# Finding the generator matrix for a Markov jump process

Vehicles in a certain country are required to be assessed every year for road-worthiness. At one vehicle assessment center, drivers wait for an average of 15 minutes before the road-worthiness assessment of their vehicle commences. The assessment takes on average 20 minutes to complete. Following the assessment, 80% of vehicles are passed as road-worthy allowing the driver to drive home. A further 15% of vehicles are categorized as a “minor fail”; these vehicles require on average 30 minutes of repair work before the driver is allowed to drive home. The remaining 5% of vehicles are categorized as a “significant fail”; these vehicles require on average three hours of repair work before the driver can go home.

A continuous-time Markov model is to be used to model the operation of the vehicle assessment centre, with states W (waiting for assessment), A (assessment taking place), M (minor repair taking place), S (significant repair taking place) and H (travelling home).

generator matrix i got:

$\space\space\space\space\space\space\space$ $\begin{matrix} W & \space\space A & M & \space\space S &\space\space\space H \end{matrix}$

$\begin{matrix} W \\ A \\ M \\ S \\ H \end{matrix} \left[ \begin{matrix} \frac{-1}{15} & \frac{1}{15} & 0 & 0 & 0 \\ 0 & \frac{-1}{20} & & & \frac{1}{25} \\ 0& 0 & \frac{-1}{30} & 0 & \frac{1}{30} \\ 0 & 0 & 0 & \frac{-1}{180} & \frac{1}{180} \\ 0 & 0 & 0 & 0 & 0 \end{matrix} \right]$

But I don't know how to find the 2 empty spaces in second row. Any help would be appreciated.

W, A, M, S, H are labels of rows and columns.

• Have you used the 80%, 15%, 5% figures anywhere yet when determining these entries? – whuber May 17 '12 at 21:27
• $\frac {80}{(100)(20)} = \frac {1}{25}$ in second row. – amul28 May 18 '12 at 1:33
• My comment was a hint to look more closely at how you can use these figures. Notice that you haven't employed the 15% or 5% information directly. – whuber May 18 '12 at 13:48
• I was calculating taking into account the waiting time in H also. As you take waiting time in H when you enter that state. But actually it has reached state H from A, so the waiting time will be only 1/20. Am I correct. – amul28 May 18 '12 at 14:59

Think of the entries as rates of flow. Drivers flow out of state $A$ at a rate of 1/20 minutes (the - sign means "out"), and therefore must flow into states $M$, $S$, and $H$ combined at a rate of << fill in the blank. >> Now, given that total flow rate into $M$, $S$, and $H$, why is the number for $A \to H$ 1/25? Obviously it must relate to a number given in the text somewhere...
• The probability 80% is given... and the implications for how to calculate the two blank fields, which have probabilities of transitioning to from $A$ of ___ and ___, are... – jbowman May 18 '12 at 1:47
• Right! You have 1 transition out of $A$ every 20 minutes, on average, and 80% go to H, so ON AVERAGE 1 every 25 minutes, and 15% to M, so ON AVERAGE 3 every 400 minutes, and 5% to S, son ON AVERAGE 1 every 400 minutes. – jbowman May 18 '12 at 2:41