Determining What Can Be Predicted: Identifiability

One of the best developed ways to predict how changing inputs to a complex system will change its probable outputs is to simulate the behavior of the system. Modern simulation modeling software environments (such as MATLAB/SIMULINK^®, or STELLA/ITHINK^® for continuous simulation, and SIMUL8^® for discrete-event simulation) make the mechanics of simulation model building and use relatively straightforward. Stochastic simulation risk models have been developed for business, engineering, biological, social, and economic systems. (Agent-based simulation models have also been developed for complex social and economic systems, but this chapter focuses on continuous simulation.)

Appendices

Appendix A: Proof of Theorem 1

The following succinct proof is due to Professor William Huber (see the acknowledgments at the start of the book).

For k > 1, the w’s satisfy recursive relationships

• (^*) w _k = a _k,k–1 w _k–1 and

• (^{* *}) 1/w _jk–1 + a _kk /w _jk = a _jj /w _jk for all j between 1 and k–1.

These are immediate from the definitions; (^{* *}) is just a rearrangement of the relationship (a _jj  – a _kk )w _jk–1 = w _jk . For k = 1, the theorem is trivial. For k > 1, we calculate

$$\begin{aligned} (\boldsymbol{A}{\bf z}(t))_k&=a_{k,k-1}z_{k-1}(t)+a_{kk}z_k(t)\\ & = a_{k,k-1}w_{k-1}[\exp(a_{11}t)/w_{1\,k-1}+\cdots+ \exp(a_{k-1,k-1}t)/w_{k-1,k-1}]\\&\hspace*{12pt}+ a_{kk}w_{k}[\exp(a_{11}t)/w_{1\,k}+\cdots+\exp(a_{kk}t)/w_{kk}].\end{aligned}$$

Apply (^*) to the first term and distribute a _kk over the second term to obtain

$$\begin{aligned} &w_{k}[\exp(a_{11}t)/w_{1\,k-1}+\cdots+\exp(a_{k-1,k-1}t)/w_{k-1,k-1}]+ w_{k}[a_{kk}\exp(a_{11}t)/w_{1\,k}\\&\hspace*{12pt}+\cdots+a_{kk}\exp(a_{kk}t)/w_{kk}]. \end{aligned}\vspace*{5pt}$$

Finally, upon factoring out w _k , collecting the coefficients for each exponential, and applying (^{* *}) to the first k–1 of them, the derivative of z _k (t) becomes recognizable:

$$\begin{aligned} &w_{k}[(1/w_{1\,k-1}+a_{kk}/w_{1\,k})\exp(a_{11}t)+\cdots+(1/w_{k-1,k-1}+a_{kk}/w_{kk})\exp(a_{k-1,k-1}t)\\ &\hspace*{12pt}+(a_{kk}/w_{1\,k})\exp(a_{kk}t)]=w_{k}[(a_{11}/w_{1\,k})\exp(a_{11}t)+\cdots +(a_{kk}/w_{kk})\exp(a_{kk}t)]\\ &\hspace*{106pt}=dz_k(t)/dt. \end{aligned}$$

This demonstrates that z(t) satisfies the system of equations. It remains to show that it also satisfies the initial condition. For k = 1, the value is z ₁(0) = 1^*[exp(a ₁₁ ^*0)] = 1, as desired. For larger values of k, we need to show that

$$\begin{aligned} 0&=z_k(0)=\exp(a_{11}^{\ \ \ast} 0)/w_{1\,k}+\cdots +\exp (a_{kk}^{\ \ \ast}0)/w_{kk}\\ &=1/w_{1\,k}+\cdots + 1/w_{kk}. \end{aligned}$$

The partial fraction expansion of 1/w _1k  + … + 1/w _kk is a sum whose terms are in the form u _ikj /(a _ii – a _jj ). By inspection, we obtain

$$u_{ikj}=1/\Pi(a_{ii}-a_{ll})-1/\Pi(a_{jj}-a_{ll}),$$

with the products extending over all l between 1 and k but skipping i and j. The products are subtracted, not added, because (a _ii  – a _jj ) appears in w _ik while (a _jj  – a _ii ) = –(a _ii  – a _jj ) appears in w _jk with the opposite sign. Evidently, the limiting value of u _ikj as a _ii and a _jj become equal is zero, because the two products approach a common finite value. Thus, (a _ii  – a _jj ) is a factor of u _ikj , implying 1/w _1k  + … + 1/w _kk has a “removable singularity” on the set a _ii  = a _jj . Since i and j were arbitrary, we conclude that z _k (0) has no singularities at all and therefore is really a polynomial. The proof is finished by observing that z _k (0) approaches zero whenever any a _ii becomes arbitrarily large, which for a polynomial can occur only when it is identically zero. QED.

Appendix B: Listing of ITHINK™ Model Equations for the Example in Figure 11.3

Compartment M

M(t) = M(t – dt) + (flow_FM) ^* dt

INIT M = 0

flow_FM = F^*(bFM + delta_FM)

Compartment F

F(t) = F(t – dt) + (flow_PF + proliferation_F – flow_FM) ^* dt

INIT F =

INFLOWS to Compartment F:

flow_PF = P^*switch?^*(bNP + deltaNP) + P^*(1 – switch?)^*(bPF + deltaPF)

proliferation_F = F^*(bF + delta_bF)

Compartment P

P(t) = P(t – dt) + (flow_NP – flow_PF) ^* dt

INIT P = 0

INFLOWS to Compartment P:

flow_NP = N^*(1 – switch?)^*(bNP + deltaNP) + N^*switch?^*(bPF + deltaPF)

OUTFLOWS:

flow_PF = P^*switch?^*(bNP + deltaNP) + P^*(1 – switch?)^*(bPF + deltaPF)

Compartment N

INFLOWS to Compartment N:

N(t) = N(t – dt) + (growth – flow_NP) ^* dt

INIT N =

growth = if (TIME < 20) then 100/20 else 0

OUTFLOWS:

flow_NP = N^*(1 – switch?)^*(bNP + deltaNP) + N^*switch?^*(bPF + deltaPF)

FORMULAS AND PARAMETERS

bF = 0.08 {0.08}

bFM = 0.00008

bNP = 0.00006

bPF = 0.05

deltaNP = bNP^*RNP^*exposed?

deltaPF = bPF^*RPF^*exposed?

delta_bF = bF^*RF^*exposed?

delta_FM = bFM^*RFM^*exposed?

end_time = 60

exposed? = if ((TIME >= start_time) and (TIME <= end_time)) then exposure_factor else 0

exposure_factor = 1 {fraction of saturation exposure intternal dose}

M_x_100 = M^*100

RF = 0.9

RFM = 2.19

RNP = 2 {2}

RPF = 0.2 {0.2}

start_time = 20

switch? = 0