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8
SYSTEM DYNAMICS
Control model of autopilot aircraft discussed in chapter one shows how various parts and their activities
are controlled by gyro, and redirects the aircraft in the desired direction. Can we apply control theory
to every day life? It was seen in the example of autopilot that main concern of the control mechanism
is to control the stability and oscillations of the system.
There are many examples in nature where we can also apply control theory. Like engineering
problems, instability and oscillations also occur in nature. Let us take the example of population
explosion. It is observed that if we do not control the population of human being or even any other
specie, it will grow exponentially. This is one of the fields in nature, where control theory is applied
to study the growth of population. In the field of medicines, multiplication of cancerous cells in
human body is another example. Similarly in market there are oscillations in prices of products,
which effect their supply and production. In Physics, decay of radioactive material can also be
modeled with the help of control theory. Although the precision applicable to engineering problems
can not be attained in such problems yet control theory can suggest changes that will improve the
performance of such systems.
In scientific literature studies connected with industrial problems are called Industrial Dynamics
(Geoffrey Gordon), where as study of urban problems is called Urban Dynamics. Similarly control
of environmental problems is called World Dynamics. In all these systems there is no difference in
the techniques to be used to study the system, therefore it is appropriate to call this field as System
Dynamics.
The principal concern of a system Dynamics study is to understand the forces operating on
a system, in order to determine their influence on the stability or growth of the system (Geoffrey
Gordon). Output of such study may suggests some reorganization, or changes in the policy, that
can solve an existing problem, or guide developments away from potentially dangerous directions.
Unlike engineering problems, in this case system dynamics may not produce certain parameters
to improve the performance of system. But this study definitely helps the system analyst to predict
the scenario so that corrective steps can be taken in time.
System Modeling and Simulation
198
8.1 EXPONENTIAL GROWTH MODELS
There are various activities in nature, in which rate of change of an entity is proportional to itself.
Such entities grow exponentially and study of such models are known as exponential growth models.
To understand this concept, let us consider the birth rate of monkeys. Unlike other animals, birth
rate of monkeys grows very fast. If not controlled, their population grows exponentially. Let us
try to model this simple natural problem mathematically. If in a region, say x is the number of
monkeys at time t, then their rate of growth at time t is proportional to their number x at that time.
Let proportionality constant be k. This type of functions can be expressed in the form of differential
equations as,
dx
= kx
dt
...(8.1)
with the condition x = x0 at t = 0.
This is first order differential equation and its solution is
kt
x = x0 e
...(8.1a)
Fig. 8.1: Exponential growth curves.
Figure 8.1 shows variation of x vs. t for different values of k. Figure shows, higher is the
value of k, more steep is the rise in x. These types of curves are called exponential growth curves
and equation (8.1a) is a mathematical expression for exponential curve. We now apply this law to
the population of monkeys.
Let us take example of population of monkeys in a city. Let their production period is six months.
Assuming that in this city, there are 500 monkeys i.e., 250 couples at time t = 0. If each couple
produces four offspring’s in time = 1 (six months), then proportionality constant k = 2. In first six
months population becomes 1850 (7.4 times) by this relation and in next six months it becomes 13684,
which is equal to 54.7 times of 250.
System Dynamics
199
We can also express function (eq. 8.1) on a semi lag graph. Equation (8.1) can be written as
ln x = ln x0 + 2t
...(8.1b)
which is equation of straight-line in ‘ln x’ (natural log of x) and t. This means that, if we plot this
equation by taking x on log axis and t on simple axis, we will get output as a straight-line, with slope
equal to 2. What are the dimensions of k? From equation (8.1a) it can be seen that dimensions of
k are nothing but 1/time. Sometimes coefficient k is written as 1/T, that is total period under study.
Thus equation (8.1a) can be written as
x = x0et/T
The constant T is said to be time constant, since it provides the measure of growth of x.
8.2 EXPONENTIAL DECAY MODELS
Radioactive materials continuously decay, because they radiate energy and thus lose mass, and ultimately
some part of the matter is changed to some other material. This is proved due to the fact that one can
not get pure radium from any ore. It is always be mixed with some impurities such as carbon. It is
thought that radium was 100% pure when earth was formed. By finding the quantum of carbon in the
radium ore, scientists have determined the life of Earth. Let us see, how this situation is modeled
Mathematically. If rate of change of variable x is proportional to its negative value, then such growth
is called negative growth and such models are called negative growth models, or exponential decay
models. The equation for such a model is
dx
= –kx
dt
...(8.2)
with the condition x = x0 at t = 0.
Solution of this equation is
x = x0e–kt
Fig. 8.2: Exponential decay curve.
...(8.2a)
System Modeling and Simulation
200
Equation 8.2a is plotted in Figure 8.2. From the figure 8.2 we can see that value of x at any
time t, decreases from its initial value and comes down to zero when t is infinity. Decay rate increases
with k.
8.3 MODIFIED EXPONENTIAL GROWTH MODEL
A production unit, which is planning to launch a new product, first problem faced by it is, how much
quantity of a product can be sold in a given period. A market model should be able to predict, rate
of selling of a product, which obviously cannot be proportional to itself. There are several other
parameters to be considered while modeling such a situation. In practice, there is a limit to which
one can sell the product. It depends on how many other brands are available in market, and what
is the probable number of customers. The exponential growth model can not give correct results as
it shows unlimited growth. Thus we have to modify this model.
Exponential growth model can be modified if we assume that rate of growth is proportional to
number of people who have yet not purchased the product. Suppose the market is limited to some
maximum value X, where X is the number of expected buyers. Let x be the number of people who
have already bought this product or some other brand of same product. The numbers of people who
have yet to buy are (X – x). Thus equation (8.1) can be modified as
dx
= k ( X − x)
dt
...(8.3)
with the condition x = 0 at t = 0. Solution of equation (8.3) is as
x = X(1 – e–kt)
Fig. 8.3: Modified exponential model.
...(8.3a)
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201
Figure 8.3 gives the plot of equation (8.3a) for various values of k. This type of curve is sometimes
referred as modified exponential curve. As it can be seen, maximum slope occurs at the origin and
the slope steadily decreases as time increases. As a result of it, the curve approaches the limit more
slowly, and never actually gets the limit. In marketing terms, the sale rate drops as the market
penetration increases. The constant k plays the same role as the growth rate constant as in Exponential
growth model. As k increases, the sale grows more rapidly. As with the growth model, k is sometimes
expressed as equal to 1/T, in which case it can be interpreted as a time constant.
Example 8.1: A builder observes that the rate at which he can sell the houses, depends directly
upon the number of families who do not have a house. As the number of families without house
diminish, the rate at which he sells the houses drops. How many houses in a year can he sell?
Solution: Let H be the potential number of households and y be the number of families with
houses. If
dy
dy
is the rate at which he can sell the houses, then
is proportional to (H – y), i.e.,
dt
dt
dy
= k ( H − y ), y = 0 at t = 0
dt
This is nothing but Modified Growth case and solution is
y = H(1 – e–kt)
where H is the potential market.
Example 8.2: Radioactive disintegration
The rate of disintegration of a radioactive element is independent of the temperature, pressure,
or its state of chemical combination. Each element thus disintegrates at a characteristic rate
independent of all external factors. In a radioactive transformation an atom breaks down to give
one or more new atoms.
If to start with t = 0, the number of atoms of A present is a (say). After time t, x atoms will
have decomposed leaving behind (a – x) atoms. If then in a small time interval dt, dx is the number
of atoms which change, the rate of disintegration
dx
can be expressed as
dt
dx
= k (a − x )
dt
...(8.4)
This is called law of mass action and k is called velocity constant or disintegration constant or
transformation constant. Equation (8.4) on integration, with initial condition, x = 0 at t = 0 gives
ln
or
a−x
= – kt
a
a – x = ae–kt
If we write y = (a – x) and a = y0, we get
y = y0e–kt
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202
which is same as equation (8.2a). If T is the time when half of the element has decayed i.e., x = a /2 ,
we get T, as
2.303
log 2
k
T = 0.693 / k
T=
This means the disintegration of an element to half of its period T depends only on k and is
independent of amount present at time t = 0.
8.4 LOGISTIC MODELS
Let us again come back to sale of a product in the market. The model of section 8.2 is some what
unrealistic because, in modified exponential model, the slope of the product in the beginning is shown
to be maximum. In fact in the beginning sale is always less and when product becomes popular in
the market, sale increases, that is, slope increases as occurs in the exponential growth model. When
for the sale of product, saturation comes, slope again decreases, making the curve of market growth
like modified exponential curve. The result is an S-shaped curve as shown in Figure 8.4.
Fig. 8.4: S-shaped growth curve.
Such curves are called logistic curves.
The logistic function is, in effect, a combination of the exponential and modified exponential
functions, that describes this process mathematically. The differential equation defining the logistic
function is
dx
= kx(X – x)
...(8.5)
dt
In this relation, initially x is very small and can be neglected as compared to X. Thus equation
(8.5) becomes
dx
= kxX
dt
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203
which is the equation for exponential growth curve with proportionality constant equal to kX. Much
later, when the market is almost saturated, the value of x becomes comparable to X, so that it changes
very little with time. The equation for the logistic curve then takes the approximate form
dx
= kX ( X – x )
dt
which is the differential equation for the modified exponential function with a constant kX.
The true differential equation is nonlinear and can be integrated numerically with the boundary
conditions x = 0, when t = 0. Exact analytic solution is tedious and has been given by (Croxton et al.,
1967). Interested students may see the reference. Apart from market trends, many other systems follow
logistic curve, for example population growth can also follow logistic curve (Forrester JW, 1969).
During initial stages, there may be ample resources for the growth of population but ultimately when
resources reduce to scarcity, rate of population comes down.
The model is also applicable to spread of diseases. Initially it spreads rapidly as many acceptors
are available but slowly people uninfected drop and thus growth rate of disease also decreases.
8.5 MULTI-SEGMENT MODELS
In market model, we can introduce more than one products, so that sale of one depends on the sale
of other and thus both are mutually related. For example in example 8.1, we have considered the
case of a builder who wants to sell houses. It was observed that rate of sale of houses depended
on the population which did not have houses. Now suppose another firm wants to launch its product,
say air conditioners in the same market. He can only sell air conditioners to people who already have
purchased houses from the builders. Otherwise they will not require air conditioners. Let us make
a model of this situation. This model can be constructed as follows:
Let at time t, H be the number of possible house holds, y, the number of houses sold and, x
the number of air conditioners installed. Then
dy
= k1 ( H – y )
dt
dx
= k2 ( y – x)
dt
...(8.6)
Equation (8.6) means the rate of sale of houses at time t, is proportional to (H – y), which is the
number of people who do not have houses. Similarly second equation of (8.6) says, the rate of sale
of air conditioners is proportional to (y – x) i.e., the number of houses which do not have air conditioners
installed so far. Both these equations are modified exponential growth models. In equation (8.6),
we have not taken into account the air conditioners which become unserviceable and require replacement. This factor can be added by modifying second equation of (8.6) as,
dx
= k 2 ( y – x ) – k3 x
dt
...(8.6a)
Equations (8.6) and (8.6a) can easily be computed numerically, when k1, k2 and k3 are known.
System Modeling and Simulation
204
8.6 MODELING OF A CHEMICAL REACTION
In a chemical reaction, reaction velocity is very much similar to that of velocity of motion in kinetics
and the term denotes the quantity of a given substance which undergoes change in unit time. In this
process, the rate of reaction is never uniform and falls of with time the reactants are used up. A case
of unimolecular reaction has already been discussed in example 8.2 of this chapter. It is known that
dx
in a chemical reaction, velocity of reaction
, where x is available reactant is proportional to
dt
(a – x) where x = a at the beginning of reaction i.e. at t = 0. This is called law of mass action in
chemistry. That is if x is a molecular concentration of a reactant at any time t and k is the proportionality
constant, then
dx
= (a – x)
dt
In a chemical reaction, the number of reacting molecules whose concentration alters as a result
of chemical changed as order of reaction.
Consider a general reaction,
nA + mB → pC + qD
where concentration of both A and B alters during the reaction. At any time t, velocity of this chemical
reaction is given by,
dx
= kAn Bm
dt
where (m + n) represents the order of the reaction. One example of reaction of first order has already
been considered in example 8.2. Let us consider another example below.
Example 8.3: Following data as obtained in a determination of the rate of decomposition of
hydrogen peroxide, when equal volumes of the decomposing mixture were titrated against standard
KMnO4 solution at regular intervals:
Time (in minutes)
Volume of KMnO4
0
25
10
16
20
10.5
30
7.08
used (in cc’s)
Show that it is a unimolecular reaction.
Solution: For a unimolecular reaction,
k =
a
2.303
log10
t
a–x
Here the volume of KMnO4 solution used at any time corresponds to undecomposed solution
i.e., (a – x) at that time. The initial reading corresponds to a. Inserting experiment values in above
equation one gets,
1
25
log 10
= 0.0194
k′ =
10
16
k′ =
1
25
log 10
20
10.5
= 0.0188
1
25
log10
30
7.08
= 0.0183
k′ = k
System Dynamics
205
The constant value of k′ shows that the decomposition of KMnO4 is a unimolecular reaction.
8.6.1 Second Order Reaction
In a second order reaction, the minimum number of molecules required for the reaction to proceed
is two. Let a be the concentration of each of the reactants to start with and (a – x) their concentration
after any time t. Then we have in this case,
dx
= k (a − x)(a − x)
dt
= k (a − x) 2
(Law of mass action)
Initial condition for integrating this equation are, at t = 0, x = 0. Thus equation becomes on integration,
 a 2 kt 
x= 

1 + akt 
If we start with different amounts a and b of the reactants, then according to law of mass action,
dx
= k (a − x)(b − x)
dt
with initial conditions, at t = 0, x = 0, one gets,
x

1 − 
a
= exp( kt ( a − b))
 x
1 − 
b
8.7 REPRESENTATION
OF
TIME DELAY
In all the models so far discussed, we considered different proportionality constants. By dimensional
analysis, we observe that these constants have dimensions of 1/time. Thus if inplace of these constants,
we take average time to complete a task, it will be more meaningful. Let us consider market model of
section 8.4. If average time required to complete the housing project of an area is T1, say 10 years and
time required to install air conditioners in available houses is T2, then equation (8.6) can be written as
1
dy
( H − y)
=
T1
dt
1
dx
( y − x)
=
T2
dt
...(8.7)
This is simplistic form of model. Determination of T1 and T2 is not that straight forward. For
example T1 depends on number of factors viz., economic conditions of the people, cost of land, speed
of housing loans available etc. Similarly, T2 depends on again, how many people owning house, can
afford air conditioners, and also it depends on weather conditions and many other factors. In order
System Modeling and Simulation
206
to model such situations, one has to take all the factors into account. In order to model such situation
correctly, feedback of information regarding market trends, is very much essential. Without this
information, air conditioner vendor, may stock air conditioners based on houses available. If he is
not able to install the air conditioners because of other conditions, he will have to stock them for
longer time and bear losses. This becomes another problem called inventory control. Thus all the
parameters have to be taken into account while constructing such models.
8.8 A BIOLOGICAL MODEL
Here we consider a biological model, an application of System Dynamics. There are many examples
in nature of parasites, that must reproduce by infesting some host animal and, in doing so, kill the
animal. As a result, the population of both, the host and parasite fluctuate. As the parasite population
grows, host population declines. Ultimately, decline in host population results in the decline in parasite
population, owing to which host population starts increasing. This process can continue to cause
oscillations indefinitely.
To construct this model of balance between host and parasite, let x be the number host and
y be the number of parasites at a given time t. Let birth rate of host over the death rate due to
natural causes is a, where a is positive. In the absence of parasites, the population of hosts should
grow as,
dx
= αx
dt
The death rate from infection by the parasites depends upon the number of encounters between
the parasites and hosts, which is assumed to be proportional to the products of the numbers of parasites
and hosts at that time. Thus the rate of growth of hosts modifies to,
dx
= αx – kxy
dt
...(8.8)
Here a simplifying assumption, that each death of a host due to parasite results in the birth of a
parasite. This is the only mean, by which parasite population can grow. It is also assumed that death
rate of parasites is δ due to natural death. Thus the equation controlling the parasite population is,
dy
= kxy – δy
dt
...(8.9)
We can solve equations (8.8) and (8.9) numerically by taking values of parameters α, k, and
δ as, α = 0.005, k = 6 × 10–6, and δ = 0.05.
Table 8.1
Day no.
0
100
200
300
Host population
Parasite population
10000
7551.74
7081.92
8794.23
1000
1303.45
570.282
446.034
Contd...