Final answer:
To calculate the total time required to transmit 10,000 bits from Host A to Host B, we sum the transmission time at Host A (100 μs), the total propagation delay (40 μs), and the retransmission delay at switch S (35 μs), resulting in a total time of 175 μs.
Step-by-step explanation:
The student's question deals with a network transfer scenario where Host A is transmitting data to Host B via a switch which uses store-and-forward transmission. The task is to calculate the total time required to transmit 10,000 bits from A to B as a single packet with given link speed, propagation delay, and retransmission delay.
First, we need to calculate the transmission time at Host A, which is the time taken to send the packet on to the network. With a link speed of 100 Mbps (megabits per second), the time to transmit 10,000 bits would be:
Transmission time = (Number of bits / Bandwidth) = 10,000 bits / 100,000,000 bits per second = 0.0001 seconds or 100 μs.
Next, we add the propagation delay, which is the time it takes for the bits to travel from Host A to the switch S, given as 20 μs. Similarly, there will be another propagation delay from switch S to Host B:
Total propagation delay = Propagation delay (A to S) + Propagation delay (S to B) = 20 μs + 20 μs = 40 μs.
Finally, we consider the retransmission delay of the switch S, which is the time the switch waits before forwarding the packet after receiving it fully. The given retransmission delay is:
Retransmission delay = 35 μs.
Adding up all the calculated times:
- Transmission time at Host A: 100 μs
- Total propagation delay: 40 μs
- Retransmission delay at switch S: 35 μs
The total time required for 10,000 bits to go from A to B is:
Total time = Transmission time + Total propagation delay + Retransmission delay = 100 μs + 40 μs + 35 μs = 175 μs.