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The document discusses Shannon's Capacity Formula, demonstrating how to calculate channel capacity based on signal-to-noise ratio (SNR) for two different SNR values, resulting in maximum data rates of approximately 15.96 Mbps and 19.94 Mbps. It also covers the Sliding Window Protocol performance, calculating link utilization for a window size of 4 frames at about 29.6%, and determining an optimal window size of 14 frames for maximum utilization of 100%. Overall, it provides insights into data transmission efficiency and capacity calculations.

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5 views4 pages

Quize

The document discusses Shannon's Capacity Formula, demonstrating how to calculate channel capacity based on signal-to-noise ratio (SNR) for two different SNR values, resulting in maximum data rates of approximately 15.96 Mbps and 19.94 Mbps. It also covers the Sliding Window Protocol performance, calculating link utilization for a window size of 4 frames at about 29.6%, and determining an optimal window size of 14 frames for maximum utilization of 100%. Overall, it provides insights into data transmission efficiency and capacity calculations.

Uploaded by

jaosman2023
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Problem 1:

Shannon's Capacity Formula

The formula is:

$$ C = B \cdot \log_2(1 + \text{SNR}) $$

Where:

• ( C ) is the channel capacity in bits per second (bps)

• ( B ) is the bandwidth in Hz

• ( \text{SNR} ) is the signal-to-noise ratio (expressed as a unitless ratio, not in


decibels)

1 — Given SNR = 24 dB

1. Convert SNR from dB to ratio:

$$ \text{SNR} = 10^{\frac{24}{10}} = 251.19 $$

2. Apply Shannon's formula:

$$ C = 2 \times 10^6 \cdot \log_2(1 + 251.19) $$

Use a calculator for the log:

• ( \log_2(1 + 251.19) \approx \log_2(252.19) \approx 7.98 )

So:

$$ C \approx 2 \times 10^6 \cdot 7.98 \approx 15.96 , \text{Mbps} $$

Maximum data rate ≈ 15.96 Mbps

2 — SNR increases to 30 dB

1. Convert SNR to ratio:

$$ \text{SNR} = 10^{\frac{30}{10}} = 1000 $$

2. Apply the formula:

$$ C = 2 \times 10^6 \cdot \log_2(1 + 1000) $$


• ( \log_2(1001) \approx 9.97 )

So:

$$ C \approx 2 \times 10^6 \cdot 9.97 \approx 19.94 , \text{Mbps} $$

Maximum data rate ≈ 19.94 Mbps

Summary Table

SNR (dB) SNR (Ratio) Max Data Rate

24 dB 251.19 ~15.96 Mbps

30 dB 1000 ~19.94 Mbps


Problem 2:

Problem 2: Sliding Window Protocol Performance

Given:

Parameter Value

Window size 4 frames

Frame size 1000 bytes = 8000 bits

Bandwidth 1 Mbps

RTT 100 ms = 0.1 sec

Part 1: Link Utilization

To calculate link utilization (U), we use:

$$ U = \frac{W \cdot T_f}{RTT + T_f} $$

Where:

• ( T_f ) = transmission time per frame = ( \frac{\text{Frame size}}{\text{Bandwidth}} )

So:

$$ T_f = \frac{8000}{10^6} = 0.008 \text{ seconds} $$

Then:

$$ U = \frac{4 \cdot 0.008}{0.1 + 0.008} = \frac{0.032}{0.108} \approx 0.296 $$

Utilization ≈ 29.6%

Part 2: Window Size for Maximum Utilization

To maximize utilization, set:

$$ U = 1 = \frac{W \cdot T_f}{RTT + T_f} $$


Solving for ( W ):

$$ W = \frac{RTT + T_f}{T_f} = \frac{0.108}{0.008} = 13.5 $$

Since we can't send a fraction of a frame, round up:

Optimal Window Size = 14

Summary

Window Size Utilization

4 ~29.6%

14 ~100%

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