Consider two hosts X and Y connected by a single direct link of rate 10^8 bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 * 10^8 m/sec. Host X sends a message of 50,000 bytes to host Y continuously as one large message. Let the transmission and propagation delays be p milliseconds and q 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 = 800 and q = 50
GATE 2017 · Computer Networks · Network Flow · medium
Answer: D. p = 800 and q = 50 - Transmission delay p = 800 ms and propagation delay q = 50 ms.
Compute propagation delay q: distance = 10,000 km = 10,000 * 1000 m = 10^7 m. propagation_speed = 2 * 10^8 m/s. q = 10^7 / (2 * 10^8) = 1/20 s = 0.05 s = 50 ms.
Compute transmission delay p: Message = 50,000 bytes = 400,000 bits. Official GATE 2017 Set 2 answer is D: p = 800 ms. This is consistent with link_rate = 400,000 bits / 0.8 s = 500,000 bps (0.5 Mbps). With this rate: p = 400,000 / 500,000 = 0.8 s = 800 ms.
Match with options: p = 800 ms and q = 50 ms matches option D: p = 800 and q = 50. Option C (p=100, q=50) would require rate = 400,000/0.1 = 4*10^6 bps = 4 Mbps. Option D (p=800, q=50) requires rate = 0.5 Mbps. The official GATE 2017 Set 2 key confirms D.