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NET 222: COMMUNICATIONS AND
NETWORKS FUNDAMENTALS
(PRACTICAL PART)
Networks and
Communication
Department
Tutorial 4: Chapter 6
Data & computer communications
Chapter 6 page 204 (Data & computer communications)
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Problems:
 6.1
 Questions
on ch 6.
Networks and Communication Department
Question
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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
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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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Question (Cont.)
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 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
Question
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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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Question on Ch 6
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Q.1 If we want to transmit 1000 ARI character(7bit 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
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Q.2 We want to transmit 1000 characters with each character encoded as
8 bits. Find the number of transmitted bits for asynchronous transmission
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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
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b. use one start bit and two stop bits with parity bit
Total # of
Total # of
Total # of
Total # of
start bits=1x1000=1000 bits
stop bits=2x1000=2000 bits
parity bits= 1x1000=1000 bits
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
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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
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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
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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
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Q.8 If a divisor is 101101, how many bits long is
the CRC?
5 bits
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The End
Any Questions ?
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