Project ForgeConsider two hosts X and Y, connected by a single direct link of rate 10⁶ bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2×10⁸ 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
- Ap=50 and q=100
- Bp=50 and q=400
- Cp=100 and q=50
- p=400 and q=50
Explanation
Transmission delay p = (50,000×8 bits)/(10⁶ bits/sec) = 400,000/1,000,000 s = 0.4 s = 400 ms. Propagation delay q = distance/speed = (10,000×1000 m)/(2×10⁸ m/s) = 10⁷/2×10⁸ s = 0.05 s = 50 ms. So p=400 and q=50.
Source: GATE CS 2017 (Session 2) (2017)
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