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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?

Short Answer

Expert verified
For (a): Relative accuracy < \frac{1}{3840}. For (b): Relative accuracy < \frac{1}{480000}.

Step by step solution

01

- Understanding SONET Synchronization

The SONET receiver resynchronizes its clock with each 1 bit received to maintain synchronization with the sender's clock. This means that as long as a 1 bit appears, the receiver can adjust its clock to match the sender’s clock.
02

- Calculate the Total Number of Bits for 480 Bytes

To find out how accurate the clock needs to be, first figure out the total number of bits being transmitted. There are 8 bits in a byte. Therefore, for 480 bytes: \[ 480 \text{ bytes} \times 8 \text{ bits/byte} = 3840 \text{ bits} \]
03

- Derive the Required Relative Accuracy for One Resynchronization

The relative accuracy needs to be such that the clock drift does not cause a bit error for 3840 bits. If a bit drift occurs, the sample may happen outside the middle of the bit's time slot. Assume the maximum drift per bit is \Delta t, and suppose the bit accuracy requirement is a fraction of the bit duration T: \[ \frac{\text{Drift}}{T} = \frac{\text{Relative Accuracy}}{480 \times 8 \times T} \]The allowable drift cannot exceed 1 bit period over 3840 bits.
04

- Simplify the Expression for Relative Accuracy

We need the drift to be less than one bit period over the entire transmission. This means: \[ \frac{\text{Drift}}{T} < \frac{1}{3840} \] Thus, the required relative accuracy is: \[ \text{Relative Accuracy} < \frac{1}{3840} \]
05

- Calculate the Frame Transition Period Accuracy for Part (b)

Forwarding station A needs to retransmit without accumulating more than one frame per minute. STS-1 has a frame every 125 µs. For one minute:\[ 1 \text{ minute} = 60 \times 10^6 \text{ µs} \]Therefore, the number of frames in one minute is: \[ \frac{60 \times 10^6 \text{ µs}}{125 \text{ µs}} = 480,000 \text{ frames} \]
06

- Compute Relative Accuracy Requirement

The relative accuracy must ensure that only one frame is lost over 480,000 frames. Hence, the allowable drift per frame should be less than one frame period:\[ \frac{\text{Drift}}{T_f} < \frac{1}{480000} \]Thus, the relative accuracy required is:\[ \text{Relative Accuracy} < \frac{1}{480000} \]

Key Concepts

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

Clock Drift
Clock drift is a crucial concept in synchronous networks like SONET (Synchronous Optical Network). It refers to the gradual displacement of the clock signal in a synchronization process. This displacement happens because every clock has a slightly different frequency. Over time, even tiny inaccuracies can add up, causing the receiver's clock to become misaligned with the sender's clock.

In the context of SONET, the receiver resynchronizes its clock each time it encounters a '1' bit. This helps to minimize the drift. However, if there are many consecutive zeros, the inability to resynchronize may cause bit errors. Understanding clock drift is essential for calculating the relative accuracy required to maintain proper synchronization over varying numbers of bits or frames.
Frame Retransmission
Frame retransmission is another vital aspect of SONET and similar networking protocols. It is concerned with the accuracy in timing between sending and receiving stations, ensuring that data frames are retransmitted correctly downstream in a process called 'forwarding'.

Consider a station A receiving frames from station B. The station needs to keep its internal clock aligned with B's clock to avoid accumulating errors. In practice, retransmitting frames without the need for frequent resynchronizations is crucial to maintain network efficiency. For example, in the given problem, station A needs a relative clock accuracy that prevents it from receiving more than one extra frame per minute. Therefore, the required accuracy would be less than one frame period over the total number of frames per minute.
Relative Clock Accuracy
Relative clock accuracy directly influences both clock drift and frame retransmission. It defines how closely the receiver's clock needs to align with the sender's clock over a given period or number of bits.

For instance, in case (a) of the problem, where the task is to receive 480 bytes worth of data without errors, the relative clock accuracy must ensure that any clock drift does not cause a bit to be sampled outside its intended time slot. This means the relative accuracy should be less than one part in 3840, based on the number of bits transmitted. Similarly, in case (b), to retransmit frames without losing synchronization over one minute (or 480,000 frames), the relative accuracy must be below one part in 480,000 for it to work effectively.

Ensuring high relative clock accuracy is essential for maintaining synchronization and avoiding data errors in high-speed communication networks like SONET.

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Most popular questions from this chapter

Describe a protocol combining the sliding window algorithm with selective ACKs. Your protocol should retransmit promptly, but not if a frame simply arrives one or two positions out of order. Your protocol should also make explicit what happens if several consecutive frames are lost.

Suppose you are designing a sliding window protocol for a 1-Mbps point-to- point link to the moon, which has a one-way latency of \(1.25\) seconds. Assuming that each frame carries \(1 \mathrm{~KB}\) of data, what is the minimum number of bits you need for the sequence number?

In stop-and-wait transmission, suppose that both sender and receiver retransmit their last frame immediately on receipt of a duplicate ACK or data frame; such a strategy is superficially reasonable because receipt of such a duplicate is most likely to mean the other side has experienced a timeout. (a) Draw a timeline showing what will happen if the first data frame is somehow duplicated, but no frame is lost. How long will the duplications continue? This situation is known as the Sorcerer's Apprentice bug. (b) Suppose that, like data, ACKs are retransmitted if there is no response within the timeout period. Suppose also that both sides use the same timeout interval. Identify a reasonably likely scenario for triggering the Sorcerer's Apprentice bug.

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 \(N\).

Suppose \(A, B\), and \(C\) all make their first carrier sense, as part of an attempt to transmit, while a fourth station D is transmitting. Draw a timeline showing one possible sequence of transmissions, attempts, collisions, and exponential backoff choices. Your timeline should also meet the following criteria: (i) initial transmission attempts should be in the order \(A, B, C\), but successful transmissions should be in the order \(\mathrm{C}, \mathrm{B}, \mathrm{A}\), and (ii) there should be at least four collisions.

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