[SOLVED] R C 1)

1)
Sample Problems (From Previous Exams)
2)
Sender wants to transmit MSG=10011010. The generator pattern is 1101. What is the FCS pattern? What is the transmitted pattern? Suppose the MSB and LSB of the transmitted pattern are flipped (i.e. in error), will the receiver be able to detect the error?

3)

Solution
!A->B:
Propagation time = 4000 x 5 sec = 20 msec
1000 = 10msec 100103
!B->C:
Propagation time = 1000 x 5 sec = 5 msec
Transmission time per frame = x = 1000/R R = data rate between B and C (unknown)
Transmission time per frame =
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Solution Continued
!A can transmit three frames to B and then must wait for the acknowledgment of the first frame before transmitting additional frames. The first frame takes 10 msec to transmit; the last bit of the first frame arrives at B 20 msec after it was transmitted and therefore 30 msec after the frame transmission began. It will take an additional 20 msec for Bs acknowledgment to return to A.
Thus, A can transmit 3 frames in 50 msec.
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Solution Continued
!BcantransmitoneframetoCatatime.Ittakes5+ x msec for the frame to be received at C and an additional 5 msec for Cs acknowledgment to return to B. Thus, B can transmit one frame every 10 + x msec, or 3 frames every 30 + 3x msec. Thus:
30 + 3x = 50
x = 6.66 msec
R = 1000/x = 150 kbps
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Sender: Example
Sender wants to transmit MSG=10011010
MSG=10011010,n=8 correspondsto M(x)=x7+x4 +x3 +x1 Divisor=1101 , k=3 corresponds to G(x) = x 3+ x 2 + 1
Multiply M(x) by x k
In this example, we get:
M(x).x3=x10+x7 +x6+x4=10011010000
Divide result by G(x) =1101 (Subtraction or addition is XOR in polynomial arithmetic)
The remainder is E(x) = x 2 +1 = 101
Send P(x) = M(x).x k + E(x) which is exactly divisible by G(x) i.e. Send 10011010000 + 101 = 10011010101, since this is exactly divisible by G(x) =1101
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