = 168000s
-- P2P: max(168, 8400, 1000×168×10^9/(10^9+1000×7×10^6)) = max(168, 8400, 2400) = 8400s
-
-#### Analysis and Commentary
-
-#### Client-Server Distribution Features:
-**Advantages:**
-- Simple implementation and management
-- Predictable performance
-- Server has complete control over distribution
-- No coordination overhead between peers
-
-**Disadvantages:**
-- Server becomes bottleneck as N increases
-- Distribution time grows linearly with number of peers
-- Inefficient use of peer upload capacity
-- Single point of failure
-
-**Scaling characteristics:** T_cs grows linearly with N when NF/U_s dominates
-
-#### P2P Distribution Features:
-**Advantages:**
-- Utilizes aggregate peer upload capacity
-- More scalable than client-server for large N
-- Distributes load across all participants
-- Self-scaling with more peers
-
-**Disadvantages:**
-- Complex protocol implementation
-- Coordination overhead
-- Dependent on peer participation and upload capacity
-- Less predictable performance
-
-**Scaling characteristics:** Better scalability as collective upload capacity grows with N
-
-#### Key Differences:
-
-1. **Scalability:** P2P scales better with large N, especially when peer upload capacity is significant
-2. **Resource utilization:** P2P utilizes all available upload bandwidth, client-server wastes peer capacity
-3. **Performance crossover:** P2P becomes superior when total peer upload capacity exceeds server limitations
-4. **Minimum distribution time:** P2P bounded by max(F/U_s, F/D_min), client-server bounded by max(NF/U_s, F/D_min)
-
-**Critical insight:** P2P effectiveness depends heavily on peer upload rates. With sufficient peer upload capacity (U ≥ 2 Mbps), P2P maintains constant distribution time regardless of N, while client-server time increases linearly.
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