Math Problem Statement

Consider a point-to-point link 50 km in length. At what bandwidth would propagation delay (at a speed of 2 × 108m/s) equal transmit delay for 100-byte packets? What about 512-byte packets?

Solution

To determine the bandwidth at which the propagation delay equals the transmission delay, we first need to calculate the propagation delay and the transmission delay separately.

Step 1: Calculate the Propagation Delay

The propagation delay is the time it takes for a signal to travel from the source to the destination. It can be calculated using the formula:

Propagation Delay=DistancePropagation Speed\text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}}

Given:

  • Distance = 50 km = 50,000 meters
  • Propagation Speed = 2×1082 \times 10^8 meters/second

Propagation Delay=50,000 meters2×108 meters/second=0.00025 seconds=250 microseconds\text{Propagation Delay} = \frac{50,000 \text{ meters}}{2 \times 10^8 \text{ meters/second}} = 0.00025 \text{ seconds} = 250 \text{ microseconds}

Step 2: Calculate the Transmission Delay

The transmission delay is the time it takes to push all the packet's bits onto the wire. It can be calculated using the formula:

Transmission Delay=Packet Size (in bits)Bandwidth\text{Transmission Delay} = \frac{\text{Packet Size (in bits)}}{\text{Bandwidth}}

Let's determine the bandwidth for which the transmission delay equals the propagation delay.

Step 3: Find Bandwidth for 100-byte Packets

First, convert the packet size from bytes to bits:

Packet Size=100 bytes=100×8=800 bits\text{Packet Size} = 100 \text{ bytes} = 100 \times 8 = 800 \text{ bits}

Set the transmission delay equal to the propagation delay:

800 bitsBandwidth=0.00025 seconds\frac{800 \text{ bits}}{\text{Bandwidth}} = 0.00025 \text{ seconds}

Solve for bandwidth:

Bandwidth=800 bits0.00025 seconds=3,200,000 bits/second=3.2 Mbps\text{Bandwidth} = \frac{800 \text{ bits}}{0.00025 \text{ seconds}} = 3,200,000 \text{ bits/second} = 3.2 \text{ Mbps}

Step 4: Find Bandwidth for 512-byte Packets

Similarly, for 512-byte packets:

Packet Size=512 bytes=512×8=4096 bits\text{Packet Size} = 512 \text{ bytes} = 512 \times 8 = 4096 \text{ bits}

Set the transmission delay equal to the propagation delay:

4096 bitsBandwidth=0.00025 seconds\frac{4096 \text{ bits}}{\text{Bandwidth}} = 0.00025 \text{ seconds}

Solve for bandwidth:

Bandwidth=4096 bits0.00025 seconds=16,384,000 bits/second=16.384 Mbps\text{Bandwidth} = \frac{4096 \text{ bits}}{0.00025 \text{ seconds}} = 16,384,000 \text{ bits/second} = 16.384 \text{ Mbps}

Summary of Results:

  • For 100-byte packets: Bandwidth = 3.2 Mbps
  • For 512-byte packets: Bandwidth = 16.384 Mbps

Would you like more details or have any questions?

Related Questions:

  1. How would the results change if the propagation speed was different, say 3×1083 \times 10^8 m/s?
  2. What is the impact of different distances on the required bandwidth for equal propagation and transmission delays?
  3. How would these calculations change if the packet size was 1500 bytes, typical for Ethernet?
  4. How does the bandwidth affect network latency in point-to-point communication?
  5. What are some real-world applications where propagation delay is critical?

Tip:

Remember that while increasing bandwidth reduces transmission delay, it does not affect the propagation delay, which depends solely on the distance and the speed of signal propagation.

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Math Problem Analysis

Mathematical Concepts

Network Communication
Propagation Delay
Transmission Delay
Bandwidth Calculation

Formulas

Propagation Delay formula
Transmission Delay formula

Theorems

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Suitable Grade Level

Advanced