4. Calculate the total time required to transfer a 1.5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent: (a) The bandwidth is 10 Mbps, and data packets can be sent continuously. (b) The bandwidth is 10 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next. (c) The link allows infinitely fast transmit, but limits bandwidth such that only 20 packets can be sent per RTT. (d) Zero transmit time as in (c), but during the first RTT we can send one packet, during the second RTT we can send two packets, during the third we can send four (2 3−1 ), etc. (A justification for such an exponential increase will be given in Chapter 6.)

To solve this problem, we'll calculate the total time required to transfer a 1.5-MB file under the given conditions. Let's first break down the problem:

- **File Size**: 1.5 MB = \(1.5 \times 1024 \times 1024\) bytes = 1,572,864 bytes.
- **RTT (Round-Trip Time)**: 80 ms = 0.08 seconds.
- **Packet Size**: 1 KB = 1024 bytes.
- **Number of Packets**: Total packets required = \( \frac{1,572,864 \text{ bytes}}{1024 \text{ bytes/packet}} \) = 1536 packets.

### (a) Bandwidth is 10 Mbps, and data packets can be sent continuously.

- **Bandwidth**: 10 Mbps = \(10 \times 10^6\) bits per second.
- **Packet Size**: 1 KB = 1024 bytes = 8192 bits.

**Time to send one packet**:
\[ \text{Transmission time per packet} = \frac{\text{Packet Size}}{\text{Bandwidth}} = \frac{8192 \text{ bits}}{10 \times 10^6 \text{ bits/sec}} = 0.0008192 \text{ seconds} \]

**Total transmission time for 1536 packets**:
\[ \text{Total Transmission Time} = 1536 \times 0.0008192 \text{ seconds} = 1.25952 \text{ seconds} \]

**Total time required**:
\[ \text{Total Time} = 2 \times \text{RTT} + \text{Total Transmission Time} = 2 \times 0.08 \text{ seconds} + 1.25952 \text{ seconds} = 1.41952 \text{ seconds} \]

### (b) Bandwidth is 10 Mbps, but after we finish sending each data packet, we must wait one RTT before sending the next.

**Time to send one packet and wait for an RTT**:
\[ \text{Time per packet} = \text{Transmission time} + \text{RTT} = 0.0008192 \text{ seconds} + 0.08 \text{ seconds} = 0.0808192 \text{ seconds} \]

**Total time required**:
\[ \text{Total Time} = 2 \times \text{RTT} + 1536 \times 0.0808192 \text{ seconds} = 0.16 \text{ seconds} + 124.456512 \text{ seconds} = 124.616512 \text{ seconds} \]

### (c) Infinitely fast transmit, but only 20 packets can be sent per RTT.

- **Number of RTTs needed**:
\[ \text{Number of RTTs} = \left\lceil \frac{1536 \text{ packets}}{20 \text{ packets/RTT}} \right\rceil = \left\lceil 76.8 \right\rceil = 77 \text{ RTTs} \]

**Total time required**:
\[ \text{Total Time} = 2 \times \text{RTT} + 77 \times 0.08 \text{ seconds} = 0.16 \text{ seconds} + 6.16 \text{ seconds} = 6.32 \text{ seconds} \]

### (d) Zero transmit time, but exponential increase in packets per RTT.

- **Packets sent in each RTT**:
  - RTT 1: 1 packet
  - RTT 2: 2 packets
  - RTT 3: 4 packets
  - ...
  - \( \text{RTT } n: 2^{n-1} \text{ packets} \)

**Sum of packets up to RTT \( n \)**:
\[ \text{Total packets sent by RTT } n = 1 + 2 + 4 + ... + 2^{n-1} = 2^n - 1 \]

**Finding the smallest \( n \) such that \( 2^n - 1 \geq 1536 \)**:
\[ 2^n - 1 \geq 1536 \Rightarrow 2^n \geq 1537 \Rightarrow n \geq \log_2{1537} \approx 10.56 \]

So \( n = 11 \) RTTs.

**Total time required**:
\[ \text{Total Time} = 2 \times \text{RTT} + 11 \times 0.08 \text{ seconds} = 0.16 \text{ seconds} + 0.88 \text{ seconds} = 1.04 \text{ seconds} \]

### Summary of Results

- **(a)**: 1.41952 seconds
- **(b)**: 124.616512 seconds
- **(c)**: 6.32 seconds
- **(d)**: 1.04 seconds

Would you like more details or have any questions? Here are some additional questions to consider:

1. How does increasing RTT affect the overall transmission time in scenario (a)?
2. What would happen to the transmission time if the packet size was doubled in scenario (b)?
3. How does reducing the number of packets per RTT in scenario (c) affect the total time?
4. Can you derive a general formula for transmission time based on the number of RTTs and bandwidth?
5. What are some practical scenarios where the exponential packet increase in scenario (d) could be used?

**Tip**: In network transmissions, minimizing RTTs or optimizing the packet size relative to bandwidth can significantly reduce total transmission time.

Explore how to calculate the total time required to transfer a 1.5-MB file under different network conditions: continuous transmission, delayed transmission, limited packets per RTT, and exponential packet increase. Understand the impact of bandwidth, RTT, and packet size on transmission time.

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