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NET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS (PRACTICAL PART)

Tutorial 4: Chapter 6 Data & computer communications

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Chapter 6 page 204 (Data & computer communications)

Problems: 6.1 Questions on ch 6.

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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=10,000bitsTotal #of stop bits=1x10,000=10,000bitsOverhead in bits=Total # of start bits + Total # of stop=20,000 bitsOverhead 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 framesOverhead in bits=# of frames x # of control bits =10x48=480 bitsOverhead in time=overhead in bits / data rate =480/2400 =0.2 second

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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

secondsFor synchronous : Overhead in bits :same as (b)Overhead in time=480/9600 =0.05 second

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Question

For P=1011 and D=1010011110 ,find the CRC

D= 1010011110K=10 bitsP=1011n=p+k-1 4+10-1=13Fcs =n-k=13-10 =3

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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

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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

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Question

Q.4 Assuming 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

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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

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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

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Question

Q.8 If a divisor is 101101, how many bits long is

the CRC?5 bits

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The End