Skip to content

GATE CS 2021 Set 2 Q34 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

Explain with AI

Consider the cyclic redundancy check (CRC)\text{(CRC)} based error detecting scheme having the generator polynomial X3+X+1X^3+X+1. Suppose the message m4m3m2m1m0=11000m_4m_3m_2m_1m_0=11000 is to be transmitted. Check bits c2c1c0c_2c_1c_0 are appended at the end of the message by the transmitter using the above CRC\text{CRC} scheme. The transmitted bit string is denoted by m4m3m2m1m0c2c1c0m_4m_3m_2m_1m_0c_2c_1c_0. The value of the checkbit sequence c2c1c0c_2c_1c_0 is

Choose one option
View worked solution

Answer: C

To find the CRC check bits c2c1c0c_2c_1c_0, we perform modulo-2 division of the appended message by the generator polynomial.

Step 1: Represent the Generator Polynomial in Binary

The generator polynomial is given by: G(X)=X3+X+1=1⋅X3+0⋅X2+1⋅X1+1⋅X0G(X) = X^3 + X + 1 = 1\cdot X^3 + 0\cdot X^2 + 1\cdot X^1 + 1\cdot X^0 In binary, this corresponds to the divisor: 10111011 (degree n=3n = 3).

Step 2: Append Zeros to the Message

The message is M=11000M = 11000. Since the degree of the generator polynomial is 33, we append 33 zeros to the message: Appended Message=11000000\text{Appended Message} = 11000000

Step 3: Perform Modulo-2 Division

We divide 1100000011000000 by 10111011 using modulo-2 division (XOR operation):

      111100  (Quotient)
    -----------
1011 | 11000000
       1011
       ----
        1110
        1011
        ----
         1010
         1011
         ----
          0010
          0000
          ----
           0100
           0000
           ----
            100  (Remainder)

The remainder of the division is 100100.

Conclusion

The check bits c2c1c0c_2c_1c_0 are the remainder, which is 100100.

Thus, the correct option is C.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging questions