GATE CS 2017 Set 2 Q.35 (CS Q.35): OSI/TCP-IP Layering, Physical Layer and Switching

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

Then the values of $p$ and $q$ are

$p = 50$ and $q = 100$

$p = 50$ and $q = 400$

$p = 100$ and $q = 50$

$p = 400$ and $q = 50$

Official answer and solution

Official answer: D
From: Official GATE 2017 answer key

Solution

Compute transmission delay using $p = \frac{L}{R}$ and propagation delay using $q = \frac{d}{v}$ .

Calculate transmission delay $p$ :

Message length $L = 50,000 \text{ bytes} = 50,000 \times 8 \text{ bits} = 400,000 \text{ bits} = 4 \times 10^5 \text{ bits}$ .
Link transmission rate $R = 10^6 \text{ bits/sec}$ .

p = \frac{L}{R} = \frac{4 \times 10^5 \text{ bits}}{10^6 \text{ bits/sec}} = 0.4 \text{ sec} = 400 \text{ milliseconds}

Calculate propagation delay $q$ :

Distance $d = 10,000 \text{ km} = 10,000 \times 10^3 \text{ m} = 10^7 \text{ m}$ .
Propagation speed $v = 2 \times 10^8 \text{ m/sec}$ .

q = \frac{d}{v} = \frac{10^7 \text{ m}}{2 \times 10^8 \text{ m/sec}} = \frac{1}{20} \text{ sec} = 0.05 \text{ sec} = 50 \text{ milliseconds}

Thus, $p = 400$ and $q = 50$ , which matches option D.

Final answer

Common trap

Forgetting to convert bytes to bits when computing transmission delay ( $50,000 / 10^6 = 50$ ms), leading to option B.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 35 of 65, printed as CS Q.35
Section: Computer Science and Information Technology

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