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CSE 434 – TTh 9:15 AM
Homework #3 Sample Solution
October 14, 2003
Please email Catherine (
cwen@asu.edu
) if you have questions.
(10 points for each question)
1.
(3.11)
A 12bit Hamming code whose hexadecimal value is 0xE4F arrives at a
receiver.
What was the original value in hexadecimal?
Assume that not more than
1 bit is in error.
0xE4F=1110 0100 1111
Position
1
2
3
4
5
6
7
8
9
10
11
12
Parity Result
Pos. as power of 2
1
2
2+1
4
4+1
4+2
4+2+1
8
8+1
8+2
8+2+1
8+4
Bitstream
1
1
1
0
0
1
0
0 1
1
1
1
Check bit #1 Parity
1
1
0
0
1
1
Even
Check bit #2 Parity
1
1
1
0
1
1
Odd
Check bit #4 Parity
0
0
1
0
1
Even
Check bit #8 Parity
0
1
1
1
1
Even
The second bit is incorrect. So the original value is 1010 0100 1111, i.e. 0xA4F.
Dada: 1010 1111, i.e. 0xAF
(3.14)
What is the remainder obtained by dividing
1
5
7
+
+
x
x
by the generator
polynomial
1
3
+
x
?
M(x) =
1
5
7
+
+
x
x
G(x) =
1
3
+
x
Since degree of G(x) is r = 3, append 3 zero bits to the lower end of M(x).
Frame:
10100001
Generator:
1001
Message after 3 zero bits are appended:
10100001000
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The remainder obtained by dividing x7 + x5 + 1 by the generator polynomial x3 + 1 is
111.
2.
(3.19)
In protocol 3, is it possible that the sender starts the timer when it is already
running?
If so, how might this occur?
If not, why is it impossible?
Yes, this can happen.
This is the case when "no positive ack arrives" at the sender in
last iteration of the while loop in sender3(void) code on page 210. So, the frame that the
sender is transmitting when the timer is running (and therefore reset), is a repeat/duplicate
frame.
(3.20)
Imagine a sliding window protocol using so many bits for sequence numbers
that wraparound never occurs.
What relations must hold among the four window
edges and the window size, which is constant and the same for both the sender and
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