Calculate packet loss with random transmission times and TTL

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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?







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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?
    – user144410
    Aug 21 at 13:18










  • I guess a sum would have been a better. I have really no idea how to approach this.
    – katakuriki
    Aug 21 at 13:21










  • The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
    – user144410
    Aug 21 at 13:22










  • I edited the question and explained it a bit more in detail.
    – katakuriki
    Aug 21 at 13:26














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?







share|cite|improve this question






















  • what do you mean by this last expression? you are taking the limit of which function? and what does "$0$ else $1$" mean?
    – user144410
    Aug 21 at 13:18










  • I guess a sum would have been a better. I have really no idea how to approach this.
    – katakuriki
    Aug 21 at 13:21










  • The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
    – user144410
    Aug 21 at 13:22










  • I edited the question and explained it a bit more in detail.
    – katakuriki
    Aug 21 at 13:26












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?







share|cite|improve this question














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?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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?
    – user144410
    Aug 21 at 13:18










  • I guess a sum would have been a better. I have really no idea how to approach this.
    – katakuriki
    Aug 21 at 13:21










  • The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
    – user144410
    Aug 21 at 13:22










  • I edited the question and explained it a bit more in detail.
    – 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?
    – user144410
    Aug 21 at 13:18










  • I guess a sum would have been a better. I have really no idea how to approach this.
    – katakuriki
    Aug 21 at 13:21










  • The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
    – user144410
    Aug 21 at 13:22










  • I edited the question and explained it a bit more in detail.
    – 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?
– 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?
– user144410
Aug 21 at 13:18












I guess a sum would have been a better. I have really no idea how to approach this.
– katakuriki
Aug 21 at 13:21




I guess a sum would have been a better. I have really no idea how to approach this.
– katakuriki
Aug 21 at 13:21












The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
– user144410
Aug 21 at 13:22




The expression you wrote doesn't make sense. Please edit your question and explain it in writing.
– user144410
Aug 21 at 13:22












I edited the question and explained it a bit more in detail.
– katakuriki
Aug 21 at 13:26




I edited the question and explained it a bit more in detail.
– katakuriki
Aug 21 at 13:26










2 Answers
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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 $frac 0.150.2=0.75$






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    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 $25%$






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      2 Answers
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      active

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      2 Answers
      2






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      active

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      active

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      up vote
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      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 $frac 0.150.2=0.75$






      share|cite|improve this answer
























        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 $frac 0.150.2=0.75$






        share|cite|improve this answer






















          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 $frac 0.150.2=0.75$






          share|cite|improve this answer












          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 $frac 0.150.2=0.75$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 21 at 13:35









          Ross Millikan

          278k21188354




          278k21188354




















              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 $25%$






              share|cite|improve this answer


























                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 $25%$






                share|cite|improve this answer
























                  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 $25%$






                  share|cite|improve this answer














                  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 $25%$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 21 at 13:40

























                  answered Aug 21 at 13:34









                  user144410

                  7341519




                  7341519






















                       

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