3. introduction to data buffering and disk scheduling for ...enpklun/eie552/buffering.pdf · –...

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DEPARTMENT OF ELECTRONIC AND INFORMATION ENGINEERING

Internet Technology for Multimedia Applications

3. Introduction to Data Buffering and Disk Scheduling for

Multimedia Servers

via Network to Clients

buffers

:

:: empty

filled

Disks

2



ReferencesD.J. Gemmell, “Disk Scheduling for Continuous Media”, Chapter 1, Multimedia Information Storage and Management, Kluwer Academic Publishers, 1996.K.K. Ramakrishnan, L. Vaitzblit, C. Gray, U. Vahalia, D. Ting, P. Tzelnic, S. Glaser, and W. Duso, “Operating System Support for a Video-on-demand File Service”, Multimedia Systems, pp.53-65, No.3, 1995.D.J. Gemmell and J. Han, “Multimedia network file servers: multichannel delay-sensitive data retrieval”, Multimedia Systems, pp.240-252, No.1, 1994.

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ContentsIntroduction to On-demand Media ServersMulti-stream Scheduling– Disk placement policy for multi-stream– Buffer conservation– Admission control– Disk scheduling– Data replication and striping

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3.1 Introduction to On-demand Media Server

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A simplified model for Internet multimedia system

Clients

bufferbuffer

bufferbuffer

bufferbuffer

buffer

bufferbuffer

bufferbuffer

bufferbuffer

DISKs

buffer

Media Server

Internet

DISKs

buffer

Media Server

Proxy

Proxy

Other networks

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On-demand multimedia servicesProvide high quality on-demand multimedia services is the goal of current Internet multimedia streaming research Critical components in the design of multimedia services are – Multimedia storage servers

Able to support efficient multimedia data retrieval and storage

– Network infrastructureEnable multimedia data delivery with quality of service

– Client support (critical only in some environments, e.g. mobile)

Playback the received multimedia data in the best quality

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Design consideration of on-demand multimedia servers

Video and audio are continuous media (CM)– Convey meaning only when presented in time

– Interaction between media streams further require accurate timing when presenting the data

– Hence multimedia servers need real-time response

On-demand multimedia servers may receive enormous requests for service at any instance of time – Servers need to cater for worst case situation

– Need effective schedule in using the resource

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Ideal on-demand multimedia servers

An ideal on-demand multimedia server should have the following attributes:– Have enough storage capacity for all on-line multimedia items– Can respond to user request with minimum latency– Can deliver multimedia services in time– Require the least amount of memory storage for buffering– Impose the lowest buffer requirement to the clients – Impose the lowest processing power requirement to the clients

Usually trade-off is required to makeTo keep a good balance is the objective of current research

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How about using the conventional file server?

Conventional file server cannot achieve the task since– does not guarantee real-time delivery of data

Retrieve and store data in a best effort manner

– does not consider the continuous nature of media dataMedia data can be fragmented in the hard disk hence unnecessary disk arm movements are introduced

– does not aware of the huge file size and enormous request of services in worst case

Can lead to starvation due to inefficient use of resource

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Disk bandwidth is one of the major concerns

Mbytes/sec Growing very fastHow about serving 1000 customers each with 2.5Mbps (MPEG-2)?⇒ 312 Mbytes/sec!

Besides, does the disk bandwidth can be fully exploited?– Decrease by seek time,

rotational latency, and othersExtracted from www.adaptec.com

Resource constraints within a server

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Seek time and rotational latencyDelay is introduced for the disk arm to locate at the required position on the diskSeek time - time to move the read-write head to the new radial positionRotational latency - time it takes for the target sector to rotate under the headFor example– Seagate Cheetah 73 : 73.4GB SCSI Ultra160

10033RPM (worst case rotational latency 5.98ms), seek time 5.85ms avg

The delay is enough to transfer 1.8Mbytes of data!

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Disk storage capacity needs to be carefully considered– Total size of video files kept on-line cannot exceed disk space

– Depend on the quality, multimedia file can require a very large disk storage

Digital Video (DV) - 10GB per hourMPEG-2 - 1.1GB per hourMPEG-1 - 450 MB per hour

– To improve the redundancy and performance, items may be replicated in a disk

– Can be expensiveSeagate Cheetah 73 : 73.4GB SCSI Ultra160 costs USD1,070

– Disk storage should be hierarchical so that only popular items are kept on-line

Other resource constraints - Storage

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Other resource constraints - Memory


buffers

:

:: empty

filled

Disks

• To support multiple clients, need buffers to read-ahead• All buffers receives a chance of being refilled in each

round of service (time = service period)• Disk overheads(seek and rotation) are incurred• Must refill buffers before under-run; must not over-run

buffers (i.e. do not read more than the buffer can store) • Need an efficient scheduling algorithm

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3.2 Multi-stream Scheduling

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Disk access model


buffers

:

:: empty

filled

Disk

• Each buffer has its turn to be read or to be filled• A server is said to complete a round of service by giving one

service to refill the buffers of each stream• Such process are repeated periodically• Depend on the disk scheduling policy, each buffer will be filled

in a specific order (i.e. may not be sequentially)

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A. Disk placement policy for multi-streamContiguous Non-Contiguous

Trackno.

0

n• Avoid excessive seek time• Can fully utilize disk storage for

write-once-read-many access• Will have fragmentation problem

if the file is frequently updated• May need off-line defragmentation

• Reduce greatly the disk bandwidth due to the increase of seek time

• Can fully utilize disk storage for write-many-read-many access

• Use in traditional file systems

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admission playing startsservice starts service ends

Buffered data

time

playing endsstarvation

d1 d2 d3

• The time to refill can change from time-to-time• Depend on the number of clients admitted and the disk

scheduling policy• To avoid starvation, the buffer size and admission control

are important

B. Buffer conservation

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pmax and dmax

p1 p2 p3

d2d1

time

• pi is the time for each round• controlled by the number of clients admitted

• di is the time for two successive read• controlled by the number of clients admitted and the disk

scheduling policy • The maximum possible p and d are called pmax and dmax

• dmax ≤ 2 pmax and pmax ≤ dmax

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A simple algorithm to avoid starvation - Buffer conservation algorithmAssumptions• Data of all streams are placed in the contiguous mode on the disk• The consumption rate of the network to the buffer is constant and

is denoted as rc

AlgorithmPre-fetch rcdmax bytes of data into the buffer before the serviceFor each round

Let the current content in the buffer is fif (rcdmax - f) > rcpmax

read rcpmax from disk and add to buffer else

read rcdmax- f bytes from the disk and add to buffer

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p1 p2 p3

d2d1time

Buffered data

time

no need

rcdmaxrcpmax

f1

f2

p4

d3

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Explanation• To the worst case, dmax = 2pmax

• Case 1

pmax

d2d1 = dmaxtime

Buffered data

time

rcdmax rcpmaxf1

pmax pmax pmax

No starvation at anytime

f2

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Explanation• To the worst case, dmax = 2pmax

• Case 2

pmax

d2d1 = dmax

time

Buffered data

time

rcdmax rcpmaxf1

pmax pmax pmax

No matter when the maximum data consumption is required, no starvation is occurred

d3 = dmax

f2

rcpmax

f3

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Since the average data consumption to the buffer is rcpmax, the refill of rcpmax in each round achieves buffer conservation

⇒ operations in a reading period never lead to a net decrease in data buffered for any channel

The only condition that the refill cannot achieve rcpmax

is when there is not enough free space in the buffer

In this case, the buffer storage is top-up to rcdmax, which is the maximum data consumption of the buffer

Hence starvation is not possible

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It seems from the result above that the lower bound of buffer size is rcdmax

Unfortunately, for most hard disks, basic unit of disk access is a sector

Assume the sector size of a particular disk is s, the minimum buffer size is

It indicates that extra space is required to store the whole sector although only 1 byte is required

Buffer size lower bound

Minimum buffer size = rcdmax + s -1

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C. Admission control

Since pmax is directly determined by the no. of clients in the system, admission control is very important

where N is the no. of clients and α is the time to serve each client

pmax = α N

α = disk overhead + disk read timedisk overhead = seek time (ts ) + rotational latency (tr )

disk read time = rc pmax / disk data rate (rd )* Here we assume that the bottleneck is on fetching data

from disk but not pushing data to the network

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It is seen that the relationship between pmax and N is

recurrent N ⇑ ⇒ pmax ⇑ ⇒ α ⇑ ⇒ N ⇓

N

Buffer size

Typical relationship between buffer size and N

When α is dominated by the disk overhead

When α is dominated by the

disk read time

d

crs r

prtt

pNmax

max

++=It can be shown

that(a)

α

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N is limited by the total memory available (M) for buffer

Here we assume that all streams have the same rate rc

It is seen that, again, it is a recurrent relationship

M ⇑ ⇒ N ⇑ ⇒ pmax ⇑ ⇒ dmax ⇑ ⇒ N ⇓

Hence increase memory does not necessarily increasenumber of clients

N(rcdmax+ s - 1) < M

M

N

Typical relationship between M and N

(b)

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Combine eqns.a and b and assume the worst cases

dmax = 2pmax, tsmax = the maximum seek time, trmax = the maximum rotational latency, we have

( ) Ms

rNr

ttNrN

d

c

rsc <

−+−

+ 11

2 maxmax(c)

Here we need to ensure that total consumption rate rcN should be smaller than the disk data rate rd

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A Simple Algorithm for Admission Control

Given M, ts , tr , rc , rd and sFor a new request from client such that the total number of clients is N

if rcN > rd

reject request else if eqn.c is satisfied

admit request else

reject request

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D. Disk schedulingThe disk overhead can greatly reduce the number of clients in the system

The disk overhead can be reduced by using a good disk access schedule

Disk scheduling ≡ ordering of service

Approaches– Control the order of service (refilling) and reduce the seek-

length incurred between reads.– Avoid seeks that can occur within a disk read.– Reduce rotational latencies between reads.

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Round robin scheduling

Append service of new stream at the end of recurrent scheduleRetrieve data from disk in the same order of requestCan lead to a great disk overhead due to many track seeking

time

Head position

RR

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SCAN schedulingRefill stream buffers according to cylinder positions on both inward and outward sweeps of the disk-armOptimized to minimize the seek timeMay require large buffer since worst case inter-service time can be 2*service period

time

Head position

SCAN

AB

C

D

E

AB

C

D

E

worst case

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CSCAN schedulingRefill stream buffers according to cylinder positions only on inward sweep (or outward sweep) of the disk-armA balance to seek time and buffer size

time

Head position

CSCAN

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Maximum seek time of schedulesIt is given that the seek time may be approximated by the following linear function

ε+−

−−+=

1)1(

max

minmaxmin S

TTstTts

whereTmax and Tmin are the maximum and minimum seek time as shown in the specification of the disk drivest is the number of tracks to be soughtSmax is the maximum number of seeking trackε is a constant depend on the drive

(d)

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Consider in each round, j streams are to be read that leads to j seeks (assume one seek per stream and the data of each stream are located at different track)Assume for the i th seek, sti tracks are to be soughtThe total seek time is

ε

ε

jjstS

TTjT

STTstTt

i

j

i

j

iistotal

+−−

−+=

+

−−

−+=

∑

∑

=

=

)(1

1)1(

1max

minmaxmin

1 max

minmaxmin

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For Round Robin scheduling method, the number of tracks to be sought is random. Consider the worst case

222)1)(2(2

)1)))(1(()1(()1(

))1((...)2()1(2max

maxmax

maxmaxmax

maxmaxmax1

−+−−=

−−−+−+−=

−−++−+−≈

∑=

SjjS

jjSSS

jSSSstj

ii

Trackno.

1

Smax

Always swing to the opposite end

time

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For SCAN, the maximum total number of tracks to be sought in each round is bounded to be Smax

1max max1

−=

∑=

Sstj

ii

For CSCAN, the maximum total number of tracks to be sought in each round is bounded to be 2Smax

)1(2max max1

−=

∑=

Sstj

ii

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εjjSjjSS

TTjTtRRs +−

−+−−−

−+= )

222)1)(2((

1maxmax

max

minmaxminmax

Hence the maximum seek time for each scheduling method is as follows:

εjjSS

TTjTtSCANs +−−

−−

+= )1(1 max

max

minmaxminmax

εjjSS

TTjTtCSCANs +−−

−−

+= )22(1 max

max

minmaxminmax

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E. Data replication and stripingWhen multiple disks are available, titles can be replicated according to their popularity

Title A Title A Title A Title A

Title C Title B Title C Title C

Title D Title E Title F Title G

Data replication increases the number of clients in the system with the cost of using up more storageRequire prior information about the title popularityShould update constantly



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Striped Retrieval

Not all clients are watching the same part of a titleTitle can be striped and placed on different drives

A1

A2

A3

A4

Provide same of amount of bandwidth and serve same amount of clientsThe storage is much reduced

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If the size of each stripe is large, it is highly possible that a large no. of clients will make request to a single disk

To achieve better load balancing, each stripe should be further divided into smaller unit, e.g. disk block and distributed to different disks, known as cyclic striping

A1

A2

A3

A4

Large stripe Small stripe

0 to 1 hr 1 to 2 hr

2 to 3 hr 3 to 4 hr

All requests in the first

hour will be sent to the first disk

0 to 10 sec10 to 20 sec

20 to 30 sec

30 to 40 sec

40 to 50 sec

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Time(A,1)

(B,1) (A,2)

(C,1) (B,2) (A,3)

(D,1) (C,2) (B,3) (A,4)

(x,y) = title x block y

=Seek time

To minimize the seek time between read, the SCAN or CSCAN scheduling may be applied to each diskThat is, data on nearby track should be read first

(A,5) (D,2) (C,3) (B,4)

1 round

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Final comments• Much research is doing to cater for different practical

situations• Streaming with variable consumption rate•Disks with variable disk data rate• Supporting hypermedia• Supporting VCR function • and many other

• Even built correctly, a streaming server can only deliver 60% to 80% of the aggregated disks’ raw bandwidth• Anything done wrong, may get as little as 10%

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