update a1

1 files changed, 42 insertions, 5 deletions

diff --git a/assignments/1/README.md b/assignments/1/README.md
index 8fe516b..6b924cd 100644
--- a/assignments/1/README.md
+++ b/assignments/1/README.md
@@ -273,6 +273,8 @@ Receiving Email:
 2. The MUA will download the mail message to the device and synchronize any
    changes via IMAP with the mail server.
 
+Browser -> HTTPS over TCP/IP -> webmail server accepts message and (via SMTP over TCP/IP) relays to recipient's mail server -> recipient's IMAP client connects to mail server (IMAP over TCP/TLS) and downloads the message.
+
 # Part 2: Long Answer Questions (70%)
 
 > I provide short, clear answers first, then 1-2 sentences of reasoning.
@@ -316,10 +318,11 @@ Later hops filtered probes, so the total hop count to the destination could not
 > c) What is the speed for each identified link based on your best calculation? Show your work.
 
 My ISP speed is 960 Mbps up and down. I don't know how to calculate the speed for each identified link using RTT data.
+Traceroute gives path and RTT but not bandwidth. The only reliable known speed here is my ISP speed anything else would be a guess.
 
 > d) Assume you start uploading the assignment at t0. At what time will the last packet be pushed into the first link?
 
-Total bits transmitted on the first link = 8,081 packets * 10,000 bits/packet = 80,810,000 bits.
+Total bits sent to the first link = 8,081 packets * 10,000 bits = 80,810,000 bits.
 
 My home ISP speed is 960 Mbps up and down.
 
@@ -327,21 +330,55 @@ My home ISP speed is 960 Mbps up and down.
 
 > e) At what time will the last packet arrive at the university server?
 
-The last packet will arrive at approximately (0.084s + 0.015s) ~ 0.099 seconds after t0.
+The last packet will arrive at approximately (0.084s + 0.018) ~ 0.102 seconds after t0.
 
-- Transmission delay (first link): 0.084s
-- One-way propagation delay: ~15ms (estimated from traceroute RTT/2 + additional hops)
-- Total: 0.084s + 0.015s = 0.099s after t0
+The traceroute showed an approximate RTT of 18 ms. Using that plus the transmission time above of ~0.084 seconds equals ~0.102 seconds after t0.
 
 ## 2.2 Propagation Delay and Bandwidth-Delay Product (20%)
 
 > (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.
 > a) Based on your best estimate and calculation, what is the total distance your assignment data will travel to reach the university server?
+
+The distance is approximately 3,600 km.
+
+Propagation delay is represented by: `d_prop = d / s`
+
+Where:
+
+- d = distance in meters
+- s = speed in meters/second
+
+So Distance is equal to `d = d_prop * s` where s is the speed of light fibre ~(2 * (10^8))
+
+Propagation time: 0.018 s
+Distance =
+  - speed fibre * `d_prop`
+  - 2×10^8 m/s × 0.018 s ~ 3,600,000 m ~ 3,600 km.
+
 > b) Suppose the propagation speed over all the links is the same 2*10^8 meters/sec. What is propagation delay T_prop from your computer to the university server?
+
+The propagation delay is 9 ms.
+
+Because the propagation delay is the RTT/2 for one-way and not the full round trip of 18 ms.
+
 > c) Further assume all the links have the same speed R bps. What is the bandwidth-delay product R*T_prop?
+
+The bandwidth-delay is ~2.16 MB for R = 960 Mbps.
+
+BDP = R × `T_prop`.
+BDP = 960,000,000 * 0.018 ~ 17280000 bits ~ 17.28 Mb ~ 2.16 MB.
+
 > d) 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?
+
+The maximum is 2.16 MB.
+
+When sender is pushing continuously, the number of bits simultaneously populating the path equals link rate × end-to-end propagation time.
+
 > e) Based on the results from c and d, what does the bandwidth-delay product imply?
 
+The bandwidth-delay product is used in TCP tuning window size and buffer sizing.
+It helps keep the pipe full.
+
 ## 2.3 Web Cache Implementation and Performance (20%)
 
 > (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.