Computer Networks | Set 12

Following questions have been asked in GATE CS 2009 exam. 

1) Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error? 
(A) G(x) contains more than two terms 
(B) G(x) does not divide 1+x^k, for any k not exceeding the frame length 
(C) 1+x is a factor of G(x) 
(D) G(x) has an odd number of terms. 

Answer (C) 
Odd number of bit errors can be detected if G(x) contains (x+1) as a factor. See this for proof. 

2) Frames of 1000 bits are sent over a 10^6 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). 
What is the minimum number of bits (i) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames. 
(A) i=2 
(B) i=3 
(C) i=4 
(D) i=5 

Answer (D) 
Transmission delay for 1 frame = 1000/(10^6) = 1 ms 
Propagation time = 25 ms 
The sender can atmost transfer 25 frames before the first frame reaches the destination. 
The number of bits needed for representing 25 different frames = 5 

3) Consider the data of previous question. Suppose that the sliding window protocol is used with the sender window size of 2^i where is the number of bits identified in the previous question and acknowledgments are always piggybacked. After sending 2^i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.) 
(A) 16ms 
(B) 18ms 
(C) 20ms 
(D) 22ms 

Answer (B) 

Propagation time is given as 25 ms.
Bandwidth = 10^6 bps.

So, to fully utilize the channel, we must send 10^6 bits into the channel in a second, which will be 1000 frames per second as each frame is 1000 bits. Now, since the propagation time is 25 ms, to fully pack the link we need to send at least:

1000×25×10-3=25 frames.

So, we need ⌈log_225⌉=5 bits.

I=5, so 2I =32 frames are sent.

Now, we need to get RTT (which is the time between which a frame is sent and its ACK is received), to determine the waiting time.

Transmission time (for a frame of size 1000 bits) = \frac{1000}{10^6} =1 ms.

So, transmission time for 32 frames = 32 ms.

RTT = Propagation time for frame + Transmission time for frame + Propagation time for ACK + Transmission time for ACK

=25 ms+1 ms+25 ms+1 ms

(ACK is piggybacked and assuming frame size for piggybacking is also 1000 bits)

=52 ms.

So, waiting time = 52−32=20 ms.
(For the 32 ms, the sender was transmitting and not waiting.)

Correct Answer: C