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Consider the following two causes of a 1 -second network delay (assume ACKs return instantaneously): One intermediate router with a 1 -second outbound per-packet bandwidth delay and no competing traffic One intermediate router with a 100-ms outbound per-packet bandwidth delay and with a steadily replenished (from another source) 10 packets in the queue (a) How might a transport protocol in general distinguish between these two cases? (b) Suppose TCP Vegas sends over the above connections, with an initial CongestionWindow of 3 packets. What will happen to CongestionWindow in each case? Assume BaseRTT \(=1\) second and \(\beta\) is 1 packet per second.

Short Answer

Expert verified
TCP Vegas will keep CongestionWindow around 3 packets for both routers but adjusting actively for the second one due to queue delays.

Step by step solution

01

- Understand the two scenarios for 1-second delay

First, identify the scenarios: 1. A router with a 1-second delay and no competing traffic. 2. A router with a 100-ms delay and 10 packets always in the queue.
02

- Analyze how a transport protocol distinguishes between the two cases

A transport protocol might distinguish the two cases by measuring Round Trip Time (RTT) or packet delays. In the first case, the delay will be consistently 1 second. In the second case, the delay will fluctuate around 1 second due to queue length fluctuations and the additional 100-ms delay.
03

- Assess the initial RTT for both cases

For the first router, the RTT would be consistently 1 second (since there's no competing traffic, RTT equals outbound delay). For the second router, initial RTT could vary but will be approximately 1 second due to the 100-ms delay plus the variable waiting time in the queue.
04

- Calculate CongestionWindow for TCP Vegas for the first router scenario

TCP Vegas measures RTT and adjusts CongestionWindow based on the difference between actual RTT and BaseRTT. Initial RTT = 1 second, BaseRTT = 1 second, so there’s no significant difference initially. CongestionWindow remains around 3 packets.
05

- Calculate CongestionWindow for TCP Vegas for the second router scenario

Initially, RTT might be slightly greater than BaseRTT due to queue delays. TCP Vegas observes this difference and would initially keep CongestionWindow around 3 packets but can adjust it based on subsequent RTT measurements since the queue delay might vary.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Round Trip Time (RTT)
Round Trip Time, or RTT, is the time it takes for a packet to travel from the sender to the receiver and back. It includes all types of delays:
  • Propagation delay
  • Transmission delay
  • Queuing delay
  • Processing delay
In network communications, RTT is a key metric for measuring the efficiency and performance. Simply put, a lower RTT indicates a faster and more responsive network. By constantly measuring RTT, transport protocols such as TCP adapt to changing network conditions. For example, if RTT increases, it might imply congestion or extended queuing delays. Conversely, a decrease could indicate improved network conditions. Understanding RTT helps protocols adjust their congestion control mechanisms to optimize data flow.
TCP Vegas
TCP Vegas is an enhancement over traditional TCP that places a greater emphasis on congestion avoidance rather than reacting to congestion. Unlike traditional protocols that rely on packet loss to signal congestion, TCP Vegas uses changes in RTT to detect congestion early. Here’s how it works:
  • **BaseRTT:** The minimum RTT ever observed on the connection. Ideally, this is measured when the network is least congested.
  • **Expected Throughput:** This is calculated based on the BaseRTT and current CongestionWindow size.
  • **Actual Throughput:** This is measured in real-time based on the observed RTT and current CongestionWindow size.
TCP Vegas compares Expected Throughput with Actual Throughput. If the Actual Throughput is significantly lower than Expected Throughput, it means there might be congestion in the network. TCP Vegas adjusts the CongestionWindow size gradually, increasing it when throughput is good and decreasing it when congestion is detected. This proactive approach helps maintain a smoother and more stable network performance.
CongestionWindow
The CongestionWindow (CWND) is an essential part of TCP's congestion control algorithm. It dictates the number of packets a sender can send into the network before needing an acknowledgment from the receiver. Here’s how it impacts data flow:
  • **Initial CWND:** TCP starts with a small CWND to probe the network's capacity.
  • **Slow Start Phase:** Initially, CWND grows exponentially to quickly discover the available bandwidth.
  • **Congestion Avoidance Phase:** Once a threshold is reached, CWND grows linearly to avoid congestion.
  • **Congestion Detection:** If packet loss is detected, CWND is reduced to ease congestion.
TCP Vegas refines this mechanism by adjusting CWND based on RTT variations rather than solely on packet loss. For example, in our scenarios:
In the **first case**, with a consistent 1-second RTT, TCP Vegas maintains the CWND around 3 packets, as there’s no significant sign of congestion.
In the **second case**, where RTT might fluctuate slightly around 1 second due to queuing delays, TCP Vegas might make small adjustments to CWND to account for varying network conditions.

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Most popular questions from this chapter

Suppose an RSVP router suddenly loses its reservation state, but otherwise remains running. (a) What will happen to the existing reserved flows if the router handles reserved and nonreserved flows via a single FIFO queue? (b) What might happen to the existing reserved flows if the router used weighted fair queuing to segregate reserved and nonreserved traffic? (c) Eventually, the receivers on these flows will request that their reservations be renewed. Give a scenario in which these requests are denied.

Consider a simple congestion-control algorithm that uses linear increase and multiplicative decrease but not slow start, that works in units of packets rather than bytes, and that starts each connection with a congestion window equal to one packet. Give a detailed sketch of this algorithm. Assume the delay is latency only, and that when a group of packets is sent, only a single ACK is returned. Plot the congestion window as a function of round-trip times for the situation in which the following packets are lost: \(9,25,30,38\), and 50 . For simplicity, assume a perfect timeout mechanism that detects a lost packet exactly 1 RTT after it is transmitted.

Suppose that between \(A\) and \(B\) there is a router \(R\). The \(A-R\) bandwidth is infinite (that is, packets are not delayed), but the R-B link introduces a bandwidth delay of 1 packet per second (that is, 2 packets take 2 seconds, etc.). Acknowledgments from \(\mathrm{B}\) to \(\mathrm{R}\), though, are sent instantaneously. \(\mathrm{A}\) sends data to \(\mathrm{B}\) over a \(\mathrm{TCP}\) connection, using slow start but with an arbitrarily large window size. R has a queue size of 1 , in addition to the packet it is sending. At each second, the sender first processes any arriving ACKs and then responds to any timeouts. (a) Assuming a fixed TimeOut period of 2 seconds, what is sent and received for \(\mathrm{T}=0,1, \ldots, 6\) seconds? Is the link ever idle due to timeouts? (b) What changes if TimeOut is 3 seconds instead?

Two users, one using Telnet and one sending files with FTP, both send their traffic out via router \(R\). The outbound link from \(R\) is slow enough that both users keep packets in R's queue at all times. Discuss the relative performance seen by the Telnet user if \(\mathrm{R}\) 's queuing policy for these two flows is (a) round-robin service (b) fair queuing (c) modified fair queuing, where we count the cost only of data bytes, and not IP or TCP headers Consider outbound traffic only. Assume Telnet packets have 1 byte of data, FTP packets have 512 bytes of data, and all packets have 40 bytes of headers.

Consider a router that is managing three flows, on which packets of constant size arrive at the following wall clock times: flow A: \(1,2,4,6,7,9,10\) flow B: \(2,6,8,11,12,15\) flow C: \(1,2,3,5,6,7,8\) All three flows share the same outbound link, on which the router can transmit one packet per time unit. Assume that there is an infinite amount of buffer space. (a) Suppose the router implements fair queuing. For each packet, give the wall clock time when it is transmitted by the router. Arrival time ties are to be resolved in order \(\mathrm{A}, \mathrm{B}, \mathrm{C}\). Note that wall clock time \(T=2\) is FQ-clock time \(A_{i}=1.5\). (b) Suppose the router implements weighted fair queuing, where flows \(\mathrm{A}\) and \(\mathrm{B}\) are given an equal share of the capacity, and flow \(\mathrm{C}\) is given twice the capacity of flow A. For each packet, give the wall clock time when it is transmitted.

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