Computer Networks Test 2

Using the Back-off algorithm:

Since both 1 and 2 transmit after the first collision.

Range (0, 2¹ -1) = Rang (0, 1)

Station whose time is less wins the or the person whose time is more back-offs

The probability that neither X nor Y wins = \(\frac{2}{{4\;}} = 0.5\)

Concept:

For any sliding window protocol,

Available Sequence Numbers ≥  Sender window size + Receiver window size

In Selective repeat Protocol, sender  window size = receiver window size = W

Calculation:

With 16 bits, the total number of sequence numbers possible = 2¹⁶.

Available Sequence Numbers ≥  Sender window size + Receiver window size

2¹⁶ ≥  W + W

∴ W ≤ \(\frac{2^{16}}{2}\) 

W_max = 2¹⁵ = 32768

The maximum window size for the data transmission is 32768.

Concept:

If the polynomial is of order n then the number bits generated by CRC generator is n + 1.

Data:

Message = (7766554433221100)8 = 0010 1010 0100 1110

CRC polynomial = x11 + x10 +x8+ x5 + x4 + x + 1

Explanation:

CRC polynomial = x11 + x10 + 0.x9 + x8 + 0.x7 + 0.x6 + x5 + x4 + 0.x3 + 0.x2 + 1.x + 1

CRC generator generates = (110 100 110 011)₂ = (6463)₈ 

1^st lost:

8^th packet is lost, and window size is 4 then packet number 8, 9, 10 ,11 is send again

Note that out of 8,9,10,11 packet 9^th packet is first packet send

2^nd lost:

∴ 12^th packet is the 8^th lost packet

3^rd lost:

∴ 16^th packet is the 8^th lost packet

∴ number of packets transmitted = 11 + 8 + 5 + 1 = 25 

Data:

Probability = p = 0.3

Throughput = T = 0.21

Number of hosts = n

Formula:

\(nT={{~}^{n}}{{C}_{1}}\left( p \right){{\left( q \right)}^{n-1}}~\)

Solution

n × 0.21 = n × (0.3) × (0.7)^n-1

0.7 = (0.7)^n-1

n – 1 = 1

∴ n = 2

Message: 11010101

CRC polynomial: x³ + x² + 0.x+1

CRC generator: 1101

Since polynomial is of order 3 append 3 0’s at the end of message

Message: 11010101000

Data:

Distance = 5000 km

Bandwidth = 500 MBps = 500 × 106  × 8 bps = 4 × 10⁹ bps

Propagation speed = 2× 107 m/s

Average packet size = 108 bits

Window size in Go-Back-N protocol = 2n – 1

Formula:

\({\rm{Transmission\;time}} = \frac{{{\rm{packet\;size}}}}{{{\rm{bandwidth}}}}\)

\({\rm{Propagation\;time}} = \frac{{{\rm{Distance}}}}{{{\rm{velocity}}}}\)

Calculation:

\({\rm{Transmission\;time}} = \frac{{{{10}^8}}}{{4 × {{10}^9}}} = 25\;ms\)

\({\rm{Propagation\;time}} = {\rm{\;}}\frac{{5000 × 1000}}{{2 × {{10}^7}}} = 250{\rm{\;ms}}\)

RTT = 2 × Propagation time = 2 × 250 = 500 seconds

Number of packets that can be transmitted =  500 ÷ 25

∴ window size = 20 

2n – 1 ≥   20

2n ≥ 21,

n > 4

n = 5

I. True: The most popular ARQ protocol is the go-back-N ARQ, where the sender sends the frames continuously without waiting for acknowledgment. That is why it is also called as continuous ARQ.

II. True: If no acknowledgement is received after sending N frames, the sender takes the help of a timer. After the time-out, it resumes retransmission.

Concept:

In stop and wait protocol, packet lost during transmission is the only packet retransmitted.

Data:

Total packet send = 750

Actual packet = 560

Error rate = p

Calculation:

Total number of packets transmitted by the sender:

570 + p× (570) + p × (p × 570) + p × (p × p× 570) … = 750

570(1 + p + p² + p³ …) = 750

\(570 × \left( {\frac{1}{{1\; - p}}} \right) = 750\)

\(1- p=\frac{{570}}{{750}}\)

p = 1 - 0.76 = 0.24

Error in link = 0.24× 100 = 24%.