Math Problem Statement

What is the maximum sending rate in Stop-and-wait technique if 1.5 Mbps link with 45ms round-trip time and frame size is 1KB

Solution

To calculate the maximum sending rate in the Stop-and-Wait ARQ technique, we use the following steps:


Formulas and Steps:

  1. Transmission Time (T_tx): Ttx=Frame Size (bits)Link Speed (bps)T_{\text{tx}} = \frac{\text{Frame Size (bits)}}{\text{Link Speed (bps)}}

  2. Round Trip Time (RTT): Given directly in the problem.

  3. Maximum Sending Rate: The Stop-and-Wait protocol allows only one frame to be sent at a time, so the maximum sending rate is: Max Sending Rate=Frame Size (bits)RTT+Ttx\text{Max Sending Rate} = \frac{\text{Frame Size (bits)}}{\text{RTT} + T_{\text{tx}}}


Given Data:

  • Link speed (RR) = 1.5Mbps=1.5×106bps1.5 \, \text{Mbps} = 1.5 \times 10^6 \, \text{bps}
  • Frame size (LL) = 1KB=1KB×8bits/byte=8192bits1 \, \text{KB} = 1 \, \text{KB} \times 8 \, \text{bits/byte} = 8192 \, \text{bits}
  • RTT = 45ms=0.045s45 \, \text{ms} = 0.045 \, \text{s}

Step-by-Step Calculation:

Step 1: Calculate Transmission Time (TtxT_{\text{tx}}):

Ttx=Frame Size (bits)Link Speed (bps)=81921.5×106sT_{\text{tx}} = \frac{\text{Frame Size (bits)}}{\text{Link Speed (bps)}} = \frac{8192}{1.5 \times 10^6} \, \text{s} Ttx=0.00546sT_{\text{tx}} = 0.00546 \, \text{s}


Step 2: Compute Maximum Sending Rate:

Max Sending Rate=LRTT+Ttx\text{Max Sending Rate} = \frac{L}{RTT + T_{\text{tx}}} Max Sending Rate=81920.045+0.00546bits/sec\text{Max Sending Rate} = \frac{8192}{0.045 + 0.00546} \, \text{bits/sec} Max Sending Rate=81920.05046bits/sec\text{Max Sending Rate} = \frac{8192}{0.05046} \, \text{bits/sec} Max Sending Rate162,387.35bps\text{Max Sending Rate} \approx 162,387.35 \, \text{bps}


Final Answer:

The maximum sending rate using the Stop-and-Wait technique is approximately: 162.39kbps\boxed{162.39 \, \text{kbps}}


Let me know if you need further clarifications! Below are some related questions for your practice:


Related Questions:

  1. How does the Stop-and-Wait protocol handle errors during transmission?
  2. What happens to efficiency as the frame size increases in Stop-and-Wait ARQ?
  3. How does RTT affect the overall throughput in Stop-and-Wait ARQ?
  4. What is the impact of using a higher link speed on the maximum sending rate?
  5. What are some real-world applications of the Stop-and-Wait protocol?

Tip:

For high-speed networks with significant RTT, Stop-and-Wait ARQ becomes inefficient. To improve performance, consider protocols like Sliding Window ARQ or Selective Repeat ARQ

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Data Transmission
Stop-and-Wait Protocol
Network Throughput

Formulas

Transmission Time: T_tx = Frame Size (bits) / Link Speed (bps)
Maximum Sending Rate: Max Rate = Frame Size (bits) / (RTT + T_tx)

Theorems

-

Suitable Grade Level

Undergraduate (Computer Science / Networking)