Question

The computer center at Dong-A University has been experiencing computer down time. Let us assume that the trials of an associated Markov process are defined as one-hour periods and that the probability of the system being in a running state or a down state is based on the state of the system in the previous period. Historical data show the following transition probabilities: From Running down Running 0.90 0.10 Down 0.30 0.70 If the system is initially running, what is the probability of the system being down in next hour of operation? At a current period, the system is in a down state. After 2 periods of time, what is the probability that the system will be in the state of running? c. What are the steady-state probabilities of system being in the running state and in the down state?

Answer 1

Answer:

(a)0.16

(b)0.588

(c)$[s_1 s_2]=[0.75, 0.25]$

Step-by-step explanation:

The matrix below shows the transition probabilities of the state of the system.

$\left(\begin{array}{c|cc}& Running& Down\\---&---&---\\ Running&0.90&0.10\\ Down&0.30&0.70\end{array}\right)$

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix $P^k$.

(a)

$P^1=\left(\begin{array}{c|cc}& Running& Down\\---&---&---\\ Running&0.90&0.10\\ Down&0.30&0.70\end{array}\right)$

$P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}$

If the system is initially running, the probability of the system being down in the next hour of operation is the $(a_{12})th entry of the P^2 matrix.$

The probability of the system being down in the next hour of operation = 0.16

(b)After two(periods) hours, the transition matrix is:

$P^3=\begin{pmatrix}0.804&0.196\\ 0.588&0.412\end{pmatrix}$

Therefore, the probability that a system initially in the down-state is running

is 0.588.

(c)The steady-state probability of a Markov Chain is a matrix S such that SP=S.

Since we have two states, $S=[s_1 s_2]$

$[s_1 s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1 s_2]$

Using a calculator to raise matrix P to large numbers, we find that the value of $P^k$ approaches [0.75 0.25]:

Furthermore,

$[0.75 0.25]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[0.75 0.25]$

The steady-state probabilities of the system being in the running state and in the down-state is therefore:

$[s_1 s_2]=[0.75 0.25]$