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

2. Presentation outline Scheduling packets in packet-switched networks Problem formulation Relative fairness bound Absolute fairness bound 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 w i – weight of flow i, i = 1, …, n L i – size of packet i, i = 1, …, n Fair sequence?

4. Relative fairness S i P (t 1, t 2 ) – service obtained by flow i in time interval (t 1, t 2 ) using discipline P t1t1 t2t2

5. Relative fairness bound t1t1 t2t2

6. Generalized Processor Sharing Policy

7. C – resource capacity (rate)

8. Absolute fairness bound

9. Apportionment problem formulation n – number of states p = [p 1, …, p n ] – vector of populations h – house size a = [a 1, …, a n ] – vector of apportionment:

10. Apportionment problem properties House monotone methods No method minimizing is house monotone. Population monotone methods Every population monotone method is also house monotone.

11. Apportionment number of states population (p i ) of state i number of seats (a i ) assigned to state i in a parliament of size h n – number of flows w i – weight of flow i x i – number of packets of flow i sent in the considered time interval of length h x i L i /C – number of time units assigned to flow i in the considered time interval Packet scheduling RFB transformation

12. Relation between RFB and the apportionment problem t1t1 t2t2 Ca j

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. Product Rate Variation 10% 15% 25% 50% 10 pcs 15 pcs 25 pcs 50 pcs 1234567891011121314151617181920 100/10=10 100/15=6.67 100/25=4 100/50=2 123456789 10 11121314151617181920

15. Product Rate Variation x ik – number of copies of product i completed by time k d i – demand for product i in the planning horizon  i – weight of product i minimize 123456789 10 11121314151617181920 k

16. Relation between AFB and the PRV problem assume L i = L t1t1 t2t2 k packets

17. PRV number of products total number of copies completed in k time units demand (d i ) of product i number of copies (x ik ) of product i completed in k time units n – number of flows k – total number of packets sent w i – weight of client i x i – number of packets of flow i sent in the sequence of k packets Packet scheduling AFB transformation

18. Relation between AFB and the PRV problem ii riri 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.