Question 1. 200 nodes are connected to a 1,500-meter length of coaxial cable. Jsing some protocol, each node can transmit 50 frames/second, where che average frame length is 2,000 bits. each node is 100 Mbps (where 1 Mbps = 1,000,000 bps). What is the efficiency of this protocol? The transmission rate at
Question 1. 200 nodes are connected to a 1,500-meter length of coaxial cable. Jsing some protocol, each node can transmit 50 frames/second, where che average frame length is 2,000 bits. each node is 100 Mbps (where 1 Mbps = 1,000,000 bps). What is the efficiency of this protocol? The transmission rate at
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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Transcribed Image Text:### Networking Efficiency Question
**Problem Statement:**
200 nodes are connected to a 1,500-meter length of coaxial cable. Using a certain protocol, each node can transmit 50 frames per second, with each frame averaging 2,000 bits in length. The transmission rate at each node is 100 Mbps (where 1 Mbps = 1,000,000 bps). What is the efficiency of this protocol?
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To determine the efficiency of the protocol described, follow these steps:
1. **Calculate the total bits transmitted by all nodes per second:**
Each node transmits 50 frames/second, and each frame is 2,000 bits long.
So, total bits per node per second = 50 frames/second * 2,000 bits/frame = 100,000 bits/second.
2. **Calculate the total bits transmitted by all 200 nodes per second:**
Total bits per second for all nodes = 100,000 bits/second * 200 nodes = 20,000,000 bits/second = 20 Mbps.
3. **Find the transmission rate at each node:**
The given transmission rate is 100 Mbps.
4. **Calculate the efficiency of the protocol:**
Efficiency = (Total bits transmitted by all nodes per second) / (Transmission rate at each node * Number of nodes).
Substituting in the values:
Efficiency = 20 Mbps / (100 Mbps * 200) = 20 Mbps / 20,000 Mbps = 0.001 or 0.1%.
Therefore, the efficiency of the protocol is 0.1%.
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This calculation illustrates how efficiently the protocol utilizes the available bandwidth on the network.
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