A simple Model of P2P
Streaming and Some
Extensions
September 29, 2008
Dah Ming Chiu
Chinese University of Hong Kong
P2P content distribution

Network multicast
 Is
pre-occupied with network efficiency
 Find an efficient tree to reach all receivers

P2P content distribution
 Tries
to maximize throughput (or minimize makespan)
Makespan of broadcasting



Various algorithms studied
Given n receivers, makespan for single message
O(log n) * T
But divide “message” into M pieces, makespan
~ (M + O(log n) * (T/M)
= T + O(log n) * T/M
~T
as M becomes large
~
The Secret of p2p content distribution

Use multiple trees
Divide content into multiple
“chunks” or “stripes”
 Let different chunks (stripes)
flow through different trees
 Every node serves as interior
node in some tree(s), hence
contribute in the distribution
 The multiple trees are
distributing content in
parallel!

source
1
2
3
5
6
4
7
8
The capacity of p2p networks


The multiple distribution trees are logical trees in an
overlay network
To study the capacity of p2p network, need a
model of how these overlay trees map to physical
network resources
The uplink sharing model - Mundinger

Assume away the network complexity
Only “uplinks” can be bottlenecks in peer communications
 Practically all models make this assumption
 Under this assumption, it is possible to derive a simple “rule
of thumb”, even if peers have different uplinks

J. Mundinger, R. Weber, and G. Weiss. Optimal Scheduling of Peer-to-Peer File
Dissemination. Journal of Scheduling, 2007.
When peers have unequal capacity


What is the minimum makespan when peers have different
uplink capacities?
It is a MILP (Mixed integer linear programming problem)



See Mundinger’s thesis for details…
Solutions can be derived for small M and N
But there is a simple asymptotic, closed-form solution when M is
very large! -> fluid approximation
The simple formula
Given the uplink sharing
model (right figure) and very
large M, the maximum
throughput is
These two values are equal when
C0 = sum(Cj) / (n-1)
Most of the time, server’s capacity is larger,
so R equals to the 2nd term
Proof step 1: R is upper bound

C0 is uplink bandwidth from server


Server must send content out at least once –
C0 is a clear upper bound
C0 + sumj(Cj) is the total uplink
bandwidth from server and all peers


Each peer must receive server’s content from
some where.
The total demand is therefore n*R, which must
equal to the total supply
Proof step 2:
Realizing R by construction




One 1-hop tree (from sender)
N 2-hop trees (from each peer)
Assign rates optimally, satisfying the
uplink constraints
Two cases:


C0
C1
C0 > C/(n-1) where C = sum Ci
C0 <= C/(n-1)
Cn
C2 C3
C0
C1
Cn
C2 C3
Maximum rate achieved

Case 1: C0 > C/(n-1) where C = sum Ci
Assign rate Ci/(n-1) to the ith 2-hop spanning
tree
The ith spanning tree can deliver Ci/(n-1) to
each other peer
 Assign rate C0 – C/(n-1) to the 1-hop spanning
tree
The server can deliver (C0-C/(n-1))/n to each
peer
Each peer receives (C0 + C)/n


C0
C1
Cn
C2 C3
C0
Case 2: C0 <= C/(n-1)
Assign rate C0*Ci/C to the ith 2-hop spanning
tree, which can then deliver to all other peers
at that rate
Each peer receives sum(C0*Ci/C) = C0

C1
Cn
C2 C3
How to build multiple distribution trees?

Two approaches:

Structured P2P network: build multiple trees explicitly


“Splitstream: High-bandwidth Multicast in Cooperative Environments”,
SOSP, 2003 (500+ citations in GS)
Unstructured P2P network: build multiple trees dynamically
and implicitly, based on who has and can provide content


“Incentives Build Robustness in BitTorrent”, Bram Cohen, 2003 (900+
citations in GS)
PPLive: very popular P2P streaming platform
Why does it work, especially for streaming? A mystery.
Approx model of unstructured P2P



Each peer gets a list of neighbors, during
bootstrapping (e.g. from tracker)
In each time slot, a peer “talks to” a random
neighbor, and select a piece to download, using a
piece selection algorithm
This model is commonly used for studying “BT-like” or
“gossip” protocols


Metrics: throughput, delay…
Our work is also based on this model

Metrics for streaming: Continuity, start-up latency…
Our recent work
1.
“A simple model for analyzing p2p streaming protocols”,
ICNP 2007 (YP Zhou, DM Chiu, J Lui)
Peer streaming model



Server “pushes” chunks to
peers sequentially, a new
chunk in each time slot
Buffer shifts ahead one
position per time slot
In a p2p system, a peer may
download one chunk from
another peer per time slot
server
t=1
t=2
t=3
playback
1
2
3
1
2
1
Buffer
……….
Simple p2p model

M homogeneous peers







Same upload capacity
Same playback buffer size
Synchronized playback
In each time slot, server
“push” a chunk to a random
peer
continuity = p(n)
Without P2P network
continuity = p(n) = 1/M
With P2P network, recursively
compute p(i). p(i) is defined
as:
p(i)=prob(position i filled)
server
playback
1/M
1
2
…………… n
2
…………… n
1/M
1
… M peers
1/M
1
2
…………… n
Sliding window



Each peer’s buffer is a
sliding window
In each time slot, each peer
downloads from a random
neighbor
q(i) = the probability Buf[i]
gets filled
p(i  1)  p(i)  q(i)
p(1)=1/M
time=t
1
2
p(n)=?
…………… n
sliding window
t+1
1
2
p(1)=1/M
…………… n
p(i+1) = p(i) + q(i)
q(i) = w(i)*h(i)*s(i)
 w(i) = peer wants to fill Buf[i]
 h(i) = the selected peer has
the content for Buf[i]
 s(i) = Buf[i] determined by chunk
P2P technology effect
selection strategy
Chunk selection strategies

Greedy
try to fill the empty buffer closest to playback

Rarest First
try to fill the empty buffer for the newest chunk
since p(i) is an increasing function, this means “Rarest First”
playback
1
Buffer map
2
3
X
RF Selection
4
5
X
X
6 7
8
X
Greedy Selection
Chunk selection strategy - cont

Greedy
S(i) = (1-1/M)∏j>I Pr[¬(W(k,j)H(h,j))]
= (1-1/M)∏j>I (p(j) + (1-p(j))2)
= (1- p(1) - p(n) + p(i+1))

Rarest first
s(i) = (1-1/M) ∏j<I (p(j) + (1-p(j))2)
= 1 – p(i)
Chunk selection strategy - cont

Greedy
p(i+1) = p(i) + (1-p(i)) p(i) (1- p(1) - p(n) + p(i+1))
w(i)

h(i)
s(i)
Rarest first
Lemma 1
p(i+1) = p(i) + (1-p(i)) p(i) (1- p(i))
w(i)
h(i)
s(i)
Lemma 2
Also studied


continuous forms for these difference equations
simulation
Which strategy is better?
What does “better” mean?

Playback continuity: p(n) as large as possible

Start-up latency:  p(i) as small as possible
Given buffer size (n) and large peer population (M)
1)
“Rarest first” is more scalable, and is often better in continuity!
2)
Greedy is better in start-up latency
Numerical result



M=1000
N=40
In simulation,


# neighbors=60
Uploads at most 2 in each
time slot
Mixed strategy




Partition the buffer into [1,m] and [m+1,n]
Use RF for [1,m] first
If no chunks available for download by RF, use Greedy for
[m+1,n]
Difference equations become
p(1) = 1/M
p(i+1) = p(i) + p(i) (1- p(i))2
for i = 1,…,m-1
p(i+1) = p(i) + p(i) (1- p(i))(1- p(m)- p(n) + p(i+1)) for i = m, … n-1

Proposition: Given peer population M, if buffer length n is large
enough, Mixed Strategy beats both Greedy and RF
Comparison
s(i) = 1 achieves
upper bound
For different buffer sizes
 Mixed achieves better continuity than both RF and Greedy
 Mixed has better start-up latency than RF
Closer look with simulation



Simulate 2000 time slots
Continuity is the average of
all peers
Continuity for Mixed is
more consistent, as well
highest
Adapting m to unknown population

Adjust m so that p(m) achieves a target probability (e.g. 0.3)

In simulation study, 100 new peers arrive every 100 slots

m adapts to a larger value as population increases
Optimality

Proposition: Mixed strategy is asymptotically
optimal
Idea:
 Upper bound case: s(i)=1or q(i) = p(i))(1-p(i))
 Derive continuity for upper bound and mixed, in terms
of M (population) and n (buffer)
 Show they are the same order
Extensions: Unsynchronized playback


This is still on-going work
Three unsynchronized cases considered so far:
1.
2.
3.
Peers in one cluster have same buffer length but
different playback offset.
Peers in one cluster have same playback offset but
different buffer length.
Peers in different clusters and there are lags among
clusters.
Unsynchronized playback

Case 1: Same buffer length different offset

Objective: maximize peers’ own buffer overlap.
Overlap
Unsynchronized playback

Case 1: Same buffer length different offset

All peers can achieve maximum overlaps only when they have same
playback offset, and it is a Nash Equilibrium.
Peer*’s overlap = 20
Peer*’s overlap = 22
By moving from g1 to g2, peer* gains more overlap
Unsynchronized playback

Case 2: Same playback offset different buffer length

If peers can change buffer length at will, what is the best strategy?
We prove that cluster with same buffer length has
better average continuity (by continuous model)
Unsynchronized playback

Case 2: Same offset different buffer length


Peers playback the same chunk at different time slot according to startup algorithm, which is equivalent to the different buffer length case.
Start-up algorithm: each peer will not start the playback until they have
downloaded N consecutive chunks
We run a simulation to determine the
peer population of different clusters.
M = 1000, n = 40, N = 10
The result is similar to a normal distr.
Unsynchronized playback

Case 2: Same latency different buffer length

Simulation results validate that homogeneous buffer length performs
better (continuity).
In these simulations, we apply the population distribution derived
by start-up algorithm to the heterogeneous case.
Unsynchronized playback

Case 3: Different clusters with lags

Can the lag among clusters improve performance?
We consider two cluster case, C1 and C2. Compare the
average continuity of the single cluster and of two small
cluster case.
Unsynchronized playback

Case 3: Different clusters with lags

Can the lag among clusters improve performance? Yes
The performance of two cluster
case is better than single cluster
More smaller clusters, better
performance.
Unsynchronized playback

Conclusions:
Given sufficient buffer size, it is better for peers to play the
video at the same time in one cluster
 It is better for peers to chose the same buffer length in one
cluster.
 If there are too many peers, it is better to divide them into
different clusters. The lag among clusters can improve
performance.

Other directions


ISP-friendly P2P algorithms (sensitive to ISP connectivity)
P2P VoD




May or may not be based on streaming
Optimal replication problem (optimize what?)
P2P traffic detection and traffic management
The economics of ISP and P2P conflicts