9104302 - CS 473(ug Combinatorial Algorithms Fall 2005...

CS 473 (ug): Combinatorial Algorithms, Fall 2005
Homework 1 Solutions
This homework was due on Sep 9, with an automatic extension to Sep 12.

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
linear-time 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 bit-rate of the first i streams, and since the streams are sorted in
increasing order of bit-rates, 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 bit-rate 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 bit-rate is ≤ r .
To show that the bit-rate 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 bit-rate is
≤ r after the end of every stream. In particular, at time t , after stream i
1
has been sent, the
average bit-rate 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
1

CS 473 (ug) Homework 1 Solutions Fall 2005
condition for a schedule to exist, while the second method actually outputs a feasible sched-
ule.