以下为面向“TCP 拥塞控制机制（Reno/CUBIC，含对比视角）”的高阶多选题（A–E，至少2个正确选项/题）。每题均为中英双语，并包含图表描述性文字。参考答案部分仅列出正确选项，符合“严谨、可验证的推理；参考答案仅提供正确答案”的要求。命题所依据的关键事实与模型可由经验证的标准与文献支持（例如 RFC 5681、RFC 9438、Padhye 模型、Jain 公平性指标等）。
题目一（状态转移与阈值更新 | State transitions and threshold updates）
中文题干：
给定一个 TCP Reno 发送端，初始条件：cwnd=1 MSS，ssthresh=32 MSS。观测到的事件序列如下：
- 阶段A（慢启动）：经过4个RTT后，cwnd 达到 16 MSS（1,2,4,8,16）。
- 点P（快速重传/快速恢复触发）：进入下一拥塞轮次时，某个分组丢失，接收方回送3个重复ACK；未发生超时。丢失被检测时的飞行报文量近似等于 cwnd=16 MSS。
- 恢复后进入拥塞避免，线性增长。
- 点Q（超时）：稍后在拥塞避免阶段发生一次超时，当时 cwnd 约为 10 MSS。
问题：下面哪些判断与 Reno 的标准行为一致（选择所有适用项）。
English stem:
A TCP Reno sender has initial cwnd=1 MSS and ssthresh=32 MSS. Events:
- Phase A (Slow Start): After 4 RTTs, cwnd reaches 16 MSS (1,2,4,8,16).
- Point P (Fast Retransmit/Fast Recovery trigger): In the next congestion round, a packet is lost; 3 duplicate ACKs are received; no timeout occurs. At loss detection, FlightSize ≈ cwnd=16 MSS.
- After recovery, the sender enters Congestion Avoidance and grows linearly.
- Point Q (Timeout): Later in Congestion Avoidance, a timeout occurs with cwnd ≈ 10 MSS.
Task: Select all statements consistent with Reno’s standard behavior.
图表描述（文字）/ Figure description (textual):
- 折线图显示 cwnd 随时间在阶段A指数增长至16 MSS；点P处出现3个重复ACK，随后进入快速恢复并回落至一半阈值；之后线性增长；点Q处由于超时，cwnd 归一并重新慢启动。
- A line plot shows cwnd doubling to 16 MSS in Phase A; at P, three duplicate ACKs trigger Fast Retransmit/Recovery with deflation to half; then linear growth; at Q, timeout resets cwnd to 1 MSS and Slow Start resumes.
选项 / Options:
A. P点触发的是快速重传/快速恢复，进入拥塞避免而非重新进入慢启动。
At P, Fast Retransmit/Fast Recovery is triggered; upon recovery the sender enters Congestion Avoidance, not Slow Start.
B. 在P点，按照 Reno 的规则，ssthresh 设为 8 MSS（即将丢包时的飞行报文量的一半）。
At P, Reno sets ssthresh to 8 MSS (half of the FlightSize at loss).
C. 在P点，进入快速恢复时临时将 cwnd 设为 ssthresh + 3 MSS；恢复完成后将 cwnd 归一（deflate）为 ssthresh。
At P, cwnd is temporarily set to ssthresh + 3 MSS during Fast Recovery and then deflated to ssthresh upon recovery.
D. 在Q点发生超时后，cwnd 被设置为 ssthresh 而非 1 MSS。
At Q, after timeout, cwnd is set to ssthresh rather than 1 MSS.
E. 在Q点发生超时后，ssthresh 设为 5 MSS（约为超时前 cwnd=10 MSS 的一半），cwnd 置为 1 MSS，进入慢启动。
At Q, after timeout, ssthresh is set to 5 MSS (about half of cwnd=10 MSS), cwnd is reset to 1 MSS, and Slow Start begins.
正确答案 / Correct answers: A, B, C, E
题目二（RTT/丢包信号对公平性与吞吐的影响；Reno vs CUBIC 对比 | Effects of RTT/loss on fairness and throughput; Reno vs CUBIC)
中文题干：
两条长流在同一路径共享单一瓶颈，MSS 相同，接收端不延迟ACK（每报文段确认一次），队列为 Drop-Tail，无ECN标记。流A的往返时延 RTT_A=50 ms，流B 的往返时延 RTT_B=100 ms。稳态时两流的分组丢失概率近似相同且较小。比较在 Reno 与 CUBIC 拥塞避免阶段的行为。选择所有正确陈述。
English stem:
Two long-lived flows share a single bottleneck on the same path with identical MSS, no delayed ACKs, Drop-Tail queue, and no ECN marking. Flow A has RTT_A=50 ms; Flow B has RTT_B=100 ms. In steady state, both see similar small loss probability. Compare Reno and CUBIC behavior in Congestion Avoidance. Select all correct statements.
图表描述（文字）/ Figure description (textual):
- 条形图示意 Reno 下流A吞吐约为流B两倍；Jain 公平指数约为 0.9（理想化 2:1 场景）。
- 折线示意 CUBIC 在高BDP下增长随“距上次丢包的时间”呈立方曲线，RTT 敏感性较低于Reno。
- A bar chart suggests Flow A’s throughput ≈ 2× Flow B’s under Reno; Jain’s index ≈ 0.9 for an idealized 2:1 split. A line sketch indicates CUBIC’s cubic-in-time growth with reduced RTT sensitivity at high BDP.
选项 / Options:
A. 根据 Padhye 等模型，在小丢包率下 Reno 吞吐近似与 1/(RTT·√p) 成正比，因而若两流丢包率相近，则 T_A/T_B ≈ RTT_B/RTT_A ≈ 2。
By the Padhye model, Reno throughput scales as ≈ 1/(RTT·√p) at small loss rates; with similar loss, T_A/T_B ≈ RTT_B/RTT_A ≈ 2.
B. CUBIC 的窗口增长在拥塞避免阶段主要取决于 ACK 计数而非时间，从而强化了 RTT 敏感性。
In Congestion Avoidance, CUBIC’s growth depends primarily on ACK counts rather than time, which increases RTT sensitivity.
C. 在上述理想化 2:1 吞吐分配下，Jain 公平性指数约为 (2+1)^2 / [2·(2^2+1^2)] = 0.9。
For an idealized 2:1 throughput split, Jain’s fairness index is ≈ 0.9.
D. 与 Reno 相比，CUBIC 的拥塞避免以“距上次丢包的时间”为自变量（立方函数），在高 BDP 场景对 RTT 的依赖较弱，因而通常更接近 RTT 公平（虽非完全公平）。
Compared with Reno, CUBIC’s avoidance phase uses a cubic function of elapsed time since last loss, reducing RTT dependence at high BDP and yielding closer-to-RTT-fair sharing (though not perfectly fair).
E. CUBIC 的乘性减小系数 β 通常约为 0.7（cwnd 降至 0.7·cwnd），而 Reno 约为 0.5。
CUBIC’s multiplicative decrease β is typically ≈ 0.7 (cwnd reduced to 0.7·cwnd), whereas Reno uses ≈ 0.5.
正确答案 / Correct answers: A, C, D, E
题目三（基于抓包摘要评估吞吐-时延权衡与改进建议 | Throughput–delay trade-off assessment and improvement proposals from a trace)
中文题干：
抓包摘要显示两条 CUBIC 长流在同一瓶颈链路（无ECN，Drop-Tail，大缓冲）并发传输：
- 初始慢启动阶段发生明显过冲，队列积压上升，队列化时延从基线 20 ms 提升至 200–250 ms。
- 进入拥塞避免后，cwnd 呈现随“距上次丢包时间”的立方增长曲线，周期性溢出导致丢包与时延振荡。
- 链路利用率高，但交互流（小流）RTT 明显恶化。
问题：在不改变应用负载的前提下，以下哪些系统性改进更可能在保持或接近现有吞吐的同时，显著改善时延（选择所有适用项）。
English stem:
A packet-trace summary shows two long-lived CUBIC flows over the same bottleneck link (no ECN, Drop-Tail, large buffers):
- Slow Start overshoot causes queue build-up; queuing delay rises from 20 ms baseline to 200–250 ms.
- In Congestion Avoidance, cwnd follows a cubic-in-time growth; periodic overflows induce loss and delay oscillations.
- Link utilization is high, but small interactive flows suffer inflated RTT.
Task: Without changing application load, which systemic changes are most likely to improve delay significantly while maintaining comparable throughput? Select all that apply.
图表描述（文字）/ Figure description (textual):
- 上图：时延曲线从 20 ms 爬升到 200+ ms，呈周期性锯齿；下图：cwnd 呈立方增长与周期性回落。
- Top: Delay rises from 20 ms to >200 ms with sawtooth oscillations; Bottom: cwnd shows cubic growth with periodic backoffs.
选项 / Options:
A. 将瓶颈队列由 Drop-Tail 替换为支持 ECN 标记的 AQM（例如 PIE/CoDel 家族）并启用 ECN，使发送端以标记而非丢包为拥塞信号从而提前退让，降低队列化时延且维持高利用率。
Replace Drop-Tail with an ECN-marking AQM (e.g., PIE/CoDel family) and enable ECN so senders back off on marks rather than loss, reducing queuing delay while keeping high utilization.
B. 在发送端启用分组节奏化（pacing）与合理的字节队列上限（BQL），以降低突发性和短时队列尖峰，从而改善时延抖动且不显著牺牲吞吐。
Enable sender-side packet pacing and Byte Queue Limits (BQL) to reduce burstiness and transient queue spikes, improving delay jitter without materially harming throughput.
C. 将两端拥塞控制从 CUBIC 切换到 Reno，可系统性降低时延并提高小流的完成时间。
Switch both endpoints from CUBIC to Reno to systematically reduce delay and improve small-flow completion times.
D. 在高BDP与缓冲胀气场景下，改用 BBR（带带宽-RTT 模式）常能显著压缩队列化时延并维持高吞吐，但若与损失型算法共存且无 AQM，可能引入跨算法不公平。
In high-BDP, bufferbloat-prone settings, switching to BBR (bandwidth–RTT model) often collapses queues and sustains high throughput; however, when coexisting with loss-based flows without AQM, it may induce inter-algorithm unfairness.
E. 将 ssthresh 设置为无限大（长期停留在“慢启动”形态）以避免模式切换，可提升稳定性并降低排队时延。
Set ssthresh to infinity (never leaving a “Slow Start”-like mode) to avoid mode switching, improving stability and reducing queuing delay.
正确答案 / Correct answers: A, B, D
参考文献（供命题依据核查，不附解析）/ References (for verification; no solutions/explanations):
- Allman, Paxson, Blanton. TCP Congestion Control. RFC 5681.
- Eddy. Transmission Control Protocol (TCP). RFC 9293.
- Rhee, Xu, Ha et al. CUBIC for Fast Long-Distance Networks. RFC 9438.
- Padhye, Firoiu, Towsley, Kurose. Modeling TCP Reno Performance: A Simple Model and its Empirical Validation. IEEE/ACM Trans. Netw., 2000.
- Jain, Chiu, Hawe. A Quantitative Measure of Fairness and Discrimination. DEC TR-301, 1984.
- Cardwell, Cheng, Gunn et al. BBR: Congestion-Based Congestion Control. ACM Queue, 2016.
- Floyd, Gummadi, Shenker (for ECN); Nichols & Jacobson (CoDel); Pan et al. (PIE). Empirical foundations for AQM/ECN improving delay–throughput trade-offs.
