1. Quality of Service in Multimedia Distribution
G. Calinescu Illinois IT
G. Fernandes Brazil
I. Măndoiu UConn
A. Olshevsky GaTech
K. Yang Georgia State
A. Zelikovsky Georgia State

2. Outline QoS for multimedia distribution
Quality of Service Steiner Tree Problem
Previous work
Maxemchuk‘s heuristic
Rounding heuristic
Naive primal-dual
Restarting primal-dual
Experimental results
Conclusions

3. QoS for Multimedia Distribution
Given:
source in the network
set of customers requesting high volume data at different rates (bandwidths)
cost of link is proportional to
link length (or data unit cost)
rate (or bandwidth)
Find:
Minimum cost tree connecting source to each customer
S.t. each customer gets data with at least requested rate 2 4 S 4 1 3 6 6 1 8 7 4 2 2 4 6

4. QoS for Multimedia Distribution
Given:
source in the network
set of customers requesting high volume data at different rates (bandwidths)
cost of link is proportional to
link length (or data unit cost)
rate (or bandwidth)
Find:
Minimum cost tree connecting source to each customer
S.t. each customer gets data with at least requested rate 2 4 S 4 1 3 6 6 1 8 7 4 2 2 4 6
cost = 68 +44= 64

5. QoS for Multimedia Distribution
Given:
source in the network
set of customers requesting high volume data at different rates (bandwidths)
cost of link is proportional to
link length (or data unit cost)
rate (or bandwidth)
Find:
Minimum cost tree connecting source to each customer
S.t. each customer gets data with at least requested rate 2 4 S 4 1 3 6 6 1 8 7 4 2 2 4 6
cost = 69 +42= 62

6. Removing Source and Directions
We may get rid of source and directions by
assigning to source the maximum rate and
introducing rates on edges: rate of edge e, r(e) is the lowest rate between the maximum node rate in two components T1 and T2 in the routing tree T-e
tree T e T1 T2
r(e)=min{r1,r2}
Max node rate r1
Max node rate r2

7. Formal QoSST Problem Given: Undirected graph G=(V,E, l, r) with
rates r: VR+ on nodes (r = 0 means Steiner point)
lengths l: ER+ on edges (l is a metric)
Find: Spanning tree T of minimum cost
cost(T) = e E r(e) l(e),
where rate of edge e, r(e), is the smaller among maximum node rates in connected components of T-e

8. Previous Work
Introduced by Current[1986] in the context of network design and Maxemchuk[1997] in the context of network routing
Known as Multi-Tier/QoS/Grade-of-Service STP
Case of a single rate = classical Steiner tree problem
Case of few rates explored in a series of papers by Mirchandani-Balakrishnan-Magnanti [1994, 1996]
Mirchandani at al [1996] and Xue et al[2001] obtained better results for the case of two and three rates
First constant-factor approximation algorithm for arbitrary number of rates given by Charikar-Naor-Schieber[2001].

9. Maxemchuk’s Heuristic2
Kruskal-like algorithm
Each node is a component
Repeat while there are disconnected components:
Connect two closest components of highest rate with the edge of this rate
When only single component has highest rate, then down-rate this component to the next closest rate 2 4 6 4 1 3 6 6 1 8 7 4 2 2 4 6
cost = 68 +44= 64

10. How Accurate is Maxemchuk’s Algorithm?
rates
1-
Maxemchuk’s
 1+1/21 + 1/42 + 1/84 + … = 1+K/2
K = #rates
1-2
1-2
1-3
1-3
1-3
1-3
1-4
1/4
1/4
1/16 1 1
Optimal  1 = length  rate
1/8
1/8
1/8
1/8
1/2
Approximation ratio
= O(K) = O(log2 n),
n = # of nodes 1

11. Rounding Algorithm cost = 69 +42= 62
Algorithm (Charikar-Naor-Shieber’2001):
Round all rates to the closest power of 2 (or better)
Compute Steiner Trees for each rate
Output the union of these Steiner Trees.
Constructs a tree that is at most 2e (5.4) times larger than optimal
Can construct a tree that is at most 4.2 times optimal at the expense of impractical running time 2 4 6 4 1 3 6 6 1 8 7 4 2 2 4 6
cost = 69 +42= 62

12. Naive Primal Dual Regular Primal Dual Naive Primal Dual
Form components out of each vertex
Grow components at an equal rate
Two components collapse into one when
touched
Naive Primal Dual
Grow each component at the rate proportional to the maximum rate of a vertex it contains 1 1
The white circles have rate 1 and the crosses have very small rate. The distance between successive crosses is is a much smaller number than rates

13. Integer Linear Program and Relaxation
Given a network with k rates ri replace each edge e with length l by k edges, each having a cost ri l.
Let x(a,r) be the boolean variable denoting whether the edge of length r emanating from a is used.
The optimal QoSS tree is the solution to the following integer program:
Minimize
Subject to:
For all cuts C r1 s r2
The dual of this program is:
Maximize
Subject To:
y_C => 0

14. Restarting Primal Dual2
1. Replace each edge of length l with R
edges of cost ri l , where i=1,…,R, where R
is the number of rates. 2.
4. Grow each component with its own rate
along edges whose rate is at least as high
as the rate of the component
5. When two components touch, freeze the
component with the lower rate.
6. When highest rate component spans all nodes
of highest rate, do
- all highest rate edges add to the tree
- remove all other edges
- collapse all highest rate nodes into a single
node of the next highest rate
- restart with the next highest rate. 4 6 4 1 3 6 6 1 8 7 4 2 2 4 6
cost = 69 +42= 62

15. Experimental Study

16. Experimental Study

17. Conclusions We give 2 new primal-dual heuristics for QoSST problem
Compare on simulated data PD heuristics with two known before heuristics showing up to 7% cost improvement
Give a provably good practical approximation algorithm which guarantees at most 4.5 worse than optimum