To solve these problems, let's break down the scenarios one by one.

Common Parameters:

• File size = 1000 KB
• RTT (Round-Trip Time) = 50 ms = 0.05 seconds
• Packet size = 1 KB
• Initial handshaking time = 2 × RTT = 100 ms = 0.1 seconds

Part (a): Bandwidth = 1.5 Mbps, Data Packets Can Be Sent Continuously

1. Calculate the time to send each packet:

   $\text{Time to send 1 KB} = \frac{\text{Packet size} \times 8}{\text{Bandwidth}} = \frac{1 \text{ KB} \times 8 \text{ bits/byte}}{1.5 \text{ Mbps}} = \frac{8 \text{ Kbits}}{1.5 \text{ Mbps}} \approx 5.33 \text{ ms}$

2. Total data transmission time:

   $\text{Total time for data transmission} = \text{Number of packets} \times \text{Time to send 1 packet} = 1000 \times 5.33 \text{ ms} = 5333.33 \text{ ms} \approx 5.33 \text{ s}$

3. Total time including handshaking:

   $\text{Total time} = 5.33 \text{ s} + 0.1 \text{ s} = 5.43 \text{ s}$

   Answer: The total time is approximately 5.43 seconds.

Part (b): Bandwidth = 1.5 Mbps, After Each Packet We Must Wait One RTT

1. Time to send each packet (same as part a): 5.33 ms

2. Total time for each packet including RTT:

   $\text{Total time per packet} = 5.33 \text{ ms} + 50 \text{ ms} = 55.33 \text{ ms}$

3. Total time for all packets:

   $\text{Total time} = \text{Handshaking} + 1000 \times 55.33 \text{ ms} = 0.1 \text{ s} + 55.33 \text{ s} = 55.43 \text{ s}$

   Answer: The total time is approximately 55.43 seconds.

Part (c): Bandwidth = Infinite, Up to 20 Packets Can Be Sent per RTT

1. Number of RTTs required to send all packets:

   $\text{Number of RTTs} = \frac{1000 \text{ packets}}{20 \text{ packets/RTT}} = 50 \text{ RTTs}$

2. Total time:

   $\text{Total time} = \text{Handshaking} + 50 \times \text{RTT} = 0.1 \text{ s} + 50 \times 0.05 \text{ s} = 0.1 \text{ s} + 2.5 \text{ s} = 2.6 \text{ s}$

   Answer: The total time is approximately 2.6 seconds.

Part (d): Bandwidth = Infinite, Exponential Increase in Packet Transmission

Here, the number of packets sent doubles each RTT.

• RTT 1: 1 packet (1 = 2^1 - 1)
• RTT 2: 2 packets (3 = 2^2 - 1)
• RTT 3: 4 packets (7 = 2^3 - 1)
• RTT 4: 8 packets (15 = 2^4 - 1)
• RTT 5: 16 packets (31 = 2^5 - 1)
• RTT 6: 32 packets (63 = 2^6 - 1)
• RTT 7: 64 packets (127 = 2^7 - 1)
• RTT 8: 128 packets (255 = 2^8 - 1)
• RTT 9: 256 packets (511 = 2^9 - 1)
• RTT 10: 489 packets (1000 - 511)

Total time:

$\text{Total time} = 10 \times \text{RTT} + \text{Handshaking} = 10 \times 0.05 \text{ s} + 0.1 \text{ s} = 0.5 \text{ s} + 0.1 \text{ s} = 0.6 \text{ s}$

Answer: The total time is approximately 0.6 seconds.

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Would you like more details on any of these scenarios?

Here are five related questions to further explore:

1. How would the total time change if the RTT were doubled in each case?
2. What would happen to the time if the packet size were increased to 10 KB?
3. How would the introduction of an additional RTT in handshaking affect the total times?
4. What if there were a fixed delay of 5 ms after each packet transmission in case (a)?
5. How would different packet loss rates affect the transmission time in each scenario?

Tip: Understanding how RTT and bandwidth interact can be crucial in optimizing data transfer protocols.