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ECC
Electronic Communications Committee
Mobile-DTT(14)xx
CPG-15 PTD
CPG-15 PTD CG Mobile-DTT
Date issued: 18 March 2014
Source: EBU Technology & Innovation
Subject: EBU Response to CPG-PTD(14)026 "Implications of adding further Time
Elements to the Monte Carlo simulation methodology"
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Summary:
Appendix 1 details the mathematical and other errors in the Motorola & al. contribution to PTD
(CPG-PTD(14)026 ) and JTG 4-5-6-7, in which it was erroneously concluded that the probability of
interference P = IP, the interference probability.
Basically, the Motorola & al. contribution confuses an average probability of DTT interference with
respect to a single random UE configuration with the cumulative probability of DTT interference
with respect to many sequential random UE configurations.
The Appendices 2 to 4 demonstrate in various ways that the probability of interference, P, for a
DTT site is P = 1 - (1 - IP)C, where IP is the interference probability (the probability that a random
DTT site is interfered with by a random UE configuration), and C is a number of such random UE
configurations occurring within a DTT viewer's viewing time, say 1 hour.
Proposal
To consider the present comments and questions expressed on document CPG-PTD(14)026 from
Motorola & al.
1. INTRODUCTION
The document in Appendix 1 was presented at the 5th meeting of CPG-PTD. It claims to show
that the Monte Carlo simulations for interference probability, IP, take into account the
multiplicity of interference effects that a DTT site might expect over a period of time in the
presence of several independent UE configurations.
A somewhat modified version of the same document was also presented to the February
meeting of JTG 4-5-6-7.
The Motorola & al. conclusion is that the probability of interference, P, due to a large number of
UE configurations, C, during a time period, TW, is IP, the same as the interference probability
for a DTT site 'exposed' to a single UE configuration; i.e. that P = IP, instead of P = 1 - (1 - IP)C.
The present contribution demonstrates the faults of the Motorola & al. document.
2. DISCUSSION
There are many errors, misinterpretations etc in the Motorola & al. document.
These are dealt with in the present contribution in the following manner.
- In Appendix 1, the Motorola & al. document is presented in its entirety. EBU comments,
corrections etc to various parts of the text are included at the relevant points within the
Motorola & al. text and marked in green. Further comments are included in Annex 3 of
Appendix 1.
The following 3 Appendices show in various ways that the probability of interference, P, for DTT
reception in the presence of C independent random UE configurations is: P = 1 - (1 - IP)C.
- In Appendix 2 is provided a brief (slightly mathematical) explanation showing that the
probability of interference is P = 1 - (1 - IP)C.
- In Appendix 3 a 'practical' (i.e., less mathematical) example is presented showing that the
probability of interference P = 1 - (1 - IP)C.
- In Appendix 4 a comparison of IP and P and their relationship is presented for a simple
'practical' example with C = 1 and C = 2, which again shows that P = 1 - (1 - IP)C.
3. CONCLUSION
Appendix 1 details the mathematical and other errors in the Motorola & al. contribution to PTD
and JTG 4-5-6-7, in which it was erroneously concluded that the probability of interference P =
IP, the interference probability.
Basically, the Motorola & al. contribution confuses an average probability of DTT interference
with respect to a single random UE configuration with the cumulative probability of DTT
interference with respect to many sequential random UE configurations.
The Appendices 2 to 4 demonstrate in various ways that the probability of interference, P, for a
DTT site is P = 1 - (1 - IP)C, where IP is the interference probability (the probability that a
random DTT site is interfered with by a random UE configuration), and C is a number of such
random UE configurations occurring within a DTT viewer's viewing time, say 1 hour.
2
CEPT
APPENDIX 1
ECC
CPG-PTD(14)026
Electronic Communications Committee
CPG-15 PTD #5
Rome, 13-17 January 2014
Mobile – DTT CG final session Jan 6-8
Date issued:
January 6, 2014
Source:
Motorola Solutions, Ericsson, EADS Cassidian, Selex ES
Subject:
Implications of adding further Time Elements to the Monte Carlo
simulation methodology
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3
Summary:
A recent proposal to ITU-R JTG4-5-6-71 argued that the Monte Carlo simulation
methodology being used to analyze coexistence between LTE and DTT is flawed and
propose to take into account the time factor in correctly calculating the probability of
interference.
However, in this study, by applying the theorem of the Strong Law of Large Numbers, we
show that the widely applied Monte Carlo simulation methodology has already taken into
account the time element and that as a result the proposal and results presented in JTG
document 326 becomes flawed.
 It has been demonstrated in the field [Reference needed. Where was this
demonstrated for LTE vs. DTTB???] that the Monte-Carlo simulation tool, albeit
simplified in the modelling, is accurate in predicting the probability of interference
and defining RF requirements.
 The Monte-Carlo simulation tool has already taken into account the time element
[which Monte Carlo simulation tool? - SEAMCAT certainly doesn't do this] in such
a way that the probability of interference obtained from the Monte-Carlo simulation
represents the average probability of interference over time [It is the probability
(which is an average) that a single random UE configuration interferes with a
single random DTT site; this has nothing to do with 'interference over time'].
Therefore this time element does not depict temporal correlation [the previous
sentence doesn't make sense].
 Post processing the results from Monte-Carlo simulation to try to incorporate
another [no, not 'another'] time element is not correct and will lead to conclusions
which are different from what can be demonstrated in the field [reference needed].
Background:
The CEPT CPG PT D and ECC PT 1 has been tasked to elaborate the European Common
Position regarding the WRC-15 Agenda Item 1.2 and the ECC PT 1 to study and answer the EU
Commission’s Mandate on the band 694 – 790 MHz. Both these tasks involve the study of the
interference probability between DTT and LTE services which is addressed by this contribution.
Proposal:
1
JTG4-5-6-7/326-E, AI 1.2: MONTE CARLO SIMULATIONS: REQUIREMENT FOR INFORMATION ON
CHANGES OF STATE IN AN IMT NETWORK of October 11, 2013
4
Using formula (1) as proposed by [1] to predict the probability of interference is not the right way
to incorporate time element into the Monte-Carlo simulation and we propose to seize further work
using this approach. It is demonstrated in the field that the well known Monte-Carlo simulation
tool is accurate in predicting the probability of interference and defining RF requirements
[reference needed for LTE vs. DTTB field results].
Post processing the results from Monte-Carlo simulation to try to incorporate time element is not
correct [this has not been shown in this Motorola contribution] and will lead to wrong conclusions.
It is therefore proposed to apply the well known Monte Carlo simulation tool as a proven
methodology to establish interference probability between LTE and DTT services.
Discussion:
In [1] the following formula is proposed to be used for calculating the probability P of interference
from LTE UE to DTV receiver (below in italic is copied from [1]. )
“If IP is the interference probability derived from the mtcMC calculations and C is the number of
network state changes then the probability P of LTE UE causing harmful interference to DTT
reception is given by:
”
This is not true, as demonstrated below [the demonstration below is incorrect. A correct
demonstration that the formula is correct is given in Appendix 5.]. In fact a very simple
demonstration that it is false is the following:
If C tends to infinity, (the observation period is very long), then P tends to 1. This would mean, if
the proposed assumption was true, that whatever is the value of the probability of interference in
one state (IP), provided that it is not zero, then the probability (P) of causing interference over an
infinite duration would be one. So this would mean that the interference would be harmful in all
[infinite duration] cases. [But nobody is discussing 'infinite duration' cases; for example,
broadcasters are proposing to calculate the probability of 1 interference during one hour...]
This is of course totally false [why is this 'totally false'? For infinite time durations it would be true]
and this demonstrates that the proposed calculation method
is not at all
representing the probability of an interference event to have a duration of CxDT [of course not an interference event does not have a duration of CxDT] but that it represents the probability that
an [at least 1] interference event, whatever its duration, occurs during CxDT [this is correct: the
intent of the calculation is to determine the probability that at least one interference occurs during
CxDT = 1 hour - in other words the formula is correct]. So the proposed method does not target
at all the right figure [yes it does!] and would just give over-estimated and absurd results [no it
wouldn't!].
Let’s first take a closer look at this formula.
The Figure below show for each independent state, the probability of causing interference to
DTV receiver is IP and there are total C independent states. The duration of each state is DT [DT
has been called the 'decorrelation time', but it is not clear that DT is also 'the duration of each
state'... in fact, "DT" has never been analytically defined] and the total time of observation is
CxDT.
5
[From the above figure, it would seem that the probability of interference should increase as the
number of 'IP intervals' increases, i.e. P ≠ IP.
For example, in the first IP rectangle at the left, the IP represents the interference probability that
a single random UE configuration causes an interference to a DTT site. The second IP rectangle
also represents the interference probability that another single random UE configuration causes
an interference to the DTT site. So, with 2 random UE configurations it is obvious that the
probability of interference, P, will increase, to about double. And this increase in P will continue
as the number of single random UE configurations, C, increases.]
For better understanding of the formula, a slightly different derivation of the formula is shown
below:
We see that the probability of
states causes interference:
, where take the values from 1 to C, and
are the binomial coefficients (C
chooses i).
From this follow that the total probability P of at least one state causing interference is
(1)
In Annex 1 we show the details on how (1) is derived.
Now, (1) does however not represent the fraction of time that interference to DTV is
occurring [we are not interested in the 'fraction of time' that interference is occurring - we
are interested in the number, C, of interference events in an hour (TW); furthermore, it is
not even clear what 'fraction of time' means]. Rather, (1) represents the probability of
interference occurring during at least one interval DT over the total time considered.
[Yes, this is what it is supposed to represent.] So using (1) to calculate the probability of
interference from LTE UE to DTV is actually over-estimating the occurrence of
interference, and results are highly unreliable. [The proposed DTTB protection criteria
is based on the probability that at least 1 interference event occurs during the
'observation time', TW. This is the result (1) delivers.]
In an LTE Network it may take 1ms or many seconds for a LTE UE to change from one
state to the next independent state. It depends on many factors including the speed of
the UE relative to the eNB, environment (multipath-reflections), eNB implementation etc.
[It should be kept in mind in the next 4 paragraphs that a very 'brief' UE interference
event (e.g. lasting only a few msec or less) may cause several seconds of interfered
DTTB viewing.]
Let’s now assume the DTV protection criterion is at least 1 second of interference
occurrence during a 20 second period [No. The proposed protection criterion is based on
the probability of one or more interference events during an hour]. Only when the
duration of each independent state is exactly 1 second, can formula (1) be used to
estimate the probability of interference (with C=20). [No. It is not the duration of each
6
independent state that is significant in the P calculation, but rather the NUMBER of such
states, C, occurring in one hour (TW).]
If the duration of each independent state is less than 1 second, applying (1) will over
estimate the probability of interference. [No. It's the number of states, C, that is
significant, not their duration. A state has a certain IP which determines, together with the
number of states, C, in the time period TW, the probability of interference, P]
If the duration of each independent state is more than 1 second, then applying (1) does
not tell you the probability of at least 1 second occurrence during every 20 second.
[Again, it is the number of interference states, C, that determines P, and not the time
duration of the states.]
In reality, the duration of each state can vary significantly. This makes it even harder to
apply (1) to calculate the probability of interference. [No. It is the number of states, C,
that is important, not their duration.]
The time element of the Monte-Carlo simulation methodology
In this section we demonstrate how this methodology already takes the time element into
account.
As has been clarified in summary, time element here merely means time average. This is
explicitly stated (underlined) after equation (2). [If anything, the 'time average' is the IP
for a DTT interference event due to a single random UE configuration, and not to a
sequence of different random UE configurations.]
Note, that we do not claim that the Monte-Carlo simulation methodology has taken into
account the temporal correlation in time, in fact, we simply claim that the Monte Carlo
simulation, (even though it is snap-shot based, and sampled through ensemble of the
same stochastic process), the time average of a particular realization (of the stochastic
process) equals the ensemble average. Annex 3 has more details on it.
Also notice that EBO’s method assumes independency between events, as our analysis
assumes the same, so there is no temporal correlation (or “correlation equals zero”) in
the modelling.
Consideration of the fraction of time
First we consider how the fraction of time is taken into consideration when deriving the
probability of interference.
When i states have caused interference with probability of PT, during the overall
observation period of CxDT, the interference occurred
of the time
[But for the purposes of calculating the probability of
interference, P, during the time interval TW, the important point is that the i states occur
during TW with 100% probability]
The probability when
states cause interference is thus given by:
[Even though the i states don't last the entire TW, they occur during TW with the probability
pt = 1 = 100%. So the interference probability due to i states during TW is
7
C 
C 
C 
  pt  IP i (1  IP )C i   1  IP i (1  IP )C i    IP i (1  IP )C i , and not the following:]
i 
i 
i 
, where i takes the values from 1 to C, and
are the binomial coefficients
(C choose i) [No, this is incorrect for the reasons given just above]
Therefore the total probability of interference is
(2)
[No. The correct formulation is (the derivation of which is shown in Annex1):
]
Annex 2 presents the details on how the sum in (2) is derived [Unfortunately, the sum in
(2) does not represent, P, the probability of interference]. It is interesting to see that the
time average over the observation period equals exactly the probability of interference
for each state. [It is also interesting to see that equation (1) and equation (2), both
calculating P, contradict each other!]
(2) is basically saying that, the probability of interference occurred during a certain
period, is actually equal to IP. [But, as shown above, (2) is incorrect in terms of
calculating P.]
If IP = 0.1%, and the observation period is 1 hour, the (2) establishes that during this 1
hour period, 0.1% of the time or around 3.6 seconds (may or may not be consecutive),
you may experience interference. [No. IP is not percentage of time. IP = 0.1% means
that if a single random UE configuration occurs, there is a 0.1% chance that a random
DTT site will be interfered.]
However, this is not happening by coincidence as The theorem of the Strong Law of
large Numbers [2] has already predicted such behaviour.
This theorem says in words:
“some time average (i.e., an average over time of a sample function of a
stochastic process) approaches the expected value of the process at a given
time.”
[Unfortunately, there is apparently a misunderstanding of the meaning of this theorem.
It is true that the average of a very large number of random events approaches the
expected value. This is true whether the random events occur in a number of 'snap
shots' or at various times. In the present application, the resulting average (expected
value) is the interference probability, IP, i.e. the probability that an interference occurs
at a DTT site due to a single random UE configuration.
However, with "P", we are interested in the probability of interference, i.e. in the
probability that at least one interference occurs at a DTT site when a number (in
general more than one) of random UE configurations are present during a time
interval, TW. Obviously, if the interference probability due to a single random UE
configuration is IP, the probability of interference will increase if the number of random
UE configurations increases from 1 to C during the observation time, TW.
8
The average of the IPs of the C individual UE configurations will of course still be IP
(which is what the strong law of large numbers is all about) but we are not interested at
this point in the average of the individual interference probabilities that 1 interference
occurs in the presence of a single UE configuration, but rather we are interested in the
probability that at least one interference occurs in the presence of C UE configurations.
This latter probability is not treated in the theorem of large numbers.]
[2] page 17.
The Monte-Carlo simulation methodology is exactly based on the theorem of the law of
large numbers. [This is true, and it gives as a result, IP, which represents the probability
that a DTT site will be interfered with in the presence of a single random UE
configuration.]
During each snap shot, UEs are randomly dropped, and statistics are collected.
Assuming each snap shot is independent, with large enough drops, the expected value
(or probability of interference) is calculated numerically (by averaging over all the snap
shots).
This expected value of any given time (or snap shot), is actually the same as the
expected value averaged over time, according to the law of large numbers. [Again, it
must be kept in mind that the 'expected value' is the probability, IP, that a DTT site
experiences interference in the presence of a single random UE configuration. The
probability of interference, P, that is to be calculated is the probability that a DTT site
experiences interference in the presence of many (C) random UE configurations during
a time window, TW = 1 hour.]
In Annex 3, the strong law of large numbers theorem is described, and how this
theorem is applied to the interference problem between LTE UE and DTT.
REFERENCES
[1]
JTG 4-5-6-7/326 ‘Requirement for Information on Changes of State in an
IMT Network’, Broadcast Network Europe.
[2]
‘Discreet Stochastic Processes’, Robert G. Gallager, Kluwer Academic
Publishers
9
Annex 1: Derivation of Formula (1)
This annex shows how formula (1) is derived.
(a.1)
(a.1) is based on binomial theorem.
Annex 2: Derivation of Formula (2)
This annex shows how formula (2) is derived. [However, "P" is not the probability of
interference.]
(a.2)
(a.2) is replacing the variables (
. Note that (binomial theorem)
(a.3)
Take derivative on both side of (a.3) with regard to
(a.4)
From (a.4), we have
(a.5)
From (a.5) and (a.2), we have
Annex 3: Strong Law of Large Numbers Theorem
Theorem of the Strong Law of Large Numbers:
10
Let
Then with probability 1,
are IID random variables with a finite mean
Given the theorem above, let’s see how this can be applied to the snap-shot based
Monte-Carlo simulation for co-existence analysis. Consider IID random variables
And
represents a sample point in a particular time [or a particular 'snap shot'] instance ,
it takes value of 1 if the event LTE UE interfering DTV receiver happens, with
probability of IP. Therefore, the expectation of
is,
is the time average [No. It is not the "time" average; it is simply the average of the
numbers Si, i = 1,… , n. The numbers Si could occur over any time interval, from 0 to
and the average would be the same] of the random process, which can be interpreted
as the total probability of interference [No. It is the interference probability, IP, as
defined by SEAMCAT, which is simply the average over a large number of 'samples'] or
the fraction of time [here the confusion is quite evident: it is the fraction which is equal
to the number of times a sample causes interference compared to the total number of
times that a sample is calculated] when the interference occurs [No.
is the
AVERAGE of the random process involving n 'trials'; temporal 'time' is not involved. For
example, if n = 100,000, then
is the average of the set {Si, i = 1,…, 100,000}; the
individual Si are simply calculated numbers and have nothing to do with temporal time,
although they might be generated over time, for example during a Monte Carlo
simulation]. According to the strong law of large numbers,
.
To better understand the meaning of the strong law of large numbers. Rewrite as
, then
can be viewed as a particular realization ( ) of the
sample space (
from a stochastic process (
,
is the time average [No. It is
simply the average. 'Time' is irrelevant] of this particular realization. The strong law of
large numbers says the time average [No. The strong law doesn't say anything about
'time'; it talks about the average] of this particular realization is equal to the expectation
of the process at any given time [in any realization; of course, the 'realizations may
'take place' over time (e.g. during a computer run)]. And this applies to all the
realizations from the sample space.
So back to the problem of interference from LTE UE to DTV receiver, the result from
Monte-Carlo simulation (probability of interference [the term used by SEAMCAT is
'interference probability'] is IP) is essentially the expected value for a particular time
(snap-shot) [realization; once again, it is the probability that a DTT site is interfered with
due to a single random UE configuration]. Based on the strong law of large numbers,
the result also means it is the probability of interference in time, or the fraction of time
11
when LTE UE interferes with DTV receiver is IP. [No. The strong law of large numbers
is a theorem about averages of large sets of random numbers and the averages of
those sets as they become larger. IP is the probability that a single random UE
configuration causes interference to a DTT site. IP is derived by averaging over a large
number of random DTT sites and single random UE configurations. There is no 'fraction
of time' involved.]
12
APPENDIX 2
Brief Explanation of P = 1 - (1 - IP)C (a bit mathematical)
In a SEAMCAT Monte Carlo simulation a large number of interference 'snap shots' are
considered, each of which defines an interference 'state' or 'event'. A state/event is defined by a
random positioning of a DTT site and a random UE configuration. A UE 'configuration' consists
of one (or more) UE(s) per sector randomly distributed throughout the cells of an IMT network.
Then, the probability, P, that an interference event occurs in a 'time window', TW, depends on
the number of states-events (i.e. not the random UE configurations), C, occurring during that
time window.
For example, if 1 state-event (1 random UE configuration) occurs during TW, then the
probability2 that 1 interference event occurs is 1P1 = IP and the probability that 0 interference
events occur is 1P0 = 1 - 1P1 = 1 - IP and
and the probability that at least one interference event occurs is
1
1P1 = 1 - 1P0 = 1 - (1 - IP) = IP.
If 2 states-events (i.e. 2 random UE configurations) occur during TW, then the probability, 2P0,
that 0 interference events occur is the probability that the first state-event causes no
interference (1P0) times the probability that the second state-event causes no interference (1P0),
i.e.
2
2P0 = 1P0 x 1P0 = (1P0) .
Then, the probability that at least one interference event occurs is
2
2P1 = 1 - 2P0 = 1 - (1 - IP) .
In the same manner, if K state-events (random UE configurations) occur during TW, the
probability that 0 interference events occur is KP0 = (1P0)K, and the probability that at least one
interference event occurs is KP1 = 1 - KP0 = 1 - (1 - IP)K.
2
The notation KPJ represents a probability of interference where K is the number of UE configurations
considered and J = 0 means the probability that no interference event has occurred at a DTT site, and J =
1 means the probability that 1 (or more) interference events has (have) occurred at a DTT site.
13
APPENDIX 3
PROBABILITY OF INTERFERENCE DURING TW (PRACTICAL EXAMPLE)
The following exposition gives a 'practical' example of how the 'cumulative' probability of
interference, P, increases as the number, C, of UE configurations during a 'time window', TW,
increases.
The 'interference probability', IP, is the probability that interference occurs at a single random
DTT site due to the occurrence of a single random UE configuration. We call 'a single UE
configuration' an 'event'.
We give in the following a numerical example deriving the probability of interference, P, from IP.
ASSUMPTIONS
- Assume that there are N = 100,000 DTT sites within the area of interest, for example
within an IMT sector.
- Assume that IP = 0.0001 has been calculated as the interference probability that a
random UE configuration cause an interference event at a DTT site.
- Assume that there are C = 10 events during the 'time window' TW = 1 hour.
-Assume that the C events take place sequentially during TW.
FIRST EVENT
When the first event occurs, the probability is IP = 0.0001 that interference occurs amongst the
N sites. This means that, on average IP x N = 0.0001 x 100,000 = 10 sites are interfered once:
we label the sites S1, …, S10. The probability of interference3 is:
1P1
= 10/100,000 = 0.0001 = IP; thus 1P1= IP = 1 - (1 - IP).
SECOND EVENT
When, thereafter, the second event occurs, again the probability is IP = 0.0001 that
interference is produced amongst the N sites. This means that, on average IP x N = 0.0001 x
100,000 = 10 sites are interfered once: we label the sites S11, …, S20.
Since the events are statistically independent and have a very small IP, the probability that Si =
Sj for i ≠ j is very small. For example, the probability that site S11 is one of the sites S1, …, S10 is
10/100,000 = 0.0001; the probability that any of the sites S11, …, S20 is the same as any of the
sites S1, …, S10 is about 10 x 0.0001 = 0.001, in other words, about 1 in a thousand, very small.
For this simplified numerical examination we ignore this small probability, but indicate it as
'approx'.
The probability of interference is (with the 20 interfered DTT sites), approximately:
= 20/100,000 = 0.0002; but, 2P1 =1 - (1 - IP)2 = 1 - (1 - .0001)2 = 1 - 0.99980001 =
0.00019999.
2P1_approx
So, 2P1_approx = 0.0002  0.00019999 = 1 - (1 - IP)2 = 2P1
THIRD EVENT
When, thereafter, the third event occurs, again the probability is IP = 0.0001 that interference is
produced amongst the N sites. This means that, on average IP x N = 0.0001 x 100,000 = 10
sites are interfered once: we label the sites S21, …, S30.
The probability of interference is (with the 30 interfered DTT sites), approximately:
= 30/100,000 = 0.0003; but, 3P1 = 1 - (1 - IP)3 = 1 - (1 - .0001)3 = 1 - 0.99970003 =
0.00029997.
3P1_approx
3
The notation KPJ represents a probability of interference where K is the number of UE configurations
considered and J = 0 means the probability that no interference event has occurred at a DTT site, and J =
1 means the probability that 1 (or more) interference events has (have) occurred at a DTT site.
14
So, 3P1_approx = 0.0003  0.00029997 = 1 - (1 - IP)3 = 3P1.
FOURTH EVENT, ETC.
We can continue to the third event and so on until the tenth event during TW where we find
TENTH EVENT
When the tenth event occurs, again the probability is IP = 0.0001 that interference is produced
amongst the N sites. This means that, on average IP x N = 0.0001 x 100,000 = 10 sites are
interfered once: we label the sites S91, …, S100.
DURING TW
This means that about 100 sites, S1, S2, …, S100 will be interfered with during TW, that is 100
sites out of 100,000 sites, which means the probability of interference is (with the 100 interfered
DTT sites):
10P1
= 100/100,000 = 0.0010; but,
0.0009996.
10P1
- (1 - IP)10 = 1 - (1 - .0001)10 = 1 - 0.9990004 =
So, 10P1_approx = 0.0010  0.0009990004 = 1 - (1 - IP)10 = 10P1
Thus, if we calculate 10P1 with C = 10 events during TW using the formula, 10P1 = 1 - (1 - IP)10,
we find 10P1 = 0.0009996  0.0010, which differs by about 0.045% from 100/100,000 = 0.001.
The small differences between nP1_approx and nP1 come from the slight inexactness of the way
we've estimated nP1_approx.
CONCLUSION
It has been shown how the probability of interference CP1 increases as C increases.
Furthermore, if there are C events during a time window, TW, then the probability of
interference during TW, is
CP1
= 1 - (1 - IP)C, and CP1 > IP for C > 1.
15
APPENDIX 4
EXPLANATION OF THE RELATIONSHIP BETWEEN IP AND P
IP ('interference probability') is the result of averaging the number of undesired outcomes
(interference to a DTT site) over the total number of outcomes for a large number of random
UE configurations. It is the probability that a single UE configuration causes interference to a
random DTT site.
In other words, once the IP has been calculated, the probability of interference4, 1P1, for a
single random UE configuration occurring during a given time period is: 1P1 = 1 - (1 - IP)1 = IP.
If, during the given time period, two random UE configurations occur, the probability of
interference due to the two configurations, 2P1, is:
2
2P1 = 1 - (1 - IP) ; expanding this, we find
2
2
2P1 = 1 - (1 - 2xIP - IP ) = 2xIP - IP = IPx(2 - IP) > IP = 1P1.
Thus we see that the probability of interference increases if 2 random UE configurations occur
during the given time period compared to only 1 random UE configuration occurring during the
time period, which is to be expected.
If, during the given time period, C random UE configurations occur, the probability of
interference, CP1, is: CP1 = 1 - (1 - IP)C.
NUMERICAL WORKED EXAMPLE
We show how this formula [i.e., CP1 = 1 - (1 - IP)C] works for C = 1 and C = 2 in the following
Table. We use a simple model where IP = 1/3.
The Table below shows the results. The following paragraphs describe the situation calculated
in the Table.
For simplicity, we consider 3 outcomes for a random single-UE configuration:
(0) for interference (referred to as 'failure', F)
(1) for no interference (referred to as 'no failure', No F)
(2) for no interference (referred to as 'no failure', No F)
Each of the outcomes, (0), (1), and (2) has probability 1/3.
Three Sets of 'random' UE trials are considered: 2 Sets of 'single-event' trials and 1 Set of
'double-event' trials.
'Single-event' trials consist of a number of 'random' single-UE configurations/events. In a singleevent, a (0) corresponds to interference ("F" for 'failure'); and a (k) (for k = 1 or 2) corresponds
to no interference ('no failure') for the random UE configuration. This situation reflects the case
where only one random UE configuration/event occurs during the time period (i.e., C = 1).
'Double-event' trials consist of a number of 'random' double-UE configurations/events. Note that
interference is considered to occur during the time period if there is an interference, (0), for
either one of the two events (or for both). In a double-event, (0 k) and (k 0) (for k = 0, 1, 2)
correspond to interference ('failure'), and (j k) (for j = 1 or 2, k = 1 or 2) corresponds to no
interference ('no failure'). This situation reflects the case where two random UE configurations
occur during the time period (i.e., C = 2).
The first two Sets (Set 1 and Set 2) consist of 9 single-event trials. The 9 trials display 3 "(0)"
results, 3 "(1)" results, and 3 "(2)" results (maintaining the 1/3 probability of each outcome).
4
The notation KPJ represents a probability of interference where K is the number of UE configurations
considered and J = 0 means the probability that no interference event has occurred at a DTT site, and J =
1 means the probability that 1 (or more) interference events has (have) occurred at a DTT site.
16
The third Set (Set 3) consists of 9 double-event trials. The 9 trials represent the 9 possible
results when 2 random events are considered5. These results reflect the individual probabilities
of each event of the double-events for the "(0)", "(1)", and "(2)" outcomes.
In the Table below, for each trial the following are indicated:
the individual results:
(n), n = 0, 1, 2 for single-events;
(n m), n = 0, 1, 2 and m = 0, 1, 2 for double-events,
the number of (0)'s for Set 1 and for Set 2,
the number of (0 k)'s and (k 0)'s for Set 3
the number of trial 'failures' ("F").
In addition, the total number of (0)s and F's are indicated, and their respective probabilities.
It can be noted that the third row from the bottom entitled 'prob' shows in the columns entitled
''# 0" that the ratio of #0s/#events = IP, which is to be expected because #0s/#events is the probability
that a single random UE event causes 'failure', F.
The penultimate line for each Set shows, P, the probability of interference (entitled "probability
for failure for the set").
The final line calculates CP1 = 1 - (1 - IP)C, for C = 1, 2. It is seen that the last line and the
penultimate line (calculation of the ratio #failures/#trials = P) give the same results.
In particular, it is seen that 2P1 ≠ IP. In general, CP1 ≠ IP, for C ≥ 2.
1 or more failures in a set of C events constitutes failure: "F"
Outcomes (0), (1) and (2) each have probability 1/3
Trials consist of single-events (Set 1 and Set 2) or double-events (Set 3)
F = failure
(Interference)
No F = no
failure
(no
Interference)
IP = 1/3
Trial #
1
2
3
4
5
6
7
8
9

Prob
Probability of
failure for the
set
Outcomes for a single-event trial:
(0) = F
(1) = no F
(2) = no F
Set 1 - single events
(C = 1)
Trial
# (0)
#F
result
(2)
0
0
(2)
0
0
(0)
1
1
(1)
0
0
(0)
1
1
(1)
0
0
(2)
0
0
(1)
0
0
(0)
1
1
9 trials;
9x1 = 9
(0)'s =
F's
UE
3
=3
events
3/9 =
3/9 =
1/3
1/3
Set 2 - single events
(C = 1)
Trial
# (0)
#F
result
(1)
0
0
(0)
1
1
(1)
0
0
(2)
0
0
(2)
0
0
(0)
1
1
(2)
0
0
(1)
0
0
(0)
1
1
9 trials;
9x1 = 9
(0)'s =
F's
UE
3
=3
events
3/9 =
3/9 =
1/3
1/3
1/3
1/3
5
Outcomes for a double-event
trial:
(0) (0) = F
(0) (1) = F
(1) (0) = F
(0) (2) = F
(2) (0) = F
(1) (1) = no F
(1) (2) = no F
(2) (1) = no F
(2) (2) = no F
Set 3 - double events
(C = 2)
Trial result
# (0)
#F
(2) (1)
(2) (0)
(0) (1)
(1) (2)
(0) (2)
(1) (0)
(2) (2)
(1) (1)
(0) (0)
0
1
1
0
1
1
0
0
2
0
1
1
0
1
1
0
0
1
9 trials;
9x2 =18
UE events
(0)'s =
6
F's =
5
6/18 =
1/3
5/9
5/9
If there are 3 possible outcomes for the first event and 3 possible outcomes for the second event, there
will be 3 x 3 = 9 possible outcomes for the double-event.
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Formula
= 1 - (1 - IP)1 =
1 - (1 - 1/3) = 1/3
= 1 - (1 - IP)1 =
1 - (1 - 1/3) = 1/3
1P1(f)
1P1(f)
2
2P1(f) = 1 - (1 - IP) =
1 - (1 - 1/3)2 = 1 - 4/9 = 5/9
APPENDIX 5
FURTHER DEMONSTRATION THAT P = 1 - (1 - IP)C
The text in Annex 3 of Appendix 1 describes how the average value of the interference
probability, IP, is calculated.
Xi  {0, 1} is a random variable representing the occurrence of interference (1) or not (0) from a
random UE configuration affected at a random DTT site. Xi is referred to as a ' sample point'.
The sums of many sample points, Sn, were considered: Sn = X1 + X2 + ... + Xn and the average
was calculated:
Sn
 E ( X )  IP .
n  n
lim
This means that any given UE configuration has a probability IP to cause interference to any
given DTT site.
However the problem at hand is, having calculated the value of IP for a single UE
configuration, how to calculate the probability of interference, P, due to a number of UE
configurations. These configurations could happen simultaneously or consecutively.
We take an example from the world of craps. For example, the probability of throwing a "1" with
a six-sided die is IP = 1/6. However if we throw two dice, the probability, P, of throwing at least
one "1" will be larger than 1/6. Of the 36 possible outcomes we find the following 11 with at
least one "1":
Throw #
outcome
1
(1 1)
2
(1 2)
3
(1 3)
4
(1 4)
5
(1 5)
6
(1 6)
7
(2 1)
8
(3 1)
9
(4 1)
10
(5 1)
11
(6 1)
That is, P = 11/36  1/3 > 1/6.
The probability, P, of throwing at least one "1" would be the same if the 2 dice were thrown
consecutively.
FURTHER EXAMPLE
We assume a simple case where C = 3 UE configurations occur during a time interval, TW.
Each configuration can cause
- interference with probability IP, or
- no interference with probability (1 - IP).
The configurations which cause interference are labelled I; the configurations which cause no
interference are labelled N.
The sample space, Sall, consists of the 2C = 23 = 8 sample points:
Sall = {(I I I), (I I N), (I N I), (I N N), (N I I), (N I N), (N N I ), (N N N)}.
The probabilities of these sample points are, respectively:
IP3, IP2(1 - IP), IP2(1 - IP), IP(1 - IP)2, IP2(1 - IP), IP(1 - IP)2, IP(1 - IP)2, (1 - IP)3.
If we assume IP = 0.1, then the probabilities of the sample points as calculated as, respectively:
0.001, 0.009, 0.009, 0.081, 0.009, 0.081, 0.081, 0.079.
We see that the sum of the probabilities of all the sample points is 1, as it should be:
Pall = 0.001 + 0.009 + 0.009 + 0.081 + 0.009 + 0.081 + 0.081 + 0.729 = 1
We are interested in the probability of the subset of sample points which have at least one
interference (I):
S≥1 = {(I I I), (I I N), (I N I), (I N N), (N I I), (N I N), (N N I )}.
The sum of the probabilities of this subset is
P≥1 = 0.001 + 0.009 + 0.009 + 0.081 + 0.009 + 0.081 + 0.081 = 0.271.
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If we calculate with the formula: P≥1 = 1 - (1 - IP)C = 1 - (1 - 0.1)3 = 0.271.
We see that the formula gives the correct result. In particular, we see that P≥1  IP.
More generally, for C ue configurations, the sample space would contain sample points of the
form:
(e1 e2 e3 ... eC), with each ei  {I, N}, i = 1, ..., C.
As before, each sample point in the sample space would have probability IPk(1-IP)C-k, where k
is the number of I's in the sample point and C-k is the number of N's in the sample point. We
call this a "k-combination".
To find the overall probability, P, of at least one ei = I, we need to add the individual
independent probabilities, taking into account all of the k-combinations for k ≥ 1. With C
C 
C!
elements, the number of k-combinations is equal to the binomial coefficient:   
.
 k  k!(C  k )!
Then P 
C
C 
  k IP
k 1
 
k
(1  IP)C  k  1  (1  IP)C as was shown in Annex 1 of Appendix 1.
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