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CS421/S03
Solutions to Homework Two
Chapter 6 (7/e)
1.
Problem 6.9
a.
We have:
Pr [single bit in error] = 10
3
Pr [single bit not in error] = 1 – 10
3
= 0.999
Pr [8 bits not in error] = (110
3
)
8
= 0.992
Pr [at least one error in frame] = 1(110
3
)
8
= 0.008
b.
Pr [at least one error in frame] = 1 (110
3
)
10
= 1(0.999)
10
= 0.01
2
.
Problem 6.12
_____ 10110110__________
110011/ 1110001100000
110011
101111
110011
111000
110011
101100
110011
111110
110011
CRC =
11010
Chapter 7
.
3.
Problem 7.2
Let L be the number of bits in a frame. Then, using Equation 7.5 of Appendix 7A:
a = (Propagation Delay)/(Transmission Time) =
)
10
4
/(
10
20
3
3
×
×

L
= 80/
L
Using equation 7.4 of Appendix 7A:
5
.
0
)
/
160
(
1
1
2
1
1
≥
+
=
+
=
L
a
U
L
≥
160
Therefore an efficiency of at least 50% requires a frame size of at least 160 bits.
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View Full Document4.
Problem 7.9
5.
Problem 7.12
Let t
1
= time to transmit a single frame
t
1
= 1024 bits/10
6
bps = 1.24 msec
The transmitting station can send 7 frames without an acknowledgment. From the
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 Spring '08
 Shen

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