8.17 Discrete Compartmental Models

INTRODUCTION

The construction of the mathematical model (3) in Section 2.9, describing the number of pounds of salt in two connected tanks in which brine is flowing into and out of the tanks, is an example of compartmental analysis. In the discussion in Section 2.9, the compartmental model was a system of differential equations. In this section we introduce the notion of a discrete mathematical model.

The General Two-Compartment Model

Suppose material flows between two tanks with volumes V1 and V2. In the diagram shown in FIGURE 8.17.1, F01, F12, F21, F10, and F20 denote flow rates. Note that the double-subscripted symbol Fij denotes the flow rate from tank i to tank j. Next, suppose a second substance, called the tracer, is infused into compartment 1 at a known rate I(t). As we did in Section 2.9, we will assume that the tracer is thoroughly mixed in both compartments at all times t. If x(t) denotes the amount of tracer in compartment 1 and y(t) the amount of tracer in compartment 2, then the concentrations are c1(t) = x(t)/V1 and c2(t) = y(t)/V2, respectively. It follows that the general two-compartment model is

(1)

A diagram consists of two boxes. The box on the left is labeled compartment 1 and the one on the right is labeled compartment 2. The inflow of material into compartment 1 from the outside is indicated by an arrow on the left edge of the box 1 and pointing to the right labeled as F subscript 01. The inflow of material from compartment 1 into compartment 2 is indicated by an arrow pointing to the right and labeled as F subscript 12. The inflow of material from compartment 2 into compartment 1 is indicated by an arrow pointing to the left and labeled as F subscript 21. The outflow of material from compartment 1 is indicated by an arrow at the bottom edge of the box 1 pointing downward and labeled as F subscript 10. The outflow of material from compartment 2 is indicated by an arrow at the bottom edge of the box 2 pointing downward and labeled as F subscript 20. A downward pointing arrow at the top edge of box 1 is labeled as I (t).

FIGURE 8.17.1 Material flowing between two compartments at specified rates

The model in (1) keeps track of the amount of tracer that flows between the compartments. The material consisting of, say, a fluid and a tracer is continually interchanged. We present next a model that keeps track of compartmental contents every Δt units of time and assumes that the system changes only at times Δt, 2Δt, . . ., nΔt, . . . . Of course, by selecting Δt very small, we can approximate the continuous case.

Discrete Compartmental Models

In constructing a compartmental model of a physical system, we conceptually separate the system into a distinct number of small components between which material is transported. Compartments need not be spatially distinct (like the tanks used in Section 2.9) but must be distinguishable on some basis. The following are a few examples:

  • Acid rain (containing strontium 90, for example) is deposited onto pastureland. Compartments might be grasses, soil, streams, and litter.
  • In studying the flow of energy through an aquatic ecosystem, we might separate the system into phytoplankton, zooplankton, plankton predators, seaweed, small carnivores, large carnivores, and decay organisms.
  • A tracer is infused into the bloodstream and is lost to the body by the metabolism of a particular organ and by excretion. Appropriate compartments might be arterial blood, venous blood, the organ, and urine.

Suppose then that a system is divided into n compartments and, after each Δt units of time, material is interchanged between compartments. We will assume that a fixed fraction τij of the contents of compartment j are passed to compartment i every Δt units of time, as depicted in FIGURE 8.17.2. This hypothesis is known as the linear donor-controlled hypothesis.

A compartmental diagram consists of two boxes. The box on the left is labeled compartment i. The box on the right is labeled compartment j. An arrow pointing right from box 1 to box 2 is labeled tau subscript j i. An arrow pointing left from box 2 to box 1 is labeled tau subscript i j.

FIGURE 8.17.2 Interchange of material between compartments

Let the entries xi in the n × 1 matrix X,

(2)

represent the amount of tracer in compartment i. We say that X specifies the state of the system. The n × 1 matrix Y is the state of the system Δt units of time later. We will show that X and Y are related by the matrix equation Y = TX, where T is an n × n matrix determined by the transfer coefficients τij. To find T, observe, for example, that

If we let τ11 = 1 − τ21 − τ31 − . . . − τn1, then τ11 is just the fraction of the contents of compartment 1 that remains in 1.

Letting τii = 1 − Σj≠iτji we have, in general,

(3)

The matrix equation in (3) is the desired equation Y = TX. The matrix T = (τij)n×n is called the transfer matrix. Note that the sum of the entries in any column, the transfer coefficients, is equal to 1.

Note: A transfer matrix is an example of a stochastic matrix. See Problem 32 in Exercises 8.8.

Discrete compartmental models are illustrated in the next two examples.

EXAMPLE 1 Transfer Matrix

In FIGURE 8.17.3 the three boxes represent three compartments. The content of each compartment at time t is indicated in each box. The transfer coefficients are shown along the arrows connecting the compartments.

A compartmental diagram consists of three boxes forming a downward pointing triangle. Starting from the left and moving in a clockwise direction the boxes are labeled as followed: Box 1, 100; box 2, 250; box 3, 80. An arrow pointing right from box 1 to box 2 is labeled 0.2 per day. An arrow pointing to the left from box 2 to box 1 is labeled 0.05 per day. An arrow pointing downward to the left from box 2 to box 3 is labeled 0.3 per day. An arrow pointing upward to the left from box 3 to box 1 is labeled 0.25 per day.

FIGURE 8.17.3 Compartments and transfer coefficients in Example 1

(a) Find the transfer matrix T.

(b) Suppose Δt = 1 day. Find the state of the system Y 1 day later.

SOLUTION

(a) The state of the system at time t = 0 is X = . Remember that τij specifies the rate of transfer to compartment i from compartment j. Hence we are given that τ21 = 0.2, τ12 = 0.05, τ32 = 0.3, τ23 = 0, τ13 = 0.25, and τ31 = 0. From these numbers we see that the matrix T is

(4)

But since the column entries must sum to 1, we can fill in the blanks in (4):

(b) The state of the system 1 day later is then

≡

If X0 denotes the initial state of the system, and Xn is the state after n(Δt) units of time, then

Because

we have in general (5)

We could, of course, use the method illustrated in Section 8.9 to compute Tn, but with the aid of a calculator or a CAS it is just as easy to use the recursion formula Xn + 1 = TXn by letting n = 0, 1, . . . .

EXAMPLE 2 States of an Ecosystem

Strontium-90 is deposited into pastureland by rainfall. To study how this material is cycled through the ecosystem, we divide the system into the compartments shown in FIGURE 8.17.4. Suppose that Δt = 1 month and the transfer coefficients (which have been estimated experimentally) shown in the figure are measured in fraction/month. (We will ignore that some strontium-90 is lost due to radioactive decay.) Suppose that rainfall has deposited the strontium-90 into the compartments so that X0 = . (Units might be grams per hectare.) Compute the states of the ecosystem over the next 12 months.

A compartmental diagram consists of four boxes. Starting from the left top and moving in a clockwise direction the boxes are labeled as followed: Box 1, Grasses; box 2, soil; box 4, streams, box 3, dead organic matter. An arrow pointing right from box 1 to box 2 is labeled (leaching) 0.05. An arrow pointing left from box 2 to box 1 is labeled 0.01 (growth). A downward arrow from box 1 to box 3 is labeled 0.1 (death). A downward arrow from box 2 to box 4 is labeled 0.01 (run-off). An upward diagonal arrow to the right from box 3 to box 2 is labeled 0.2 (decomposition).

FIGURE 8.17.4 Ecosystem in Example 2

SOLUTION

From the data in Figure 8.17.4 we see that transfer matrix T is

.

We must compute X1, X2, ..., X12. The state of the ecosystem after the first month is

.

The remaining states, computed with the aid of a CAS and the recursion formula Xn + 1 = TXn with n = 1, 2, . . ., 11, are given in Table 8.17.1. ≡

TABLE 8.17.1
A table has 5 columns and 13 rows. The column headings from left to right are month, grasses, soil, dead organic matter, streams. The row entries are as follows. Row 1. Month, 0; grasses, 20.00; soil, 60.00; dead organic matter, 15.00; streams, 20.00. Row 2. Month, 1; grasses, 17.60; soil, 62.80; dead organic matter, 14.00; streams, 20.60. Row 3. Month, 2; grasses, 15.59; soil, 65.22; dead organic matter, 12.96; streams, 21.23. Row 4. Month, 3; grasses, 13.90; soil, 67.29; dead organic matter, 11.93; streams, 21.88. Row 5. Month, 4; grasses, 12.49; soil, 69.03; dead organic matter, 10.93; streams, 22.55. Row 6. Month, 5; grasses, 11.31; soil, 70.46; dead organic matter, 9.99; streams, 23.24. Row 7. Month, 6; grasses, 10.32; soil, 71.61; dead organic matter, 9.13; streams, 23.95. Row 8. Month, 7; grasses, 9.48; soil, 72.52; dead organic matter, 8.33; streams, 24.66. Row 9. Month, 8; grasses, 8.79; soil, 73.21; dead organic matter, 7.61; streams, 25.39. Row 10. Month, 9; grasses, 8.20; soil, 73.71; dead organic matter, 6.97; streams, 26.12. Row 11. Month, 10; grasses, 7.71; soil, 74.04; dead organic matter, 6.40; streams, 26.86. Row 12. Month, 11; grasses, 7.29; soil, 74.22; dead organic matter, 5.89; streams, 27.60. Row 13. Month, 12; grasses, 6.94; soil, 74.28; dead organic matter, 5.44; streams, 28.34.

8.17 Exercises Answers to selected odd-numbered problems begin on page ANS-22.

    1. Use the data in the compartmental diagram in FIGURE 8.17.5 to determine the appropriate transfer matrix T and the initial state of the system X0.
    2. Find the state of the system after 1 day. After 2 days.
    3. Eventually, the system will reach an equilibrium state that satisfies . Find . [Hint: x1 + x2 = 150.]
      A compartmental diagram consists of two boxes. The box on the left is labeled 1 and has 90 written in it. The box on the right is labeled 2 and has 60 written in it. An arrow pointing right from box 1 to box 2 is labeled 0.2 per day. An arrow pointing left from box 2 to box 1 is labeled 0.4 per day.

      FIGURE 8.17.5 Compartments for Problem 1

    1. Use the data in the compartmental diagram in FIGURE 8.17.6 to determine the appropriate transfer matrix T and the initial state of the system X0.
    2. Find the state of the system after 1 day. After 2 days.
    3. Find the equilibrium state that satisfies . [Hint: What is the analogue of the hint in part (c) of Problem 1?]
      A compartmental diagram consists of three boxes forming a downward pointing triangle. Starting from the left and moving in a clockwise direction the boxes are labeled as followed: Box 1, 100; box 2, 200; box 3, 150. An arrow pointing right from box 1 to box 2 is labeled 0.3 per day. An arrow pointing downward to the left from box 2 to box 3 is labeled 0.2 per day. An arrow pointing upward to the left from box 3 to box 1 is labeled 0.5 per day.

      FIGURE 8.17.6 Compartments for Problem 2

    1. Use the data in the compartmental diagram in FIGURE 8.17.7 to determine the appropriate transfer matrix T and the initial state of the system X0.
    2. Find the state of the system after 1 day. After 2 days.
    3. Find the equilibrium state that satisfies .
      A compartmental diagram consists of three boxes forming a downward pointing triangle. Starting from the left and moving in a clockwise direction the boxes are labeled as followed: Box 1, 100; box 2, 0; box 3, 0. An arrow pointing right from box 1 to box 2 is labeled 0.3 per day. An arrow pointing to the left from box 2 to box 1 is labeled 0.5 per day. An arrow pointing downward to the left from box 2 to box 3 is labeled 0.4 per day. An arrow pointing downward to the right from box 1 to box 3 is labeled 0.5 per day.

      FIGURE 8.17.7 Compartments for Problem 3

  4. A field has been completely devastated by fire. Two types of vegetation, grasses, and small shrubs will first begin to grow, but the small shrubs can take over an area only if preceded by the grasses. In FIGURE 8.17.8, the transfer coefficient of 0.3 indicates that, by the end of the summer, 30% of the prior bare space in the field becomes occupied by grasses.
    1. Find the transfer matrix T.
    2. Suppose X = and that area is measured in acres. Use the recursion formula Xn + 1 = TXn, along with a calculator or a CAS, to determine the ground cover in each of the next 6 years.
      A compartmental diagram consists of three boxes forming an upward pointing triangle. Starting from the top and moving in a counter-clockwise direction the boxes are labeled as followed: Box 1, bare space; box 2, grasses; box 3, small shrubs. An arrow pointing downward to the left from box 1 to box 2 is labeled 0.3 per year. An arrow pointing upward to the right from box 2 to box 1 is labeled 0.05 per year (death). An arrow pointing to the right from box 2 to box 3 is labeled 0.2 per year. An arrow pointing upward to the left from box 3 to box 1 is labeled 0.15 per year (death).

      FIGURE 8.17.8 Compartments for Problem 4

Discussion Problem

  1. Characterize the vector in part (c) of Problems 1–3 in terms of one of the principal concepts in Section 8.8.

Computer Lab Assignment

  1. Radioisotopes (such as phosphorus-32 and carbon-14) have been used to study the transfer of nutrients in food chains. FIGURE 8.17.9 is a compartmental representation of a simple aquatic food chain. One hundred units (for example, microcuries) of tracer are dissolved in the water of an aquarium containing a species of phytoplankton and a species of zooplankton.
    1. Find the transfer matrix T and the initial state of the system X0.
    2. Instead of the recursion formula, use Xn = TnX0, n = 1, 2, . . ., 12, to predict the state of the system for the next 12 hours. Use a CAS and the command to compute powers of matrices (in Mathematica it is MatrixPower[T, n]) to find T2, T3, . . ., T12.
      A compartmental diagram consists of three boxes forming an upward pointing triangle. Starting from the top and moving in a counter-clockwise direction the boxes are labeled as followed: Box 1, Phytoplankton; box 2, Water; box 3, Zooplankton. An arrow pointing downward to the left from box 1 to box 2 is labeled 0.06 per hour (respiration). An arrow pointing upward to the right from box 2 to box 1 is labeled 0.02 per hour (uptake of dissolved tracer). An arrow pointing to the right from box 2 to box 3 is labeled 0.01 per hour. An arrow pointing to the left from box 3 to box 2 is labeled 0.05 per hour (excretion). An arrow pointing downward to the right from box 3 to box 1 is labeled 0.06 per hour (grazing).

      FIGURE 8.17.9 Compartments for Problem 6