Post on 13-Dec-2015
transcript
Contents
1. Informal Introduction
2. Palm Calculus
3. Other Palm Calculus Formulae
4. Application to RWP
5. Other Examples
6. Perfect Simulation
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1. Event versus Time Averages
Consider a simulation, state St
Assume simulation has a stationary regime
Consider an Event Clock: times Tn at which some specific changes of state occur
Ex: arrival of job; Ex. queue becomes empty
Event average statistic
Time average statistic
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Sampling Bias
Ws and Wc are different
A metric definition should mention the sampling method (viewpoint)
Different sampling methods may provide different values: this is the sampling bias
Palm Calculus is a set of formulas for relating different viewpoints
Can often be obtained by means of the Large Time Heuristic
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The Large Time Heuristic
We will show later that this is formally correct if the simulation is stationary
It is a robust method, i.e. independent of assumptions on distributions (and on independence)
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Impact of Cross-Correlation
Sn = 90, 10, 90, 10, 90
Xn = 5000, 1000, 5000, 1000, 5000
Correlation is >0
Wc > Ws
When do the two viewpoints coincide ? 10
0 90 100 190 200 290 300
50001000
t (ms)
job arrival
50001000
50001000
Two Event Clocks
Stop and Go protocol
Clock 0: new packets; Clock a: all transmissions
Obtain throughput as a function of t0, t1 and loss rate
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t (ms)
timeoutt0 t1 t0
0,a 0,a a 0,a
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Load Sensitive Routing of Long-Lived IP FlowsAnees Shaikh, Jennifer Rexford and Kang G. Shin
Proceedings of Sigcomm'99
ECDF, per flow viewpoint
ECDF, per packet viewpoint
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2. Palm Calculus : Framework
A stationary process (simulation) with state St.
Some quantity Xt measured at time t. Assume that
(St;Xt) is jointly stationary
I.e., St is in a stationary regime and Xt depends on the past, present and future state of the simulation in a way that is invariant by shift of time origin.
ExamplesSt = current position of mobile, speed, and next waypoint
Jointly stationary with St: Xt = current speed at time t; Xt = time to be run until next waypoint
Not jointly stationary with St: Xt = time at which last waypoint occurred
Stationary Point Process
Consider some selected transitions of the simulation, occurring at times Tn.
Example: Tn = time of nth trip end
Tn is a called a stationary point process associated to St
Stationary because St is stationary
Jointly stationary with St
Time 0 is the arbitrary point in time
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Palm ExpectationAssume: Xt, St are jointly stationary, Tn is a stationary point process associated with St
Definition : the Palm Expectation is
Et(Xt) = E(Xt | a selected transition occurred at time t)
By stationarity:
Et(Xt) = E0(X0)
Example: Tn = time of nth trip end, Xt = instant speed at time t
Et(Xt) = E0(X0) = average speed observed at a waypoint
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E(Xt) = E(X0) expresses the time average viewpoint.
Et(Xt) = E0(X0) expresses the event average viewpoint.
Example for random waypoint: Tn = time of nth trip end, Xt = instant speed at time t
Et(Xt) = E0(X0) = average speed observed at trip end
E(Xt)=E(X0) = average speed observed at an arbitrary point in time
Xn
Xn+1
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Intensity of a Stationary Point Process Intensity of selected transitions: := expected number of transitions per time unit
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Two Palm Calculus Formulae Intensity Formula:
where by convention T0 ≤ 0 < T1
Inversion Formula
The proofs are simple in discrete time – see lecture notes
Campbell’s Formula
Shot noise model: customer n adds a load h(t-Tn,Zn) where Zn is some attribute and Tn is arrival time
Example: TCP flow: L = λV with L = bits per second, V = total bits per flow and λ= flows per sec
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t
Total load
T1 T2 T3
Two Event Clocks
Two event clocks, A and B, intensities λ(A) and λ(B)
We can measure the intensity of process B with A’s clock
λA(B) = number of B-points per tick of A clock
Same as inversion formula but with A replacing the standard clock
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Is the previous simulation stationary ?
Seems like a superfluous question, however there is a difference in viewpoint between the epoch n and time
Let Sn be the length of the nth epoch
If there is a stationary regime, then by the inversion formula
so the mean of Sn must be finite
This is in fact sufficient (and necessary)
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Time Average Speed, Averaged over n independent mobiles
Blue line is one sample
Red line is estimate of E(V(t))
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A Random waypoint model that has no stationary regime !
Assume that at trip transitions, node speed is sampled uniformly on [vmin,vmax]
Take vmin = 0 and vmax > 0
Mean trip duration = (mean trip distance)
Mean trip duration is infinite !
Was often used in practice
Speed decay: “considered harmful” [YLN03]
max
0max
1v
v
dv
v
Closed Form Assume a stationary regime exists and simulation is run long enough
Apply inversion formula and obtain distribution of instantaneous speed V(t)
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Removing Transient MattersA. In the mobile case, the nodes are more often towards the center, distance between nodes is shorter, performance is betterThe comparison is flawed. Should use for static case the same distribution of node location as random waypoint. Is there such a distribution to compare against ?
Random waypoint
Static
A (true) example: Compare impact of mobility on a protocol:
Experimenter places nodes uniformly for static case, according to random waypoint for mobile case
Finds that static is better
Q. Find the bug !
A Fair Comparison
We revisit the comparison by sampling the static case from the stationary regime of the random waypoint
Random waypoint
Static, from uniform
Static, same node location as RWP
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5. PASTAThere is an important case where Event average = Time average
“Poisson Arrivals See Time Averages”More exactly, should be: Poisson Arrivals independent of simulation state See Time Averages
6. Perfect Simulation
An alternative to removing transients
Possible when inversion formula is tractable
Example : random waypointSame applies to a large class of mobility models
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Removing Transients May Take Long
If model is stable and initial state is drawn from distribution other than time-stationary distribution
The distribution of node state converges to the time-stationary distribution
Naïve: so, let’s simply truncate an initial simulation duration
The problem is that initial transience can last very long
Example [space graph]: node speed = 1.25 m/sbounding area = 1km x 1km
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Perfect simulation is highly desirable (2)
Distribution of path:
Time = 100s
Time = 50s
Time = 300s
Time = 500s
Time = 1000s
Time = 2000s
Solution: Perfect Simulation
Def: a simulation that starts with stationary distribution
Usually difficult except for specific models
Possible if we know the stationary distribution
Sample Prev and Next waypoints from their joint stationary distributionSample M uniformly on segment [Prev,Next]Sample speed V from stationary distribution
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Perfect Simulation Algorithm
Sample a speed V(t) from the time stationary distributionHow ?
A: inversion of cdf
Sample Prev(t), Next(t)How ?
Sample M(t)
QuestionsQ1: A node receives messages from 2 sensors S1 and S2. Each of the sources sends messages independently of each other. The sequence of time intervals between messages sent by source S1 is iid, with a Gaussian distribution with mean m1 and variance v1 and similarly for S2. The node works as follows. It waits for the next message from S1. When it has received one message from S1, it waits for the next message from S2, and sends a message of its own.
How much time does, in average, the node spends waiting for the second message after the first is received ?
Q2: A sensor detects the occurrence of an event and sends a message when it occurs. However, the sensing system needs some relaxation time and cannot sense during T milliseconds after an event was sensed. There are l events per millisecond. Can you find the probability that an event is not sensed, as a function of T?
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QuestionsQ3: Consider the random waypoint model, where the distribution of the speed drawn at a random waypoint has a density f(v) over the interval [0, vmax]. Is it possible to find f() such that (1) the model has a stationary regime and (2) the time stationary distribution of speed is uniform over [0, vmax] ?
Q4: A distributed protocol establishes consensus by periodically having one host send a message to n other hosts and wait for an acknowledgement. Assume the times to send and receive an acknowledgement are iid, with distribution F(t). What is the number of consensus per time unit achieved by the protocol ? Give an approximation when the distribution is Pareto, using the fact that the mean of the kth order statistic in a sample of n is approximated by F−1( k/ n+1).
Q5: We measure the distribution of flows transferred from a web server. We find that the distribution of the size in packets of an arbitrary flow is Pareto. What is the probability that, for an arbitrary packet, it belongs to a flow of length x ?
Q6: A node receives messages from 2 sensors S1 and S2. Each of the sources sends messages independently of each other. The sequence of time intervals between messages sent by source S1 is iid, with a Gaussian distribution with mean m1 and variance s1
2 and similarly for S2. The node works as follows. It waits for the next message from S1. When it has received one message from S1, it waits for the next message from S2, and sends a message of its own. How do you implement a simulator for this system ?How much time does, in average, the node spends waiting for the second message after the first is received ? 64