Exponential Decay

    Figure 1 below is a system dynamics representation of a linear first order negative feedback loop system with an implicit goal of zero. The figure includes the model's equations, a causal loop diagram, and a simulation of the model's structure. The simulation is an example of exponential decay behavior that corresponds to the exact analytical solution to the system presented in Equation 2 of Goal Seeking Behavior.

    (Click on image to run simulation)

    Figure 1: First Order, Linear, Negative Feedback Loop System with an Implicit Goal of Zero

    Measures of Exponential Decay: Half Life

    One measure of how long it will take for a first order, linear, negative feedback loop system to reach its goal of zero is its half life. The half life of such a system is the time it takes to drain half of the contents from the system's stock. Analogous to the case of the doubling time for a first order positive feedback loop system above, the system's time constant must be determined before its half life can be found.

    As was previously shown, the analytical solutions to Equations 1 of Exponential Growth and Equation 2 of Goal Seeking Behavior differ only by sign. It should not be surprising therefore that the equation for the outflow in Figure 1:

    Flow = Stock / TimeCoef

    or:

    Flow = Stock * (1/TimeCoef)

    is very similar to the equation for the inflow in the Figure 20 of Exponential Growth:

    Flow = Stock * Coef

    Given that the time constant for the positive feedback loop system of Figure 20 of Exponential Growth is Tc = 1/Coef, and applying the same logic to the negative feedback loop system presented in Figure 1, yields a time constant of Tc = 1/(1/TimeCoef) = TimeCoef. This result can be used to answer a question similar to that asked about the positive feedback loop system: How much progress toward its goal will the negative feedback loop system have made after one time constant has elasped? The answer can be found by substituting the time constant into the Equation 2 of Goal Seeking Behavior:

    Stockt = Stock0 * e-(1 / TimeCoef) *Tc

    Stockt = Stock0 * e-(1/TimeCoef)*TimeCoef

    Stockt = Stock0 * e-1

    Stockt = Stock0 * (1/e)

    Stockt = Stock0 * .37

    This result indicates that the system loses 63% of the contents of its stock after one time constant has elapsed. Extending this logic reveals that the system loses 86% of the contents of its stock after two time constants have elasped (Stockt = Stock0 * .37 * .37 = Stock0 * .14) and 95% of the contents of its stock after three time constants have elapsed (Stockt = Stock0 * .37 * .37 * .37 = Stock0 * .05). Figure 2 repeats the simulation from Figure 1 and confirms this result graphically.

    As in the case of a first order positive feedback loop system, the negative loop system's half life is 70% of its time constant. That is: Th = .7 * Tc. Once again, the derivation of ths result is presented in the shaded box below for the more methematically inclined reader. A graphical presentation of the result is also shown in Figure 2.

    (Click on image to run simulation)

    Figure 2: Behavior of a First Order, Linear, Negative Feedback Loop System after Three Time Constants have Elapsed

    The analysis of the first order negative feedback loop system with an implicit goal of zero can be extended to the more general negative feedack loop model originally presented in Figure 2 of Goal Seeking Behavior, giving the following insight: A first order, negative, feedback loop system reaches approximately 95% of its goal after three time constants have elapsed.

    Derivation of the Half Life for a First Order, Linear, Negative Feedback Loop System

    The half life of a first order, linear, negative feeback loop system is calculated using Equation 2 of Goal Seeking Behavior. If the half life, Th, is defined to be the time it takes to cut the initial value of a stock in half, then:

    1/2 * Stock0 = Stock0 * e-1/Coef * Th

    Since the system's time constant, Tc, is 1/Coef, then the system's Coef = 1/Tc. Thus:

    1/2 * Stock0 = Stock0 * e-Tc * Th

    Dividing both sides by Stock0 yields:

    1/2 = e - Th/Tc

    Taking the natural logarithm of both sides yields:

    ln(1/2) = ln(e -Th/ Tc)

    Or:

    ln(1) -ln(2) = ln(e-Th/Tc)

    Or:

    0 - .69 = - Th / Tc

    Or:

    .7 = Th / Tc

    Or:

    Th = .7 * Tc

    Thus, the half life of a first order, linear, negative feeback loop system is seventy percent of its time constant.