Link layer Link layer description, Exam questions for Computer Networks. A.P. University of Law
shahtab
shahtab

Link layer Link layer description, Exam questions for Computer Networks. A.P. University of Law

PDF (299 KB)
10 pages
2Number of visits
Description
Link layer description Link layer description Link layer description Link layer description Link layer description
20 points
Download points needed to download
this document
Download the document
Preview3 pages / 10
This is only a preview
3 shown on 10 pages
Download the document
This is only a preview
3 shown on 10 pages
Download the document
This is only a preview
3 shown on 10 pages
Download the document
This is only a preview
3 shown on 10 pages
Download the document
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

no comments were posted
This is only a preview
3 shown on 10 pages
Download the document