GATE CS 2022 Set 1 — Question 59
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Computer Networks → Network Fundamentals
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Question
Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3x10⁸ m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is __________.
Correct answer
7.07 to 7.09
Solution
The total time for the receiver to completely receive the packet is the sum of the transmission time () and the propagation time ().Given data:
This is the time required to push all the bits of the packet onto the link.
seconds
To convert to milliseconds, multiply by 1000:
ms2. Calculate Propagation Time ():
This is the time it takes for the first bit to travel from the sender to the receiver.
seconds
To convert to milliseconds, multiply by 1000:
ms3. Calculate Total Time:
The time for the receiver to completely receive the packet is the time from when the first bit is sent until the last bit is received. This is the sum of transmission time and propagation time.
Total Time =
Total Time = msThe question asks to round off to two decimal places, which gives 7.08.
- Bandwidth (B) = 100 Mbps = bits per second
- Packet Size (L) = 1000 bytes = bits
- Distance (d) = 2100 km = m = m
- Propagation Speed (v) = m/s
This is the time required to push all the bits of the packet onto the link.
seconds
To convert to milliseconds, multiply by 1000:
ms2. Calculate Propagation Time ():
This is the time it takes for the first bit to travel from the sender to the receiver.
seconds
To convert to milliseconds, multiply by 1000:
ms3. Calculate Total Time:
The time for the receiver to completely receive the packet is the time from when the first bit is sent until the last bit is received. This is the sum of transmission time and propagation time.
Total Time =
Total Time = msThe question asks to round off to two decimal places, which gives 7.08.
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