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Consider a broadcast channel with \(N\) nodes and a transmission rate of \(R\) bps. Suppose the broadcast channel uses polling (with an additional polling node) for multiple access. Suppose the amount of time from when a node completes transmission until the subsequent node is permitted to transmit (that is, the polling delay) is \(d_{\text {poll }}\). Suppose that within a polling round, a given node is allowed to transmit at most \(Q\) bits. What is the maximum throughput of the broadcast channel?

Short Answer

Expert verified
The maximum throughput is \( \frac{R}{1 + d_{\text{poll}}\frac{R}{Q}} \).

Step by step solution

01

Understand the Problem

We are considering a broadcast channel with \(N\) nodes and one extra polling node. Each node is allowed to transmit up to \(Q\) bits in one round after the polling delay \(d_{\text{poll}}\). We want to determine the maximum throughput of this system.
02

Identify Key Elements

Identify the maximum number of bits a node can transmit, which is \(Q\), and the time it takes to transmit these bits, which is \(\frac{Q}{R}\) seconds. Also consider the polling delay \(d_{\text{poll}}\). This delay occurs every time the transmission moves from one node to the next.
03

Compute the Cycle Time

A complete cycle consists of one transmission opportunity for each node in addition to the polling delay for all nodes except the last one. Thus, the total cycle time \(T_{\text{cycle}}\) is given by \( T_{\text{cycle}} = N \left( \frac{Q}{R} + d_{\text{poll}} \right) \).
04

Determine Maximum Throughput

The throughput is the total number of bits transmitted in each cycle divided by the duration of the cycle. The maximum number of bits transmitted in a cycle is \(NQ\), so the maximum throughput \(\text{Throughput}_{\text{max}}\) is given by: \( \text{Throughput}_{\text{max}} = \frac{NQ}{T_{\text{cycle}}} = \frac{NQ}{N \left( \frac{Q}{R} + d_{\text{poll}} \right)} = \frac{Q}{\frac{Q}{R} + d_{\text{poll}}} \).
05

Simplify the Result

By simplifying the expression, we see that the maximum throughput of the broadcast channel is determined solely by the bit rate, the maximum number of transmittable bits per node, and the polling delay: \( \text{Throughput}_{\text{max}} = \frac{R}{1 + d_{\text{poll}}\frac{R}{Q}} \). This shows that the throughput decreases as the polling delay increases in proportion to \(Q\).

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Key Concepts

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

Polling Delay
Polling delay is a crucial concept in network communication, especially within broadcast channels that involve polling mechanisms. It refers to the time interval from when one node finishes transmitting its data until the next node is given permission to start its transmission. Understanding polling delay is important because it significantly impacts the efficiency and overall speed of data transmission in a network. During this delay, no data is being sent, leading to brief periods of inactivity, which can add up and reduce system performance.

In the context of the exercise, the polling delay is denoted as \( d_{\text{poll}} \). When considering a full transmission cycle for \( N \) nodes, each node experiences this delay, except the last one. Consequently, the total cycle time includes multiple polling delays, which adds to the time required for all the nodes to complete a transmission round.
  • Polling delays slow down data transfer since they introduce idle time between transmissions.
  • A longer polling delay means that the channel spends more time inactive, decreasing the channel’s overall efficiency.
  • Optimizing polling mechanisms can reduce these delays, enhancing throughput performance.
Maximum Throughput
Maximum throughput is a pivotal metric in network communications that indicates the highest data transfer rate achievable over a network channel within a specific period. It measures the efficiency of the network, showing how much data is successfully transmitted without errors. In the case of broadcast channels using polling for multiple access, achieving maximum throughput involves fine-tuning various parameters like transmission rate, data packet size, and polling delays.

This exercise showcases how to calculate maximum throughput by using the formula \( \text{Throughput}_{\text{max}} = \frac{R}{1 + d_{\text{poll}} \frac{R}{Q}} \), where \( R \) is the transmission rate and \( Q \) is the maximum number of transmittable bits in a round.
  • Throughput is affected by transmission rate \( R \), the polling delay \( d_{\text{poll}} \), and bit quota per cycle \( Q \).
  • High throughput ensures faster data transmission and better channel utilization.
  • Efforts to maximize throughput often aim to minimize idle times and optimize data flow rates.
Multiple Access
In communication networks, multiple access refers to the ability of several nodes or devices to share and use the same communication channel without interference. This approach is essential for maximizing resource utilization and ensuring efficient data transmission in shared environments. In the exercise, multiple access is managed through polling, which ensures that each node gets a turn to transmit data in sequence without collisions.

The polling method implemented allows each node to transmit a predefined number of bits \( Q \) and involves organizing and coordinating when each node can access the channel. This systematic access helps prevent data collisions, which can occur when multiple nodes attempt to transmit simultaneously on the same channel.
  • Multiple access strategies include Time-Division Multiple Access (TDMA) and Frequency-Division Multiple Access (FDMA), among others.
  • Polling is a type of controlled access method where nodes patiently wait for their turn, reducing competition over the channel.
  • Efficient multiple access ensures all nodes utilize the channel effectively, improving overall network throughput.
Transmission Rate
The transmission rate is a fundamental aspect of network communications, indicating how fast data can be sent over a network channel. Measured in bits per second (bps), this rate determines how quickly nodes can push data out and is crucial in calculating the maximum throughput of a network. In scenarios like the one presented, a high transmission rate \( R \) is desired to ensure optimal data flow and network performance.

As seen in the exercise, the transmission rate is directly tied to the calculation of maximum throughput. The given formula \( \text{Throughput}_{\text{max}} = \frac{R}{1 + d_{\text{poll}} \frac{R}{Q}} \) vividly shows the relationship between the transmission rate and throughput.
  • Transmission rate \( R \) is a measure of the channel's capacity to move data swiftly.
  • A higher transmission rate helps in reducing the time required to transmit a specific amount of data.
  • Balancing transmission rate with other parameters like polling delay and bit quota is essential to maximize throughput.

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Most popular questions from this chapter

Suppose nodes A and B are on the same \(10 \mathrm{Mbps}\) broadcast channel, and the propagation delay between the two nodes is 245 bit times. Suppose A and \(B\) send Ethernet frames at the same time, the frames collide, and then \(\mathrm{A}\) and \(\mathrm{B}\) choose different values of \(K\) in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from \(\mathrm{A}\) and \(\mathrm{B}\) collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at \(t=0\) bit times. They both detect collisions at \(t=245\) bit times. Suppose \(K_{A}=0\) and \(K_{B}=1\). At what time does B schedule its retransmission? At what time does A begin transmission? (Note: The nodes must wait for an idle channel after returning to Step 2 -see protocol.) At what time does A's signal reach B? Does B refrain from transmitting at its scheduled time?

Suppose nodes \(\mathrm{A}\) and \(\mathrm{B}\) are on the same \(10 \mathrm{Mbps}\) broadcast channel, and the propagation delay between the two nodes is 325 bit times. Suppose CSMA/CD and Ethernet packets are used for this broadcast channel. Suppose node A begins transmitting a frame and, before it finishes, node B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted without a collision. Hint: Suppose at time \(t=0\) bits, A begins transmitting a frame. In the worst case, A transmits a minimum-sized frame of \(512+64\) bit times. So A would finish transmitting the frame at \(t=512+64\) bit times. Thus, the answer is no, if B's signal reaches A before bit time \(t=512+64\) bits. In the worst case, when does B's signal reach A?

What are some of the possible services that a link-layer protocol can offer to the network layer? Which of these link-layer services have corresponding services in IP? In TCP?

How big is the MAC address space? The IPv4 address space? The IPv6 address space?

Suppose two nodes start to transmit at the same time a packet of length \(L\) over a broadcast channel of rate \(R\). Denote the propagation delay between the two nodes as \(d_{\text {prop. }}\). Will there be a collision if \(d_{\text {prop }}

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