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Chapter 2: Problem 23

Consider an ARQ algorithm running over a \(20-\mathrm{km}\) point-to-point fiber link. (a) Compute the propagation delay for this link, assuming that the speed of light is \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in the fiber. (b) Suggest a suitable timeout value for the ARQ algorithm to use. (c) Why might it still be possible for the ARQ algorithm to time out and retransmit a frame, given this timeout value?

Short Answer

Expert verified
Propagation delay is 100 ms. Suitable timeout value is 250 ms. Timeouts can occur due to congestion and processing delays.

Step by step solution

01

Calculate the propagation delay

To compute the propagation delay, use the formula: \[ \text{Propagation Delay} = \frac{\text{Distance}}{\text{Speed}} \] Given the distance is 20 km (or 20,000 meters) and the speed of light in the fiber is \( 2 \times 10^8 \text{ m/s} \), we can plug these values into the formula: \[ \text{Propagation Delay} = \frac{20{,}000 \text{ meters}}{2 \times 10^8 \text{ m/s}} = 0.0001 \text{ s} = 100 \text{ ms}\]
02

Suggest a suitable timeout value

A suitable timeout value for the ARQ algorithm should be slightly more than the round-trip time to account for processing and potential delays. First, calculate the round-trip time:\[ \text{RTT} = 2 \times \text{Propagation Delay} = 2 \times 100 \text{ ms} = 200 \text{ ms}\] To account for additional delays and processing time, it's good to add a margin to the timeout value. For example, adding 50 ms would make the timeout value:\[ \text{Timeout Value} = 200 \text{ ms} + 50 \text{ ms} = 250 \text{ ms} \]
03

Explain why the ARQ algorithm might time out

Even with the calculated timeout value, the ARQ algorithm might still time out and retransmit a frame. This can happen due to various unforeseen factors such as network congestion, errors in the data link causing retransmissions, or additional delays in the processing time at either end of the communication.

Key Concepts

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

Propagation Delay
Propagation delay refers to the time it takes for a signal to travel from the sender to the receiver. It's crucial in networking because it affects the overall latency. In our example, we use the formula: \( \text{Propagation Delay} = \frac{\text{Distance}}{\text{Speed}} \) Given a 20 km (20,000 m) fiber link and the speed of light in fiber at \( 2 \times 10^8 \text{ m/s} \), the propagation delay is calculated as 100 milliseconds. This delay directly influences network performance.
Timeout Value
The timeout value in an ARQ algorithm determines how long the sender waits for an acknowledgment before retransmitting the frame. It's essential to set this value correctly to avoid unnecessary retransmissions. Based on our computed propagation delay, we calculate the round-trip time (RTT) and suggest a slightly higher timeout value to account for processing and potential delays. If the RTT is 200 milliseconds, adding 50 milliseconds as a safety margin results in a 250 milliseconds timeout value. This helps ensure that the network can handle minor delays without unnecessary retransmissions.
Round-Trip Time (RTT)
Round-trip time, or RTT, is the time it takes for a signal to go from the sender to the receiver and back again. It's a critical factor in determining the efficiency of the ARQ algorithm. To find the RTT, we multiply the one-way propagation delay by two: \( \text{RTT} = 2 \times \text{Propagation Delay} \) For a propagation delay of 100 milliseconds, the RTT is 200 milliseconds. Adding a buffer to account for additional delays means that our final timeout value should be a bit more than the RTT, like 250 milliseconds.
Network Congestion
Network congestion occurs when the network is overloaded with more data than it can handle efficiently. This congestion can cause increased delays, packet loss, and retransmissions. Even with an optimized timeout value, congestion can lead to timeouts because packets or acknowledgments are delayed beyond their expected arrival times. This is why it's essential to consider typical network traffic patterns and potential congestion when setting timeout values for ARQ algorithms.
Data Link Errors
Data link errors happen when there are problems in transmitting data over a network link, causing frames to be corrupted or lost. These errors can trigger retransmissions in ARQ algorithms, leading to increased delays. Factors like physical defects in the transmission medium, interference, or weak signals can cause such errors. Even with a well-calculated timeout value, these errors can lead to timeouts if the frame doesn't arrive correctly or if the acknowledgment from the receiver doesn't get back in time. Therefore, robust error detection and correction mechanisms are vital to minimize the impact of data link errors.

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Most popular questions from this chapter

For a 100-Mbps token ring network with a token rotation time of \(200 \mu\) s that allows each station to transmit one 1-KB packet each time it possesses the token, calculate the maximum effective throughput rate that any one host can achieve. Do this assuming (a) immediate release and (b) delayed release.

Assume that a SONET receiver resynchronizes its clock whenever a 1 bit appears; otherwise, the receiver samples the signal in the middle of what it believes is the bit's time slot. (a) What relative accuracy of the sender's and receiver's clocks is required in order to receive correctly 480 bytes (one ATM AAL. 5 cell's worth) in a row? (b) Consider a forwarding station A on a SONET STS-1 line, receiving frames from the downstream end \(B\) and retransmitting them upstream. What relative accuracy of A's and B's clocks is required to keep A from accumulating more than one extra frame per minute?

Suppose a \(100-\mathrm{Mbps}\) delayed-release token ring has 10 stations, a ring latency of \(30 \mu \mathrm{s}\), and an agreed-upon TTRT of \(350 \mu \mathrm{s}\). (a) How many synchronous frame bytes could each station send, assuming all are allocated the same amount? (b) Assume stations \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in increasing order on the ring. Due to uniform synchronous traffic, the TRT without asynchronous data is \(300 \mu \mathrm{s}\). B sends a \(200-\mu \mathrm{s}(2.5-\mathrm{Kb})\) asynchronous frame. What TRT will \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) then see on their next measurement? Who may transmit such a frame next?

With 1 parity bit we can detect all 1-bit errors. Show that at least one generalization fails, as follows: (a) Show that if messages \(m\) are 8 bits long, then there is no error detection code \(e=e(m)\) of size 2 bits that can detect all 2-bit errors. Hint: Consider the set \(M\) of all 8-bit messages with a single 1 bit; note that any message from \(M\) can be transmuted into any other with a 2 -bit error, and show that some pair of messages \(m_{1}\) and \(m_{2}\) in \(M\) must have the same error code \(e\). (b) Find an \(N\) (not necessarily minimal) such that no 32 -bit error detection code applied to N-bit blocks can detect all errors altering up to 8 bits.

Suppose that \(N\) Ethernet stations, all trying to send at the same time, require \(N / 2\) slot times to sort out who transmits next. Assuming the average packet size is 5 slot times, express the available bandwidth as a function of \(N\).
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