PREVIOUS YEAR SOLVED PAPERS - August 2016 Paper 3
- Option : A
- Explanation :
- High speed Ethernet works on optic fiber Correct
- A point to point protocol over Ethernet is a network protocol for encapsulating PPP frames inside Ethernet frames not for encapsulating Ethernet frames inside PPP frames.Correct.
So, option (A) is correct.
- Option : A
- Explanation : Given frame for transmission is 1101011011 and generator polynomial is G(x) = x4 + x + 1 i.e. 10011. We have to append 4 0's (1 0's less then divisor according to CRC):
We have to append 1110 to the frame. Now our frame for transmission is 11010110111110. So, option (A) is correct.
- Option : C
- Explanation : For pure aloha Throughput(S) = Ge-2G. For Slotted aloha Throughput(S) = Ge-G. (it is maximum at G = 1). i.e. S = 1 * e-1. S = 1 / e. S = 1 / 2.71 S = 0.368. So, option (C) is correct.
- Option : A
- Explanation : Analog signal has a bit rate of 8000 bps and a baud rate of 1000. So, each signal will clearly carry bit rate / baud rate bits. i.e. 8000 / 1000 = 8 bits and 28Â = 256 signal. So, option (A) is correct.
- Option : B
- Explanation : RSA algorithm using e = 3, d = 7 and n = 33 and text message BAHI(using the values 00 to 25 → A(00), B(01), C(02),............). for B → [m = cd mod n] i.e. m = 17 mod 33 m = 1 mod 33. m = 1 → B A → 07 mod 33 m = 0 mod 33. m = 0 → A m = 77 mod 33 m = 13 → N m = 87 mod 33 m = 17 → R Message will be BANR (using the values 00 to 25 → A(00), B(01), C(02),............). No option is matching. But if we use the values 01 to 26 → A(01), B(02), C(03)....... Then B → [m = cd mod n] i.e. m = 27 mod 33 m = 8 mod 33. m = 8 → H A → 17 mod 33 m = 1 mod 33. m = 1 → A m = 87 mod 33 m = 17 → Q m = 97 mod 33 m = 3 → C Message will be HAQC. So, option (B) is correct.
