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Suppose you want to send some data using the BISYNC framing protocol, and the last 2 bytes of your data are DLE and ETX. What sequence of bytes would be transmitted immediately prior to the CRC?

Short Answer

Expert verified
DLE, DLE, ETX

Step by step solution

01

- Understand BISYNC Protocol

The BISYNC (Binary Synchronous) protocol uses special characters for communication control. The DLE (Data Link Escape) character is used to indicate that the following character is a control character rather than data.
02

- Identify Control Characters Sequence

When DLE is used in a data frame, it indicates that the next character should be interpreted differently. In the BISYNC protocol, if DLE is followed by ETX, it signifies the end of the text frame.
03

- Apply BISYNC Framing Rules

According to BISYNC rules, if DLE itself appears in the data, it must be doubled to distinguish it from a DLE used to indicate control sequences. Therefore, if the last two bytes of the data are DLE and ETX, we need to encode DLE by doubling it.
04

- Determine the Transmitted Sequence

Given that our last bytes are DLE and ETX, the transmitted sequence immediately prior to the CRC will have DLE doubled to indicate a data DLE. Thus, the sequence will be DLE, DLE, ETX.

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Key Concepts

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

Data Link Escape (DLE) character
The Data Link Escape (DLE) character is vital in communication protocols like BISYNC. It signals that the next character holds special importance and should be treated differently from regular data characters.
DLE is primarily used to signify that the subsequent character is a control character. This way, the receiving end can appropriately interpret the sequence.
For example, if you have a data sequence where DLE appears followed by another character, you should anticipate a specific control sequence rather than typical data. This ensures that the transmitted data maintains integrity and follows the protocol's rules.

Here are some key points to remember about DLE:
  • DLE indicates that the following character is a control character.
  • It helps differentiate between control sequences and regular data.
  • In BISYNC, DLE, when followed by specific characters, holds special meanings.
End of Text (ETX) character
The End of Text (ETX) character is another significant control character in framing protocols. It denotes the end of a text or data frame.
In the context of BISYNC, when DLE is followed by ETX, it indicates the conclusion of the data sequence or frame. This helps the receiver understand where the data ends, making the communication more structured and error-free.
Understanding ETX is crucial in ensuring proper data interpretation and transmission. When sending data, especially using BISYNC, recognizing ETX helps in identifying the end of meaningful data.

Important aspects of ETX include:
  • Often used to signify the end of a text frame.
  • Interpreted specially when preceded by DLE.
  • Ensures proper framing and termination of data sequences.
CRC (Cyclic Redundancy Check)
The Cyclic Redundancy Check (CRC) is a powerful error-detecting mechanism used to ensure data integrity. In BISYNC and other communication protocols, CRC is appended at the end of data frames.
CRC works by performing polynomial division on the data to create a remainder. This remainder, or CRC code, is then transmitted with the data. The receiver performs the same CRC calculation and checks if the remainder matches the sent CRC. If they match, it means the data likely arrived without errors.
CRC is crucial in identifying and correcting errors that might occur during data transmission due to noise, interference, or other issues.

Key points about CRC include:
  • Used for error detection.
  • Calculates a remainder using polynomial division of data.
  • Ensures data integrity by verifying if transmitted and received data match.

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Most popular questions from this chapter

Suppose Ethernet physical addresses are chosen at random (using true random bits). (a) What is the probability that on a 1024-host network, two addresses will be the same? (b) What is the probability that the above event will occur on some one or more of \(2^{20}\) networks? (c) What is the probability that of the \(2^{30}\) hosts in all the networks of (b), some pair has the same address? Hint: The calculation for (a) and (c) is a variant of that used in solving the socalled Birthday Problem: Given \(N\) people, what is the probability that two of their birthdays (addresses) will be the same? The second person has probability \(1-\frac{1}{365}\) of having a different birthday from the first, the third has probability \(1-\frac{2}{365}\) of having a different birthday from the first two, and so on. The probability all birthdays are different is thus $$ \left(1-\frac{1}{365}\right) \times\left(1-\frac{2}{365}\right) \times \cdots \times\left(1-\frac{N-1}{365}\right) $$ which for smallish \(N\) is about $$ 1-\frac{1+2+\cdots+(N-1)}{365} $$

What kinds of problems can arise when two hosts on the same Ethernet share the same hardware address? Describe what happens and why that behavior is a problem.

Suppose we want to transmit the message 1011001001001011 and protect it from errors using the CRC-8 polynomial \(x^{8}+x^{2}+x^{1}+1\). (a) Use polynomial long division to determine the message that should be transmitted. (b) Suppose the leftmost bit of the message is inverted due to noise on the transmission link. What is the result of the receiver's CRC calculation? How does the receiver know that an error has occurred?

The text suggests that the sliding window protocol can be used to implement flow control. We can imagine doing this by having the receiver delay ACKs, that is, not send the ACK until there is free buffer space to hold the next frame. In doing so, each ACK would simultaneously acknowledge the receipt of the last frame and tell the source that there is now free buffer space available to hold the next frame. Explain why implementing flow control in this way is not a good idea.

Suppose you are designing a sliding window protocol for a 1-Mbps point-to- point link to a stationary satellite revolving around the earth at \(3 \times 10^{4} \mathrm{~km}\) altitude. Assuming that each frame carries \(1 \mathrm{~KB}\) of data, what is the minimum number of bits you need for the sequence number in the following cases? Assume the speed of light is \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) \(\mathrm{RWS}=1\) (b) RWS = SWS

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