GATE CSE 2017 Set 2 | Question: 35

Question

Consider two hosts $X$ and $Y$, connected by a single direct link of rate $10^6\; \text{bits/sec}$. The distance between the two hosts is $10,000\;\text{km}$ and the propagation speed along the link is $2 \times 10^8 \;\text{m/sec}$. Host $X$ sends a file of $50,000\;\text{bytes}$ as one large message to host $Y$ continuously. Let the transmission and propagation delays be $p\; \text{milliseconds}$ and $q \; \text{milliseconds}$ respectively. Then the value of $p$ and $q$ are

• A. $p=50$ and $q=100$
• B. $p=50$ and $q=400$
• C. $p=100$ and $q=50$
• D. $p=400$ and $q=50$

Comments

This question seems so trivial to solve, like marks for free, am i the only one to think like this or is there any deep concept to apply in this question?

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@UltraRadiantX
You are thinking right. This question is really easy to solve.

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@prithatiti this question is part of the easier section of the paper:)

Answer 1

Propagation time=Length/bandwidth & Transmission time=distance/speed gives option D.

Comments

I answered it as B) 50 and 400!!! such a silly mistake!