The number of packets arriving in one microsecond at a router is well-modeled as
ID: 3230762 • Letter: T
Question
The number of packets arriving in one microsecond at a router is well-modeled as a Poisson random variable. For this problem, we will model the number of packets X arriving within a given microsecond as a Poisson (3) random variable. Using Markov's inequality, give an upper bound on the probability that 6 or more packets arrive in a microsecond, P[X greaterthanorequalto 6]. Comment on how this compares to the exact value, which you can calculate directly in this case. Using Chebyshev's inequality, give an upper bound on the probability that X deviates from its mean by 2 or more, P[|X - E[X]| greaterthanorequalto 2]. Comment on how this compares to the exact value, which you can calculate directly in this case. The total number of packets that arrive over 100 consecutive time intervals can be expressed as W_100 = X_1 + ... X_100 where X_1, ..., X_100 are i.i.d. Poisson(3) random variables. Using the Weak Law of Large Numbers, determine the probability that W_100 deviates from its mean by 60 or more, P[|W_100 - 300| greaterthanorequalto 60]. Assuming the Central Limit Theorem can be applied, approximate the probability that Wi00 deviates from its mean by 60 or more, P[|W_100 - 300 | greaterthanorequalto 60]. Comment on the difference between this approximation and your upper bound from part (c).Explanation / Answer
a) Markov inequality
P(X >=6) is less than or equal to E(x)/6
The upper bound is E(x)/6.
E(x) for 1 microsecond is 3 therefore this is equal to 3/6 or 0.5
Exact value P(x>=6) = 1- ( P(x=0)+ P(x=1) + ... P(x=5))
= 1- 0.916082 = 0.08
As we see this is very far away from value estimated by Markov method
b)
Chebysev's rule P( [x - E(x)] >=r ) <= V(x)/r^2 ;
Since V(x) = 3
3/ (2^2) = 3/4 = 0.75
Exact value P ( [x-3] >= 2) this implies P ( x >= 5 ) or P ( x < = 1 )
P(x>=5) = 1 - ( P(x=0)+ P(x=1) + ... P(x=4)) = 1-0.815263 = 0.184
and P(x <=1) = P(x=0) + P(x=1) = 0.199
Adding P ( x >= 5 ) + P ( x < = 1 ) = 0.383
This is still far away, but better than Markov
c) Weak law population Mean approaches sample mean
Chebysev's inequality will imply P( [x-E(x)] >= r) <= Var(x)/ (r^2)
Var(x) = 300= E(x) , r = 60
P( [x -E(x)] >= 60) <= 300/(60^2) =0.0833
d) Central Limit implies
W100 is a Normal with mean 300 and Var= 100*3 = 300
The P([ W100-300] >= 60 ])
P( W100 > 360 ) + P ( W100 < 240) = 0.000266 + 0.000266 = 0.000532
The values can be looked up from cumulative normal distribution tables