Applied Course | Mock GATE | Test 1 | Question: 32

Question

Alice and Bob are staying in Guntur and Amaravathi, and the distance between them is $30$ kms.  Suppose Alice want to send an image of size $4KB$ to Bob and connected using a LAN and link works at a speed of light in fiber. Then what data rate does the round-trip delay equal to the transmission delay of the packet?

• A. $32768 \times 10^4 \text{bits/sec}$
• B. $16384 \times 10^4 \text{bits/sec}$
• C. $8192 \times 10^4 \text{bits/sec}$
• D. None of these

Answer 1

The speed of light in fiber is $3 \times 10^8 \: m/sec$
For $25$ kms distance the propagation delay is $10^{-4} \: sec$
$\begin{array}{ll} \text{Round Trip time} & = 2 \times \text{propagation delay} \\ & = 2 \times 10^{-4} \: sec \end{array}$
Given
$\text{Image size }= 4KB = 32768  \text{ bits}$
Transmission delay = Round trip time
$\begin{array}{lll} \Rightarrow & \frac{L}{B}  & = 2 \times 10^{-4} \text{ sec} \\ \Rightarrow & B  & = L/(2 \times 10^{-4}) \text{bits/sec} \\ \Rightarrow & B  & = 32768 /(2  \times 10^{-4}) \text{ bits/sec} \\ \Rightarrow & B & = 16384 \times 10^4 \text{ bits/sec} \end{array}$

Comments

I think the distance should be 30Km instead of 25Km...calculations are correct ..am I missing something?

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Round trip time is transmission delay plus two times propagation delay

Why have you considered only $2*T_p$?

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Transmission delay plus two times propagation delay is the total delay, not just the round trip...that is why while calculating efficiency we take the whole term in the denominator and useful time is only transmission time. So it is kept in the numerator.

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@amanverma==

You're wrong. Context is everything. RTT = $T_t+T_p+T_{t'}+T_p$

But when the distance between the machines is large enough, $T_t$ becomes negligible compared to $T_p$, and then we approximate RTT = $T_p+T_p$