Problem 5. Consider a tram at a tourist attraction. The tram is on a fixed loop

Question

Problem 5. Consider a tram at a tourist attraction. The tram is on a fixed loop stopping at a number of sites. At each site passengers queue up to board the tram and passengers on the tram may elect to get off. Assume the capacity of the tram is 5. The probability that any passenger on the tram will get off at any site is 0.8. The number of passengers waiting to board at a site has the following probability mass function. PO customers waiting to board ] = 0.4 PU customer waiting to board]-0.4 P[2 customers waiting to board] = 0.2. Each site is assumed to be mathematically equivalent. Assume at each stop any passengers who wish to get off will do so before anyone waiting to board. Also, anyone who does not find a seat will simply walk to their desired destination. Formulate the problem as a Markov process Assume the state of the system will be the number of passengers on the bus (just after boarding) Compute the one-step transition probability matrix, P and determine the steady state probabilities

Explanation / Answer

A stochastic process X = {Xn : n 0} on a countable set S is a Markov Chain if, for any i, j S and n 0,

P{Xn+1 = j|X0,...,Xn} = P{Xn+1 = j|Xn}

P{Xn+1 = j|Xn = i} = pij .

The pij is the probability that the Markov chain jumps from state i to state j. These transition probabilities satisfy jS pij = 1, i S, and the matrix P = (pij ) is the transition matrix of the chain.

The one-step transition probability is the probability of transitioning from one state to another in a single step. The Markov chain is said to be time homogeneous if the transition probabilities from one state to another are independent of time index n

pij = Pr{Xn = j | Xn-1 = i}

The transition probability matrix, P, is the matrix consisting of the one-step transition probabilities, Pij.

According to our problem,

P0=0.4

P1=0.4

P2=0.2

P(passenger getting down from tram at a particular stop) = 0..8

P(person staying back in tram) = 1- P(person getting down)

                                                       =1-0.8 =0.2

The one-step transition matrix for a Markov chain with states S = { 0, 1, 2 } is P=

P00

P01

P02

P10

P11

P12

P20

P21

P22

Pij= Pr{X1 = j | X0 = i}

So,

Transition probability matrix:

(1 r) 2

2r(1 r)

r 2

p(1 r)

pr + (1 p)(1 r)

r(1 p)

p 2

2p(1 p)

(1 p) 2

Steady-state probabilitiesare: 0 = [r/(p+r)]2 , 1 = 2[r/(p+r)][p/(p+ r)] and 2 = [p/(p + r)]2 .

Each tram remains full with average time of 1/p and the average time of 1/r for filling.

Thus, the proportion of time it is filled is (1/p)/(1/p + 1/r) = r/(p + r).

That is, each time, independently of each other, each component is functioning with probability r/(p + r) and not functioning with probability p/(p + r).

Hence, the limiting distribution of Xn is Binomial with parameters 2 and r/(p + r)..

P00

P01

P02

P10

P11

P12

P20

P21

P22

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