fix calculations and replace unicode with ascii characters

2 files changed, 88 insertions, 91 deletions

diff --git a/Brewfile b/assignments/1/Brewfile
index 580ce3e..580ce3e 100644
--- a/Brewfile
+++ b/assignments/1/Brewfile
diff --git a/assignments/1/README.md b/assignments/1/README.md
index ab0c6a9..8ea75bc 100644
--- a/assignments/1/README.md
+++ b/assignments/1/README.md
@@ -397,8 +397,8 @@ Scenario: Web-based email (Gmail) sent to Alice who accesses email via IMAP on s
 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)
+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
@@ -411,12 +411,12 @@ Scenario: Web-based email (Gmail) sent to Alice who accesses email via IMAP on s
 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)
+#### 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)
+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
@@ -430,8 +430,8 @@ Scenario: Web-based email (Gmail) sent to Alice who accesses email via IMAP on s
 
 Key Protocol Stack Traversals:
 
-- Downward: Each device processes from Application → Physical when sending
-- Upward: Each device processes from Physical → Application when receiving
+- 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%)
@@ -464,9 +464,9 @@ Number of packets = Total file bits ÷ Payload bits per packet
 
 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)
+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
 
@@ -476,26 +476,26 @@ Total identified links: 14 hops
 
 Based on traceroute RTT measurements:
 
-| Link  | From → To       | RTT (ms)   | Estimated Speed                    |
+| 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)   |
+| 1     | Home -> Router  | 16         | ~100 Mbps (Local Ethernet/Wi-Fi)   |
+| 2     | Router -> ISP   | 16         | ~960 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)
+First link speed estimation: ~960 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
+= 10,000 bits ÷ (960 × 10^6 bps)
+= 0.0104 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
+= 8,081 packets × 0.0104 ms/packet
+= 84.04 ms = 0.084 seconds
 
-At t₀ + 1.62 seconds, the last packet will be pushed into the first link.
+At t₀ + 0.084 seconds, the last packet will be pushed into the first link.
 
 ### e) Time for last packet to arrive at university server
 
@@ -506,16 +506,16 @@ Total end-to-end delay consists of:
 - 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  
+- Time for last packet to enter first link: 0.084 s
+- Transmission time through all links: ~0.0104 ms × 14 = 0.146 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
+Total time = 0.084 s + 0.000146 s + 0.079 s + 0.014 s + 0.070 s
+= 0.247 seconds
 
-The last packet will arrive at the university server at t₀ + 1.79 seconds.
+The last packet will arrive at the university server at t₀ + 0.25 seconds.
 
 ## 2.2 Propagation Delay and Bandwidth-Delay Product (20%)
 
@@ -531,15 +531,15 @@ 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)
+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
+Total distance ~ 1,775,000 meters
 
 ### b) Propagation delay calculation
 
@@ -548,31 +548,39 @@ Given:
 - Propagation speed: 2 × 10^8 m/s
 - Total distance: 1,775,000 m
 
+```
 Propagation delay (T_prop):
-T_prop = distance ÷ speed
+T_prop = distance / speed
 = 1,775,000 m ÷ (2 × 10^8 m/s)
 = 0.008875 seconds
 = 8.875 ms
+```
 
-### c) Bandwidth-delay product
+### c) Bandwidth-delay
 
-Assuming all links have the same speed R bps:
+Assuming all links have the same speed `R` bps:
 
-From our analysis, the bottleneck is likely the residential connection at ~50 Mbps.
+From our analysis, the bottleneck is likely the residential connection at 960 Mbps.
 
+```
 Bandwidth-delay product:
 
-R × T_prop = 50 × 10^6 bps × 8.875 × 10^-3 s
-= 443,750 bits
-= 443.75 Kbits
+R × T_prop = 960 × 10^6 bps × 8.875 × 10^-3 s
+= 8,520,000 bits
+= 8.52 Mbits
+```
 
 ### 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.
+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
+Maximum bits in links = `R × T_prop` = 8,520,000 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.
+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
 
@@ -580,9 +588,9 @@ The bandwidth-delay product represents the network pipe capacity and has several
 
 1. Buffer Requirements:
   - Minimum buffer size needed for optimal performance
-  - TCP window size should be at least R × T_prop for maximum throughput
+  - 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
+  - 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
@@ -606,32 +614,14 @@ The bandwidth-delay product represents the network pipe capacity and has several
 
 ### 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
+I implemented an Apache HTTP proxy server with disk caching capabilities on the home network:
+
+- Installation: `brew bundle` to install Apache HTTP Server
+- Configuration: Forward proxy with disk caching (see `etc/apache/apache.conf`)
+- Management: Server startup/shutdown scripts in `bin/server` and `bin/client`
+- Network Setup: Server runs on port 8081, family devices configured to use proxy at `192.168.x.x:8081`
+
+The proxy server enables all family video requests to flow through the cache, providing the foundation for the performance analysis below.
 
 ### Performance Analysis
 
@@ -641,21 +631,21 @@ Given parameters:
 - 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)
+- Home internet connection: 960 Mbps
 
 ### 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
+- Link speed: 960 Mbps (download)
+- Transmission time = 800 Mb ÷ 960 Mbps = 0.833 seconds
 
 ```
 Total time (α):
 
 α = ISP processing time + transmission time
-α = 5 seconds + 16 seconds = 21 seconds
+α = 5 seconds + 0.833 seconds = 5.833 seconds
 ```
 
 ### b) Traffic intensity (μ) without caching
@@ -663,13 +653,13 @@ Total time (α):
 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
+- Service time per request: 5.833 seconds (from part a)
+- Service rate (μ): 1/5.833 = 0.171 requests/second
 
 ```
 Traffic intensity (ρ):
 
-ρ = λ/μ = 0.3 ÷ 0.0476 = 6.3
+ρ = λ/μ = 0.3 ÷ 0.171 = 1.75
 
 Note: ρ > 1 indicates system overload - requests arrive faster than they can be served.
 ```
@@ -677,19 +667,19 @@ Note: ρ > 1 indicates system overload - requests arrive faster than they can be
 ### c) Average access delay (β)
 
 ```
-Given formula: β = α/(μ-1)
+Given formula: β = α/(μ-λ)
 
-Since μ = 0.0476 and λ = 0.3:
-μ - λ = 0.0476 - 0.3 = -0.2524
+Since μ = 0.171 and λ = 0.3:
+μ - λ = 0.171 - 0.3 = -0.129
 
 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) 
+β = 5.833/(0.171 - 0.3) = 5.833/(-0.129) = -45.2 seconds
 
 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.
+Corrected interpretation: With ρ = 1.75, average delay approaches infinity due to queue buildup.
 ```
 
 ### d) Total average response time with proxy caching
@@ -706,32 +696,39 @@ Cache hit scenario:
 
 Cache miss scenario:
 
-- Same as no caching: 5 + 16 = 21 seconds
+- Same as no caching: 5.833 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
+= (0.5 × 0.1) + (0.5 × 5.833)
+= 0.05 + 2.917
+= 2.967 seconds
 ```
 
-With proxy caching, total average response time = 10.55 seconds
+With proxy caching, total average response time = 2.97 seconds
+
+### Summary of Results
+
+**Assignment Question Answers:**
 
-### Benefits Analysis
+- **a) Average time (α)**: 5.833 seconds for home router to receive video object
+- **b) Traffic intensity (ρ)**: 1.75 (system overload - requests arrive faster than service)
+- **c) Access delay (β)**: System unstable with infinite delays due to ρ > 1
+- **d) Total response time with caching**: 2.97 seconds (50% cache hit rate)
 
-* Without cache: System unstable, infinite delays
-* With 50% cache hit rate: 10.55 seconds average response time
+**Web Cache Performance Benefits:**
 
-Performance improvement:
+The analysis demonstrates the critical importance of web caching in home networks:
 
-- 50% reduction in external bandwidth usage
-- Eliminates system instability
-- Significant improvement in user experience
-- Reduces ISP bandwidth costs
+1. **System Stability**: Without caching, the traffic intensity of 1.75 creates an unstable system where requests queue indefinitely
+2. **Performance Transformation**: A 50% cache hit rate reduces average response time from infinite to 2.97 seconds
+3. **Bandwidth Efficiency**: Caching reduces external bandwidth usage by 50%, alleviating the bottleneck
+4. **User Experience**: Family members experience predictable, reasonable response times instead of system overload
+5. **Modern Internet Reality**: Even with gigabit internet (960 Mbps), the system is still overloaded due to ISP processing delays and request patterns
 
-Key insight: Even a modest 50% cache hit rate transforms an unstable system into a functional one with reasonable response times.
+**Key Insight**: Web caching remains essential even with high-speed internet connections. The bottleneck shifts from pure bandwidth to service capacity and ISP processing time, demonstrating that caching benefits extend beyond just bandwidth limitations.
 
 ## 2.4 File Distribution Comparison (10%)