Answer to Question #228293 in Statistics and Probability for Mohamed

Question

(1) State and prove the Markovian property of interarrival times.

(2) In a railway marshaling yard, goods trains arrive at a rate of 30 trains per day. Assuming that the inter-arrival

time follows an exponential distribution and the service time (the time taken to hump a train) distribution is also

exponential with an average of 36 minutes. Calculate:

(i) Expected queue size (line length)

(ii) Probability that the queue size exceeds 10

(iii) If the input of trains increases to an average of 33 per day, what will be the change in (i) and (ii)

Answer 1

1)

The Markovian Property

Theorem

A time series of random variables is said to be a Markov chain if it has the following property

P(X_t=x_t | X₀ =x₀, X₁= x₁, ... , X_t-1=x_t-1) = P(X_t=x_t |X_t-1=x_t-1)

It is said to be homogenous a stationary if it also satisfies

P(X_t=z| X_t-1=y) = P(X_t+k=z |X_t+k-1=y) at any k $\ge$ 0

In words

The value of random variable at a time in a future. The fact this value will be X of T depends only on the present of the random variable. This present value is X of minus 1.

2)

i) Expected queue size (line length)

$\lambda$ = 30/(60*24) = 1/48 trains per minute

$\mu$ = 1/36 trains per minute

The traffic intensity $\rho = \lambda / \mu$ = 0.75

Expected queue size (line length)

L_s = $\lambda / (\lambda - \mu)$ or $\rho/(1-\rho)$ = 0.75/(1-0.75) = 3 trains

ii) Probability that the queue size exceeds 10

P[n $\ge$ 10] = $\rho$ ¹⁰ = 0.75¹⁰=0.06

iii)If the input of trains increases to an average of 33 per day, what will be the change in (i) and (ii)

Now, if the input of trains increases to an average of 33 per day, then we have

$\lambda$ = 33/(60*24) = 11/480 trains per minute

$\mu$ = 1/36 trains per minute

The traffic intensity $\rho = \lambda / \mu$ = 0.83

Hence, recalculating the value for (i) and (ii)

L_s = $\rho/(1-\rho)$ = 5 trains approximately

P[n $\ge$ 10] = $\rho$ ¹⁰ = 0.83¹⁰=0.16 approximately