1. Memory Management Algorithms Huan Liu, Damon Mosk-Aoyama
for DSM Switches
Huan Liu, Damon Mosk-Aoyama

2. Distributed Shared Memory Switch
What is known (for FIFO)
2N is necessary
3N-1 is sufficient
Questions:
Can we close the gap?
Possible to have an implementable algorithm?

3. Our main results On the gap On practical algorithms
Counter example to show 2.25N is necessary
A heuristic that uses 2.5N memories
On practical algorithms
Simulation results on simple algorithms under Bernoulli i.i.d. traffic

4. … … The counter example Consider 4x4 switch
Can generalize to arbitrary N 3 4 1 2 1 2 … 5 5
Output 1 6 … 6 3 4 ? 7 7 8 8
Output 4

5. If we have N/4 more 1 1 1 6 6
Output 1 … 5 2 2 7 … 7 3 5 3 8 8 2 4 4 9 9
Output 4

6. Observation
Greedily minimizing the number of memories used can lead to trouble
Need to reuse memories later as time slot fills up 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 4 5 6 7 4

7. Heuristics with 2.5N memories
Minimize intersection between adjacent time slots
Minimize intersection between neighboring pairs
After N/2 cells arrived in a time slot, reuse memories already assigned to the adjacent time slot.
Simulation has been running for 100M+ cycles with no problem
Minimize intersection 5 6 1 2 3 4 1 2 1 2 1 6 2 4 3 4

8. …… Random algorithm Assign memories to arriving cells randomly
Drop if another cell using the memory is
departing now
departing in the future in the same time slot Si …… S2 S1

9. Upper bound on Drop Rate
Suppose there are memories. The drop probability is
The drop rate can now be computed as:
Use Si distribution from M/M/1

10. Fixed Arrival Rate

11. Fixed Number of Memories

12. Distributed random algorithm
Each packet makes independent decision
Pick a random memory that is NOT
departing now
departing in the same time slot in the future
If two arriving packets pick the same memory, we drop one

13. Distributed random algorithm - simulation

14. Centralized random algorithm
Assign each packet in turn
Randomly pick a memory that is NOT
departing now
departing in the same time slot in the future
assigned for other packets arriving at the same time

15. Centralized random algorithm - simulation

16. Conclusion Still gap  more work
Better counter example?
Prove 2.5N is sufficient
Also gap between theory and practical algorithm