Net 222: Communications and networks fundamentals ( Practical Part)

1. Net 222: Communications and networks fundamentals (Practical Part) Tutorial 4: Chapter 6 Data &computer communications Networks and Communication Department

2. Chapter 6 page 204 (Data & computer communications) • Problems: • 6.1 • Questions on ch 6. Networks and Communication Department

3. Question • 6.1 Suppose a file of 10,000 bytes is to sent over a line at 2400 bps. • a) Calculate the overhead in bits and time in using asynchronous communication. Assume 1 start bit and a stop element of length 1 bit, and 8 bits to send the byte itself for each character. The 8-bit character consists of all data bits, with no parity bit. Total # of start bits=1x10,000=10,000bits Total #of stop bits=1x10,000=10,000bits Overhead in bits=Total # of start bits + Total # of stop=20,000 bits Overhead in time=overhead in bits / data rate = 20,000/2400=8.33 seconds • b) Calculate the overhead in bits and time using synchronous communication. Assume that the data are sent in frames. Each frame consists of 1000 character= 8000 bits and an overhead of 48 control bits per frame. Overhead=48 bits/frame # of frames =10,000/1000 =10 frames Overhead in bits=# of frames x # of control bits =10x48=480 bits Overhead in time=overhead in bits / data rate =480/2400 =0.2 second Networks and Communication Department

4. Question (Cont.) • d) What would the answer to parts (a) and (b) for the original file of 10,000 character except the data rate of 9600 bps? For asynchronous : Overhead in bits :same as (a) Overhead in time=20,000/9600=2.08 seconds For synchronous : Overhead in bits :same as (b) Overhead in time=480/9600 =0.05 second Networks and Communication Department

5. Question • For P=1011 and D=1010011110 ,find the CRC D= 1010011110 K=10 bits P=1011 n=p+k-1 4+10-1=13 Fcs =n-k=13-10 =3 Networks and Communication Department

6. Question on Ch 6 • Q.1 If we want to transmit 1000 ARI character(7- bit character) asynchronously, what is the minimum number of extra bits needed? Assume parity bit is used. Total # of start bits=1x1000=1000 bits Total # of stop bits=1x1000=1000 bits Total # of parity bits=1x1000=1000 bits Minimum # of extra bits = 1000+1000+1000=3000 bits Networks and Communication Department

7. Question • Q.2 We want to transmit 1000 characters with each character encoded as 8 bits. Find the number of transmitted bits for asynchronous transmission • a. use one start bit and two stop bits Total # of start bits=1x1000=1000 bits Total # of stop bits=2x1000=2000 bits Total # of data bits=8x1000=8000 bits Total # of bits to be transmitted = # of data bits + # of extra bits = 8000+(1000 + 2000)=11000 bits • b. use one start bit and two stop bits with parity bit Total # of start bits=1x1000=1000 bits Total # of stop bits=2x1000=2000 bits Total # of parity bits= 1x1000=1000 bits Total # of data bits=8x1000=8000 bits Total # of bits to be transmitted = # of data bits + # of extra bits = 8000+(1000 + 2000 +1000)=12000 bits Networks and Communication Department

8. Question • Q.4Assuming even parity, find the parity bit for each of the following data units: • a) 1001011 1001011 0 • b) 0001100 0001100 0 • c) 1000000 1000000 1 • d) 1110111 1110111 0 Networks and Communication Department

9. Question • Q.5 A receiver receives the bit pattern 01101011. if the system is using even parity, is the pattern in error? yes because the total number of 1’s is odd Networks and Communication Department

10. Question • Q.7 Given a remainder of 111, a data unit of 10110011, and a divisor of 1001, is there an error in the data unit? No error because the reminder is 0 Networks and Communication Department

11. Question • Q.8 If a divisor is 101101, how many bits long is • the CRC? 5 bits Networks and Communication Department

12. The End Any Questions ? Networks and Communication Department