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CS 414 – Multimedia Systems Design
Lecture 30 –
Buffer Management (Part 3)
Klara Nahrstedt
Spring 2012
CS 414 - Spring 2012
Administrative


MP2 posted
MP2 Deadline – April 7, Saturday, 5pm.
CS 414 - Spring 2012
Covered Aspects of Multimedia
Image/Video
Capture
Audio/Video
Perception/
Playback
Audio/Video
Presentation
Playback
Image/Video Information
Representation
Transmission
Audio
Capture
Transmission
Compression
Processing
Audio Information
Representation
Media
Server
Storage
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A/V
Playback
Outline

VoD Client Buffering Techniques
 MinBuf
 MaxBuf
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Client/Server Video-on-Demand
System
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Buffering Strategies at VOD
Client

Minbuf - minimum buffering strategy
 Minbuf
minimizes buffering requirements at
VOD client, but makes more demands on
network (throughput and delay guarantees)

Maxbuf – maximum buffering strategy
 Maxbuf
buffers more than one unit of
information and eases QoS guarantees
demands on network
 Buffering only up to a limit (Bufmax)
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Buffering Model
ZOi(t) – amount of bits in multimedia object
(e.g., Video frame) Oi displayed at time t
 Cr(t) – amount of bits received at receiver
at time t
 Two buffer states in each buffering
strategy:

 Starvation

if Cr(t) ≤ ZOi(t)
If Cr(t) > ZOi(t), no starvation at VOD client
 Overflow
if Cr(t) ≥ ZOi(t) + Bufmax
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Client Buffer Constraints
(Received amount of data Cr(t) should always be
between ZOi (t) and ZOi (t) + Bufmax)
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Implications of Minbuf Strategy


With minbuf strategy – delivery time of a unit of
information is the display time of previous unit
Minimum Throughput (Th) in bits per time unit
required is
min buf
min
Th

Z Oi

ti  ti 1
Delivery time instant of the first unit before start
Z Oi
of display: t 
x
min buf
Thmin
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Example (1)

Question: Given Video Stream with
 320x240
pixels, 20 fps, 8bits per pixel, uncompressed
 What is the minimal buffer size if Minbuf technique is
used?

Answer:


Buffer Size is of the application frame Z = 76800 bytes if
receiving and display operations are done sequentially
One can also deploy a dual buffering scheme of 2* Z if
receiving and display operations are done concurrently
Display Thread
Receiving Thread Display Thread
Phase 1
ti
ti-1
ti-1
ti-2
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ZOi+1
ZOi
Phase 2
ZOi-1
ZOi
Receiving Thread
ti
ti+1
Example (2)


Question: What is the throughput under the
above given video characteristics and minbuf
strategy?
Answer: MinBuf Throughput needed in end-toend fashion
Z
= 76800 bytes, 20 fps (50 ms period)
 ti – ti-1 = 50 ms = 50/1000 sec
 Throughput = Z/ti – ti-1 = 76800 bytes/ (50/1000 sec) =
1536000 bytes/sec = 12288000 bits/sec = 12000 kbps =
11.71875 Mbps
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Implication of Maxbuf Strategy

Delivery schedule of VOD server and network
may cause that Bufmax bits will be delivered
every K seconds, where
K

= Bufmax/Th,
Th – throughput of network
Minimum Throughput required
 ∑ZOi
– total size in bits of all objects that will be
presented in stream:
n
max buf
Thmin

Z
i 1
Oi
 Buf max
Stream _ Display _ Duration
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Example (1)

Question: Given Video frame 320x240 pixels, 8bits/pixel;
Video frame rate 20 fps, uncompressed, and Network
throughput 8 Mbps


Answer: arrival time every k = Bufmax/Throughput =
5.8599375 Mbits/8 Mbps = 0.83242 seconds


What is the worst case arrival time of frames if Bufmax is 10
frames (holds ½ second of video frames)?
What is the Bufmax size if frames arrive on average every k = 1
second?
Answer: Bufmax = k*Throughput = 8 Mbits = 8388608
bits = 1048576 bytes = 13.653 video frames ; round up
to 14 frames; Bufmax should be 14 frames if throughput
is 8 Mbps and arrival of frames is approximately every 1
CS 414 - Spring 2012
second
Example (2)

Question: Given video characteristics of
320x240 pixels, 8 bpp, 20 fps,
uncompressed, length of the move is 1
minute, Bufmax holds 1 second of frames,
what is the minimum throughput we would
need if we use MaxBuf strategy?
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Example (3)

Given video characteristics of 320x240 pixels,
8bpp, 20fps, MJPEG compressed with I frames
50% compression (1:2) of each frame;
 Question:
what would be the arrival time (k) of frames
if we have buffer size, holding 1 second frames, and
network throughput of 10 Mbps?
 Question: what would be the minimal throughput
needed from the network under above conditions if
the video clip would be 30 seconds long?
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Example (4)
Given MPEG-2 compressed video with
GOP of IPBBPBBPBBI, average I frame
size of 10KB, average P frame size of 6
KB, and average B frame size of 2 KBytes.
 Question: What would be the buffer size
(express it in bytes and time duration) if
network throughput is 4 Mbps and arrival
time of each compressed frame is on
average every 500 ms?

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Standard Video Buffering

Video delivered over “deterministic” channel (satellite,
cable, digital TV broadcasting) simple
 Small buffer (minbuf technique)


Data arrives to the media player with deterministic delay and rate and
infrequent data drops
Video delivered over IP networks problematic

Big buffers (maxbuf technique)


No guarantees in IP networks
Streaming servers manage simple buffering

Data sent as quickly as possible; when buffer full, (reached predefined
threshold), video playback starts and server continues to send data
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Implementation Issues
 Buffers for Uncompressed Periodic Streams


Buffers for Compresses Periodic Streams



Use circular buffers, but carefully consider the size of buffer
unit (depending on MPJEG or MPEG) with maxbuf strategy
For MPEG may consider dynamic buffer allocation
Buffers for Control Information


Use circular buffers as prefetch buffers with maxbuf strategy
Use priority queues or FIFO
Buffers for Non-RT Data

Use maxbuf strategy with static buffer allocation
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Implementation Issues

Dual-Threshold Buffering (in Flash Media
Server)
 Buffer
has two thresholds:
When buffer filled with data up to first threshold,
start playback
 After that continue buffering until second, higher
threshold

 Advantage:
assures fast start, and at the same time
 provides fairly high resilience to bandwidth
fluctuation

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Conclusion

Will talk about media server in next
lectures
CS 414 - Spring 2012