Each time a machine is repaired it remains up for an exponentially distributed time with mean 20 days. It then fails, and its failure is either of two types. If it is a type 1 failure, then the time to repair the machine is exponential with mean time 1/2 day; if it is a type 2 failure, then the repair time is exponential with mean 2 days. Each failure is, independently of the time it took the machine to fail, a type 1 failure with probability 0.8 and a type 2 failure with probability 0.2. What proportion of time is the machine down due to a type 1 failure? What proportion of time is it down due to a type 2 failure? What proportion of time is it up?
We can model the state of the machine according to a continuous time markov chain X ( t ) , with state space { 0 , 1 , 2 } where
state 0 =machine is up and working
state 1 = type 1 failure
state 2= type 2 failure
Q1;
Find the transition rates:
q 0 = per day
q 10 = per day
q 20 = per day
Recall that q i j = q i p i j for i ≠ j .
q 01 = q 0 p 01 where p 01 =
q 02 = q 0 p 02 where p 02 =
q 1 =
q 2 =
Q2;
Find the steady-state probabilities of the machine being down due to type 1 failure ( π 1 ), machine being down due to type 2 failure ( π 2 ) and machine being up ( π 0 ). Either write your answers as decimals up to 4 decimal places or write them as fractions.
π 0 =
π 1 =
π 2 =
please solve this problem ASAP.
q0=1/20 per day
q10=2 per day
q20=1/2 per day
p01=.8
p02=.2
q1=2
q2=1/2
π0=25/26
π1=1/52
π2=1/52
Q1:
The transition rates can be found using the exponential mean times provided.
- q0 is the rate at which the machine transitions from state 0 (up and working) to another state. The mean time it remains in state 0 is 20 days, so q0 = 1/20 per day.
- q10 is the rate at which the machine transitions from state 1 (type 1 failure) to state 0. The mean repair time for type 1 failure is 1/2 day, so q10 = 1/(1/2) = 2 per day.
- q20 is the rate at which the machine transitions from state 2 (type 2 failure) to state 0. The mean repair time for type 2 failure is 2 days, so q20 = 1/2 = 0.5 per day.
To find q01 and q02, we use the fact that qij = qipij for i ≠ j.
- q01 = q0p01, where p01 is the probability of transitioning from state 0 to state 1. The probability of a type 1 failure given that the machine is up is 0.8, so p01 = 0.8. Therefore, q01 = q0 * 0.8 = 1/20 * 0.8 = 0.04 per day.
- q02 = q0p02, where p02 is the probability of transitioning from state 0 to state 2. The probability of a type 2 failure given that the machine is up is 0.2, so p02 = 0.2. Therefore, q02 = q0 * 0.2 = 1/20 * 0.2 = 0.01 per day.
The rates q1 and q2 represent the rate at which the machine transitions from states 1 and 2 to any other state. Since the repair times for type 1 and type 2 failures are exponential, the rates are simply the inverse of the mean times.
- q1 = 1/(1/2) = 2 per day
- q2 = 1/2 = 0.5 per day
Q2:
To find the steady-state probabilities π0, π1, and π2, we solve the system of equations πQ = 0, where π is a row vector of the steady-state probabilities and Q is the matrix of transition rates.
The transition rate matrix Q can be written as:
Q = [-(q01 + q02), q10, q20;
q01 , -q10, 0;
q02, 0, -q20]
Solving the system of equations πQ = 0, we get:
-(q01 + q02)π0 + q01π1 + q02π2 = 0
q10π0 - q10π1 = 0
q20π0 + 0π1 - q20π2 = 0
Plugging in the values of the transition rates, the system becomes:
-0.05π0 + 0.04π1 + 0.01π2 = 0
2π0 - 2π1 = 0
0.5π0 - 0.5π2 = 0
Simplifying the system, we have:
-0.05π0 + 0.04π1 + 0.01π2 = 0
π0 - π1 = 0
0.5π0 - 0.5π2 = 0
Using the second equation, we can express π0 in terms of π1:
π0 = π1
Substituting this into the third equation, we get:
0.5π1 - 0.5π2 = 0
Now, substituting π0 = π1 into the first equation, we have:
-0.05π1 + 0.04π1 + 0.01π2 = 0
Simplifying, we get:
-0.01π1 + 0.01π2 = 0
This equation tells us that π1 = π2, so the steady-state probabilities become:
π0 = π1
π1 = π2
Now, we can solve for the probabilities by normalizing them such that they sum to 1:
π0 + π1 + π2 = 1
Substituting π0 = π1 = π2, we get:
π0 + π0 + π0 = 1
3π0 = 1
π0 = 1/3
Therefore, the steady-state probabilities are:
π0 = 1/3
π1 = 1/3
π2 = 1/3
To find the proportion of time the machine is down due to each type of failure, we need to multiply the steady-state probabilities by the repair times for each failure type.
For type 1 failure, the repair time is 1/2 day, so the proportion of time the machine is down due to type 1 failure is:
π1 * (1/2) = (1/3) * (1/2) = 1/6
For type 2 failure, the repair time is 2 days, so the proportion of time the machine is down due to type 2 failure is:
π2 * 2 = (1/3) * 2 = 2/3
The proportion of time the machine is up is given by 1 minus the sum of the proportions of time the machine is down due to each failure type:
1 - (1/6 + 2/3) = 1 - (1/6 + 4/6) = 1 - (5/6) = 1/6
Therefore, the proportions of time are:
Type 1 failure: 1/6
Type 2 failure: 2/3
Machine up: 1/6
To find the transition rates, we need to calculate the probabilities of transitioning from one state to another. Recall that the transition rate (qij) is equal to the transition probability (pij) multiplied by a constant rate (qi).
Q1:
The transition rates are as follows:
q0 = per day (transition rate from state 0 to any other state)
q10 = per day (transition rate from state 1 to state 0)
q20 = per day (transition rate from state 2 to state 0)
To find q01 and q02, we need to calculate the transition probabilities (p01 and p02):
p01 = Probability of transitioning from state 0 to state 1 = Probability of a type 1 failure = 0.8
q01 = q0 * p01 = q0 * 0.8
p02 = Probability of transitioning from state 0 to state 2 = Probability of a type 2 failure = 0.2
q02 = q0 * p02 = q0 * 0.2
q1 = per day (transition rate for type 1 failure)
q2 = per day (transition rate for type 2 failure)
Q2:
To find the steady-state probabilities, we need to solve the following equations:
π0 * q0 = π1 * q10 + π2 * q20
π1 * q1 = π0 * q01
π2 * q2 = π0 * q02
Since the machine is working for an exponentially distributed time with mean 20 days, we know that the rate of being up (λ) is equal to 1/20 per day. Thus, q0 = λ = 1/20.
Substituting this value into the equations, we have:
π0 * (1/20) = π1 * q10 + π2 * q20
π1 * q1 = π0 * (q0 * p01)
π2 * q2 = π0 * (q0 * p02)
Simplifying further, we get:
π0/20 = π1 * q10 + π2 * q20
π1 * q1 = π0 * (1/20 * 0.8)
π2 * q2 = π0 * (1/20 * 0.2)
Now, we need to find the steady-state probabilities by solving the system of equations:
π0 + π1 + π2 = 1 (since the sum of probabilities must be 1)
π0/20 = π1 * q10 + π2 * q20
π1 * q1 = π0/20 * 0.8
π2 * q2 = π0/20 * 0.2
We can solve this system of equations to find the steady-state probabilities (π0, π1, π2).