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Answer 1

Given CSMA/CD:

$\frac{\text{Frame Size}}{\text{Data Rate}} \ge 2 \times \frac{\text{Segment Length}}{\text{Propagation Speed}}$

Drop the constants

• Propagation speed
• Frame size
• The coefficient 2

 

The Inequality Boils down to:

$L \propto \frac{1}{R}$

 

R (data rate) increases 10 fold from 1 to 10 Gbps

Then you expect L (segment length) to decrease 10 fold

 

L' = L/10

L' = 200/10 = 20 metres

 

Hence, Option A is Correct

 

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Disclaimer:

Although these kinds of proportionalities are easy to setup and solve, they are double-edged swords. So be careful while selecting your constants. With that said, it would probably be a good idea to refer to the official, completely solved answer by the OP. 

 

 

Answer 2

For CSMA/CD:

$$
\frac{\text { Frame Size }}{\text { Data Rate }} \geq 2 \times \frac{\text { Segment Length }}{\text { Propagation Speed }}
$$

Let:

• $F=$ frame size (bits)
   
• $R=$ data rate (bps)
   
• $L=$ segment length (m)
   
• $v=$ propagation speed (m/s)

So:
$$
\begin{aligned}
& \frac{F}{R} \geq 2 \times \frac{L}{v} \\\\
& L \leq \frac{F \times v}{2 \times R}
\end{aligned}
$$

$F=512 \text { bytes }=512 \times 8=4096 \text { bits }$

$R_1=1 \mathrm{Gbps}=10^9 \mathrm{bps}$

$L_1=200 \mathrm{~m}$

\begin{aligned}
& \qquad 200=\frac{4096 \times v}{2 \times 10^9} \\\\
& 200 \times 2 \times 10^9=4096 \times v \\\\
& 4 \times 10^{11}=4096 \times v \\\\
& v=\frac{4 \times 10^{11}}{4096} \\\\
& v=\frac{4 \times 10^{11}}{4.096 \times 10^3} \\\\
& v \approx 9.765625 \times 10^7 \mathrm{~m} / \mathrm{s}
\end{aligned}

$F \text { same }=4096 \text { bits }$

$R_2=10 \mathrm{Gbps}=10^{10} \mathrm{bps}$

$v \text { same }=9.765625 \times 10^7 \mathrm{~m} / \mathrm{s}$

\begin{aligned}
& \qquad L_2 \leq \frac{4096 \times 9.765625 \times 10^7}{2 \times 10^{10}} \\\\
& \qquad L_2 \leq 20 \mathrm{~m}
\end{aligned}