Question

Consider that only a single TCP (Reno) connection uses one \(10 \mathrm{Mbps}\) 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 receive buffer is much larger than the congestion window. We also make the following assumptions: each TCP segment size is 1,500 bytes; the two-way propagation delay of this connection is \(150 \mathrm{msec}\); and this TCP connection is always in congestion avoidance phase, that is, ignore slow start. a. What is the maximum window size (in segments) that this TCP connection can achieve? b. What is the average window size (in segments) and average throughput (in bps) of this TCP connection? c. How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?

Short answer

a) 125 segments; b) Average window: 93.75 segments, throughput: 7.5 Mbps; c) 9.375 seconds.

Step-by-step solution

Calculate the Maximum Window Size

To determine the maximum window size in segments, we use the bandwidth-delay product (BDP). The BDP is calculated as: \[ BDP = ext{Bandwidth} \times ext{Round-Trip Time (RTT)} \]Given the bandwidth is 10 Mbps (or \(10 \times 10^6\) bps) and the RTT is 150 ms (or 0.15 seconds), the bandwidth-delay product is:\[ BDP = 10 \times 10^6 \times 0.15 = 1.5 \times 10^6 \text{ bits} \]Since each TCP segment is 1,500 bytes, which is \(1,500 \times 8 = 12,000\) bits, the maximum window size in segments is:\[ \frac{1.5 \times 10^6}{12,000} = 125 \text{ segments} \]

Calculate the Average Window Size

In the congestion avoidance phase, TCP Reno incrementally increases the congestion window by one every round-trip time (RTT) until packet loss occurs, leading to a sawtooth pattern. The window size goes from \(W/2\) to \(W_{max}\), averaging to:\[ \text{Average window size} = \frac{3}{4} \times W_{max} = \frac{3}{4} \times 125 = 93.75 \text{ segments} \]

Calculate the Average Throughput

The average throughput can be calculated based on the average window size and round-trip time (RTT):\[ \text{Throughput} = \frac{\text{Average window size} \times ext{Segment size}}{\text{RTT}} \]Substituting the values, \(93.75\) segments, \(1,500 \times 8\) bits/segment, and \(0.15\) seconds for the RTT, gives:\[ \text{Throughput} = \frac{93.75 \times 12,000}{0.15} = 7.5 \times 10^6 \text{ bps or } 7.5 \text{ Mbps} \]

Determine the Time to Regain Maximum Window Size

After a packet loss, TCP Reno reduces the congestion window to half, i.e., \( \frac{W_{max}}{2} = 62.5 \). To return to \(W_{max}\), TCP Reno increases the window by 1 segment per RTT. Therefore, it takes \( \frac{62.5}{1} = 62.5 \text{ RTTs} \) to regain the maximum window size. The time taken is:\[ 62.5 \times 0.15 \text{ seconds} = 9.375 \text{ seconds} \]

Key concepts

TCP Reno

TCP Reno is a popular version of the Transmission Control Protocol that helps manage data transmission over networks. It is known for its approach to handling congestion, a key issue in maintaining smooth and efficient data communication. When a TCP Reno sender detects a packet loss through duplicate acknowledgments, it invokes Fast Recovery, reducing its congestion window size by half instead of falling back to the initial slow start entirely. This allows the network to recover more quickly following packet loss, reducing the time taken to resume maximum throughput. TCP Reno primarily operates in the congestion avoidance phase once the initial slow start is over, ensuring ongoing data flow stability without overwhelming the network.

Bandwidth-Delay Product

The Bandwidth-Delay Product (BDP) is a crucial metric in understanding network performance, calculated by multiplying a network's bandwidth with its round-trip time (RTT). This product gives the maximum amount of unacknowledged data, or 'in-flight' data, that can be maintained in the network without causing delays. In our example, the link has a bandwidth of 10 Mbps and a RTT of 150 ms. The BDP therefore is \(10 \times 10^6 \times 0.15 = 1.5 \times 10^6\) bits. This figure reflects how much data the link can handle at any given moment, essentially representing the network pipeline's capacity. Understanding BDP helps in fine-tuning the maximum window size, as it defines how full this 'pipe' can be, balance between delay and utilization.

Congestion Avoidance

Congestion Avoidance is a key phase for TCP Reno, crucial in preventing network congestion before it becomes a major bottleneck. During this phase, TCP Reno increases its congestion window size gradually rather than exponentially. This ensures that the bandwidth is used efficiently without overwhelming the network resources, which could lead to packet loss. The window size increment happens at each round-trip time (RTT), rising linearly. This careful control marks TCP Reno's signature sawtooth pattern where window size increases steadily until packet loss is detected. At this point, the window size is halved, and the process begins again, striking a balance between high throughput and minimal congestion.

Round-Trip Time

Round-Trip Time (RTT) is an essential factor in evaluating network performance, representing the time it takes for a data packet to travel from the sender to the receiver and back. RTT is affected by numerous elements, including physical distance and the type of link. In our scenario, the RTT is 150 ms, an average figure that influences calculations like the bandwidth-delay product and the timing for flow control mechanisms in TCP Reno. Understanding RTT helps in setting the congestion window size appropriately. It's crucial for determining how quickly TCP can increase its flow or how soon it detects packet loss, thereby maintaining efficient data flow while managing congestion.