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14.6 Absorbing States and the Fundamental Matrix: Accounts Receivable Application

In the examples discussed thus far, we assume that it is possible for the process or system to go from one state to any other state between any two periods. In some cases, however, if you are in a state, you cannot go to another state in the future. In other words, when you are in a given state, you are “absorbed” by it, and you will remain in that state. Any state that has this property is called an absorbing state. An example of this is the accounts receivable application.

An accounts receivable system normally places debts or receivables from its customers into one of several categories or states, depending on how overdue the oldest unpaid bill is. Of course, the exact categories or states depend on the policy set by each company. Four typical states or categories for an accounts receivable application follow:

State 1 $(π_{1})$ $(π_{1})$ : paid, all bills
State 2 $(π_{2})$ $(π_{2})$ : bad debt, overdue more than 3 months
State 3 $(π_{3})$ $(π_{3})$ : overdue less than 1 month
State 4 $(π_{4})$ $(π_{4})$ : overdue between 1 and 3 months

At any given period—in this case, 1 month—a customer can be in one of these four states.² For this example, it will be assumed that if the oldest unpaid bill is more then 3 months overdue, it is automatically placed in the bad debt category. Therefore, a customer can be paid in full (state 1), have the oldest unpaid bill overdue less than 1 month (state 3), have the oldest unpaid bill overdue between 1 and 3 months inclusive (state 4), or have the oldest unpaid bill overdue more than 3 months, which is a bad debt (state 2).

As in any other Markov process, we can set up a matrix of transition probabilities for these four states. This matrix will reflect the propensity of customers to move among the four accounts receivable categories from one month to the next. The probability of being in the paid category for any item or bill in a future month, given that a customer is in the paid category for a purchased item this month, is 100% or 1. It is impossible for a customer to completely pay for a product one month and to owe money on it in a future month. Another absorbing state is the bad debts state. If a bill is not paid in 3 months, we are assuming that the company will completely write it off and not try to collect it in the future. Thus, once a person is in the bad debt category, that person will remain in that category forever. For any absorbing state, the probability that a customer will be in this state in the future is 1, and the probability that a customer will be in any other state is 0.

These values will be placed in the matrix of transition probabilities. But before we construct this matrix, we need to know the future probabilities for the other two states—a debt that is less than 1 month old and a debt that is between 1 and 3 months old. For a person in the less-than-1-month category, there is a 0.60 probability of being in the paid category, a 0 probability of being in the bad debt category, a 0.20 probability of remaining in the less-than-1-month category, and a probability of 0.20 of being in the 1- to 3-month category in the next month. Note that there is a 0 probability of being in the bad debt category the next month because it is impossible to get from state 3, less than 1 month, to state 2, more than 3 months overdue, in just 1 month. For a person in the 1- to 3-month category, there is a 0.40 probability of being in the paid category, a 0.10 probability of being in the bad debt category, a 0.30 probability of being in the less-than-1-month category, and a 0.20 probability of remaining in the 1- to 3-month category in the next month.

How can we get a probability of 0.30 of being in the 1- to 3-month category for one month and in the less-than-1-month category in the next month? Because these categories are determined by the oldest unpaid bill, it is possible to pay one bill that is 1 to 3 months old and still have another bill that is less than 1 month old. In other words, any customer may have more than one outstanding bill at any point in time. With this information, it is then possible to construct the matrix of transition probabilities of the problem.

	NEXT MONTH
THIS MONTH	PAID	BAD DEBT	<1 MONTH	1 TO 3 MONTHS
Paid	1	0	0	0
Bad debt	0	1	0	0
Less than 1 month	0.6	0	0.2	0.2
1 to 3 months	0.4	0.1	0.3	0.2

Thus,

P = [\begin{array}{l} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0.6 & 0 & 0.2 & 0.2 \\ 0.4 & 0.1 & 0.3 & 0.2 \end{array}]

$P = [\begin{array}{l} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0.6 & 0 & 0.2 & 0.2 \\ 0.4 & 0.1 & 0.3 & 0.2 \end{array}]$

If we know the fraction of the people in each of the four categories or states for any given period, we can determine the fraction of the people in each of these four states or categories for any future period. These fractions are placed in a vector of state probabilities and multiplied times the matrix of transition probabilities. This procedure was described in Section 14.4.

Even more interesting are the equilibrium conditions. Of course, in the long run, everyone will be in either the paid or the bad debt category. This is because the categories are absorbing states. But how many people, or how much money, will be in each of these categories? Knowing the total amount of money that will be in either the paid or the bad debt category will help a company manage its bad debts and cash flow. This analysis requires the use of the fundamental matrix.

To obtain the fundamental matrix, it is necessary to partition the matrix of transition probabilities, P. This can be done as follows:

A table shows the data on the probability of debts.

14.6-3 Full Alternative Text

where

\begin{array}{l} I & = & an identity matrix (i .e ., a matrix with 1s on the diagonal and 0s everyplace else) \\ 0 & = & a matrix with all 0s \end{array}

$\begin{array}{l} I & = & an identity matrix (i .e ., a matrix with 1s on the diagonal and 0s everyplace else) \\ 0 & = & a matrix with all 0s \end{array}$

The fundamental matrix can be computed as follows:

F = {(I - B)}^{- 1}

$F = {(I - B)}^{- 1}$ (14-8)

In Equation 14-8, $(I - B)$ $(I - B)$ means that we subtract the B matrix from the I matrix. The superscript $- 1$ $- 1$ means that we take the inverse of the result of $(I - B) .$ $(I - B) .$ Here is how we can compute the fundamental matrix for the accounts receivable application:

F = {(I - B)}^{- 1}

$F = {(I - B)}^{- 1}$

F = {([\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 0.2 & 0.2 \\ 0.3 & 0.2 \end{matrix}])}^{- 1}

$F = {([\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 0.2 & 0.2 \\ 0.3 & 0.2 \end{matrix}])}^{- 1}$

Subtracting B from I, we get

F = [\begin{matrix} 0.8 & - 0.2 \\ - 0.3 & 0.8 \end{matrix}]^{- 1}

$F = [\begin{matrix} 0.8 & - 0.2 \\ - 0.3 & 0.8 \end{matrix}]^{- 1}$

Taking the inverse of a large matrix involves several steps, as described in Module 5 on the Companion Website for this book. Appendix 14.2 shows how this inverse can be found using Excel. However, for a matrix with two rows and two columns, the computations are relatively simple, as shown here.

The inverse of the matrix $[\begin{matrix} a & b \\ c & d \end{matrix}]$ $[\begin{matrix} a & b \\ c & d \end{matrix}]$ is

[\begin{matrix} a & b \\ c & d \end{matrix}]^{- 1} = [\begin{matrix} \frac{d}{r} & \frac{- b}{r} \\ \frac{- c}{r} & \frac{a}{r} \end{matrix}]

$[\begin{matrix} a & b \\ c & d \end{matrix}]^{- 1} = [\begin{matrix} \frac{d}{r} & \frac{- b}{r} \\ \frac{- c}{r} & \frac{a}{r} \end{matrix}]$ (14-9)

where

r = a d - b c

$r = a d - b c$

To find the F matrix in the accounts receivable example, we first compute

r = a d - b c = (0.8) (0.8) - (2 0.2) (2 0.3) = 0.64 - 0.06 = 0.58

$r = a d - b c = (0.8) (0.8) - (2 0.2) (2 0.3) = 0.64 - 0.06 = 0.58$

With this, we have

F = [\begin{matrix} 0.8 & - 0.2 \\ - 0.3 & 0.8 \end{matrix}]^{- 1} = [\begin{matrix} \frac{0.8}{0.58} & \frac{- (- 0.2)}{0.58} \\ \frac{- (- 0.3)}{0.58} & \frac{0.8}{0.58} \end{matrix}] = [\begin{matrix} 1.38 & 0.34 \\ 0.52 & 1.38 \end{matrix}]

$F = [\begin{matrix} 0.8 & - 0.2 \\ - 0.3 & 0.8 \end{matrix}]^{- 1} = [\begin{matrix} \frac{0.8}{0.58} & \frac{- (- 0.2)}{0.58} \\ \frac{- (- 0.3)}{0.58} & \frac{0.8}{0.58} \end{matrix}] = [\begin{matrix} 1.38 & 0.34 \\ 0.52 & 1.38 \end{matrix}]$

Now we are in a position to use the fundamental matrix in computing the amount of bad debt money that we could expect in the long run. First, we need to multiply the fundamental matrix, F, times the A matrix. This is accomplished as follows:

F A = [\begin{matrix} 1.38 & 0.34 \\ 0.52 & 1.38 \end{matrix}] \times [\begin{matrix} 0.6 & 0 \\ 0.4 & 0.1 \end{matrix}]

$F A = [\begin{matrix} 1.38 & 0.34 \\ 0.52 & 1.38 \end{matrix}] \times [\begin{matrix} 0.6 & 0 \\ 0.4 & 0.1 \end{matrix}]$

F A = [\begin{matrix} 0.97 & 0.03 \\ 0.86 & 0.14 \end{matrix}]

$F A = [\begin{matrix} 0.97 & 0.03 \\ 0.86 & 0.14 \end{matrix}]$

The new FA matrix has an important meaning. It indicates the probability that an amount in one of the nonabsorbing states will end up in one of the absorbing states. The top row of this matrix indicates the probabilities that an amount in the less-than-1-month category will end up in the paid and the bad debt categories. The probability that an amount that is less than 1 month overdue will be paid is 0.97, and the probability that an amount that is less than 1 month overdue will end up as a bad debt is 0.03. The second row has a similar interpretation for the other nonabsorbing state, which is the 1- to 3-month category. Therefore, 0.86 is the probability that an amount that is 1 to 3 months overdue will eventually be paid, and 0.14 is the probability that an amount that is 1 to 3 months overdue will never be paid but will become a bad debt.

In Action Markovian Volleyball!

Volleyball coaches turned to Markovian analysis to rank five basic volleyball skills (passing, setting, digging, serving, and attacking) in high-level volleyball. In other words, the coaches wanted to know which skill had the highest probability of leading to a point on any given play, which skill had the second-highest probability of leading to a point on any given play, and so on.

For an entire season, data were collected from all matches, and all individual skill performances (e.g., a serve) in each match were rated based on a simple rubric. For example, a set that was in perfect position (3–5 feet away from the net) was given 3 performance points, while a set that was 5–8 feet away from the net was given just 1 performance point. The end result of each skill (a point for the visiting team, the rally continued, or a point for the home team) was also recorded.

A rather large $36 \times 36$ $36 \times 36$ transition matrix was developed, and a mix of empirical and Bayesian methods (see Chapter 2) were employed to arrive at the individual transition probabilities, $P_{i j}$ $P_{i j}$ s. State transitions that were impossible (e.g., a perfectly positioned set followed by a service error) were assigned a probability of zero.

The ensuing Markovian analysis led analysts to one over-arching conclusion: whereas the most important skills in men’s volleyball are (by a large margin over the others) serving and attacking, the most important skills in women’s volleyball are (again by a large margin over the others) passing, setting, and digging. Can you dig it?

Source: M. A. Miskin, G. W. Fellingham, and L. W. Florence, “Skill Importance in Women’s Volleyball,” Journal of Quantitative Analysis in Sport 6 (2010): Article 5.

This matrix can be used in a number of ways. If we know the amounts of the less-than-1-month category and the 1- to 3-month category, we can determine the amount of money that will be paid and the amount of money that will become bad debt. We let the M matrix represent the amount of money that is in each of the nonabsorbing states:

M = (M_{1}, M_{2}, M_{3}, \dots, M_{n})

$M = (M_{1}, M_{2}, M_{3}, \dots, M_{n})$

where

$n =$ $n =$ number of nonabsorbing states
$M_{1} =$ $M_{1} =$ amount in the first state or category
$M_{2} =$ $M_{2} =$ amount in the second state or category
$M_{n} =$ $M_{n} =$ amount in the nth state or category

Assume that there is $2,000 in the less-than-1-month category and $5,000 in the 1- to 3-month category. Then M would be represented as follows:

M = (2, 000, 5, 000)

$M = (2, 000, 5, 000)$

The amount of money that will end up as being paid and the amount that will end up as bad debt can be computed by multiplying the matrix M times the FA matrix that was computed previously. Here are the computations:

\begin{array}{l} Amount paid and amount in bad debt & = & M F A \\ = & (2, 000, 5, 000) [\begin{array}{l} 0.97 & 0.03 \\ 0.86 & 0.14 \end{array}] \\ = & (6, 240, 760) \end{array}

$\begin{array}{l} Amount paid and amount in bad debt & = & M F A \\ = & (2, 000, 5, 000) [\begin{array}{l} 0.97 & 0.03 \\ 0.86 & 0.14 \end{array}] \\ = & (6, 240, 760) \end{array}$

Thus, out of the total of $7,000 ($2,000 in the less-than-1-month category and $5,000 in the 1- to 3-month category), $6,240 will be eventually paid, and $760 will end up as bad debts.

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Table of Contents for 14.6 Absorbing States and the Fundamental Matrix: Accounts Receivable Application

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Table of Contents for
14.6 Absorbing States and the Fundamental Matrix: Accounts Receivable Application