Flow Control and Error Control MCQ Quiz - Objective Question with Answer for Flow Control and Error Control - Download Free PDF

The correct answer is Connect.

Key Points

• The connect system call is responsible for establishing a connection to a remote host.
  • When the connect system call is executed, it initiates the process of setting up a TCP connection.
  • As part of this process, the client sends a SYN (synchronize) packet to the server.
  • The server responds with a SYN-ACK (synchronize-acknowledge) packet if it is ready to accept the connection.
  • The client then sends an ACK (acknowledge) packet, completing the three-way handshake and establishing the connection.
• The other system calls mentioned (Socket, Bind, Listen) are involved in different stages of socket programming but do not directly result in sending SYN packets.
  • Socket: Creates a new socket.
  • Bind: Associates a socket with a local address and port.
  • Listen: Prepares the socket to accept incoming connections.

Additional Information

• The three-way handshake is a fundamental process in establishing a reliable TCP connection.
• Understanding the role of each system call in network programming is essential for developing network applications.
• Proper handling of socket connections ensures the robustness and security of network communication.

The correct answer is Option 2.

Key Points

• The Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.
• To calculate the Hamming distance between the binary strings 10101 and 11110, compare each corresponding position:

• Position 1: 1 vs 1 (no difference)
• Position 2: 0 vs 1 (difference)
• Position 3: 1 vs 1 (no difference)
• Position 4: 0 vs 1 (difference)
• Position 5: 1 vs 0 (difference)

• There are 3 positions where the corresponding bits differ.

Hence, the Hamming distance between 10101 and 11110 is 3.

Additional Information

• The Hamming distance is used in error detection and correction in data transmission.
• It is named after Richard Hamming, an American mathematician and computer scientist.
• It is particularly useful in coding theory for designing error-correcting codes.
• The concept can be extended to other types of data, such as text strings and DNA sequences.

The correct answer is Option 4.

Key Points

• The sliding window protocol for flow control involves the following stages:
  • B. Sender sends a window of frames.
  • C. Receiver acknowledges the received frames.
  • A. Sender moves the window to the next set of frames.
  • D. If acknowledgment is not received, Sender retransmits the frame.
• Therefore, the correct sequence is Option 1: B, C, A,  D.

Additional Information

• In the sliding window protocol, the sender can send multiple frames before needing an acknowledgment for the first frame.
• The window size determines the number of frames the sender can send without receiving an acknowledgment.
• This protocol helps in efficient use of network bandwidth and reduces the waiting time for acknowledgments.

The correct answer is 2) 1000 bits.

Key Points

• The network uses CSMA/CD, which stands for Carrier Sense Multiple Access with Collision Detection.
• To determine the minimum frame size in a CSMA/CD network, we need to ensure that the transmission time is at least twice the maximum propagation delay. This is to detect collisions before the entire frame is transmitted.
• Given:
  • Bandwidth (B) = 20 Mbps
  • Maximum Propagation Time (T) = 25 μs
• The minimum frame size (S) can be calculated using the formula: S = 2 * B * T
• Calculating the minimum frame size:
  • S = 2 * 20 Mbps * 25 μs
  • Converting units: 1 Mbps = 10^6 bits per second and 1 μs = 10^-6 seconds
  • S = 2 * 20 * 10^6 bits/second * 25 * 10^-6 seconds
  • S = 2 * 20 * 25 bits
  • S = 1000 bits
• However, the actual minimum frame size to avoid collision detection issues is often standardized, and in this case, it is 1000 bits.

Additional Information

• CSMA/CD is used in Ethernet networks to manage the transmission of data and avoid collisions. The calculation ensures that the round-trip time for a signal to propagate across the network and back is taken into account.
• Understanding frame size calculations is crucial for network design and ensuring efficient data transmission.

The correct answer is 160 bits

Key Points

• Channel Bit Rate and Propagation Delay:
  • The channel bit rate is 4 kbps, indicating the maximum bits transmitted in one second is 4,000.
  • The one-way propagation delay is 20 ms, denoting the time for a bit to travel from sender to receiver.
• Stop and Wait Protocol:
  • In this protocol, the sender transmits a frame and awaits an acknowledgement from the receiver.
  • The sender proceeds to send the next frame only upon receiving the acknowledgement from the receiver.
• Channel Efficiency:
  • Channel efficiency is the ratio of useful data transmitted to the total time taken for transmission.
  • For a minimum efficiency of 50%, the useful data must be at least half of the total transmission time.
• Calculation of Minimum Frame Size:
  • The minimum frame size is determined using the formula:
    • Minimum Frame Size = 2 Propagation Delay Channel Bit Rate.
    • Substituting the given values, the minimum frame size is calculated as 2 20 ms 4 kbps, resulting in 160 bits.

Therefore, to achieve a channel efficiency of at least 50%, the minimum frame size should be 160 bits.

• There are many networks that exist often with different hardware and software
• It is needed to connected devices on one network with devices on another network
• To connect such incompatible networks, necessary translation in terms of hardware and software is required, the gateway does this task
• A file server is a computer attached to a network that provides a location for shared disk access that can be accessed by the workstations through a shared link
• Printer server is a device that connects printers to client computers over a network
• The Utility Server runs a number of utility applications that support or enhance the component applications facilitating a complete single box solution

B = 10⁸ bits/sec

d = 1 km

Round trip distance = 2 km

To detect a collision, the sender must be transmitting when it receives the collision signal so that it can transmit again.

\(\begin{array}{l} \therefore \frac{{1250 \times 8}}{{{{10}^8}}} = \frac{2}{{speed}}\\ \therefore speed = \frac{{2 \times {{10}^8}}}{{{{10}^4}}} = 2 \times {10^4 } \end{array}\)

The correct option is (2)

Checksum

Concept:-

The error detection method used by the higher layer protocol is called checksum. A checksum is based on the concept of redundancy.

Key Points

• Any time a message is transmitted, noise or faulty data have the potential to scramble it.
• To prevent this, they employ error-detecting codes, which are pieces of extra information that are appended to a digital message to aid in the detection of errors that may have happened during transmission.
• The data is separated into k segments, each with a bit size of m, for the checksum error detection system.
• The segments are summed using 1's complement math at the sender's end to obtain the sum. This complements the total.

Additional InformationError bit:- The number of wrong bits that can be anticipated in a certain number of bits in a serial stream is measured by the bit error rate.

A modem (modulator-demodulator) is a network hardware device that modulates one or more carrier wave signals to encode digital information for transmission and demodulates signals to decode the transmitted information. The goal is to produce a signal that can be transmitted easily and decoded to reproduce the original digital data.

The correct option is (2)

Store and forward

Key Points

• Message switching is a switching mechanism where a message is routed to intermediary nodes where it is stored and transmitted as a single unit.
• The message is broken into smaller pieces when using packet switching. Each of these discrete components of the communication, known as packets, has a unique source and destination address. Only based on these addresses is each packet sent forward through the network.
• Some computer network protocols use packet switching as a method of transferring data over a local or remote connection. Protocols used for packet switching include Frame Relay, IP, and X.
• One of the most widely utilised computer networks is a packet switching network. On both local networks and the internet, it is commonly used.
• Circuit switching, packet switching, and message switching are the three most popular switching methods.

The correct answer is Transmission speed

Key Points

• The rating of a modem, such as 56K, refers to its data transmission capacity, which is also often understood as its speed. Essentially, 56K represents the maximum rate of data transfer that the modem is technologically capable of handling under ideal conditions.
• However, a clarification point is necessary here. While at a glance this transmission capacity might look like analogous to 'speed', it's a bit more nuanced. This "56K" stands for 56 kilobits per second (Kbps), signifying the quantity of data (in kilobits) that the modem can send or receive per second. It's the rate of data transfer rather than the literal speed at which data travels (i.e., the speed of light or electrons, depending on medium).

Additional Information

• A modem (short for modulator-demodulator) is a device that converts data between its digital form (as understood by computers and digital devices) and analog form (compatible with telephone lines or cable systems).

STEP:

Append no. of 0’s equal to the degree of the generator polynomial to the original data and then divide the message with the generator polynomial. Remainder obtained from this will be added with the original message and transmitted as CRC.

Calculation:

Generator is given as x³ +x +1

Degree of the generator is 3.

So, 3 0’s is added to the data and new data will be 01011011000 and it is divided by 1011.

Diagram

Here remainder is 101. Append this to the original message. The message will become 01011011101. So, the original message 01011011 will be transmitted as 01011011101.

The correct answer is n².

Key Points

• In this flow chart value of A should be the output.
• Value of A is described in processing box that is, n*n.
• It means the output of the above flowchart is n².

Additional Information

4 Basic Flowchart Symbols for Creating a Flowchart:

• The Oval -  An End or Beginning While Creating a Flowchart. 
• The Rectangle -  It represents any step in the process you’re diagramming and is the workhorse of the flowchart diagram. 
• The Arrow - It is used to guide the viewer along their flowcharting path. 
• The Diamond - It symbolizes that a decision is required to move forward.

An Ethernet network bridge is a device which connects two different local area networks together into one extended Ethernet. For this both networks must connect using the same Ethernet protocol.

The answer to the question is 15 Kbps

• It uses the sliding window protocol for the transmission of data. 
• The question takes into consideration the variant of sliding window protocol namely GO BACK N ARQ. In this protocol, the sender can have up to N packets unacknowledged that are still remaining in the pipeline.
• The receiver only sends cumulative acknowledgments. In case of encountering an error, the sender has to resend all the data frames following the error.
   

According to the question:

• The data rate of the link is 20 Kbps and the propagation delay = 300 ms

So, the time required to transmit 500 bytes long data will be given by

• Transmission Time \( t = {Number \,of \,bits \,to\, be\, transmitted \over data\, rate\, of\, the\, link }\)

                                             = \((500~* ~8 ~bits) \over 20 Kbps\) =  200 ms

• Now, the propagation delay is given as d = 300 ms

So the efficiency of the link is given by:

Efficiency

Where N = window size

• \(E = {3 ~* ~200ms \over (200~ms~+~2~*~300ms)}\) = 0.75

So, the maximum data rate attainable = 0.75 * 20 Kbps = 15 Kbps
So, the answer will be 15Kbps.