Chapter 1: Problem 7
Consider a point-to-point link \(2 \mathrm{~km}\) in length. At what bandwidth would propagation delay (at a speed of \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) equal transmit delay for 100 -byte packets? What about 512 -byte packets?
Short Answer
Expert verified
The required bandwidth is 80 Mbps for 100-byte packets and 409.6 Mbps for 512-byte packets.
Step by step solution
01
- Understand the problem
We need to find the bandwidth where the propagation delay equals the transmission delay for two different packet sizes on a 2 km link with the propagation speed of \(2 \times 10^8 \) m/s.
02
- Calculate the propagation delay
Propagation delay is calculated using the formula: \[ \text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}} \] Given: \[ \text{Distance} = 2 \text{ km} = 2000 \text{ m} \] \[ \text{Propagation Speed} = 2 \times 10^8 \text{ m/s} \] So, the propagation delay is: \[ \frac{2000}{2 \times 10^8} = 10^{-5} \text{ s} \]
03
- Establish the relationship between propagation delay and transmission delay
The problem states that propagation delay should be equal to transmission delay. This means: \[ \text{Propagation Delay} = \text{Transmission Delay} \] From Step 2, we know the propagation delay is \(10^{-5} \text{ s}\). Hence, we need to find the bandwidth where the transmission delay for specific packet sizes equals \(10^{-5} \text{ s}\).
04
- Calculate the required bandwidth for 100-byte packets
Transmission delay is calculated using the formula: \[ \text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Bandwidth}} \] For 100-byte packets (Note: 1 byte = 8 bits): \[ \text{Packet Size} = 100 \text{ bytes} = 100 \times 8 = 800 \text{ bits} \] Since the propagation delay is \(10^{-5} \text{ s}\), set the transmission delay equal to the propagation delay: \[ \frac{800 \text{ bits}}{\text{Bandwidth}} = 10^{-5} \text{ s} \] So, \[ \text{Bandwidth} = \frac{800}{10^{-5}} = 8 \times 10^7 \text{ bits/s} = 80 \text{ Mbps} \]
05
- Calculate the required bandwidth for 512-byte packets
Using the same transmission delay formula: \[ \text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Bandwidth}} \] For 512-byte packets: \[ \text{Packet Size} = 512 \text{ bytes} = 512 \times 8 = 4096 \text{ bits} \] Set the transmission delay equal to the propagation delay: \[ \frac{4096 \text{ bits}}{\text{Bandwidth}} = 10^{-5} \text{ s} \] So, \[ \text{Bandwidth} = \frac{4096}{10^{-5}} = 4.096 \times 10^8 \text{ bits/s} = 409.6 \text{ Mbps} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Propagation Delay
Propagation delay is the time it takes for a signal to travel from the sender to the receiver. This delay is primarily dependent on the distance between the two points and the speed at which the signal travels (propagation speed).
To calculate propagation delay, you can use the formula:
\(\text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}} \).
For example, if the distance is 2 km (2000 meters) and the propagation speed is 2 × 10⁸ meters per second, the propagation delay will be:
\( \frac{2000}{2 \times 10^8} = 10^{-5} \text{ seconds} \).
This means it takes 10 microseconds for the signal to travel 2 km.
To calculate propagation delay, you can use the formula:
\(\text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}} \).
For example, if the distance is 2 km (2000 meters) and the propagation speed is 2 × 10⁸ meters per second, the propagation delay will be:
\( \frac{2000}{2 \times 10^8} = 10^{-5} \text{ seconds} \).
This means it takes 10 microseconds for the signal to travel 2 km.
Transmission Delay
Transmission delay is the time it takes to push all the packet's bits onto the wire. It depends on the packet size and the bandwidth of the network. The formula to calculate transmission delay is:
\( \text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Bandwidth}} \).
For instance, if our packet size is 100 bytes (which is 800 bits, as 1 byte = 8 bits), and the bandwidth needed is such that the transmission delay equals the propagation delay of 10 microseconds, you need to set up the equation:
\(\frac{800 \text{ bits}}{\text{Bandwidth}} = 10^{-5} \text{ seconds} \).
Solving this, we find the bandwidth is 80 Mbps for 100-byte packets.
\( \text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Bandwidth}} \).
For instance, if our packet size is 100 bytes (which is 800 bits, as 1 byte = 8 bits), and the bandwidth needed is such that the transmission delay equals the propagation delay of 10 microseconds, you need to set up the equation:
\(\frac{800 \text{ bits}}{\text{Bandwidth}} = 10^{-5} \text{ seconds} \).
Solving this, we find the bandwidth is 80 Mbps for 100-byte packets.
Bandwidth Calculation
Bandwidth refers to the maximum rate at which data can be transferred over a network path. It is usually measured in bits per second (bps), megabits per second (Mbps), or gigabits per second (Gbps).
In the context of our problem, we calculated the required bandwidth for both 100-byte and 512-byte packets. For a 100-byte packet, with a transmission delay requirement of 10 microseconds, the bandwidth needed was found to be 80 Mbps. Similarly, for a 512-byte packet, the required bandwidth is much higher at 409.6 Mbps, to ensure the same propagation delay of 10 microseconds.
In the context of our problem, we calculated the required bandwidth for both 100-byte and 512-byte packets. For a 100-byte packet, with a transmission delay requirement of 10 microseconds, the bandwidth needed was found to be 80 Mbps. Similarly, for a 512-byte packet, the required bandwidth is much higher at 409.6 Mbps, to ensure the same propagation delay of 10 microseconds.
Packet Size
Packet size is the amount of data that is sent in a single packet over the network. It’s important because it affects transmission delay. Packet sizes are typically measured in bytes.
In our example, we considered two packet sizes: 100 bytes and 512 bytes. For the calculation, we converted these into bits (because bandwidth is usually measured in bits per second). So, 100 bytes became 800 bits, and 512 bytes became 4096 bits.
The packet size directly impacts the required bandwidth for maintaining a specific transmission delay.
In our example, we considered two packet sizes: 100 bytes and 512 bytes. For the calculation, we converted these into bits (because bandwidth is usually measured in bits per second). So, 100 bytes became 800 bits, and 512 bytes became 4096 bits.
The packet size directly impacts the required bandwidth for maintaining a specific transmission delay.
Data Transmission Speed
Data transmission speed is how fast data can be sent from one point to another in a network. It's often referred to in terms of bandwidth and is usually measured in bits per second (bps).
In this exercise, the speed of data transmission had to match a specific delay requirement. For 100-byte packets, the necessary speed (bandwidth) was determined to be 80 Mbps. For the larger 512-byte packets, the required transmission speed increased to 409.6 Mbps to maintain equal propagation and transmission delays.
In this exercise, the speed of data transmission had to match a specific delay requirement. For 100-byte packets, the necessary speed (bandwidth) was determined to be 80 Mbps. For the larger 512-byte packets, the required transmission speed increased to 409.6 Mbps to maintain equal propagation and transmission delays.