UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2008
1
EE 550: Problem Set
#
4 Solution
I. S
TOP AND
W
AIT
E
FFICIENCY
a) Remaining time =
n
·
RTT
if, starting at the time the sender gets the “ACK” with an error, the sender gets the first correct
“ACK” after
n
retransmissions, which happens with probability
q
n

1
(1

q
)
. Obviously the remaining time is
RTT
times
a geometric random variable with parameter
1

q
. Thus the expected remaining time equals to
RTT/
(1

q
)
.
b) Define
X
to be the time between first deliveries of two consecutive packets. The expectation of X equals to
E
{
X
}
=
RTT
+
p
E
{
X
}
+ (1

p
)
q
RTT
1

q
+ (1

p
)(1

q
)
·
0
On the right hand side, the first term is the time needed for one transmission. The second term represents, with prob.
p
,
there are errors in the forward transmission; and after the first round trip time, the remaining time until the first delivery
of the next packet is still
X
. The third term represents, with prob.
(1

p
)
q
, after the first round trip time during which a
packets has been successfully delivered but there are errors in the ACK, by part a) the expected remaining time until the
first delivery of the next packet is
RTT/
(1

q
)
. The last term represents, with prob. (1p)(1q), a packet is successfully
received and acked during the first round trip time. The above equation yields
E
{
X
}
=
1

pq
(1

p
)(1

q
)
RTT,
and the throughput equals to
1
E
{
X
}
=
(1

p
)(1

q
)
(1

pq
)
RTT
II. S
TOP AND
W
AIT WITH
E
NERGY
C
ONSIDERATION
a) The throughput equals to
1

p
RTT
=
1

1
/
2
16
=
1
32
b) Let
X
be the time taken by a particular packet. Then
E
{
X
}
=
E
{
X

first transmission is lossfree, errorfree
}
(
3
4
)(
3
4
)
+
E
{
X

error occurs, no loss
}
(
1
4
)(
3
4
)
+
E
{
X

first transmission is lost
}
(
1
4
)
=
9
16
RTT
+
3
16
(
RTT
+
E
{
X
}
) +
1
4
(
TO
+
E
{
X
}
)
,
where
RTT
= 16
and
TO
= 20
yield
E
{
X
}
= 272
/
9
, and throughput equals to
9
/
272
.
c) The throughput is low because the time between retransmissions is at least
RTT
. The throughput is maximized by setting
TO
= 1
, i.e., retransmit a packet every timeslot without waiting for ACK/NACKs until receiving the first ACK of the
packet. Let
X
be the time until receiving the first ACK of a packet. Then
X
=
RTT
w.p.
(
3
4
)(
3
4
) =
9
16
,
q
RTT
+ 1
w.p.
(1

q
)
q
RTT
+ 2
w.p.
(1

q
)
2
q
. . . . . .
,
which yields
E
{
X
}
=
RTT
+ (1

q
)
/q
= 151
/
9
. Thus the throughput equals to
9
/
151
.
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