## Modelling Average Unit Maintenance Cost

From\To | Good | Fair | Poor | Broken |

Good | 0 | 0.8 | 0.2 | 0 |

Fair | 0 | 0.6 | 0.4 | 0 |

Poor | 0 | 0 | 0.5 | 0.5 |

4 | 1 | 0 | 0 | 0 |

The transition matrix is

\[A= \left( \begin{array}{cccc} 0 & 0.8 & 0.2 & 0 \\ 0 & 0.6 & 0.4 & 0 \\ 0 & 0 & 0.5 & 0.5 \\ 1 & 0 & 0 & 0 \end{array} \right)\]

Now use, for a smoothly running operation

\[\mathbf{x}B=\mathbf{x}\]

to get\[x_1=x_4\]

\[x_2=0.8x_1+0.6x_2\]

\[x_3=0.2x_1+0.4x_2+0.5x_3\]

\[x_4=0.5x_3\]

Where

\[x_1, \; x_2, \; x_3, \; x_4\]

are the probabilities of a randomly selected box being in the category good, fair, poor or broken.Solving these equations gives

\[x_1=1/6, \; x_2=1/3, \; x_3=1/3, \; x_4=1/6\]

.The average weekly cost of maintaining a crate and lost production is is

\[1/6(2.50+1.85)=£0.725\]

.