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Chapter 6: Problem 16

Assume that TCP implements an extension that allows window sizes much larger than \(64 \mathrm{~KB}\). Suppose that you are using this extended TCP over a 1-Gbps link with a latency of \(100 \mathrm{~ms}\) to transfer a \(10-\mathrm{MB}\) file, and the TCP receive window is \(1 \mathrm{MB}\). If TCP sends 1-KB packets (assuming no congestion and no lost packets): (a) How many RTTs does it take until slow start opens the send window to \(1 \mathrm{MB}\) ? (b) How many RTTs does it take to send the file? (c) If the time to send the file is given by the number of required RTTs multiplied by the link latency, what is the effective throughput for the transfer? What percentage of the link bandwidth is utilized?

Short Answer

Expert verified
10 RTTs for slow start, 10 RTTs to send the file, effective throughput is 5 MB/s, and 4% utilization of the link bandwidth.

Step by step solution

01

- Calculate RTT

First, calculate the Round-Trip Time (RTT). Given that latency is 100 ms, the RTT is the round trip of this latency. Thus, RTT = 100 ms = 0.1 seconds.
02

- Determine slow start to reach 1 MB

In slow start, the congestion window (cwnd) size doubles every RTT. Starting with a 1-KB packet: - After 1st RTT: cwnd = 2 KB - After 2nd RTT: cwnd = 4 KB - After 3rd RTT: cwnd = 8 KB Continue this until cwnd = 1 MBgeneral form of cwnd after n RTTs is cwnd = 1 KB * 2^nSet cwnd to 1 MB and solve for n:1 MB = 1 KB * 2^n 1024 KB = 1 KB * 2^n 1024 = 2^n solving for n, n = log2(1024) n = 10 RTTs.
03

- Calculate RTTs to send file

During each RTT, TCP sends up to the window size allowed. To send a 10-MB file:Each RTT, sending window is 1 MB after slow start.Number of RTTs needed to send 10 MB = 10 MB / 1 MB per RTTNumber of RTTs = 10.
04

- Total Time Calculation

Total time to send the file is the sum of the RTTs for slow start and sending the file:Total RTTs = 10 RTTs (slow start) + 10 RTTs (sending file) = 20 RTTs.
05

- Calculate Effective Throughput

Effective throughput is the total file size divided by the total time:Total time = 20 RTTs * 100 msTotal time = 20 * 0.1 s = 2 secondsThroughput = file size / total timeThroughput = 10 MB / 2 secondsThroughput = 5 MB/s (Megabytes per second).
06

- Calculate Bandwidth Utilization

The link bandwidth is 1 Gbps. Convert throughput to the same unit:Throughput = 5 MB/s = 5 * 8 = 40 MbpsLink bandwidth = 1 Gbps = 1000 MbpsPercentage utilized = (throughput / link bandwidth) * 100Percentage utilized = (40 / 1000) * 100 = 4%.

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Key Concepts

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

RTT calculation
The Round-Trip Time (RTT) is crucial for understanding TCP's performance. RTT measures the time it takes for a signal to travel from the sender to the receiver and back again. In this exercise, latency is 100 milliseconds (ms), so the RTT equals twice the latency: 100 ms each way means an RTT of 200 ms or 0.2 seconds. Calculating RTT accurately ensures we can predict TCP behavior under different network conditions.
congestion window
The congestion window (cwnd) is a TCP state variable that limits the amount of data a TCP sender can send into the network before receiving an acknowledgment (ACK). During slow start, cwnd starts at a size of one Maximum Segment Size (MSS) and doubles each RTT. This exponential growth continues until the cwnd reaches the receiver window size (rwnd) or encounters packet loss. For this exercise, we calculated that the cwnd reaches 1 MB after 10 RTTs. Understanding cwnd's behavior helps predict TCP performance and efficiency.
effective throughput
Effective throughput measures the actual data transfer rate over the network, considering all protocol overheads. To calculate effective throughput, divide the total file size by the total transmission time. For our 10-MB file over 20 RTTs (each RTT being 0.2 seconds), the total time is 4 seconds. Therefore, the throughput is 10 MB / 4 seconds = 2.5 MB/s. This metric is essential to gauge the efficiency of a TCP connection in real-world scenarios.
bandwidth utilization
Bandwidth utilization quantifies the percentage of the available bandwidth actually used by the data transmissions. First, convert effective throughput to a corresponding unit: 2.5 MB/s = 20 Mbps (since 1 byte = 8 bits). Given a link bandwidth of 1 Gbps (1000 Mbps), the percentage utilized is (20 / 1000) * 100 = 2%. This low utilization indicates the transmission isn't using the full potential of the available bandwidth due to TCP's slow start and other factors.
TCP performance
TCP's performance in this scenario is influenced by several factors: RTT, cwnd evolution, and bandwidth utilization. The slow start mechanism, designed to prevent congestion, delays reaching maximum throughput. Effective throughput and bandwidth utilization provide insights into the efficiency and effectiveness of the TCP transmission. By analyzing these metrics, we can understand the trade-offs and behavior of TCP under various network conditions, ultimately aiming to optimize performance and reduce transmission times.

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Most popular questions from this chapter

Give an argument why the congestion-control problem is better managed at the internet level than the ATM level, at least when only part of the internet is ATM. In an exclusively IP-over-ATM network, is congestion better managed at the cell level or at the TCP level? Why?

It is possible to define flows on either a host-to-host basis or a process-to- process basis. (a) Discuss the implications of each approach to application programs. (b) IPv6 includes a FlowLabel field, for supplying hints to routers about individual flows. The originating host is to put here a pseudorandom hash of all the other fields serving to identify the flow; the router can thus use any subset of these bits as a hash value for fast lookup of the flow. What exactly should the FlowLabel be based on, for each of these two approaches?

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.

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.

Suppose host A reaches host B via routers R1 and R2: A-R1-R2-B. Fast retransmit is not used, and A calculates TimeOut as \(2 \times\) EstimatedRTT. Assume that the A-R1 and \(R 2-B\) links have infinite bandwidth; the \(R 1 \longrightarrow R 2\) link, however, introduces a 1 -second-per-packet bandwidth delay for data packets (though not ACKs). Describe a scenario in which the R1-R2 link is not \(100 \%\) utilized, even though A always has data ready to send. Hint: Suppose A's CongestionWindow increases from \(N\) to \(N+1\), where \(N\) is R1's queue size.
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