Efficient Streaming for

Delay-tolerant Multimedia Applications

Saraswathi KrithivasanAdvisor: Prof. Sridhar Iyer

Problem Definition

Given:• A multicast tree with source S serving clients Cis• Resources such as transcoders, layer encoders, and

streaming servers;• Each Ci specifies two requirements:

• a minimum acceptable rate αmin and

• δi, the time it is willing to wait, once connection is established, for start of playout of content;

• Content has playout duration T and base encoding rate α

Assumptions: The topology is known with link bandwidths

remaining constant over a given prediction interval P.

Problem Definition

• Leverage delay tolerance δi, specified by each Ci to provide (for each Ci):

• continuous playout at ri such that ris across all Cis are maximized and

• each ri is greater than or equal to αmin .

Problem space

Our context

Key assumptions• The network topology is known and static

• a subscription based network • different streaming sessions may have different clients

participating

• Buffers are available at all network nodes• unlimited buffers for the analysis of the cases considered • Buffer management issues including the case of memory

constrained client devices are addressed

• Client requirements are known. • The minimum rate requirement αi of Ci, is used mainly as an

admission control parameter • When αmin is not defined by the client, a minimum rate (say

128 kbps) is assumed to be required.

• End-to-end delays are negligible.

Service types

Bandwidth

Resource used

Scheduled On-demand

Varying

Layered encoder /streamer

Static

Transcoder/streamer

Parameters defining dimensions of the problem

Contributions: Scheduled streaming, Static bandwidth

case

Details

Details

Details

Details

Example

Example

Contributions: Scheduled streaming, Varying

bandwidth case

Example

Example

Example

Details

Details

Contributions: On-demand streaming, Static bandwidth

case

Details

Details

Overview

Node architectures

• Scheduled streaming• Source Node architecture• Relay Node architecture• Client Node architecture

• On-demand streaming• Source Node – with data transfer

capabilities architecture• Streaming Relay Node – with streaming

capabilities architecture

Buffer: Resource management issues

Buffer requirements• At Source/ relay node

• B ni max = ΣP(bi

p − bjp ) × P,

where P is the prediction interval duration

• At a client node• When the last link to the client supports the optimal

delivered rate at a client, buffers can be moved up to the upstream relay node.

• When the last link is the weakest link in the client’s path, it is necessary that adequate buffers are available at the client to support play out at the optimal rate.

• Size and duration over which buffer is maintained depends on δ value of client

Effect of δ on buffer requirementWhen there is no sharing link constraint,

• increasing δj value of a client Cj, improves delivered rate• δj value for which client receives stream at Γ

is denoted by đj • Beyond đj no improvement in delivered rate; stream is

buffered at client till playout• When δj takes values between đ j and (đj +T ), mechanism

moves from streaming to partial downloading to complete downloading at (đj + T).

• When δj takes values greater than (đj + T), it is equivalent to complete download and play back at a convenient time. entire content is stored at the client for a time

(δ j − (đj +T)). • Leads to the interesting notion of “residual delay

tolerance” đR = (δ j − đj).

Service types

Bandwidth

Resource used

Scheduled On-demand

Varying

Layered encoder /streamer

Static

Transcoder/streamer

To summarize ……..

Other interesting research problems…

• In VoD scenario, the following problems pose interesting challenges:

• handling new requests during a scheduled streaming session

• handling multiple servers with independent or partially overlapping contents.

• On-demand streaming, varying bandwidths case needs to be studied in depth.

• In-depth analysis of running delay-tolerant applications on wireless devices

• Exploring possibilities of extending declarative networking frameworks to facilitate the implementation of our proposed algorithms

• Analyzing the problem in a business perspective

Future work: A planning tool for

CSPs

Selected publications• S. Krithivasan, S. Iyer,

Strategies for Efficient Streaming in Delay-tolerant Multimedia Applications , IEEE-ISM 2006, Sandiego, USA, December 2006.

• A. Th. Rath, S. Krithivasan, S. Iyer, HSM: A Hybrid Streaming Mechanism for Delay-tolerant Multimedia Applications , MoMM 2006, Yogyakarta, Indonesia, December 2006.

• S. Krithivasan, S. Iyer, Enhancing QoS for Delay-tolerant Multimedia Applications: Resource Utilization and Scheduling from a Service Provider's Perspective, Infocom 2006, Barcelona, Spain, April 2006.

• S. Krithivasan, S. Iyer, Maximizing Revenue through Resource Provisioning and Scheduling in Delay-tolerant Multimedia Applications: A Service Provider's Prespective, Infocom 2006, Barcelona, Spain, April 2006.

• S. Krithivasan, S. Iyer, Enhancing Quality of Service by Exploiting Delay Tolerance in Multimedia Applications, ACM Multimedia 2005, Singapore, November 2005.

• S. Krithivasan, S. Iyer, To Beam or to Stream: Satellite-based vs. Streaming-based Infrastructure for Distance Education, Edmedia, World Conference on Educational Multimedia, Hypermedia & Telecommunications 2004, Lugano, Switzerland, June 2004.

• S. Krithivasan, S. Iyer, Mechanisms for Effective and Efficient Dissemination of Multimedia , Technical Report, Kanwal Rekhi School of Information Technology (KReSIT), IIT Bombay, September 2004.

Thank you

Extra slides

Overview of HSM

• Components• Selection of Streaming Point (SP) - the

relay node at which the streaming server is placed.

• Delivered stream rate calculation• Cache management at SP

ClientSP

PSMHSM Contents

cached

Back

Thumb rules for placement of streaming servers

• Find bandwidth of the weakest link in the distribution network, bmin

When rj <= bmin

• Weakest link occurs in the access network• Choose node having most out going links in the

distribution network in the path of Cj

When rj > bmin

• bmin may be the weakest link in clients’ paths

• Choose node at the end of bmin, future requests for the same content can be served without incurring delays

Back

Algorithm: find_max_clients

• The algorithm involves the following two steps:

1. Choosing an appropriate streaming point

2. Finding the number of clients serviced and the delivered rates at the clients.

Back

5121

R12562

4483

R2

1284

C1

C2

S

Illustration of impact of δ =512 kbps• T=60 min.

Case 1δ1 = 0;

δ2= 0;

192kbps

Case 2δ1 =

30;δ2= 0;

128kbpsCase 3

δ1 = 30;δ2= 0;

Transcoder at R1

448

192

448

Back

3841

2562

1284

R1

R2

C1

C2

S

Illustration of impact of shared link

=512 kbps• T=60 min.Transcoder at R1

δ1 = 15;δ2= 60;

2563

480

160

480

C2’s delivered rate is compromised from 512 to 480 due to shared link constraint Back

Problem 1(a): Finding delivered rates

• Algorithm find_opt_rates_I• 2 pass iterative algorithm• Proven optimal• Complexity: O(DC), D-

number of levels, C-Number of clients

• For given transcoder placement option, finds

• Stream rates through links• Delivered rates at clients

• Optimization approach• Insights from analysis:

• Theorem 4.1: rmaxi = bi (1 + (δp/T ))• Theorem 4.2:

γmaxk = bw (1 + (δk/T )) if bw (1 + (δk/T )) < Γ

= Γ otherwise;

Back

Algorithm: find_opt_rates_I

• Two iterations:• Bottom to top – to find maximum stream rate

that can flow through each link• Top to bottom – to find the actual stream

rate flowing through each link

findoptratesI finds the optimal delivered rates at clients for any given transcoder placement option

function find opt rates I (transinfo); Initialize stream rate [numlinks] % stream rate is a vector of dimension numlinks % Initialize optim rates [numclients] % optim rates is the output vector of dimension numclients %

% start pass 1 % for each link lj % Starting from the leaf nodes of the tree % if link is directly attached to a client max stream rate = find max deliverable rate (Ci)

% Calculated using Theorem 4.2 % else max rate thru link = linkbandwidth(1+ bj/T); if there is a transcoder at node ni

% at the terminating end of the link % max stream rate = max(stream rates flowing through the outgoing links from ni); else

max stream rate = min(stream rates flowing through the outgoing links from ni); end end stream rate [li] = min(max rate thru link, max stream rate); % Apply Theorem 4.1 to assign appropriate rate to link %end

Algorithm: find_opt_rates_IPass 1

Next

% start pass 2% for each client C for link L in C’s path starting from S if L is the first link in path max rate = stream rate [L]

% Traverse path from source to client and assign streamrate of first link to max rate %

else if stream rate [L] is greater than max rate actual rate [L] = max rate; else actual rate [L] = stream rate [L]; max rate = stream rate [L]; end end optim rates [C] = actual rate [last link in C’s path] end

Algorithm: find_opt_rates_IPass 2

Back

n1 nm

Redundancy rule 1

512 256

384

384

Redundancy rule 2

If transcoders are needed in a string n1, n2, . . . nm, it is sufficient to place a transcoder at n1.

For each link li

if li is serving equi-deliverable rate clients,

assign equi-deliverable rate to li Else

assign maximum rate supported by link without loss

If l1 supports 512 kbps, transcoder at

R2 is redundant

Even if l1 supports 512 kbps, assigning it to l1 would introduce redundant transcoder transcoder at R2

Algorithm: find_opt_rates_NRBack

Optimality claims

We have the following two claims with respect to the optimal placement of transcoders:

1.Optimality of O: O is the minimal set that is required for providing the best delivered rates to clients. (the redundant rules identify all instances of redundant transcoders)

2.Optimality of find_opt_rates_NR: The set of transcoders enabled by find_opt_rates_NR when AT placement option is given as input, is the optimal set and hence is the optimal placement option for providing the best rates to clients. (the algorithm applies all the redundancy rules)

Back

Properties of O-set• O provides the best delivered

rates to clients• removing any transcoder from

O results in reduction of the delivered rate for at least one client.

• placing transcoders at nodes not in O in addition to O, does not improve the delivered rate at any client.

• removing one or more transcoders from O and placing an equal number at any node not in O does not improve the delivered rate for any client. i.e., even if O is not unique, any other combination of nodes would not provide higher delivered rates at the clients. Back

for each client C for each link L in C’s path

if link L serves equi-deliverable rates clients % Apply redundancy rule due to Lemma 5.3 %

if link L is first link in client’s path stream rate [L] = delivered rate at client; else stream rate [L] = min(stream rate of previous link, delivered rate at client) end else % Apply redundancy rule Corollary 5.2 % max deli rate = stream rate [L1]

%Traverse path from source to client and assign stream rate of first link to max deli rate if stream rate [L] is greater than max deli rate actual rate [L] = max deli rate; else actual rate [L] = stream rate [L]; max deli rate = stream rate [L]; end end end optim rates [C] = actual rate [last link in C’s path] end

Algorithm: find_opt_rates_NR

Pass 2

Back

Optimal set O = {4, 5, 6}

q = 2

Optimal combination {2, 6}

Node 2 is the super node

In Case (ii), dependent set is {4, 5, 6}

Useful set is: {2, 4, 5, 6}

Back

More definitions• Super node: A common ancestor (other

than S) to two or more nodes in the optimal set that does not have a transcoder.

• Eligible set: This set contains all the relay nodes in the network.

• Useful set: The set of nodes that includes the optimal set and the super nodes. This set is denoted by U.

Next

•Optimization based approach

• Optimal algorithm considering • combinations from the eligible set• combinations from the useful set

Comparison of optimal placement algorithms

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Topology

00.511.522.5

CP

U ti

me

(in s

ecs.

)

No. of combinations -Eligible setNo. of combintions - Useful setCPU time -Eligible setCPU time - Useful set

Back

• Greedy algorithms• Max-gain algorithm

• Given a limited number of transcoders, where can each of these be inserted (among the useful nodes) in turn such that the gain from each inserted transcoder is maximized?

• Min-loss algorithm• Given transcoders at all the useful nodes, which

of these transcoders can be removed (to use just the given number of transcoders) such that the loss is minimized?

• Performance evaluation

Comparison of greedy algorithms with optimal-U algorithm

Next

Greedy algorithms perform close to the optimal algorithm

Number of affected clients vary

Thumb rules:

For very small values of n or very large values of n, opt_u algorithm can be used

when q< |O|/2, findplacementMax gain is preferredElse Min loss is preferred

Comparison of greedy algorithms when n = |O|/4 and n= |O|3/4

number of transcoders placed q = |O|/4.

number of transcoders placed q = |O|3/4.

Back

With transcoder only at S, run find opt rates I to find the base deli rates at clients if num trans == 1 % when only one transcoder is available % for each node in useful set place a transcoder and find the delivered rates at clients; find the node that maximizes the improvement in delivered rate compared to the base deli rates end return trans placement, deli rates else % when multiple transcoders are available % if option considered == 1 max num combns = (factorial(E)/ factorial(n) factorial(E-n)); else max num combns = (factorial(U)/ factorial(n) factorial(U-n)); end for count = 1: max num combns for each combination of nodes find delivered rates at the clients for each client find the improvement compared to the base delivered rates choose combination that provides maximum improvement new transinfo = chosen placement of transcoders end trans placement = new transinfo; deli rates = delivered rates with new transinfo; num combns = combinations consideredend

Algorithm: find_opt_placement

Back

Minimize Σk, k = 1,2,... m, Σ P (Γ − rlk )2, where Γ is the base encoding rate

and rlk is the rate flowing through the last link ll for each client k over each

prediction interval in P. Constraints: • Rate constraint:

rij <= Γ ,Ұi,i = 1,2,... n, Ұj, j = 1,2,... P.

• Layering constraint: r i-1

j >= rij, Ұ i, i =1,2,... n, Ұ j, j =1,2,... P.

• Delay tolerance constraint:1. Constraints for loss-free transmission over a link :

• For a given link at the end of its active period, buffer should be empty.• At any prediction interval over the active period of a link, the available data capacity should be greater than or equal to the required data capacity.

2. Constraint for loss-free transmission to a client:• For a given client, for each prediction interval over the playout duration, for each link in its path, the available data capacity should be greater than or equal to the required data capacity.

Optimization approach for varying bandwidth case

Back

Back

Total data sent from S over T = Sent data capacity = S1

5 = (ra1 + ra

2 . . . + ra5)*P

Total data available over the active period of l1 = Available data capacity =A1

6 = (b11+ b1

2. . . + b16)*P

Total data available over the active period of l2 = Available data capacity =A2

7 = (b21+ b2

2. . . + b27)*P

Total data required over the active period of l1 and l2 = Required data capacity =D1

5 = D25 = S15 = (ra

1 + ra2. . . + ra

5)*P

Example to illustrate various solution approaches

Number of nodes N = 7; Number of links L = 6; Number of clients C = 3; Play out duration T = 100 seconds; Prediction interval duration P = 20 seconds; Delay tolerance 1 of C1 = 20 seconds; Delay tolerance 2 of C2 = 40 secondsDelay tolerance 3 of C3 = 20 seconds; Number of prediction intervals N = (T + max (i)) /P = = 140/20 = 7;

Link Bandwidths over prediction intervals 1 2 3 4 5 6 7

Link1 384 384 384 256 448 512 512 Link2 128 256 448 512 256 128 384Link3 256 128 128 192 128 256 128 Link4 384 384 320 192 320 128 128 Link5 128 192 128 384 128 128 192 Link6 384 256 384 128 192 256 192

1 2 3 4Back

Illustration of using minimum bandwidth

• Bandwidths available on the links are:• Link 1: 256 kbps;• Link 2 - Link 6: 128 kbps.

• Using algorithm find_opt_rates_I we find the delivered rates at the clients: • C1 : 153.6 kbps; C2 : 153.6 kbps; C3 : 153.6 kbps;

Back

Links are under-utilized over most prediction intervals

Illustration of using average bandwidth

• Consider three instances of predicted bandwidths on link 2.

• Predicted bandwidths on link 2 :• Case 1: 128, 256, 448, 512, 256, 128 over prediction

intervals 1, 2, 3, 4, 5, and 6 respectively• Case 2: 512, 128, 256, 448, 256, 128 over prediction

intervals 1, 2, 3, 4, 5, and 6 respectively.• Case 3: 128, 256, 128, 256, 448, 512 over prediction

intervals 1, 2, 3, 4, 5, and 6 respectively.

• The average rate is the same in all three cases, 288 kbps.

• Using Theorem 4.2, the stream rate that flows through the link is computed as: (288 + (288 * 20/100)) = 345.6 kbps.

Illustration of using average bandwidth

1 345.6 128 128 217.6 128 0 2 345.6 256 (217.6 + 38.4) 307.2 384 345.6 3 345.6 448 (307.2 + 140.8) 204.8 832 691.2 4 345.6 512 (204.8 + 307.2) 38.4 1344 1036.8 5 345.6 256 (38.4 + 217.6) 128 1600 1382.4 6 0 128 128 0 1728 1728

1 345.6 512 345.6 0 345.6(UU) 0 2 345.6 128 128 217.6 473.6 345.6 3 345.6 256 (217.6 + 38.4) 307.2 729.6 691.2 4 345.6 448 (307.2 + 140.8) 204.8 1177.6 1036.8 5 345.6 256 (204.8 + 51.2) 294.4 1433.6 1382.4 6 0 128 128 166.4 1561.6 1728

345.6 128 128 217.6 128 0 345.6 256 (217.6 + 38.4) 307.2 384 345.6

3 345.6 128 128 (179.2 + 345.6) 512 (BP) 691.2 = 524.6

4 345.6 256 256 (268.8 + 345.6) 768 1036.8 = 614.4

5 345.6 448 448 (166.4 + 345.6) 1216 1382.4 = 512

6 0 512 512 0 1728 1728

(i)

(ii)

(iii)

PI Stream rate b/w Data sent Data buffered Aij Di

j

BackLoss-free delivery is not guaranteed

Illustration of I-by-I approach

The values of the clients are applied equally across prediction intervals spanning the playout duration:

• delta applied for C1 = (20/5) = 4 seconds.

• delta applied for C2 = (40/5) = 8 seconds.

• delta applied for C3 = (20/5) = 4 seconds.

PI Delivered ratesC1 C2 C3

1 133.12 133.12 133.12 2 266.14 133.12 133.123 322.8 133.12 133.124 199.68 199.68 133.125 266.24 133.12 133.12Average 237.6 146.4 133.1

I-by-I algorithm

Initialize AT option for transcoder placement; for each PI read the predicted bandwidths into matrix M, representing the topology; setup the data structures; for each client delta applied = delta of client / number of PIs in playout duration end playout duration = prediction interval duration; [stream rates, opt deli rates] =

find_opt_ rates (transinfo); delivered rates(:, PI) = opt deli rates; end

Disadvantages of I-by-I approach

• Consideration of each prediction interval in isolation

• When the weakest link changes over the prediction intervals

• does not leverage the available bandwidths

• Lack of consideration of the predicted bandwidths over all intervals spanning the active period of a link.

• When the weakest link remains constant over the prediction intervals

• may over-estimate delivered rates and hence may not provide loss-free playout

Back

L-by-L approach

• Considers each link in isolation and finds loss-free rates supported by the link over each PI• For a given link at the end of its active period, buffer

should be empty.• Algorithm find_base _stream_rate Example

• At any prediction interval over the active period of a link, the available data capacity should be less than or equal to the required data capacity, i.e., Ai

j <= Dij .

• Algorithm find_link_stream_rates Example

• Considers links in the path of each client, to find the loss-free delivered rates at client over each PI.

• Algorithm find_delivered_rates Example

Back

Comparison of L-by-L algorithm with Optimal algorithm

• L-by-L algorithm performs close to the optimal algorithm• Even for small topologies having 10 nodes, for 7 out of 22 runs, (about 32%) the optimization function didnot converge having taken about 1200 times the time taken by the L-by-L algorithm.

Effect of number of Prediction intervals: L-by-L algorithm

Average delivered rates are almost the same when number of PIs are increased without changing bandwidth values

CPU time increases as more iterations are required by algorithms: find_base_stream_rate and find_link_stream_rates

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n prediction intervals

2xn prediction intervals

4xn prediction intervals

Effect of fluctuating bandwidths: L-by-L algorithm

Average delivered rates are almost the same when bandwidths fluctuate (while bandwidth over the original PI remains constant)

Epsilon value = 32.

When bandwidth fluctuates such that higher bandwidth is available on link in the initial part of PIs, algorithm converges faster than when lower bandwidth is available in the initial part. (both cases have same available bandwidth over 2 consecutive intervals)

Algorithm: find_adjusted_stream _rates

• Start with advance estimate of predicted bandwidths • Stream rates calculated using L-by-L algorithm

• At the beginning of each prediction interval, get refined estimate for bandwidth available on link

• Using the refined estimate, if the bandwidth available at a prediction interval (after sending any data in the buffer) is:• higher than the original available bandwidth using the

advance estimate, • increase the stream rate by the difference ( upper bound

is base encoding rate)• else

• decrease the stream rate by the difference.

Example

1 469.3 Advance 384 384 384 85.3

Revised 384 384 384 85.3

2 469.3 Advance 384 298.7 298.7 170.6

Revised 320 234.7 64 (-) 405.3 234.7 170.6

3 469.3 Advance 384 213.4 213.4 255.9

Revised 256 85.4 128 (-) 341.3 85.4 255.9

4 448 Advance 256 0 0 448

Revised 448 192 192 (+) 512 192 320

5 512 Advance 448 0 0 512

Revised 448 128 128 (+) 512 128 384

6 0 Advance 512 512 512 0

a Revised 384 384 384 384 0 b Revised 448 448 384(UU) 384 0 c Revised 256 256(buffer not empty) 256 256 128

PI Stream rate Estimate Pred. b/w Available b/w Diff. in b/w Ad. rate data sent data buffered

Back

function find base stream rate (link id, start point, num pred intervals, delta);

under utilized == 1; while under utilized == 1 available data = 0; data to be sent = 0; counter = 0; start rate = find rate (link id, start point, num pred intervals, delta); stream rate = start rate; for PI = 1: TRANS INT counter = counter + 1; curr pred rate = pred table(link id, PI); available data = available data + curr pred rate; data to be sent = data to be sent + stream rate; if available data is less than or equal to data to be sent if counter == TRANS INT % when no under-utilized link under utilized = 0; end else when under-utilization of a link is detected pred table(link id, PI) = start rate; break; end end end

Algorithm: find_base_stream_rate

Back

Algorithm: find_link_stream_ratesfunction find link stream rates (linkid, start point, num PI, delta); last PI = 0; st rates = base st rates (linkid, 1); while last PI = = 0 avail data vol = 0; req data vol = 0; counter = 0; break point = 0; while break point == 0 For PI = 1: num PI counter = counter + 1; link bandwidth = pred table[counter]; avail data vol = avail data vol + link bandwidth; if counter is greater than delta % after collecting data for delta % req data vol = req data vol + current rate if req data vol is greater than avail data vol % when the rate can not be sustained % current rate = find rate(....); for interval = (start point + delta) to counter st rates[interval] = current rate end % find the rate for the intervals up to break-point % if the last PI is reached break point = 1; last PI = 1; else start from the next PI with, num delta = 0; curr rate = find rate(....); break point = 1; end else if last PI is reached assign curr rate to stream rates for all intervals; end break point = 1; last PI = 1; end end end end end

Back

for each client C for each prediction interval P

% traverse path from source to client % for each link L in its path if first link in client’s path min rate = stream˙rates(C,P); else min rate = min(min rate[L], stream˙rates(C,P)) end end delivered rates [C, P] = min rate end end

Algorithm: find_delivered_rates

Back

Illustration of find_base_stream_rate• Consider Link 4

• data available over 6 prediction intervals is consumed over 5 intervals; the start rate computed is: 345.6 kbps.

• For the first interval, Aij = 384 kbps and Si

j = 345.6 kbps; Link is underutilized.

• Reduce predicted bandwidth in PI 1 is reduced to 345.6 and start rate recomputed.

• Repeat till start rate is equal to reduced predicted bandwidth in that interval.

• Repeat same check for every interval in active period of link

Base stream rate computed for Link 4 is: 320 kbps.

Back

PI Predicted b/w

Available datacapacity

Required data capacity

Effective stream rate

1 384 384 0

2 384 768 473.6 469.3

3 384 1152 947.2 469.3

4 256 1408 1420.8 469.3

5 448 1856 (448) (480) 448

6 512 2368 512

2368 / 5 = 473.6

1408 / 3 = 469.3

(448 + 512) / 2 = 480; can not be sustained in PI 5.Back

Illustration of find_link_stream_rates

Consider Link 1:

Illustration of find_delivered_ratesLink Stream rates over prediction interval

1 2 3 4 5 Link1 469.3 469.3 469.3 448 512 Link2 345.6 345.6 345.6 345.6 345.6 Link3 208 208 208 208 208 Link4 320 320 320 320 320 Link5 256 256 256 256 256 Link6 284 284 284 284 284

PI Delivered rates C1 C2 C3

1 320 208 2082 320 208 2083 320 208 2084 320 208 2085 320 208 208Average 320 208 208

Back

findplacement-Max-gain

Find minimum set of transcoders U needed to provide best delivered rates to clients• Start with no transcoders and find delivered rates at clients.

• Place transcoder at each node in U one at a time and find the delivered rates at clients for each of these additions.

• Choose the option that maximizes the gain. • Remove the node selected for placement from

set U. With the chosen placement, repeat the process for next transcoder placing it at each node in U.

• Note that for each round of selection, the delivered rates with the chosen placement in the previous round would be used for comparison to find the additional gain.

Findplacement-Min-loss

Find minimum set of transcoders U needed to provide best delivered rates to clients

• Start with transcoders at all nodes in U and find the delivered rates at clients.

• Remove transcoders one at a time and find the delivered rates at clients for each of these removals.

• Choose that transcoder that minimizes the loss from the removal of the transcoder at the chosen location.

• Remove selected node from U.• With the chosen placement, repeat the process

for next transcoder, removing it from each node in U.

• Note that for each round of selection, the delivered rates with the chosen placement in the previous round would be used to find the cumulative minimal loss.

Solution Approach: Optimization-based

Design variables:Stream Rates xm flowing through links lm

Objective function: Maximize playout rates at the clients:

Minimize Σi ( – xi)2

is base encoding rate, xi is the rate flowing through li

Optimization: Constraints

Rate constraint:• xm<=

where is the base encoding rate,

the best possible rate that can be delivered at any client.,

Delay tolerance constraint:• Li <= di, where Li = Lb+ Lt

Lb is the latency incurred in the path from S to Ci, due to buffering, Lb = ((ri- bw)/ bw)*T

Lt is the latency due to transcoding

Optimization: Constraints…

Transcoder constraint:• Source transcoding

• for every client Ci the rates flowing through links in its path from source are equal

• Anywhere transcoding • for every Ci, the rates flowing through links in its

path are non-increasing

Back

Optimization-based approach

Design variables:Stream Rates xm flowing through links lm

Objective function: Maximize playout rates at the clients:

Minimize Σi ( – xi)2

is base encoding rate, xi is the rate flowing through li

Optimization: Constraints

Rate constraint:• xm<=

• where is the base encoding rate, the best possible rate that can be delivered at any

client.,

Delay tolerance constraint:• Li <= di, where Li = Lb+ Lt

• Lb is the latency incurred in the path from S to Ci, due to buffering, Lb = ((ri- bw)/ bw)*T

• Lt is the latency due to transcoding

• Number of Transcoder constraint: |k: node k has an outgoing link with stream rate smaller

than rate flowing into it| <= q.• where q is the number of transcoders to be placed.

Complexity of optimization approach:

• Given n transcoders, where n < |O|. E is the eligible set.

• Number of possible unique placements is given by:

ECn = E!/n!(E-n)!

• The worst case complexity is exponential if

each option is considered in turn.

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function find max client (client arrivals) num serviced client = 0; for each arriving client if this is the first client in that subtree place streaming server

% returns the streaming server node and therate delivered at the first client %

deli rates = maximum deliverable rate at client % calculated using Theorem 4.2 %

num serviced client = num serviced client + 1; elseif the links are busy if content available in cache start streaming from the streaming server

num serviced client = num serviced client + 1;update TTL deli rates = maximum deliverable rate at client % considering path of client from the streaming point %

elseif constraints can be satisfied when the links become free deli rates = maximum deliverable rate at client % considering the remaining delta of client, when the link becomes free % num serviced client = num serviced client + 1;

else reject request

end end end

Algorithm: find_max_clients

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Proof sketch • We use induction.

• Considering the base case, let l1 be a link connecting S to node n1. Let b1 be the bandwidth available on l1, where b1>>α. Ŧ, the time to transfer file of size Z from S to n1 is given by:

Ŧ = Z/ b1, as trivially, min(b1) = b1

• We consider consecutive links l1, l2, …, lm connecting S to node n having bandwidths b1, b2, …, bm which are greater than α. Let min(b1,b2..,bm) = bmin.

• If (2) holds for these consecutive links, we prove that it also holds for links l1, l2, …, lm+1 to node n+1 having bandwidths b1, b2, …, bm+1.

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Proof sketch

• By Theorem 1, if the weakest link in the path of Cj occurs in the core distribution network bmin has to be < rj when δj >0. Hence the weakest link in Cj’s path occurs in the access network. □

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Proof sketch• Let bw be the weakest link in the path of Cj.

• Suppose bw< bmin. This implies that the weakest link occurs in the access network as bmin is the weakest link in the core distribution network. By Theorem 1, rk depends on bw and δk. For large values of δk, rk can be >= bmin. In this case, even though bmin is not the weakest link in Ck’s path, rk > bmin.

• Now we consider the case when bw= bmin. In this case, by Theorem 1, rj > bmin when δj >0. Note that here the weakest link occurs in the core distribution network.

• Hence if rj > bmin, the weakest link in Cj’s path may occur in the core distribution network, when δj >0 □

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• Enhancing QoS for Delay-tolerant Multimedia Applications: Resource utilization and Scheduling from a Service Provider’s Perspective,

Short-paper: Doctoral symposium -- Presented at Infocom 2006, Barcelona, Spain, April 2006.

• Maximizing Revenue through Resource Provisioning and Scheduling in Delay-Tolerant Multimedia Applications: A Service Provider’s Perspective

Short-poster paper -- Presented at Infocom 2006, Barcelona, Spain, April 2006.

• Network Models for Distance Education Dissemination -- Full paper, co-authored with Sridhar Iyer, Presented at International Conference on Distance Education (ICDE) 2005, IGNOU, Delhi, December 2005.

• Achieving Stable Transmission on VSAT Networks -Case Study: Distance Education Program (DEP), IIT Bombay, -- Full paper, co-authored with Sameer Sahasrabuddhe and Malati Baru , Presented at ICDE 2005, IGNOU, Delhi, December 2005.

Source node architecture

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Relay node architecture

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Client node architecture

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Source node architecture (for on-demand streaming)

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Relay node with streaming server

(for on-demand streaming)

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• either of the resources - transcoders or layer encoders - can be used to adapt the encoding rate

• When bandwidth is static• Can deploy fixed number of transcoders to

deliver optimal rates to clients• Same results can be obtained with layered

encoding also; However, number of layers required may be too high for the scheme to be feasible

• When bandwidth is varying• For transcoding to be effective, every node needs

to be transcoding capable• With little additional functionality at the relay

nodes, Source adaptive layering can be implemented efficiently

Transcoding vs. Layering