EE 284
F. Tobagi
Autumn 20172018
EE284 Homework Assignment No. 3
SOLUTIONS
Total Points: 56
Problem 1:
Error recovery by means of error detection and/or
correction and retransmission (9 points)
(a) [2 points] Since A continues to retransmit the frame indefinitely, the number of
failed
transmissions to transmit a frame can be expressed as a geometric random variable
X
with probability of success 1

FER
. We can express the probability of
X
=
k
for any
integer
k >
= 0 as follows:
P
[
X
=
k
] = (
FER
)
k
(1

FER
)
,
k
= 0
,
1
,
2
, . . .
Finding the expected value for the geometric random variable, we find that the expected
value for the number of failed transmissions is:
E
[
X
] =
FER
1

FER
=
1
1

FER

1
Additionally, each unsuccessful transmission occupies the link for
B
W
+
T
0
sec., while
the (last) successful transmission occupies
B
+
C
W
+2
τ
sec. We can thus write the average
time for a single frame to be successfully transmitted and acknowledge as:
(
1
1

FER

1)(
B
W
+
T
0
) +
B
+
C
W
+ 2
τ
Hence, the throughput is 1 divided by this average time, given by:
S
=
1
(
1
1

FER

1)(
B
W
+
T
0
) +
B
+
C
W
+ 2
τ
See the diagram in Figure 1 for an illustration of this transmission process.
1
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Figure 1: Problem 1a
(b) [1 point] With the addition of a forward error correction scheme, our error rate is now
FER
(
r
), but we have also increased the number of bits in each frame.
Recall that
the coding rate
r
=
number of data bits
number of coded bits
. Thus, the error correction scheme increases the
number of bits in each frame by a factor of 1
/r
. Our new expression of the throughput
becomes:
S
=
1
(
1
1

FER
(
r
)

1)(
B
rW
+
T
0
) +
B
+
C
rW
+ 2
τ
(c) [1 point] Since there is a round trip time of 2
τ
between the stations, and the transmis
sion time of the ACK frame is
C/rW
, the minimum timeout period is given by the
following:
T
min
0
= 2
τ
+
C
rW
(d) [1 point] With
T
0
=
T
min
0
, the throughput with forward error correction becomes:
S
=
1
(
1
1

FER
(
r
)

1)(
B
rW
+ 2
τ
+
C
rW
) +
B
+
C
rW
+ 2
τ
This simplifies to:
S
=
1
(
1
1

FER
(
r
)
)(
B
+
C
rW
+ 2
τ
)
2
With
τ
= 0, we can write throughput as:
S
=
1
(
1
1

FER
(
r
)
)(
B
+
C
rW
)
The throughput for when we don’t use forward error correction is the same as the
above, except with
r
= 1 and the frame error rate as
FER
:
S
=
1
(
1
1

FER
)(
B
+
C
W
)
Forward error correction is desirable when using it can achieve greater throughput.
Thus, we can write the condition on
FER
as:
1
(
1
1

FER
(
r
)
)(
B
+
C
rW
)
≥
1
(
1
1

FER
)(
B
+
C
W
)
Simplifying this expression, we arrive at the following condition:
1

FER
≤
r
(1

FER
(
r
))
Or equivalently:
FER
≥
1

r
(1

FER
(
r
))
(e) [1 point] As
τ
increases, the propagation time of the frame becomes a larger and larger
proportion of the cycle time of a single transmission. This means it is even more fa
vorable to minimize the number of transmissions, even at the cost of a larger frame size.
Thus, as
τ
increases and becomes nonnegligible, the condition on
FER
relaxes, and
it becomes favorable to use forward error correction for even lower failure rates than
1

r
(1

FER
(
r
)).
Equivalently, we can say that as
τ
increases, the right side of
following condition on
FER
decreases:
FER
≥
1

r
(1

FER
(
r
))
.
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 Spring '09

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