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A Generic Quantitative Approach to the Scheduling of Synchronous Packets in a Shared Medium

Wireless Access network

Reuven Cohen & Liran Katzir

Dept. of Computer Science

Technion

Haifa, ISAREL

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The considered network model

We consider a shared medium wireless channel with a centralized node (base-station) that monopolizes the downstream channel and administers the access to the shared upstream channel.

Applicability– 802.16 wireless MAN– 802.11 (WiFi)– Packet switched cellular networks– other emerging standard for short range

communications, like 802.15.3

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Outline

Introduction to the problem of scheduling in a shared Introduction to the problem of scheduling in a shared medium networkmedium network

Scheduling in a wireless channel The proposed quantitative-based approach How to handle re-transmissions Simulations Conclusions

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MAC mechanisms for controlling the access to the upstream channel

The base-station can employ several scheduling services for controlling the access to the upstream channel:

– ALOHA-like contention-based access mainly for best-effort applications.

– Unsolicited Grant Service– Real-time poling

Synchronous application = an application that generates traffic periodically, and requires some upper bound on the end-to-end delay.

for synchronousfor synchronousapplicationsapplications

Packetized voice

Streaming (one way) video

periodicity packet size access delay

fixed intervals(e.g. 20 ms.)

constant packet size(e.g. 100 bytes)

very strict(e.g. 10 ms.)

fixed intervals(e.g. 40 ms.)

bounded, but not strict

variable packet size

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Using one of the scheduling services described before, the base station acquires the following information:

– the time when every packet is ready for transmission– the size of every packet– the due time of each packet

this information is determined based on the tolerated scheduling jitter parameter, which is negotiated when every synchronous connection is established between a host and the base-station.

Using this information, the base-station needs to determine when to allocate bandwidth to each packet, such that the overall channel profit is maximized.

The base-station scheduling logic

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Scheduling example

Suppose that each of the packet requires 2 slots for transmission

1 2 3 4

call-1 (voice)

call-2 (voice)

call-3 (video)

call-4 (voice)

call-5 (video)

….. ….. … …

….. …..

…..

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Scheduling algorithms: some results

If all the packets are of 1-slot, and the jitter is fixed, EDF is an optimal solution.

– EDF (Earliest Deadline First): the packet with the earliest deadline is scheduled first.

– “optimal” = no other algorithm may schedule more packets on time. When all the packets are of K-slots (for a fixed value of K):

– EDF with or without preemption is NOT optimal but in practice it yields excellent results

– the problem can be solved using dynamic programming in O(n7), if preemption is allowed

or in O(n10) when preemption is not allowed. A new result: Finding a schedule that minimizes the number of “correlated

losses” is NP-complete– specifically, minimizing the number of time we loss M packets from any

window of consecutive K>M packets is NP-complete even if all the packets are of 1-slot.

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How likely is it to get scheduling conflicts due to high load of synchronous traffic?

Not at all… We can see that if

the tolerated jitter is large enough, we get no loss even if the load is very close to 100% !!!

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Outline

Introduction to the problem of scheduling in a shared medium network

Scheduling in a wireless channelScheduling in a wireless channel The proposed quantitative-based approach How to handle re-transmissions Simulations Conclusions

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Scheduling in a shared wireless channel

The main task of the scheduler at the base-station: maximizing the number of packets transmitted on time and not lost due to transmission errors.

Scheduling considerations:1. In order to avoid consecutive losses, the scheduler should defer the

transmission of a packet over a lossy channel as much as possible.

2. If packet re-transmission is possible in the MAC layer, the scheduler should schedule every packet as early as possible in order to allow a possible re-transmission.

3. Not all packets are equal. A packet of a session that has experienced a recent loss is probably more important than a packet of a call that has not experienced a recent loss.

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How can a good scheduler contribute to the performance of the network?

A

B

load of synch.

channel condition

length oftolerated

jitter

highdon’tcare

short(< 5ms.)

don’tcare

not all the packetscan be transmitted

on time

Scenario

B’

Cdon’tcare

higherror rate

higherror rate

shorter thanan error burst

channelproblem

schedulertask

select the mostimportant packets

longer thanan error burst

many packets canbe lost due to

errors

select the best timefor transmitting

each packet

the same as B above, but re-transmissionin Layer-2 is supported by the MAC/DLC

select the best timefor transmitting

each packet many packets can

be lost due to errors

determine which packets should not be transmitted at all

The proposed scheme is said to be “generic” because it is

supposed to give a solution for each of these scenarios.

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What is the difference between Scenario B and B’?

Suppose that the channel is known to be in error when a packet P is released, and the packet has a relatively long tolerated jitter.

– in case B, the packet should be transmitted as close to its due time as possible, in order to increase the probability for success.

– in case B’, it would be better to transmit the packet earlier than the due time, in order to enable a possible re-transmission.

Release(P) tolerated jitter

Bdon’tcare

B’

higherror rate

longer thanan error burst

many packets canbe lost due to

errors

select the best timefor transmitting

each packet

the same as B above, but re-transmissionin Layer-2 is supported by the MAC/DLC

select the best timefor transmitting

each packet

Deadline(P)

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…And what is the difference between Scenario B and C?

B

load of synch.

channel condition

length oftolerated

jitter

don’tcare

Scenario

Cdon’tcare

higherror rate

higherror rate

shortershorter thanan error burst

channelproblem

schedulertask

longerlonger thanan error burst

many packets canbe lost due to

errors

select the best timefor transmitting

each packet many packets can

be lost due to errors

determine which packets should not be transmitted at all

In B we can wait until the channel is likely to be clean In C we cannot wait, so we better decide not to transmit the

packet at all. expected length of an error burst

release(P)the channel is known to be bad

Deadline(P)

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Outline

Introduction to the problem of scheduling in a shared medium network

Scheduling in a wireless channel The proposed quantitative-based approachThe proposed quantitative-based approach How to handle re-transmissions Simulations Conclusions

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The quantitative-based approach

Main idea: the scheduler maintains a profit matrix M for each scheduling interval.

Each pending packet is represented by a row. Each column represents one time slot. M[i,j] indicates the profit from transmitting packet- i starting from

slot number j1 2 3 4

12 6 6 6

12 12 12 0

6 6 12 12

0 0 4 4

packet-1

packet-2

packet-3

packet-4

5 6

6 0

0 0

12 12

4 4

2 2 2 2packet-5 0 0

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The quantitative-based approach (cont.)

The scheduler searches for a schedule that maximizes the overall benefit. e.g. assuming that each packet needs two slots for transmission, the

following schedule has the maximum benefit (of 36):

The problem of finding an optimal schedule once the matrix is given is NP-complete.

– in the paper we present efficient approximation algorithms for this problem

1 2 3 4

12 6 6 6

12 12 12 0

6 6 12 12

0 0 4 4

packet-1

packet-2

packet-3

packet-4

5 6

6 0

0 0

12 12

4 4

2 2 2 2packet-5 0 0

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The biggest question: how should the matrix be filled ???

Example (1):– Let’s the value of every packet P be proportional to the packet length

L(P). – Suppose that we know the probability e(P,s) for an error in packet P

when it is transmitted in slot s– Therefore:

the profit for transmitting packet P in slot s is L(P)(1-e(P,s)). Example (2);

– Suppose now that every packet P contains also a low-bit encoding of the previous packet P-1.

i.e. *L(P) bits are for frame P-1 and (1- )L(P) bits are for P the profit is (1- ) *L(P)*(1-e(P,s)) if P-1 was received on time,

and L(P)*(1-e(P,s)) if P-1 was not received on time.

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How to compute the value of e(P,s)?

We model the loss process using a 2-state Gilbert model:

Let S(n) be the state during slot n (S(n) = 0 or S(n) = 1).– Prob[S(n+1)=0 | S(n)=0] = p and– Prob[S(n+1)=1 | S(n)=1] = q

00goodgood

11badbad

1-q1-q

1-p1-p

pp qq

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How to compute the value of e(P,s)? (cont.)

Prob[S(n+1)=1]=Prob[S(n)=1]*q+Prob[S(n)=0](1-p)

Using this equation recursively, we determine– Prob[S(n)=1 | S(0)=1] = (p+q-1)^n +(1-p) * ((p+q-1)^n-1)/ (p+q-2)– Prob[S(n)=1 | S(0)=0] =(1-p) * ((p+q-1)^n-1) / (p+q-2)

Therefore: we know the probability of the channel to be in a good/bad condition given the outcome of the last transmission by the same node.

00goodgood

11badbad

1-q1-q

1-p1-p

pp qq

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Assuming no knowledge of p and q

The previous equations require the scheduler to know the value of p and q for every channel.

Due to the mobility of users, computing these parameters might be difficult.

It is clear that for practical values of p and q:– the value of Prob[S(n)=0 | S(0)=1] is maximum when n is maximum– the value of Prob[S(n)=0 | S(0)=0] is maximum when n is minimum– (practical values of p + q > 1)

00GG

11BB

1-q1-q

1-p1-ppp qq

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Assuming no knowledge of p and q (cont.)

A good approximation:a) if last transmission of ANY packet in the channel was good, use a

decreasing linear function whose maximum is Release(P)

b) if last transmission of ANY packet in the channel was bad, use an increasing linear function whose maximum is Deadline(P).

This approach does not work for Scenario C– For this scenario, we need to have good estimates for p and q in

order to compute Prob[S(n)=0 | S(0)=1] and Prob[S(n)=0 | S(0)=0].

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Prob.Prob.

timetime

Release(P)Release(P) Deadline(P)Deadline(P)

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ProbProb..

timetime

Release(P)Release(P) Deadline(P)Deadline(P)

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Outline

Introduction to the problem of scheduling in a shared medium network

Scheduling in a wireless channel The proposed quantitative-based approach How to handle re-transmissionsHow to handle re-transmissions Simulations Conclusions

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How to handle re-transmissions (scenario B’)

Consider the case where 1 re-transmission is possible. when should the first transmission take place if the channel is

known to be good?good?– obviously as soon as the packet is released

– When should the packet be re-transmitted in case of a bad transmission? obviously as late as possible

Release(P) tolerated jitter Deadline(P)

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How to handle re-transmissions (cont.)

When should the first transmission take place if the channel is known to be bad?bad?– on one hand, we want to postpone the transmission as much as

possible– but on the other hand we want to leave enough time for a possible re-

transmission.

– When should the packet be re-transmitted in case of a bad transmission? obviously as late as possible

Release(P) tolerated jitter Deadline(P)

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How to handle re-transmissions (cont.)

Let i and j be the time when the first transmission and the second transmission should take place, respectively. – Hence, Release(P) i < j Deadline(P).

Let Fg(i,j) be the probability for success (either in the first or in the second transmissions), provided that at slot 0 the channel is goodgood::

Fg (i,j)=Prob[S(i)=0|S(0)=0]+Prob[S(i)=1 | S(0)=0]* Prob[S(j)=0 |S(i)=1]

– as expected, Fg (i,j) is maximum for i=Release(P) and j=Deadline(P) Let Fb(i,j) be the probability for success (either in the first or in the

second transmissions), provided that at slot 0 the channel is badbad::

Fb (i,j)=Prob[S(i)=0|S(0)=1]+Prob[S(i)=1 | S(0)=1]* Prob[S(j)=0 |S(i)=1]

first transmission is good first transmission is bad, but second is good

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How to handle re-transmissions (cont.)

Fb (i,j) is maximized when j=Deadline(P) and

i=max(N/2, Release(P))

where N is the number of slots between the previous faulty transmission of any packet (sent by the same station) and Deadline(P)

Example:

Release(P) tolerated jitter Deadline(P)previous faulty transmission

N=36 slots

18 slotsi (first transmission) j (re-transmission, if needed)

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Extending this result to multiple re-transmissions

By induction on the number X of possible transmissions we show that – if the last transmission in the channel was good, then P should be

scheduled as close as possible to Release(P), regardless of X.– if the last transmission (at time t) was bad, then

if only one more transmission is allowed, this transmission should occur as close as possible to Deadline(P).

if X>1 additional transmissions are allowed, then the next transmission should take place at: i=max{Release(P),t+((Deadline(P)-t)/X}

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Extending this result to multiple re-transmissions: example

Previous faulty transmission at 1, Deadline(P)=36 Suppose that 4 transmissions are possible 36/4=9, but since the packet is only available at 13, first

transmission is recommended to take place at 13. Suppose that due to other constraints the first transmission occurs

at slot 15, and it fails. Second transmission should take place at 15+(36-15)/3=22.

Release(P) tolerated jitter Deadline(P)previous faulty transmission

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 20 1 2 3 4 5 6 7 8 9 30 1 2 3 4 5 6 7 8 9 40 1 2 3 4 5 6

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We generalize the concept with a linear function as follows:a) if last transmission of ANY packet in the channel was good, use a

decreasing linear function whose maximum is when the packet is available, i.e. MAX(Release(P), current-time)

b) if last transmission of ANY packet in the channel was bad, use an increasing linear function between the current time and the optimal time, and then a decreasing linear function until Deadline(P).– the optimal time is determined based on the number of re-

transmissions left for this packet.

How to handle multiple re-transmissions without knowing the value of p and q

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Prob.Prob.

timetime

current timecurrent time Deadline(P)Deadline(P)

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Prob.Prob.

timetime

current timecurrent time Deadline(P)Deadline(P)

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Outline

Introduction to the problem of scheduling in a shared medium network

Scheduling in a wireless channel The proposed quantitative-based approach How to handle re-transmissions Simulations ConclusionsConclusions

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Simulations – Percent bandwidth gained by our algorithm compared to the strawman alg.

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Simulations – Percent bandwidth gained by our algorithm compared to the strawman alg.

The longer average error burst does not allow our algorithm to

schedule the transmission of a packet in an erroneous channel

far enough from the previous erroneous transmission.

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Outline

Introduction to the problem of scheduling in a shared medium network

Scheduling in a wireless channel The proposed quantitative-based approach How to handle re-transmissions Simulations ConclusionsConclusions

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Conclusions

We presented a generic quantitative-based approach for the scheduling of synchronous packets in a wireless access network

– it is generic because it is applicable for different scenarios: e.g. when tolerated jitter is large (video) or small (voice) when the channel is congested or not when the error rate is high or low etc…

– it is quantitative-based because it assigns a profit to the transmission of every packet at every slot, and maximizes the total profit.

Most important part of the scheme – how to fill up the profit matrix– this is easy when we know the error distribution of each channel– more difficult when such information is not available– more difficult if the same packet is allowed to be transmitted several

times.