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## Homework Statement

there are 13,200 components that must be regularly maintained or else they fail.

At the end of month “t”, there are is a state vector describing the components given by V

_{t}= [f

_{t}, m

_{t}, a

_{t}]

^{T}

where f

_{t}is the number of failed components, m

_{t}is the number of components out for maintenance, and a

_{t}is the number of components in active production.

Of the active components one month ( a

_{t}), p is the fraction that fail the next

month, and q is the fraction that have been taken off-line for maintenance. The rest

remain active. All those in maintenance one month ( m

_{t}) are active the next month. All

those that have failed one month ( f

_{t}) are under maintenance the next.

**Describe in words why the equations that drive this system are:**

F

m

a

WRITE OUT A MATRIX, A, if this is written as a dynamical system, v

F

_{t+1}= pa_{t}m

_{t+1}= qa_{t}a

_{t+1}= m_{t}+ (1-p-q)a_{t}WRITE OUT A MATRIX, A, if this is written as a dynamical system, v

_{t+1}= AV_{t}## Homework Equations

/

## The Attempt at a Solution

F

_{t+1}= pa

_{t}: This equation states that the number of failed components in the coming month (t + 1, meaning the month after the current month), can be represented as the number of active components during the current month, multiplied by the fraction of components that fail in the proceeding month (t+1); or ANY month, t.

i.e.

if there are 12 active components during the second month (at = 12). And 25% of those fail by the end of the month (p = 0.25). The number of failed components in the third month is 12*0.25 = 3. Hence ft+1 = 3 (the number of failed components in the coming month).

m

_{t+1}= qa

_{t}: This equation states that the number of components under maintenance in the coming months (t + 1), can be represented as the number of failed components in the current month, added to the fraction of components removed for maintenance (q) multiplied by the number of active components (at) in the current month.

I.e.

The number of components under maintenance in the UPCOMING month will be the number of failed components by the end of the CURRENT month, ft + the fraction that have been REMOVED for maintenance, qat in the CURRENT month, giving the total number of components needing repair AND under repair FOR THE UPCOMING MONTH.

a

_{t+1}= m

_{t}+ (1-p-q)a

_{t}: this equation states that the number of active components in the up coming months, (t+1) is equivalent to the number of components under maintenance in the current month (since these will be available for use during the proceeding month), in addition to 1 subtract the percent that fail in the current month, and subtract the fraction taken off-line for maintenance in the current month multiplied by all active components, at; yielding the available active components for use during the proceeding month!

Now for the juicy stublack

It is stated that V

_{t}= [f

_{t}, m

_{t}, a

_{t}]

^{T}, which represents the state vector at the end of the month, t. We can assume that during the very first month, all units are functional, and hence, initially, f

_{t}= 0, m

_{t}= 0 and a

_{t}= 13200.

We are given the individual components of the state vector at any month with the equations above.

We can use this to form a matrix which represents the state vector at any month, Vt+1.

\begin{bmatrix}

&p &q &(1-p-q) \\

&1 &0 &0 \\

&1 &1 &0 \\

&1 &1 &1

\end{bmatrix}\begin{bmatrix}

&vt \\

&0\\

&0\\

&13,200

\end{bmatrix}

(not sure how to make matrices side by side, so please pretend that these two are being multiplied)

am I doing this right o____o

it looks a bit wrong,

Q

A system has a “steady state” vector if there is a state vector v*

that never changes over time, so that Av* = v*

What eigenvalue must the matrix possess if there is

to be a steady state? Show this is an eigenvalue of matrix A for any value of p

and q , so that a steady state must always exist for this system

..

I have no idea how to approach this part =[

PLEASE HELP MEEEE =)

<3 <3