Answer:
885 μs
Explanation:
Given that:
Switch via = 100 Mbps links
The propagation delay for each link = 25μs.
Retransmitting a received packet = 35μs
To determine:
The total time required to transmit 40,000 bits from A to B.
Considering the fact as a single packet:
Transmit Delay / link = size/bandwith
= 4×10⁴ bits / 100 × 10⁶ bits/sec
= 400 μs
The total transmission time = ( 2 × 400 + 2 × 25 + 35) μs
= (800 + 50 + 35) μs
= 885 μs