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(a) Suppose \(N\) packets arrive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length \(L\) and the link has transmission rate \(R\). What is the average queuing delay for the \(N\) packets? (b) Now suppose that \(N\) such packets arrive to the link every \(L N / R\) seconds. What is the average queuing delay of a packet?

Short Answer

Expert verified
(a) Average queuing delay is \( \frac{(N-1) L}{2R} \). (b) Average queuing delay is 0.

Step by step solution

01

Define Terms and Goals

To find the average queuing delay, first we need to understand what contributes to queuing delay. Each packet not being transmitted must wait its turn in the queue until the previous packets are transmitted.
02

Understand Problem (a) Conditions

For part (a), we have all packets arriving simultaneously. No packets are being transmitted or are in the queue prior. Single packet transmission takes time \(\frac{L}{R}\), where \(L\) is the packet length and \(R\) is the transmission rate.
03

Calculate Total Delay for N Packets

Each of the \(N\) packets must wait for the previous packets to be transmitted. The first packet experiences no queuing delay, the second waits \(\frac{L}{R}\), the third waits \(2\frac{L}{R}\), continuing up to \((N-1)\frac{L}{R}\) for the last packet. Thus the total wait time is the sum for the sequence: \(0 + \frac{L}{R} + 2\frac{L}{R} + \ldots + (N-1)\frac{L}{R}\).
04

Use Arithmetic Series Formula

The sum of an arithmetic series can be calculated as \(S = \frac{n(n-1)}{2}\) where \(n = N\) for the sequence of delays. The total delay for all packets is thus \(\frac{(N-1)N}{2}\frac{L}{R}\).
05

Calculate Average Delay for (a)

To find the average queuing delay, divide the total delay by \(N\). Thus the average delay is \(\frac{(N-1) L}{2R}\).
06

Understand Problem (b) Conditions

In part (b), packets arrive such that their arrival frequency is every \(\frac{LN}{R}\) seconds. This is equal to the time needed to send \(N\) packets without delay, meaning packets do not queue behind each other.
07

Calculate Average Delay for (b)

As arrival and service rate are synchronized and no queue forms due to one packet finishing as the next arrives, the average queuing delay is zero.

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Key Concepts

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

Packet Transmission
Packet transmission is the process by which data is sent from one point to another over a network. Each packet is a small chunk of data that is a part of a larger message.
The transmission of each packet uses a defined time depending on its length and the rate of the network.
- **Packet Length (L):** This is the size of the individual packet and is usually measured in bits.
- **Transmission Rate (R):** This measures how fast data can be sent across a network link and is typically given in bits per second (bps).
To calculate how long it takes to transmit one packet, you use the formula for transmission time, which is \(|\frac{L}{R}|\).
Here, a packet of length \(L\) is transmitted over a network at rate \(R\). Understanding this concept is critical to determine other metrics in networking such as delays and throughput.
Arithmetic Series
An arithmetic series is a sequence of numbers, each differing by a constant amount. It's relevant in our context because queue delays form such a series.
When packets arrive simultaneously, each one must wait in line to be transmitted after the ones before it.
The delay for the transmission forms a sequence:
  • The first packet has no delay.
  • The second packet waits while the first is transmitted, which takes \(\frac{L}{R}\).
  • The third waits while the first two are transmitted, which is \(2\frac{L}{R}\), and so on.
This pattern continues, so the total wait time is an arithmetic series sum:\[0 + \frac{L}{R} + 2\frac{L}{R} + \ldots + (N-1)\frac{L}{R}\].
To find the total delay for all packets, you calculate this sum, using the arithmetic series formula: \(S = \frac{n(n-1)}{2}\frac{L}{R}\) where \(n = N\).
The significance of using an arithmetic series helps simplify calculation steps in solving for total and average queuing delays.
Transmission Rate
Transmission rate, denoted as \(R\), is a crucial aspect of understanding how networks operate.
It describes how fast data can be transmitted over a network and is fundamental in calculating transmission times and delays.
For any length of packet \(L\), the transmission rate provides how quickly data moves through the network. It is important for prioritizing packets and managing traffic levels.
Networks aim to maximize \(R\) to improve efficiency. However, practical limitations like bandwidth impact the achievable transmission rate.
In our exercise, the transmission rate \(R\) determines the time taken to send packets entirely over the link, affecting how packets queue and their eventual delays.
Packet Arrival
Packet arrival refers to how packets are received at a network node. Imagine packets as cars arriving at a toll booth.
How these packets arrive drastically affects network performance, including queuing delay.
In part (a) of our exercise, packets arrive all at once. This means they must wait in line to be transmitted, contributing to queuing delays since succeeding packets depend on predecessors finishing.
In part (b), packets arrive exactly timed with a duration of \(\frac{LN}{R}\). This structured arrival means each arrives as the previous packet is leaving the link.
When synchronized precisely like this, no backlog or queue occurs, and thus, the average queuing delay can become zero.
Efficient network design aims to achieve optimal packet arrival patterns to prevent overload and delays.

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Most popular questions from this chapter

Consider the discussion in Section \(1.3\) of packet switching versus circuit switching in which an example is provided with a \(1 \mathrm{Mbps}\) link. Users are generating data at a rate of \(100 \mathrm{kbps}\) when busy, but are busy generating data only with probability \(p=0.1\). Suppose that the \(1 \mathrm{Mbps}\) link is replaced by a 1 Gbps link. a. What is \(N\), the maximum number of users that can be supported simultaneously under circuit switching? b. Now consider packet switching and a user population of \(M\) users. Give a formula (in terms of \(p, M, N\) ) for the probability that more than \(N\) users are sending data.

Suppose users share a 3 Mbps link. Also suppose each user requires \(150 \mathrm{kbps}\) when transmitting, but each user transmits only 10 percent of the time. (See the discussion of packet switching versus circuit switching in Section 1.3.) a. When circuit switching is used, how many users can be supported? b. For the remainder of this problem, suppose packet switching is used. Find the probability that a given user is transmitting. c. Suppose there are 120 users. Find the probability that at any given time, exactly \(n\) users are transmitting simultaneously. (Hint: Use the binomial distribution.) d. Find the probability that there are 21 or more users transmitting simultaneously.

Consider sending a large file of \(F\) bits from Host A to Host B. There are three links (and two switches) between \(A\) and \(B\), and the links are uncongested (that is, no queuing delays). Host A segments the file into segments of \(S\) bits each and adds 80 bits of header to each segment, forming packets of \(L=80+S\) bits. Each link has a transmission rate of \(R\) bps. Find the value of \(S\) that minimizes the delay of moving the file from Host A to Host B. Disregard propagation delay.

What is the transmission rate of Ethernet LANs?

Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are \(R_{1}\) and \(R_{2}\), respectively. Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send a packet of length \(L ?\) (Ignore queuing, propagation delay, and processing delay.)

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