CS 473 (ug)
Homework 1 Solutions
Fall 2005
1.
Problem 4.12:
a) The statement is false. As a counterexample, consider the streams (
b
1
, t
1
) = (1000
,
1) and
(
b
2
, t
2
) = (100
,
100) with
r
= 500.
The streams can be scheduled (do stream 2 first), but
b
1
> rt
1
.
b) Notice that the question does not ask for an algorithm that produces a schedule if one
exists; you only need an algorithm that outputs whether a schedule exists or not. A simple
lineartime solution is to calculate
B
=
∑
n
i
=1
b
i
and
T
=
∑
n
i
=1
t
i
. If
B > rT
, no schedule
exists. (This follows because after all streams have been scheduled, the total bandwidth used
will be
B
and the time elapsed will be
T
, and since
B > rT
, the constraint is violated.) If
B
≤
rT
, we claim that a valid schedule exists.
Proof:
Consider the schedule in which streams are sorted in increasing order of
b/t
. After stream
i
has completed, the total number of bits sent so far is
∑
i
j
=1
b
j
(call this
B
i
), and the total
time elapsed is
∑
i
j
=1
t
j
(call this
T
i
). We can see that
B
1
T
1
≤
B
2
T
2
≤
. . .
≤
B
n
T
n
=
B
T
≤
r
, because
B
i
T
i
is simply the
average
bitrate of the first
i
streams, and since the streams are sorted in
increasing order of bitrates, the average of the first
i
+ 1 streams must be greater than that
of the first
i
streams. But the average after all the streams is
≤
r
, and so for all
i
,
B
i
T
i
≤
r
.
We show that
B
i
+1
T
i
+1
≥
B
i
T
i
more formally below:
B
i
+1
T
i
+1
=
B
i
+
b
i
+1
T
i
+
t
i
+1
≥
B
i
+
t
i
+1
B
i
T
i
T
i
+
t
i
+1
Because
b
i
+1
t
i
+1
≥
B
i
T
i
=
B
i
T
i
+
t
i
+1
T
i
T
i
+
t
i
+1
=
B
i
T
i
We have shown that if
B
≤
rT
(we do not violate the constraint at the end), then there is a
schedule such that the constraint is not violated after each stream is completed. What about
the time while a stream is being sent? If the average bitrate is
≤
r
at the beginning and
end of the stream, then can it exceed
r
in the middle of the stream? It’s easy to show that
it cannot; here’s an interesting idea that can be converted into a proof with very little effort.
Suppose stream
i
runs from
t
1
to
t
2
, and at both times
t
1
and
t
2
, the average bitrate is
≤
r
.
To show that the bitrate is
≤
r
at any time
t
,
t
1
< t < t
2
, simply break stream
i
into two
streams, one called
i
1
of length
t

t
1
and the other,
i
2
, of length
t
2

t
. Now we have a new
schedule with 1 extra stream; by the argument above we know that the average bitrate is
≤
r
after the end of every stream. In particular, at time
t
, after stream
i
1
has been sent, the
average bitrate is
≤
r
.
Another solution to this problem is to sort the streams in increasing order of
b/t
, and show
that if any schedule does not violate the constraint, then this one does not. The argument is
essentially similar to the one above; the main difference is that the first method just gives a
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