1. The FTP client sends data to the FTP server, detailing its working process. The connection between the two machines is illustrated below:
The detailed explanation is as follows:
1.1. Assume that the initial settings are as follows: The client's FTP port number is 32768, and the server's FTP port number is 21.
1.2. The communication process between two computers on different network segments through the TCP/IP protocol is as follows: the protocol is horizontal, while the service is vertical.
At the physical layer, data is transmitted via electrical signals, ensuring a transparent bit stream. At the link layer, data is sent in frames, ensuring error-free transmission between adjacent nodes. The network layer handles packet switching, providing communication across different hosts. The transport layer manages end-to-end communication between processes, using segments as the unit of data transfer. While the network layer focuses on point-to-point transmission (between hosts or routers), the transport layer ensures reliable delivery from the source to the destination host.
1.3. Data Packet Encapsulation Process
Different layers of the protocol stack use different terms for data packets: segments at the transport layer, datagrams at the network layer, and frames at the link layer. After encapsulation into frames, the data is sent over the transmission medium. Upon reaching the destination host, each layer strips the corresponding header, and the application layer finally receives the data. When two computers are on different network segments, data passes through one or more routers.
1.4. Working Process
(1) On PC1, the original data is encapsulated into a frame and sent to Switch 1’s port 1 via a physical link. The resulting frame is as follows:
Note: How does the sender determine if the destination is on the same network segment? Each IP address has a network prefix. The sender compares the destination IP's network prefix with its own. If they match, the data can be sent directly. In this case, PC1 and PC2 are on different network segments.
(2) After receiving the data, Switch 1 checks the destination MAC address and forwards it to the router's S0 port based on its MAC address table.
(3) The router receives the data, analyzes the IP datagram, re-encapsulates it, and sends it out through S1 after consulting the routing table. The new data frame is shown below:
Note: The destination and source IP addresses remain unchanged, but the MAC address is updated. The router replaces the source MAC address with the next hop’s MAC address before sending.
(4) Switch 2 receives the data from the router and forwards it to PC2’s port 1 based on its MAC address table.
(5) PC2 verifies the data, decapsulates the TCP segment, identifies the destination port as 21, and delivers the data to the FTP application for processing.
Second, what problems may occur during data transmission, and how to solve them?
1. At the data link layer, data may be corrupted or lost, and there may be differences in transmission speed between both ends. How to solve these issues?
A: Assuming Host A sends data to Host B:
(1) Error control methods include Automatic Repeat Request (ARQ) and Forward Error Correction (FEC). ARQ allows the receiver to request retransmission upon detecting an error, while FEC enables the receiver to correct errors without retransmission. Common codes include parity check, cyclic redundancy check (CRC), and Hamming code. For example, CRC is added to the data frame so that Host B can detect errors. If an error is found, Host B may send a Negative Acknowledgment (NAK) to request retransmission.
(2) To handle data loss, timeout retransmission is used. If Host A does not receive an acknowledgment (ACK) within the set time, it retransmits the data.
(3) Flow control: One method is stop-and-wait, where the sender waits for an acknowledgment before sending the next frame. Another is sliding window, where the sender can transmit multiple frames based on the window size. The receiver also uses a sliding window to manage incoming data.
2. If an IP datagram is 5000 bytes long and the MTU is 1500 bytes, what should be done?
A: The solution is to fragment and reassemble the IP datagram. Here's the process:
2.1. IPv4 Datagram Format: An IPv4 datagram consists of a header (20 bytes fixed) and data. The header includes control information. Optional fields vary in length.
2.2. MTU values for different data link layer protocols vary. Routers may connect to networks with different MTUs.
2.3. When an IP datagram is fragmented, each fragment gets a new header with modified fields like identifier, flag, and offset. These fields help in reassembling the fragments at the destination.
2.4. Reassembly occurs at the final destination, where all fragments are combined into the original datagram using the identifier, flag, and offset fields.
2.5. Detailed Explanation: With a 5000-byte data portion and a 20-byte header, each fragment must be ≤ 1480 bytes. This results in four fragments: 1480, 1480, 1480, and 560 bytes. The fragmentation result is shown below:
3. During the application process, errors and losses may occur, and the cache at both ends may differ. How to control traffic and match port numbers?
A: The transport layer provides logical communication between application processes. TCP guarantees full-duplex, reliable delivery. Below are details on error control and flow control:
3.1. The transport layer provides logical communication, as shown in the following figure:
3.2. The structure of a TCP segment is as follows:
3.3. TCP uses sequence numbers to track each byte of data. The acknowledgment number indicates the next expected byte. This ensures reliable delivery and order.
3.4. Flow Control and Congestion Control:
(1) Sliding Window: TCP uses a variable-size sliding window for flow control. The window size determines how much data can be sent before an acknowledgment is received. The receiver dynamically adjusts the window size based on its resources.
(2) Congestion Control: Techniques such as slow start, multiplicative decrease, and congestion avoidance are used to manage network congestion. The congestion window grows linearly to prevent overload.
3.5. Port Number Identification:
(1) A port is a transport layer service access point (TSAP). It allows applications to pass data to the transport layer and receive it back.
(2) Ports are used to identify which application process should receive the data. They are 16-bit numbers, with well-known ports (0–1023) reserved for standard services.
(3) A socket combines an IP address and a port number, forming a unique endpoint for communication. It allows the transport layer to correctly route data to the appropriate application.
(4) The relationship between sockets, ports, and IP addresses is essential for establishing and maintaining connections between hosts.
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