Question

The round trip propagation delay for a 100mbps ethernet having 48-bit jamming signal is 64 microsecs. What is the minimum frame size (in bytes)?

Answer 1

Answer:

2 * propagation delay  + transmission time for jamming signal <= transmission time for frame

64 μs + 48/100 μs <= transmission time for frame

64.48 μs <= l/100 μs

l >= 6448 bits >= 806 bytes.

Explanation:

Answer 2

800 Bytes

The round trip propagation delay for a 100mbps ethernet having 48-bit jamming signal is 64 microsecs

L>= 2* Tp *B

L ->Length of frame

Tp -> Propagation time = RTT/2 = 64 * 10- 6/2 = 32 * 10-6

B -> Bandwidth = 108 bps

So, L>= 2 * 32 * 10-6 * 108

    L>= 6400 b

Converting in Bytes

=6400/8

= 800 Bytes

• Round-trip delay (RTD) or round-trip time (RTT) in telecommunications refers to the time it takes for a signal to be sent plus the time it takes for the signal to be acknowledged as having been received.
• The propagation periods for the pathways between the two communication endpoints are included in this time delay.
• The amount of time it takes latency a request to propagate to its destination. By using the straightforward formula RTT = 2 x Propagation delay, you may obtain a decent estimate of the propagation delay, which is typically the dominating component in RTT.

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