Comments on the relative and absolute fairness measures

1. Comments on the relative and absolute fairness measures Joanna Józefowska*) Łukasz Józefowski*) Wiesław Kubiak**) *)Poznań University of Technology **)Memorial University

2. Presentationoutline • Scheduling packets in packet-switched networks • Problem formulation • Relativefairnessbound • Absolutefairnessbound • Apportionment problem • Formulation • Properties • RFB transformation • PRV problem • Formulation • AFB transformation • Conclusions

3. Scheduling packets in a packet switched network • n flows • single output from the buffer of packets • wi – weight of flow i, i = 1, …, n • Li – size of packet i, i = 1, …, n Fair sequence?

4. Relative fairness • SiP(t1, t2) – service obtained by flow i in time interval (t1, t2) using discipline P t1 t2

5. Relative fairness bound t1 t2

6. GeneralizedProcessorSharing Policy

7. GeneralizedProcessorSharing Policy C – resource capacity (rate)

8. Absolute fairness bound

9. Apportionment problemformulation • n – number of states • p = [p1, …, pn] – vector of populations • h – house size • a = [a1, …, an] – vector of apportionment:

10. Apportionment problemproperties • House monotone methods • No methodminimizingis house monotone. • Population monotone methods • Every population monotone method is also house monotone.

11. Apportionment RFB transformation Packet scheduling number of states population (pi) of state i number of seats (ai) assigned to state iin a parliament of size h • n – number of flows • wi– weight of flow i • xi – number of packets of flow i sent in the considered time interval of length h • xiLi /C – number of time units assigned to flow i in the considered time interval

12. Relationbetween RFB and theapportionment problem Caj t1 t2

13. Comments Theorem There exists no house monotone method minimizing the RFB measure. Conclusion There exists no population monotone method minimizing the RFB measure.

14. ProductRateVariation 10% 10 pcs 100/10=10 15% 15 pcs 100/15=6.67 25% 25 pcs 100/25=4 50% 50 pcs 100/50=2 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 18 19 20

15. ProductRateVariation k minimize xik – number of copies of product i completed by time k di– demand for product i in the planning horizon i –weight of producti 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

16. Relation between AFB and the PRV problem assumeLi = L k packets t1 t2

17. PRV AFB transformation Packet scheduling number of products total number of copies completedink time units demand (di) of product i number of copies (xik) of product i completed inktime units • n – number of flows • k – total number ofpackets sent • wi– weight of client i • xi – number of packets of flow i sent in the sequence of k packets

18. Relation between AFB and the PRV problem ri Fi k

19. Comments • AFB with identical packet length can be transformed to the PRV problem in the min-max version. • PRV and thus AFB can be effectively solved as a linear bottleneck assignment problem.

20. Further research • Transformation of the AFB for the problem with arbitrary packet length. • Analysis of properties of schedules and algorithms minimizing the AFB.