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Microsoft Word - Link Layer Discussion.docx

Single Bit Parity (Slide 5) Data: 0111000110101011 If you use even parity: 01110001101010111 If you use odd parity: 01110001101010110

Two-dimensional Parity (Slide 5) Data: 101011111001110 Let’s say, you use even parity: 1 0 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 1 0 0 1 0 1 Let’s say, there is a single bit error: 1 0 1 0 1 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 1 0 1 You can detect error bit at the intersection.

Tutorial Question 1 1 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 1 0 0

Tutorial Question 2 Without errors: 0 0 0 0 0 1 1 1 1 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 With errors: 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0

Cyclic Redundancy Check (Slide 6) At sender, Data = 101110 G = 1001 r = 3 1 0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 0

1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 1 0 0 1 0 1 1

So, sender sends 101110011 to receiver

At receiver, Received data = 101110011 G = 1001 r = 3 If no error occurs at receiver,

1 0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1

1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 0 0 1 1 0 0 1

0

No remainder!

If error occurs at receiver, 1 0 1 1 1 1

1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 1 1 0 0

Error detected! Remainder is 100.

Tutorial Question 3

At sender, Data = 1010101010 G = 10011 r = 4 1 0 1 1 0 1 1 1 0 0 1 0 0 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 0 1 0 0 1 0 0 1 1 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 1 0 0 1 1 1 0 0 1 0 1 0 0 1 1 0 1 0 0 So, sender sends 10101010100100 to receiver

Tutorial Question 4 4a. A’s avg throughput = PA (1 – PB) B’s avg throughput = PB (1 – PA) Total throughput = A’s avg throughput + B’s avg throughput = PA (1 – PB) + PB (1 – PA) 4b. PA (1 – PB) = 2 PB (1 – PA) PA – PA PB = 2 PB – 2 PA PB PA = 2 PB – PA PB PA = PB ( 2 – PA ) PB = – No. PB should be – rather than . 4c. Node A’s retransmission probability = 2 p Other nodes’ probability of no transmission = (1 – p) N-1 Node A’s average throughput = 2 p (1 – p) N-1 One other node retransmission probability = p Node A’s probability of no retransmission = 1 – 2 p All other nodes’ probability of no retransmission = (1 – p) N-2 Other node’s average throughput = p (1 – 2 p) (1 – p) N-2

Tutorial Question 5 5a. Probability A success in a slot = p (1 - p)3 Probability A fail in a slot = 1 – [p (1 - p)3] Probability A fails in first four slots, and success in fifth slot = (1 – [ p ( 1 – p )3 ] )4 p(1 - p)3 5b. Probability a node success in a slot = p (1 – p )3 Probability that either one node success in a slot = p (1 – p )3 + p (1 – p )3 + p (1 – p )3 + p (1 – p )3 = 4 p (1 – p )3 5c. Probability that first success occurs in slot 3 = ( 1 – [ 4 p (1 – p )3 ] )2 p (1 – p )3 5d. Probability a node success in a slot = 4 p (1 – p )3