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GPS Example 2: Arrivals

0. 10. 20. Time. GPS Example 2: Arrivals. GPS Example 2: Arrivals. Eleven Sources. First source gets 0.5. Other 10 sources get 0.05 each. First source sends 11 cells. 2-11 send one each at t=0. S1. S2. p 11 1. S3. S4. S5. S6. S7. S8. S9. S10. S11. GPS Example 2: Service.

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GPS Example 2: Arrivals

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  1. 0 10 20 Time GPS Example 2: Arrivals GPS Example 2: Arrivals • Eleven Sources. First source gets 0.5. Other 10 sources get 0.05 each. First source sends 11 cells. 2-11 send one each at t=0. S1 S2 p111 S3 S4 S5 S6 S7 S8 S9 S10 S11

  2. GPS Example 2: Service • Each cell of the first source takes 2 units of time. Sources 2-11 take 20 units each. S1 p111 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 0 10 20 Time

  3. WFQ: Service • Packets finish at the same time or earlier than GPS. Some packets finish much earlier. Long period of no service Unfair. S1 S2 p111 S3 S4 S5 S6 S7 S8 S9 S10 S11 0 10 20 Time

  4. Worst Case Fair Weighted Fair Queuing (WF2Q ) • WF2Q fixes the unfairness problem in WFQ. • WFQ: Among packets waiting in the system, pick one that will finish service first under GPS. • WF2Q: Among packets waiting in the system that have started service under GPS, select one that will finish first under GPS. • WF2Q provides service close to GPS (difference in packet service time bounded by max. packet size). • WF2Q+ is an simpler implementation of WF2Q • Refs: Jon Bennett, Hui Zhang.

  5. WF2Q: Service S1 p111 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 0 10 20 Time

  6. Stochastic Fair Queueing (SFQ) • Allows # of queues less than the number of flows • Hash to queues  O(1) time for enqueuing • Packets are serviced in round-robin • If buffers full, packet in the longest queue is pushed out • Known Results: O(1) operation. • # of queues needs to be a small multiple of # of ACTIVE flows rather than total number of flows • Known Problems: Flows hashing to the same queue not isolated • Refs: McKenney, Shreedhar and Varghese

  7. Deficit Round Robin (DRR) • Use hashing to assign flows to queues • Per-Queue Quota, Per-Queue Credit (called deficit) • Send packet if Credit  packet size • After service: Credit = Credit-packet size • Known Results: Fair. O(1). • Latency depends upon frame size = sum of max packet sizes = fn of # of flows • Ref: Shreedhar and Varghese

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