Buffering problems Moran Feldman Technion Based on joint

Buffering problems Moran Feldman Technion Based on joint work with Seffi Naor

Outline • Introduction to Buffering Problems • Known Results – Results for standard buffering problems – A few proofs • Our Contribution – The problem we consider – Our results – A few proofs 2

Buffering Problems • A router in a communication network receives, buffers and transmits packets. • Often the router has to drop some of the packets arriving: – Not enough memory to keep waiting packets. – Packets have deadline, and they expire before transmission. • A buffering problem: – Describes a model of a router. – Asks for a policy for managing the packets (how to decide which packets to transmit/accept/drop). – The objective is to maximize (or minimize) some function over set of packets transmitted. 3

Standard Buffering Problems Common Properties • • Packets arrive continuously. One packet can be transmitted at every given time unit. Every packet d has an internal value w(d). The objective is to maximize the total value of transmitted packets (throughput). Bounded Delay • Each packet has a deadline. • A packet can be transmitted only before its deadline. • Any buffered packet can be transmitted. Preemptive FIFO Non‐Preemptive FIFO • Packets are buffered in a limited size FIFO queue. • The first packet in the queue is transmitted. • At any time, packets can be preempted from the queue. • Packets are buffered in a limited size FIFO queue. • The first packet in the queue is transmitted. • A packet must be either accepted to the queue or rejected upon arrival. 4

Online Problems Buffering problems are online problems. What makes a problem online? • The input arrives on the fly. • An algorithm for the problem has to make irrevocable decisions before “seeing” all the input. Game against an adversary • Online problems are often viewed as a game between the algorithm and a malicious adversary. The adversary gives the input step by step. The algorithm responds to every piece of input. 5

Competitive Ratio • Standard performance measure for online algorithms. Notation • I – An instance of an online problem. • ALG(I) – The value of the online algorithm ALG on I. • OPT(I) – The value of an optimal offline algorithm on I. Deterministic Algorithms • The competitive ratio is: • Intuitively: the adversary inspects the algorithm, and chooses the worst input for it. Remark: Usually exponential time does not improve the achievable competitive ratio. Time complexity is less important in online settings. 6

Competitive Ratio (cont. ) Randomized Algorithms • There are three adversary types with different powers. • Most works assume an oblivious adversary, i. e. , an adversary that can inspect the algorithm, but not its random coins. • The competitive ratio against oblivious adversary is: • Intuitively: the adversary choose the entire input beforehand, and then feeds it to the algorithm step by step. • Randomization often improves the achievable competitive ratio. 7

Outline • Introduction to Buffering Problems • Known Results – Results for standard buffering problems – A few proofs • Our Contribution – The problem we consider – Our results – A few proofs 8

Known Results – Bounded Delay Deterministic Best Competitive Ratio Lower Bound 1. 828 [EW 07] φ ≈ 1. 618 [AMZ 03] Randomized e/(e – 1) ≈ 1. 582 [CCFJST 06] 5/4 = 1. 25 [CF 03] • “Agreeable deadlines” assumption: the packets arrive in a non‐decreasing deadline order. Competitive ratio of φ [SS 05]. • Other assumptions yielding improved competitive ratios: – s – bounded: at most s time units between packet’s arrival and deadline. – s – uniform: exactly s time units between packet’s arrival and deadline. 9

Known Results – Preemptive FIFO • Results for this model tend to be better than for the previous one. Best Competitive Ratio Lower Bound Deterministic Randomized 3 ≈ 1. 732 [EW 06] ‐ 1. 419 [KMS 05] ‐ • If only two packet values are allowed: 1 and α. Best Competitive Ratio Lower Bound Deterministic Randomized 1. 303 [LP 02] 1. 25 [A 05] 1. 281 [KLMP 04] 1. 197 [A 05] • The worst case is usually a buffer of size 2. Improved competitive ratios are known for larger buffers. 10

Known Results – Non‐Preemptive FIFO • The optimal algorithm never preempts. Therefore, results in this model are much worse. • If packets take values from the set {1, α} only. – Tight competitive ratio of (2α – 1) / α for both deterministic and randomized algorithms [AMZ 03, AMRR 00]. • If packets take values from the range [1, α]: – Deterministic competitive ratio of e ∙ ln α [AMZ 03]. – Deterministic lower bound of 1 + ln α [AMZ 03]. Remark: Our results are for a variant of this problem allowing better results. 11

Other Buffering Problems • Most buffering problems are variants of the standard problems. Variants of the model Variants of the objective Multiple input queues Minimize the maximum delay General output bandwidth: W packets transmitted per time unit Minimize the total delay 12

Greedy Algorithm for Bounded Delay Algorithm [KLMPSS 01] At every time unit: 1. Discard packets that expired. 2. Transmit the remaining packet with maximum value. Notation • ALGt ‐ the value ALG gains till time t. • OPTt ‐ the value OPT gains starting from time t if given ALG’s configuration at time t (also denotes the corresponding schedule). • t’ ‐ a time after the deadline of the last packet. Goal Prove 2 ALGt’ ≥ OPT 0. 13

Greedy Algorithm for Bounded Delay (cont. ) Lemma For every time t, 2(ALGt+1 – ALGt) ≥ OPTt – OPTt+1. Proof • Let d. OPT (d. ALG) be the packet OPTt (respectively, ALG) transmits at time t. • OPTt+1 can use the schedule of OPTt without d. ALG and d. OPT. • OPTt – OPTt+1 ≤ w(d. OPT) + w(d. ALG) ≤ 2 w(d. ALG) = 2(ALGt+1 – ALGt). Corollary 2(ALGt’ – ALG 0) ≥ OPT 0 – OPTt’. . D E. Q. 14

Algorithm for Agreeable Deadlines Algorithm [LSS 05] Every time a packet arrives • Discard all but a schedulable set of packets of maximum value. • Sort the packets in non‐decreasing deadline order. Definitions • e – The most valuable packet among the packets with the earliest deadline. • h – The packet with the earliest deadline among the most valuable packets. Decide which packet to transmit • If w(e) ≥ w(h) / φ, transmit e. • Otherwise, transmit the first packet f with w(f) ≥ w(h) / φ and w(f) ≥ w(e) ∙ φ. 15

Algorithm Analysis • We want to prove that ALG is φ competitive. • We keep the buffers of OPT and ALG identical, at the cost of increasing OPT’s value. • Assume OPT transmits packets in non‐decreasing deadline order. Events to Consider • A packet is transmitted. • A packet arrives. Packet Transmission - (A Bit Tedious) Case Analysis 1. If both OPT and ALG transmit the same packet, their buffers remain identical. In the other cases, we use the following notation: • d. ALG – The packet ALG transmits. • d. OPT – The packet OPT transmits. 16

Packet Transmission (cont. ) 2. If d. ALG comes before d. OPT. – The deadline of d. ALG is before that of d. OPT. – OPT does not transmit d. ALG, therefore, w(d. OPT) ≥ w(d. ALG). – On the other hand, w(d. ALG) ≥ w(h) / φ ≥ w(d. OPT) / φ. We let OPT transmit d. OPT, and then replace d. ALG in its buffer with d. OPT. – ALG’s gain is at least φ‐ 1 times OPT’s gain. – The future gain of OPT only improves by the replacement. 3. If d. OPT = e. – w(d. ALG) ≥ w(e) ∙ φ Let OPT transmit both e and d. ALG, and keep e in its buffer. – Clearly, we just helped OPT. – ALG’s gain is w(d. ALG). – OPT’s gain is w(e) + w(d. ALG) ≤ (1 + φ‐ 1) w(d. ALG) = w(d. ALG) ∙ φ. The common buffer: d. OPT d. ALG d. OPT 17

Packet Transmission (cont. ) 4. If d. OPT ≠ e appears before d. ALG. – d. ALG is better than d. OPT: later deadline and higher value. – OPT is guaranteed to transmit d. ALG in the future. – OPT is not going to transmit e. We make OPT transmit d. ALG now, and then continue as before. – Only the transmission times (by OPT) of packets appearing before d. ALG are effected. – Each such packet d is still transmittable even after transmitting all earlier packets in the range e…d. – The modified OPT sends d. ALG and a subset of e…d excluding e. Hence, d remains transmittable. The common buffer: e d. OPT d d. ALG 18

Packet Arrival What Happens When a Packet Arrives? • The algorithm discards all but a schedulable set of packets of maximum value. • Can be reduced to bipartite matching (nodes on the left side are the packets, and on the right side are the transmission slots). Our Objective • Let D be the set of packets dismissed. • We want to show that: – There is a schedule for OPT that does not contain any packet of D. – Hence, OPT can dismiss D, and keep the same buffer as ALG. Proof • Let B be the set of packets ALG keeps, and let S be a schedule for transmitting all of them. • Assume the next transmission time is 1, and OPT first transmits a packet outside of B D in time t. 19

Packet Arrival (cont. ) • Observe that OPT transmits: – Till time t-1: only packets of B D. – Starting from time t: only packets outside B D. Case 1: t > |B| Replace OPT with the following schedule: • Mimic S till time |B|. • Mimic the original OPT starting from time t. Case 2: t |B| Consider the schedules of OPT and S for packets of B D. OPT: S: 1 t |B| 20

Packet Arrival (cont. ) • Assume OPT transmits a packet d D. • S transmits on the same time a packet d’ B such that: – w(d’) w(d) because S could transmit d instead. • If OPT does not transmit d’, we just replace d with d’. • Otherwise, during the time when OPT transmits d’, S transmits a packet d’’ B such that: – w(d’’) + w(d’) + w(d) because S could transmit d instead of d’ and d’ instead of d’’. • If OPT does not transit d’’, we do these replacemts. • Otherwise, continue in the same way till d is removed. OPT: S: 1 d d’ d’ d'’ t |B| 21

Outline • Introduction to Buffering Problems • Known Results – Results for standard buffering problems – A few proofs • Our Contribution – The problem we consider – Our results – A few proofs 22

Motivation • Consider a path in a communication network: Router • A packet going through the path often has a deadline. • The deadline refers to the arrival of the packet to its destination. • An online model of a router must deal with these deadlines: ‐ The standard models assume a deadline for leaving the router. ‐ A new model (proposed by Fiat, Mansour and Nadav [SODA 2008]), associates each packet with a value that diminishes over time. 23

The Model Our model is based on the model of Fiat et. al. 3 4 2 Router’s queue Time Packets Unlimited size • Arrive Continues, starting at time 0 online, each has an associated value w(d). • Packets at fractional times and are transmitted at integral Must bearrive rejected or accepted immediately. ones. • At each integral time the first packet of the queue is transmitted. Its revenue is w(d) minus the delay it suffered in the queue. 24

Results for this Model Sub models • Unrestricted model – A packet can have any positive value. • Real valued model – A packet can have any value ≥ 1. • Integral valued model – A packet can have any positive integral value. Deterministic Randomized Unrestricted Model Real Valued Model Integral Valued Model Lower bound ∞ 3(3) φ3 ≈ 4. 236 3 (3) 4 3(3) Upper bound ‐ 4. 24 4 4. 24 (4. 24) Lower bound φ∞ ≈ 1. 618 (φ) φ ≈4 1. 618 (φ) Upper bound ‐ 4 4. 24 (4. 24) Known New results appear inferred in red. from Fiat et. al. 25

Unrestricted Model – Lower Bound • Consider any algorithm ALG and a constant c. • Our job: Show that ALG is not c‐competitive. Plan of Adversary 1. For i = 1 to 2 c do: 2. Give ALG a packet of value (1/2 c)2 c+1‐i. 3. If the probability that ALG accepted any packet, so far, is less than i/2 c, terminate. 4. Give ALG a packet of value 1. Proof Idea • Packets have exponentially increasing values, but all values are at most 1, so only one packet gives revenue. • ALG must accept each packet with some probability. • OPT accepts with probability 1 the last packet. 26

The Dual‐Fitting Method Offline Primal Program • y(d, t) – Indicates whether packet d was transmitted at time t. • w(d) – The value of packet d. • a(d) – The time when packet d arrived. Remark The LP contains no FIFO constraint because this is no constraint from the offline point of view. 27

The Dual‐Fitting Method (cont. ) Offline Dual Program Method • Every dual solution is an upper bound on OPT. • To show that an algorithm is c‐competitive we do the following: § Consider any input sequence σ. § Calculate the revenue Rσ of the algorithm on σ. § Find a dual solution of value Dσ for σ. § Prove that Dσ/Rσ ≤ c. 28

Deterministic Reduction to Simple Inputs Definition An input sequence is simple if all packets in it arrive before time 1. Reduction If there is a deterministic R‐competitive algorithm A on simple input sequences, then there exists a deterministic R‐competitive algorithm B on all input sequences. Definition of B B’s Input: A’s Input: 5 5 3 3 7 8 0 1 2 3 4 5 1. Reset 2. 3. B simulates Map each the algorithm packet A andto maps when timeeach before theinput queue time packets empties. 1, andtoincrease A’s input. its value by the number of transmissions so far. 29

Deterministic Reduction to Simple Inputs (cont. ) Why Does It Work? • Let an interval be the time between two consecutive resets. • B (and A) transmit packets at all integral times in an interval. • Moving packets to earlier time: – Does not effect the transmission times. – Increase the delay that the packets suffer. • Increasing the value of the packets counters the additional delay. 5 5 0 3 3 7 8 1 2 3 30

Deterministic Algorithm for Integral Simple Inputs Algorithm (NDT) 1. Let Q be the number of packets in the router’s queue. 2. Accept a new packet d if and only if w(d) ≥ 2 Q + 1. Remark The algorithm also works for general integral inputs, although it is more difficult to show that. Reduction Assume that if a packet d is accepted, then its value is exactly 2 Q+1. Proof This is the lowest value that will get the packet accepted. 31

Analyzing NDT • Assume k packets have been received: Value for the Algorithm Assignment to the Dual LP zd = 0 Packet Value Delay The competitive ratio of NDT is: xt = max{2 k ‐ t, 0} Value of solution: 32

Randomized Reduction to Simple Inputs Difficulty The previous reduction does not work for randomized algorithms: • If we restart each time the queue empties, the simulated algorithm A will face a non‐oblivious adversary. • If we do not restart, the algorithm will not necessarily have a packet to send at each integral time of an interval. Definition A randomized algorithm is good if at all times an external observer can give a number P such that: • The algorithm has either P or P+1 packets in its queue. • P does not depend on the random decisions of the algorithm. Remark In a sense, a good algorithm is a randomized algorithm with minimal randomness. 33

Randomized Reduction to Simple Inputs (cont. ) Objective Make the previous reduction work for good randomized algorithms. Idea • We reset A’s simulation after every integral time in which we might not have a packet to transmit. • This ensures that the queue is empty after every reset. Either: – The queue was empty before the integral time. – The last packet was sent during the integral time. • A faces an oblivious adversary. – The reduction does not depend on A’s randomized decisions. • The algorithm transmits at each integral time between resets. 34

Randomized Algorithm for Real Valued Simple Inputs Algorithm (RNDT) 1. Let Q be a random variable of the size of the queue. 2. Keep the invariant Q = E(Q) or Q = E(Q) . 3. Accept a new packet d with probability: d first packet w(d) ≤ 2 E(Q)+0. 5 ½ 0 2 E(Q)+0. 5 ≤ w(d) ≤ 2 E(Q)+2. 5 w(d) ≥ 2 E(Q)+2. 5 w(d) / 2 – 0. 25 – E(Q) 1 To keep the invariant, true acceptances probability depends on the actual value of the random variable Q. Reduction Let Ef be the final value of E[Q]. No packet had value over 2 Ef + ½. Proof • For rejected packets: implied by the acceptance probabilities. • For packets d accepted with probability p: • We can assume w(d) 2 E[Q] + 0. 5 + 2 p. • E[Q] increases following the arrival by p. • 35

Analyzing RNDT Value for the Algorithm Assignment to the Dual LP zd = 0 First Packet The packet which caused E[Q] to exceed x must have value of at least 2 x + 0. 5. xt = max{2 Ef + 1. 5 ‐ t, 0} Value of solution: The competitive ratio of RNDT is: 36

Open Problems • Adversary that draws from a malicious distribution. – Competitive analysis provides too pessimistic a view. – Classic queuing theory analyzes Poisson distributed source. – Provides a middle ground between the two approaches. • A sub‐model where all values above some constant c are allowed. – Generalizes the unrestricted and real‐valued models. – The constant c represent the ratio between the shortest deadlines and the speed of the network. – The main question is finding the relation between c and the competitive ratio. 37

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