Complete assignment 2

1 files changed, 29 insertions, 10 deletions

diff --git a/assignments/2/README.md b/assignments/2/README.md
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--- a/assignments/2/README.md
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@@ -351,30 +351,49 @@ d. 192.53.56.7 -> no /23 match -> forward using Default -> Router 3.
 > (20%) Consider that only a single TCP connection uses a 1 Gbps link,
 which does not buffer any data. Suppose that this link is the only
 congested link between the sending and receiving hosts. Assume that the
-TCP sender has a huge file to send to the receiver and the receiver’s
+TCP sender has a huge file to send to the receiver and the receiver's
 receive buffer is much larger than the congestion window. Further assume
 that each TCP segment size is 1,500 bytes; the two-way propagation delay
 of this connection is 15 msec; and this TCP connection is always in the
 congestion avoidance phase (ignore slow start).
 
-Given: Single TCP flow over a 1 Gbps link, no buffering at the link; MSS=1500 B (12,000 bits); two-way propagation delay (RTT) = 15 ms; always in congestion avoidance.
+Given: 
+
+* Single TCP flow over a 1 Gbps link, no buffering at the link
+* segment size = 1500 B (12,000 bits)
+* two-way propagation delay (RTT) = 15 ms
+* always in congestion avoidance
 
 > a) What is the maximum window size (in segments) that this TCP connection can achieve?
 
+Maximum window is equal to the bandwidth delay product in segments.
+
+- BDP: 1,000,000,000 bps * 0.015 s = 15,000,000 bits.
+- segment size: = 1,500 B * 8 = 12,000 bits.
+- Maximum window = 15,000,000 / 12,000 = 1,250 segments.
+
 > b) What is the average window size (in segments) and average throughput (in bps) of this TCP connection?
 
+The window will go and up down between the window maximum and half the window maximum.
+
+- Average window: 0.75 * maximum window = 937.5 segments.
+- Average throughput: = 
+  - (average window * segment size) / RTT
+  - (937.5 * 12,000) / 0.015 = 750,000,000 bps (0.75 Gbps).
+
 > c) How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?
 
+After a loss, congestion window equal window max / 2 = 625 segments and then grows by 1 segment per RTT.
+
+- 1,250 / 2 = 625 segments
+- 625 * 0.15 s = 9.375 s
+
+Time to return to the maximum window is 9.375 s.
+
 > d) Assume we want the 1 Gbps link to buffer a finite number of segments and always keep the link busy sending data. How would you choose a buffer size? Justify your answer.
 
-Answer:
-- a) Maximum window without queuing equals the bandwidth-delay product (BDP) in segments.
-  - BDP(bits) = 1e9 bps * 0.015 s = 15,000,000 bits.
-  - MSS(bits) = 1,500 B * 8 = 12,000 bits.
-  - Wmax = 15,000,000 / 12,000 = 1,250 segments.
-- b) In AIMD congestion avoidance (Reno-like), cwnd sawtooths between Wmax and Wmax/2. Average window Wavg = 0.75 * Wmax = 937.5 segments. Average throughput = Wavg * MSS / RTT = 937.5 * 12,000 / 0.015 = 750,000,000 bps (0.75 Gbps).
-- c) After a loss, cwnd halves to Wmax/2 = 625 segments and then grows by ~1 segment per RTT. Time to return to Wmax is 625 RTTs = 625 * 0.015 s = 9.375 s.
-- d) To keep the link busy right after a multiplicative decrease, the queue should supply the missing in-flight data between BDP and cwnd. Provision about BDP/2 worth of buffering: ~625 segments. Then cwnd (625) + queue (625) = BDP, so the link remains fully utilized while cwnd ramps up (with some extra headroom and AQM/ECN recommended in practice).
+To keep the link busy right after a drop, the queue should supply the missing in-flight data between BDP and the congestion window.
+Provision about BDP/2 worth of buffering. Then congestion window + queue so that the link can be fully utilized while the congestion window ramps up.
 
 # References