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

2. Chapter 6 page 204 (Data & computer communications)
Problems:
6.1
Questions on ch 6.
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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
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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
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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
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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 = =3000 bits
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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+( )= bits
b. use one start bit and two stop bits with parity bit
Total # of parity bits= 1x1000=1000 bits
Total # of bits to be transmitted = # of data bits + # of extra bits = 8000+( )= bits
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8. Question
Q.4 Assuming even parity, find the parity bit for each of the following data units: a) b) c) d)
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9. Question
Q.5 A receiver receives the bit pattern 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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10. Question
Q.7 Given a remainder of 111, a data unit of , and a divisor of 1001, is there an error in the data unit?
No error because the reminder is 0
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11. Question Q.8 If a divisor is 101101, how many bits long is the CRC?
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12. The End
Any Questions ?
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