If each packet carries 1000 bits of data, the distance between the sender and re

Question

If each packet carries 1000 bits of data, the distance between the sender and receiver is 7200km. link speed is 1Mbps, and the propagation speed is 2 times 10^8 m/s. Assume no data or control frame is lost or damaged. Ignore transmission time for ACK. How long does it take to send 1 million bits of data if using Stop-and-Wait ARQ Protocol? Repeat for Go-back-N ARQ Protocol if using 3-bit sequence number. Use selective reject ARQ if using 3-bit sequence number. Sliding window flow control, frames are numbered modulo-8 and the window size is 4. What frames are in the window of the transmitter when the last frame transmitted is 3 and the transmitter just received RR-2 from the destination? What frames that are buffered at the transmitter? Find the frame check sequence (FCS) when the data frame is 1101 1001 0101 1001 is to be encoded using CRC error detection method with a (generator) pattern x^8 + x^6 + x^3 + 1. Show all your steps What is the pattern sequence (in bits) How many bits will the frame check sequence be? What is the transmitted sequence (message +FCS)?

Explanation / Answer

Here is the solution of given problems, please go through it throughly..

Problem1:-

(a)    Packet size = 103 bits (1000 bits)

Total data to be transmitted = 106 bits (1 million bits)
distance between the sender and receiver = 7200 Km

Therefore,

  [number of packets to be transmitted=Total data to be transmitted / Packet size]

Number of packets to be transmitted = 106 / 103 = 1000

Data frame transmission time = (1,000 bits/frame) / (1 Mbps) = 1 ms

Propagation delay for one packet = Distance / Propagation speed

= (7.2 * 106) / (2*108) = 0.036 s

Propagation delay for one ACK = Distance / Propagation speed = (7.2 * 106) / (2*108) = 0.036 s

Here, transmission delay is 0. (It is usually negligible)

So, Total time to transmit one packet and receive its ACK =2 * Propagation delay for one ACK

=2 * 0.036 = 0.072 s

Therefore,

total time to transmit 1000 packets = 1000 * 0.072 = 72 s3

(b) As per the Go-Back-N ARQ protocol,

  Sequence number, N=3
Window-size = 2n-1 =22= 4
In the worst case, we send the full window of size 4 and then wait for the acknowledgement of the whole window.
We need to send 1000 frames / 4 250 windows.

(c) Selective Reject (or Selective Repeat) protocol is one of the automatic repeat-request (ARQ) techniques used for communications.

In SR protocol the window size of the receiver and sender must be (N+1)/2, where N is the maximum sequence number.

If N is the maximum available sequence numbers then, the window size of both sender and receiver must be N/2.

If n is the number of bits in the frame sequence field then, the window size of both sender and receiver must be  2n-1.

As per the given criteria,

Here,

Sequence Number, N = 3 bit

So, the window size of both sender and receiver must be  2n-1 = 22 = 4

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