Calculate packet loss with random transmission times and TTL
Clash Royale CLAN TAG#URR8PPP
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-1
down vote
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If I have random lag times from a=0.1
to b=0.3
and a time to live (TTL) of x=0.25
, what would be the packet loss in per cent?
Ok so basically I have packets that arrive in Random [a,b]
time, if that random value is greater than x
the packet gets lost and doesn't arrive.
What is the probability of a packet to arrive?
probability percentages
add a comment |Â
up vote
-1
down vote
favorite
If I have random lag times from a=0.1
to b=0.3
and a time to live (TTL) of x=0.25
, what would be the packet loss in per cent?
Ok so basically I have packets that arrive in Random [a,b]
time, if that random value is greater than x
the packet gets lost and doesn't arrive.
What is the probability of a packet to arrive?
probability percentages
what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
â user144410
Aug 21 at 13:18
I guess a sum would have been a better. I have really no idea how to approach this.
â katakuriki
Aug 21 at 13:21
The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
â user144410
Aug 21 at 13:22
I edited the question and explained it a bit more in detail.
â katakuriki
Aug 21 at 13:26
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
If I have random lag times from a=0.1
to b=0.3
and a time to live (TTL) of x=0.25
, what would be the packet loss in per cent?
Ok so basically I have packets that arrive in Random [a,b]
time, if that random value is greater than x
the packet gets lost and doesn't arrive.
What is the probability of a packet to arrive?
probability percentages
If I have random lag times from a=0.1
to b=0.3
and a time to live (TTL) of x=0.25
, what would be the packet loss in per cent?
Ok so basically I have packets that arrive in Random [a,b]
time, if that random value is greater than x
the packet gets lost and doesn't arrive.
What is the probability of a packet to arrive?
probability percentages
edited Aug 21 at 14:03
user144410
7341519
7341519
asked Aug 21 at 13:05
katakuriki
33
33
what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
â user144410
Aug 21 at 13:18
I guess a sum would have been a better. I have really no idea how to approach this.
â katakuriki
Aug 21 at 13:21
The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
â user144410
Aug 21 at 13:22
I edited the question and explained it a bit more in detail.
â katakuriki
Aug 21 at 13:26
add a comment |Â
what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
â user144410
Aug 21 at 13:18
I guess a sum would have been a better. I have really no idea how to approach this.
â katakuriki
Aug 21 at 13:21
The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
â user144410
Aug 21 at 13:22
I edited the question and explained it a bit more in detail.
â katakuriki
Aug 21 at 13:26
what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
â user144410
Aug 21 at 13:18
what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
â user144410
Aug 21 at 13:18
I guess a sum would have been a better. I have really no idea how to approach this.
â katakuriki
Aug 21 at 13:21
I guess a sum would have been a better. I have really no idea how to approach this.
â katakuriki
Aug 21 at 13:21
The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
â user144410
Aug 21 at 13:22
The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
â user144410
Aug 21 at 13:22
I edited the question and explained it a bit more in detail.
â katakuriki
Aug 21 at 13:26
I edited the question and explained it a bit more in detail.
â katakuriki
Aug 21 at 13:26
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
You are taking a uniform random number in the range $[0.1,0.3]$ and asking the chance that it is less than $0.25$. You win in $0.25-0.1=0.15$ of the range of $0.3-0.1=0.2$, so the chance is $frac 0.150.2=0.75$
add a comment |Â
up vote
0
down vote
Assume that the time delay has a uniform distribution, then the probability of packet loss is given by the probability of the time delay being larger than 0.25.
This can be easily computed as $$(0.25-0.1)* frac10.3-0.1 = 0.15*5 = 0.75$$
i.e, probability of packet arrival is $75%$, and the probability of packet loss is $25%$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You are taking a uniform random number in the range $[0.1,0.3]$ and asking the chance that it is less than $0.25$. You win in $0.25-0.1=0.15$ of the range of $0.3-0.1=0.2$, so the chance is $frac 0.150.2=0.75$
add a comment |Â
up vote
0
down vote
accepted
You are taking a uniform random number in the range $[0.1,0.3]$ and asking the chance that it is less than $0.25$. You win in $0.25-0.1=0.15$ of the range of $0.3-0.1=0.2$, so the chance is $frac 0.150.2=0.75$
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You are taking a uniform random number in the range $[0.1,0.3]$ and asking the chance that it is less than $0.25$. You win in $0.25-0.1=0.15$ of the range of $0.3-0.1=0.2$, so the chance is $frac 0.150.2=0.75$
You are taking a uniform random number in the range $[0.1,0.3]$ and asking the chance that it is less than $0.25$. You win in $0.25-0.1=0.15$ of the range of $0.3-0.1=0.2$, so the chance is $frac 0.150.2=0.75$
answered Aug 21 at 13:35
Ross Millikan
278k21188354
278k21188354
add a comment |Â
add a comment |Â
up vote
0
down vote
Assume that the time delay has a uniform distribution, then the probability of packet loss is given by the probability of the time delay being larger than 0.25.
This can be easily computed as $$(0.25-0.1)* frac10.3-0.1 = 0.15*5 = 0.75$$
i.e, probability of packet arrival is $75%$, and the probability of packet loss is $25%$
add a comment |Â
up vote
0
down vote
Assume that the time delay has a uniform distribution, then the probability of packet loss is given by the probability of the time delay being larger than 0.25.
This can be easily computed as $$(0.25-0.1)* frac10.3-0.1 = 0.15*5 = 0.75$$
i.e, probability of packet arrival is $75%$, and the probability of packet loss is $25%$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume that the time delay has a uniform distribution, then the probability of packet loss is given by the probability of the time delay being larger than 0.25.
This can be easily computed as $$(0.25-0.1)* frac10.3-0.1 = 0.15*5 = 0.75$$
i.e, probability of packet arrival is $75%$, and the probability of packet loss is $25%$
Assume that the time delay has a uniform distribution, then the probability of packet loss is given by the probability of the time delay being larger than 0.25.
This can be easily computed as $$(0.25-0.1)* frac10.3-0.1 = 0.15*5 = 0.75$$
i.e, probability of packet arrival is $75%$, and the probability of packet loss is $25%$
edited Aug 21 at 13:40
answered Aug 21 at 13:34
user144410
7341519
7341519
add a comment |Â
add a comment |Â
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what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
â user144410
Aug 21 at 13:18
I guess a sum would have been a better. I have really no idea how to approach this.
â katakuriki
Aug 21 at 13:21
The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
â user144410
Aug 21 at 13:22
I edited the question and explained it a bit more in detail.
â katakuriki
Aug 21 at 13:26