Upload
imran
View
259
Download
3
Embed Size (px)
DESCRIPTION
solution
Citation preview
Practice Problem
Stochastic Systems
Instructor: Dr. Fahd Ahmed Khan
Prepared by: Sami Ur Rehman
Problem:
Bahawalpur Toll Plaza 1 Toll Plaza 2 Toll Plaza 3 Multan
Haider is travelling from Bahawalpur to Multan. On his route, there are a total of
three Toll Plazas. Because of exceptionally long queues, a vehicle on average
spends 0.7 hours inside each toll plaza. Given that the interconnecting road is a
motorway, we ignore the delay caused by travelling on the roads (i.e. all the delay
is caused by waiting in the queues at each Toll Plaza). Find the probability that
Haider reaches Multan in less than 0.5 hours.
Try to model this situation on your own before consulting the solution on the next
page.
Solution:
The delay caused by each Toll plaza can be modelled as an Exponential random
variable (RV), with parameter ʎ = 0.7.
Assuming that the interconnecting roads do not cause any further delay, all the
delay is caused by the three toll plazas that lie on the route. The situation is
depicted in the diagram below.
As you might have guessed by now, the scenario can be modelled as an r-stage
Erlang distribution, with r=3.
CDF of Erlang: FE (t) = 1 –∑ʎ𝒕𝒌𝒆𝒙𝒑(−ʎ𝒕)
𝒌!𝒓𝒌=𝟏
So, Pr{E<0.5} = FE (0.5)
= 1 – (𝟎.𝟕)(𝟎.𝟓)𝟏𝒆𝒙𝒑(−𝟎.𝟕×𝟎.𝟓)
𝟏! –
(𝟎.𝟕)(𝟎.𝟓)𝟐𝒆𝒙𝒑(−𝟎.𝟕×𝟎.𝟓)
𝟐! – (𝟎.𝟕)(𝟎.𝟓)𝟑𝒆𝒙𝒑(−𝟎.𝟕×𝟎.𝟓)
𝟑!
Pr {E<0.5} = 0.681 = ANSWER
Practice Problems
Topics:
1. Markov inequality
2. Chebyshev inequality
Consider a coin that comes up with head with probability 0.2 . Let us toss it n times. Now we can use Markov inequality to bound the probability that we got atleast 80% of heads.
Let X be the random variable indicating the number of heads we got in n tosses. Clearly, X is non negative. Using linearity of expectation, we know that E[X] is 0.2n. We want to bound the probability P(X >= 0.8n). Using Markov inequality , we get
Of course we can estimate a finer value using the Binomial distribution, but the core idea here is that we do not need to know it !
If the average weight of a Maine black bear is 500 pounds with standard deviation equal to 100 pounds, we can use the Chebyshev inequality to upper bound the probability that a randomly chosen bear will be more then 200 pounds away from the average.
Suppose we know that the number of items produced in a factory during a week is a random variable with mean 500. (a) What can be said about the probability that this week’s production will be at least 1000? (b) If the variance of a week’s production is known to equal 100, then what can be said about the probability that this week’s production will be between 400 and 600?
Let X be the number of items that will be produced in a week.
and so the probability that this week’s production will be between 400 and 600 is at least 0.99.