📝 Network management introduction and areas (10 MCQs)

📖 From Data Communication and Networks • 27. Network Management • 10 questions available

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What is Network management introduction and areas?

Network management is the process of monitoring, controlling, and maintaining a computer network to ensure it operates efficiently, and it is divided into five key functional areas: configuration, fault, performance, security, and accounting management.

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📝 All Network management introduction and areas MCQs

Q1. In a TCP echo scenario, if the server sends the response in multiple segments, what must the client do to guarantee it receives the entire echoed string?

A.Terminate after the first recv call

B.Close the socket immediately

C.Repeatedly call recv until no more data is available ✅

D.Send a new request for the missing parts

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The client cannot assume a single recv will retrieve the whole message because TCP may split data into several segments. By repeatedly calling recv until the call returns zero bytes (or an error), the client ensures it collects all fragments, reconstructing the complete echoed string.

Q2. When the server increments the pointer variable ptr after each recv, which logical inference about the buffer’s state is correct?

A.The pointer now points to the start of the buffer ✅

B.The pointer points to the next free byte for incoming data

C.The pointer is reset to zero after each iteration

D.The pointer indicates the total number of bytes received

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Each recv writes data at the location indicated by ptr. After storing the received bytes, the program moves ptr forward by the number of bytes just read, so it points to the next free position in the buffer where subsequent data will be placed.

Q3. If the variable maxLen is decreased by the number of bytes received in each loop, what can be deduced about the program’s behavior when maxLen reaches zero?

A.The program will allocate a larger buffer automatically

B.The server will stop receiving further data and exit the loop

C.The client will resend the entire message

D.The pointer ptr will be reset to the buffer start ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: When maxLen becomes zero, the condition that checks for remaining space fails, causing the receive loop to terminate. This prevents writing beyond the allocated buffer, ensuring that no overflow occurs and that the program stops reading once the buffer is full.

Q4. Which analytical statement best compares using a single recv call versus a loop of recv calls for receiving an echo message?

A.A single recv is always faster than a loop

B.A loop guarantees complete reception regardless of segment size, while a single recv may miss data ✅

C.Both approaches produce identical results for any message length

D.A loop consumes more memory than a single recv

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A loop of recv calls continues reading until all bytes have been transferred, handling the case where TCP splits the message. A single recv may return only a portion of the data, leading to incomplete reception if the message exceeds the size of one segment.

Q5. How does changing the waitSize parameter from 16 to 64 most likely affect the server’s ability to handle incoming connections?

A.It reduces the maximum number of simultaneous connections

B.It increases the backlog of pending connections the server can queue

C.It decreases the time needed to accept each connection ✅

D.It has no effect on connection handling

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The waitSize value determines the length of the queue that holds connections waiting to be accepted. Raising it from 16 to 64 allows more pending connections to be stored, improving the server’s capacity to handle bursts of client requests without rejecting them immediately.

Q6. In contrast to the TCP echo client, which of the following best describes a UDP echo client’s reliability characteristics?

A.UDP guarantees ordered delivery of packets ✅

B.UDP ensures that every sent packet is received exactly once

C.UDP may lose, duplicate, or reorder packets, requiring the application to handle these issues

D.UDP automatically retransmits lost packets

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: TCP provides reliable, ordered delivery, while UDP is connectionless and does not guarantee delivery, ordering, or uniqueness. Therefore, a UDP echo client must implement its own mechanisms to detect loss, duplication, or reordering, unlike a TCP client which benefits from the protocol’s built‑in reliability.

Q7. Why must the program update the len variable after each successful recv call?

A.To keep track of how many bytes have been sent to the client

B.To indicate the total number of bytes stored in the buffer so far

C.To reset the socket descriptor for the next operation

D.To change the server’s IP address dynamically ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The len variable accumulates the number of bytes received across multiple recv calls. Updating len ensures the program knows exactly how much data has been collected, which is essential for correctly forming the response and for any subsequent processing that depends on the total byte count.

Q8. If the echo server must handle messages larger than the static 256‑byte buffer, which conceptual modification would be most effective?

A.Increase the buffer size to 1024 bytes without changing any logic

B.Replace the static array with a dynamically allocated buffer that grows as needed ✅

C.Ignore any data beyond the first 256 bytes

D.Use UDP instead of TCP to avoid segmentation

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A dynamic allocation strategy allows the server to request additional memory when the incoming message exceeds the current buffer capacity. By reallocating a larger buffer and adjusting ptr, len, and maxLen accordingly, the server can safely accommodate arbitrarily large messages while preserving the existing receive loop logic.

Q9. What is the relationship between pointer arithmetic on ptr and the risk of buffer overflow in the echo server code?

A.Pointer arithmetic automatically prevents overflow by wrapping around ✅

B.Moving ptr forward without adjusting maxLen can cause writes beyond the allocated buffer, leading to overflow

C.Pointer arithmetic has no impact on buffer safety

D.Using ptr instead of array indexing eliminates overflow risk

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When ptr is advanced by the number of bytes just received, the program must also reduce maxLen by the same amount. If ptr is moved without updating maxLen, subsequent recv calls may write past the end of the allocated array, causing a buffer overflow. Proper synchronization of pointer movement and remaining length is essential for safety.

Q10. What is the primary purpose of the listen socket descriptor ls in the echo server program?

A.To store the data received from the client

B.To hold the address of the remote client

C.To accept incoming connection requests and create new sockets for communication ✅

D.To encrypt the transmitted data

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The listen socket descriptor ls is created, bound to a local address, and set to listen for connection attempts. When a client initiates a connection, the server calls accept on ls, which returns a new socket descriptor for the actual data exchange, while ls continues listening for additional clients.

🔗 Related Topics (MCQs)

📝 Configuration management in networking MCQs 📝 Fault management in computer networks MCQs 📝 Performance management in networking MCQs 📝 Network security management MCQs 📝 Accounting management in networking MCQs

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