Markov processes (Weight Suspended equally by n Cables)

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A weight of L tons is suspended by n cables that share the load equally, and if any cables break, the remaining cables redistribute the load. The discussion emphasizes the Markov property of the process, where the future state depends only on the current state of unbroken cables, not on past states. The generator matrix for this process is defined by transition rates between states of unbroken cables, with specific attention to how many cables can break before restoration occurs. The failure rate of cables is given as 0.2 per year per ton, leading to calculations on the probability of the system lasting two years and the number of cables needed to ensure a 0.999 probability of success. The conversation focuses on clarifying the state space and the transition rates necessary for constructing the generator matrix.
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Summary:: Markov processes (A weight of L tons is suspended by n cables which share the load equally)

A weight of L tons is suspended by n cables which share the load equally. If k, for 1 ≤ k ≤ n − 1, of the cables have broken, then the remaining (n − k) cables share the load equally. As soon as (n − 1) cables have failed, new cables will be installed instantly to restore the number of cables to n. Let X(t) be the number of unbroken cables at time t. The instantaneous failure rate of a single cable which carries M tons is αM, where α > 0 is a constant. The remaining time to failure of any cable operating at time t is independent of the past history of the process, conditional on X(t).

(a) Specify an appropriate state space for (X(t) : t ≥ 0), and explain why this stochastic process satisfies the Markov property.

(b) Give the generator matrix for (X(t)).

(c) Suppose that individual cables have a failure rate of 0.2 per year per ton of load.

(i) If 4 such cables are used to support 20 tons, find the probability that the system lasts for at least 2 years before reinstalling the cables. Hint: determine the number of states that the process must go through to get to the first reinstallation, and then express the total duration as the sum of the hold times in each of the states to be visited.

(ii) How many cables should be used to ensure with probability 0.999 that the system lasts for at least 2 years before reinstalling the cables?

Note: The sum of n independent Exp(µ) random variables has a Gamma distribution with shape parameter n and rate parameter µ.

My effort/wok so far:

My effort so far:

a-) An appropriate state space is: S = {0,1} where 0 means the cable is broken, and 1 means the cable is unbroken

Now, this stochastic process satisfies a markov property because it independent of the previous history of states, except being conditionally indpendent on the latest state only. That is:

P{X(t+u) = j | H(u), X(u) = i} = P(x(t+u) = j | x(u) = i)

Where it was given that the remaining time to failre of any cable operating at time (t) is independent of he history of the process, conditional on X(t). Thus, our markov process satisfies the markov property.

b-) The generator matrix (Q) has these elements:

qij = {(ri)*(pij) if i not equal to j ; or it is -ri if i = j

and since it is given that the instantaneous failure rate of a single cable which carries M tons is αM, where α > 0 is a constant, then we have r0 = αM
Now, could you please correct any of my mistakes?

Also, I don't know to continue from here? How to derive the values of pij and r1, etc. for Q matrix?

Thank you very much in advance for your help.
 
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Your answer to (a) is incorrect. The state space is the set of all possible values the Markov process can take. Since the process is X(t), the number of unbroken cables, what is the set of all possible values that X(t) can take, given it starts with n unbroken cables?
In answering this, consider whether the system can ever have 0 or 1 unbroken cables, given the statement in the second sentence of your OP. Note the word instantly, meaning the system spends no time having n-1 failed cables, meaning it jumps directly from the state of having n-2 failed cables to one of having no failed cables.
For the next bit, you use terms ri and pij (presumably ##r_i## and ##p_{ij}##) but do not say what they represent. Those are not any standard notation I recognise so I assume they are just used in your course text or notes. You'll need to explain what they represent...
OR
... just write the matrix elements directly. The element ##q_{ij}## is the instantaneous rate of migration from state ##i## (##i## unbroken cables) to state ##j## (##j## unbroken cables). Since the probability of more than one cable breaking at once is zero we know that ##q_{ij}## must be zero if ##j<i-1##. What about for ##j>i##? Hint: there is only one ##ij## combination with ##j>i## for which ##q_{ij}>0##. What is it?
 
andrewkirk said:
Your answer to (a) is incorrect. The state space is the set of all possible values the Markov process can take. Since the process is X(t), the number of unbroken cables, what is the set of all possible values that X(t) can take, given it starts with n unbroken cables?
In answering this, consider whether the system can ever have 0 or 1 unbroken cables, given the statement in the second sentence of your OP. Note the word instantly, meaning the system spends no time having n-1 failed cables, meaning it jumps directly from the state of having n-2 failed cables to one of having no failed cables.
For the next bit, you use terms ri and pij (presumably ##r_i## and ##p_{ij}##) but do not say what they represent. Those are not any standard notation I recognise so I assume they are just used in your course text or notes. You'll need to explain what they represent...
OR
... just write the matrix elements directly. The element ##q_{ij}## is the instantaneous rate of migration from state ##i## (##i## unbroken cables) to state ##j## (##j## unbroken cables). Since the probability of more than one cable breaking at once is zero we know that ##q_{ij}## must be zero if ##j<i-1##. What about for ##j>i##? Hint: there is only one ##ij## combination with ##j>i## for which ##q_{ij}>0##. What is it?
Hello Andrewkirk, thanks for your reply.

- So after thinking about what you wrote, would the state space be: S = {2,2,3,...,n-1, n}? The reason there is no 0 and 1 is because as you said, the system can never have 0 or 1 unbroken cables because once we reach n-1 unbroken cables, then it instantly gets restored to n unbroken cables.

- r is the transition rate out of a state, and p[/j] is the transition probability from state i to j, so we have:
qij =ri * pij ; for i not equal to j (i.e., j>i) (sorry, I am no sure how to write ri and p(ij) as you did)
or qij = -r ; for i = j

That's just straight from our lecture notes.

Now my question is, what is the value of r? How to calculate it? Because I need that value before being able to work out the matrix elements.

Thanks for your help, it is much appreciate it.
 
From the previous post:
CTK said:
- r[ i] is the transition rate out of a state, and p[ i][/j] is the transition probability from state i to j, so we have:
qij[ i] =ri * pij[ i][ i] ; for i not equal to j (i.e., j>i) (sorry, I am no sure how to write ri and p(ij) as you did)
or qij[ i] = -r[ i] ; for i = j

That's just straight from our lecture notes.

Now my question is, what is the value of r[ i]?
Your previous post was mangled due to the inclusion of bracketed expressions with i. Such expressions get rendered by browsers as the start of italicized expressions.

Here is what I think you intended:
##r_i## is the transition rate out of a state, and ##p_{i, j}##
is the transition probability from state i to j, so we have:
##q_{i, j}[ i] =r_i * p_{i, j}[ i][ i]## ; for i not equal to j (i.e., j>i)
or ##q_{i, j}[ i] = -r[ i]## ; for i = j

I'm not sure what you intended with what I wrote as ## p_{i, j}[ i][ i]##
CTK said:
(sorry, I am no sure how to write ri and p(ij) as you did)
Click on any of my TeX expressions to see what I wrote. Also note that there is a link in the lower left corner to our tutorial on LaTeX.
 
Mark44 said:
From the previous post:
Your previous post was mangled due to the inclusion of bracketed expressions with i. Such expressions get rendered by browsers as the start of italicized expressions.

Here is what I think you intended:
##r_i## is the transition rate out of a state, and ##p_{i, j}##
is the transition probability from state i to j, so we have:
##q_{i, j}[ i] =r_i * p_{i, j}[ i][ i]## ; for i not equal to j (i.e., j>i)
or ##q_{i, j}[ i] = -r[ i]## ; for i = j

I'm not sure what you intended with what I wrote as ## p_{i, j}[ i][ i]##

Click on any of my TeX expressions to see what I wrote. Also note that there is a link in the lower left corner to our tutorial on LaTeX.
Ops, sorry I meant to say the following:

##r_i## is the transition rate out of a state, and ##p_{i, j}##
is the transition probability from state i to j, so we have:
##q_{i, j} = r_i * p_{i, j}##
; for i not equal to j (i.e., j>i)
or ##q_{i, j} = -r[ i]## ; for i = j

So just ignore the ## p_{i, j}[ i][ i]##

So having said that, can anyone please help me with correcting my state space if it is wrong, and especially with figuring out the elements of the matrix ##q_{i, j}##?
Thanks for your help.
 

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