Assume hosts A and B are each connected to a switch Svia 100-Mbps links. The propagation delay on each link is 25μs. The switch Sis a store-and-forward device and it requires a delay of 35μs to process a packet after is has received the last bit in the packet. Calculate the total time required to transmit 40,000 bits from Ato B in the following scenarios. (The total time is measured from the start of the transmission of the first bit at A, until the last bit is received at B. We always assume that links are bi-directional with the same transmission rate and propagation delay in each direction unless specifically instructed otherwise.)

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

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