| 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
\[1/6(2.50+1.85)=£0.725\]
.