ELE4NET Communications Networks For Multilevel and Multitransition

Answer:

Line encoding is use for secure data. Line coding scheme have five types according to different levels, such as unipolar, polar, bipolar, multilevel, and multitransition. In this question, Non-return to zero is a type of unipolar, Manchester encoding is considered in polar coding. Bipolar line encoding has alternate mark invert and pseuddoternary coding. In this given bit stream 01001110, we applied different line encoding. I sketch the waveform for the following codecs:

• NRZ-L
• NRZ-I
• Bipolar AMI
• Pseudo ternary
• Manchester

I take positive voltage for zero and zero voltage for signal 1. 

Cyclic Redundancy Code is an error detection scheme, in which we take a pattern for sending data with some edition bits. Given pattern is P(x) = X⁴ + X + 1.

We used that pattern for dividing polynomial.

A message bit is this 10010011011, and it is our dataset.

We have to find Frame Check Sequence bits to be added in the end of dataset to make the message complete for transmission of message in the channel.

Cyclic Redundancy Code is an error detection scheme.

For finding added bit divide X¹⁰ + X⁷ + X⁴ + X³ + X + 1 by X⁴ + X + 1.

The remainder is X³ + X².

Additional bits are 1100.

So finally, after the attachment of 1100 into 10010011011 final dataset will be send in the channel is 100100110111100. 

When we transmit the signal over the channel then because of attenuation, distortion, and noise, some bits are changed and that was an erroneous signal. As per question, we received a bit stream that is 000110110111100.

According to received bits  X¹¹ + X¹⁰ + X⁸ + X⁷ +X⁵ + X⁴ + X³ + X².

We divide this polynomial by  X⁴ + X + 1.

The remainder after division is X³ + X² + X.

Reminder is not zero. Therefore, it means error is detected in the received dataset. 

Each character has 2 bits extra for start and stop bit as an overhead.

Given character, type is 7-bit IRA. For transferring 10,000 IRA characters, we need more bits. There are 20,000 extra bits are used for data transfer.

It takes extra time to transfer a frame, according to extra bits,

20000/2400 = 8.33 seconds.

It means more 8.33 seconds required for transfer the message.

For transmission of additional bit it required extra time, so additional bits are divided by speed of data transfer.

 480/2400 = 0.2 seconds

Total overhead bits are remaining same as previous but time would be changed by factor of four.

9600/2400=4.   

Dijkstra’s algorithm is a Greedy algorithm and it is used for negative weight.   

 Consider the network in below Figure and calculate systematic cost change according to bellman-ford algorithm.

Table shows least-cost paths of given graph.

Bellman-ford algorithm is used for finding shortest path from a vertex to all other vertices of a weighted graph. 

Bibliography

A Book

Comer, Douglas, and Ralph E. Droms, Computer Networks and Internets (Prentice-Hall, 2003)

Forouzan A. Behrouz, Data Communications and Networking (Tata McGraw-Hill Education, 2006)

Oliver C Ibe, Fundamentals of Data Communication Networks (John Wiley & Sons, 2017)

Salomon, David, Coding for Data and Computer Communications (Springer Science & Business Media, 2006)

Stallings, William, Data and Computer Communication (Pearson Education, 8th ed, 2007)

Tanenbaum, Andrew, Structured Computer Organization (Pearson Education India, 2016)

Tomasi, Wayne, Introduction to Data Communications and Networking, (Pearson Prentice Hall, 2005)

Wang, Jie, Computer network security, (Springer, 2009)