practice problems 2

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.