COMPUTER NETWORKS (MCA) – QUIZ 1

August 10, 2013


  1. What is MTU?

    (Ans: Maximum Transfer (or Transmission) Unit (MTU) is the maximum size of the frame)

  2. Name two services provided by the Network layer.

    (Ans: Unique Addressing in the world, Routing, i.e., finding the path and Forwarding, i.e., transferring data on the discovered path)

  3. Name two services provided by the Data Link layer.

    (Ans: Framing, Error Control and sometimes Flow Control)

  4. How is transport layer different from network layer?

    (Ans: Transport layer provides end-to-end service whereas network layer provides host-to-host service, i.e., transport layer allows us to reach the specific process within the destination host from a specific process in the source host)

  5. What is multiplexing and demultiplexing?

    (Ans: Multiplexing is the combination of multiple data streams onto the same channel and Demultiplexing is the separation of the data streams on the receiver side)

  6. What is the advantage of the layering principle?

    (Ans: Layering leads to separation of functionality, removes redundancy of implementing the same functionality in multiple applications/layers, leads to less complexity and therefore better reliability of software)

  7. If the generator polynomial in a CRC is x^16+x^12+x^3+x+1, what is the size of the CRC?

    (Ans: 16 bits or 2 bytes)

  8. If the generator polynomial in a CRC is x^8+x^5+x^3+x^2+x+1,what is the generator?

    (Ans: 100101111)

  9. When a channel is unreliable, which would you choose – error detection or correction? Why?

    (Ans: We choose error correction because it is not guaranteed that retransmissions will be transmitted successfully either. Overall, the bandwidth utilization is better with correction. This does not mean that detection with retransmission is not followed by some schemes. An example of this method is WiFi transmissions)

  10. What is the primary disadvantage of character count method of framing?

    (Ans: If the character count field is in error, in the worst case, all subsequent frames will be lost)

  11. What is the advantage of byte stuffing over character count?

    (Ans: Byte stuffing leads to better recovery from errors as compared to character count)

  12. If the data to be transmitted is as follows: 120, D1, 400, D2, 950, D3 and the MTU of the channel is 512B, how many frames are transmitted, assuming that only one byte is used for character count and you can pack more than one data item in a single frame? The data is also allowed to be split across multiple frames.

    (Ans: The first data uses 121B, second 401B and the third 951B. This comes to a total of 1473B. This takes overall 3 frames when the MTU is 512B)

  13. If the capacity of a channel is 1024KB/sec and the MTU is 512B, what is the minimum number of frames that can be transmitted in a second?

    (Ans: (1024*1024)/512 = 2048 frames)


  1. What is the purpose of the presentation layer in the OSI stack?

    (Ans: To take care of syntax and semantics of data – to have a common interpretation of data representation that can be different in different operating systems/processors etc.)

  2. Give an example of message switching.

    (Ans: Telegram)

  3. What is the advantage of packet switching over circuit switching?

    (Ans: Packet switching leads to a better utilization of bandwidth/resources because resources are not dedicated to a single connection as in circuit switching. The resources are shared amongst many different connections)

  4. A code consists of the following valid codewords: 0000, 1100, 0011, 1111. How many errors can be detected and corrected in this code?

    (Ans: The hamming distance of this code d = 2. To detect x errors, the hamming distance must be x+1 and to correct it must be 2x+1. Thus, only single-bit errors can be detected)

  5. If m = 5, what is the number of check bits needed for the Hamming code?

    (Ans: Using the formula for Hamming code check bits, r = 4)

  6. Given that * is used for byte stuffing. Which framing scheme has more overhead of transmitted data if the given data string is “********” - character count or byte stuffing?

    (Ans: We need an escape character for every data byte in byte stuffing for this data; thus, the total size of data transmitted = 17. With character count, it is 10 (2 + 8). Therefore, overhead is higher for byte stuffing.

  7. Is CRC capable of detecting all errors?

    (Ans: No. If the received polynomial is exactly divisible by the generator polynomial, no error is detected. Hence, the generator polynomial is usually selected to minimize this probability)

  8. What is encapsulation and decapsulation?

    (Ans: Encapsulation is the process of creating an envelope around the given data and decapsulation is the process of removing this envelope – in other words, adding a new header at each layer is encapsulation on transmitter and removing them is decapsulation on receiver)

  9. How many errors can a single parity bit correct?

    (Ans: None. It can detect single-bit errors)

  10. A code consists of the following codewords: 001100, 110011, 111000, 000111. What is the hamming distance of this code?

    (Ans: 3)

  11. Give an example of a system that uses FDM.

    (Ans: TV, Radio)

  12. What is TDM?

    (Ans: Time Division Multiplexing is the concept of dividing time into slots, where each slot is used by a different node to transmit using the same frequency. This is done when the frequencies transmitted do not change but all nodes do not transmit all the time)