Simple growth models

In the preceding chapters we have defined dynamic models as iterative models, where

each next step is based on the preceding step. In this chapter we will introduce and

discuss growth models, a type of iterative, dynamic model where the next level of a

variable is a function of its preceding value and a growth rate. We will also see that

growth processes depend on specific and limited resources. The models we will

introduce are classical, general growth models that have been used for explaining

phenomena ranging from the growth of insect populations to the replacement of steam

by diesel engines. We will explain how growth models can be applied to

developmental and learning processes. We will help the reader build his or her own

growth models of developmental phenomena.

1

Models discussed in this chapterProportional growthRestricted growthGompertz modelLogistic growthImportant conceptsGrowth rateResourcesCarrying capacityInitial stateFinal statep-value

The first classical model: proportional growth

Proportional growth is as based on an iterative (or recursive) model, in which the

change is a proportion of the current state. A simple example is the growth of

someone’s capital on a bank account. The bank gives an interest, e.g. 5%, and that is

the proportion with which the person’s capital increases. An initial capital of $ 100,-

increases with $ 5 the first year, which gives a new capital of $ 105,-, which increases

with $ 105*0.05 = $ 5.25 the second year, and so forth. Note the difference with the

random walk model that we discussed in the preceding chapter. The random walk

model involved an iterative change, just like the current growth model, but the change

consisted of a simple (random) addition to the preceding state of the model. In the

random walk, the amount of change did not depend on the preceding level (for

instance, the number added to the level does not get bigger as the level gets bigger). In

the proportional model, the amount of change depends on the level already attained:

5% of a capital of $ 100,- gives an interest of $ 5, but the same 5% gives an interest of

$ 50 on a capital of $ 1000,-

In general, we are not used to think of psychological variables or developmental

levels in terms of a specific kind of increase, e.g. either simply additional or

proportional. For instance, if we think about a child’s increasing social skill, we

should ask ourselves if the increase is likely to be additional or proportional. If we

assume that it is easier for a child to learn new things about the social world if the

child already has a considerable knowledge of that social world, we are implicitly

assuming a proportional form of growth for social skills. If we assume that the child

increases his or her social skill with some (on average) constant factor, we are

assuming an additive model. It is remarkable that, in general, developmental

psychologists have so little knowledge of the nature of change in children, in spite of

the fact that such change is the core subject of their discipline. The reason for this

ignorance is probably that most instances of development are investigated by

comparing groups of children of different ages, or by associating the variable under

scrutiny with some other variable (for instance family characteristics such as child

rearing style). By doing so, we miss a very important aspect of change, namely the

way change depends on the level already attained (which amounts to the iterative or

2

recursive nature of change and development that we discussed in the introductory

chapter).

The equation for Proportional growth

The equation for proportional growth is very simple

L/t = r.L equation 2.1

As the reader will recall from the preceding chapter, this equation can be read as

follows. The increase in a level, L, over some amount of time, t, equals the level

already attained, L, times a ratio or proportion, r. In order to calculate the level at

some later time, for instance at time t + t, we simply add the proportional increase to

the level already attained

Lt+t = Lt + r.Lt equation 2.2

In the following exercise, we will model a simple proportional growth process and

study its properties. We shall assume that our model describes the growth of a child’s

lexicon, i.e. the increase in the number of words a child uses and/or understands.

The growth of the lexicon is a well-studied field in developmental psycholinguistics.

Although a discussion of lexical growth far extends the scope of the present

discussion, we will review some general points. To begin with, the (relatively well-

educated) adult’s lexicon is conservatively estimated as comprising about 50,000

words (see for instance Aitchsion, 1994). Let us assume that an 18-year-old person

has 50000 words. He or she learned al these words in 18 times 365 days, which is

6570 days. 50,000 divided by 6570 is approximately 7.6. Hence, our 18-year-old has

learned between 7 and 8 words a day. It is rather unlikely, however, that lexical

growth can be modeled by a simple process of constant addition. We know, for

instance, that many young children show a significant spurt in their lexical growth,

around the age of 15 to 16 months (see for instance Gillis, 1984, Dromi, 1987). A

comparable increase in the rate of word learning is demonstrated by measurements

with the McArthur Communicative Development Inventories (Fenson et al., 1993;

Fenson et al., 2000). In short, lexical growth is far from a simple linear increase. Let

us take a closer look at the data from Esther Dromi’s daughter Keren. Open the file

Keren’s lexical data.xls (see also figure Keren’s lexical growth). The data suggest

that, the more words Keren knows, the more new ones she learns. It pretty much looks

like the growth of a capital on a bank account.

3

Insert figure Keren’s lexical growth about here.

Building a model of proportional growth

2.1 Open a new Excel file and save it under the name Simple growth models.xls.

Rename the first worksheet: replace the name Sheet1 by Proportional growth

see Exceltip rename Worksheet

2.2 Introduce the variables Ini_prop and Rate_prop by means of the Insert Name

procedure. Put numerical values in the cells that contain the Ini_prop and rate_prop

values (e.g. 1 and 0.1 respectively)

see Exceltip Introduce Named Variables

2.3. The model will be built in the following steps.

2.3.1 In cell D1, write MODEL

2.3.2 In cell D2, write =Ini_prop (see Exceltip Write Equation by selection)

2.3.3. In cell E1, INCREASE

2.3.4. In cell E2 write the equation that specifies the increase part,

=D2*Rate_prop.

2.3.5. In cell D3 you will add the preceding state of the model (which is in cell

D2) and the increase, which is in cell E2. Thus, in cell D3 write =D2+E2.

Copy D3 to D4:D101. First copy E2 to E3 and then copy E3 to E4:E101 (see

chapter Random Walk, Step 4; see Exceltip Quick Copy Procedure)

2.3.6. By pasting the model equation to 100 consecutive cells, you have

defined a model that contains 100 steps. The meaning of the step can be

defined as you wish. For instance, if you interpret each step as a day, your

model specifies lexical growth over a period of 100 days.

2.4 Make a line graph of your model, which is comprised in the range D1:D100 (the

procedure for making a line graph is described in the Exceltip Make Graphs). If the

growth rate is high, your model will produce VERY high levels towards the end, i.e.

very high numbers of words in the (imaginary) lexicon. Excel will specify those

4

numbers in scientific notation, for instance 3E+59. This represents a number

consisting of a 3 followed by 59 zeroes (which is probably more than the number of

atoms in your body…) Experiment with different values for Ini_prop and Rate_prop,

but change only one parameter at a time. Press F9 to recalculate your model with the

new parameter values.

Assignment: Describe the form of the growth curve. Take a look at the final state of

the curve (last point of the model range). Is this a realistic final state, taking into

account that the model is supposed to describe lexical growth? What happens if the

growth goes on for a considerably longer time?

2.5. The preceding model was a purely deterministic model. In reality, however, the

parameters represent variables, mechanisms and factors that fluctuate over time in a

random way. For instance, the growth rate of the lexicon will vary from day to day,

depending on factors such as motivation, the time the parents spend with the child,

and so forth. It is even possible that at some times the growth rate is negative, i.e. that

the child forgets more words than it learns. What will be the effect of such random

fluctuation of the parameters? We will assume that the parameters – in fact, there is

only one parameter that really matters, namely Rate_prop – fluctuate around an

average value, with a characteristic standard deviation, and that the fluctuation is

symmetric around the average. To put it differently, we will assume that the

fluctuation has a normal distribution, with a specified mean and standard deviation.

We will assume that the growth rate, Rate_prop, is the mean of the fluctuating set of

rates. Each step (day, week, …) we randomly pick a growth rate from the

probabilistic distribution of growth rates.

2.5.1 Introduce a third variable, namely Rate_prop_sd, which will specify the

standard deviation of the fluctuating growth rate. In cell A3, write

Rate_prop_sd and insert a name according to the procedure specified earlier,

see Exceltip Introduce Named Variables). Write a numerical value for this

standard deviation in cell B3 (for instance 1/3d of the value of the growth rate

itself) (see Exceltip Automatically Expand Columns)

2.5.2. Use column F to assign an arbitrary value to the growth rate for each

time step. In cell F1, write VARIABLE GROWTH RATE. In cell F2 we will

enter the equation that picks a variable growth rate from a distribution with the

5

predefined mean and standard deviation. We will use a function from

Poptools. In cell F2, write =dNormaldev(Rate_prop, rate_prop_sd) (the

variable names can be typed directly, or entered by clicking on the cells that

contain their values, B2 and B3). If you wish to use an equivalent Excel

formula, write =Norminv(Rand(),Rate_prop, Rate_prop_sd). Copy to

F3:F101. Note that your model still refers to the constant Rate_prop: you have

to change the equations so that they refer to the variable growth rate, which

appears in the F-column. To do that, select cell E2, then put the cursor in the

formula bar (under the menu). The formula bar works like a small word

processor: delete the word Rate_prop, type F2 instead (or select cell F2), then

Enter. Copy the altered formula, then paste to E3:E100. Each time you press

the F9 button, Excel will recalculate your model with values for rate_prop that

are randomly changed at each step of your model (that is, within the

specifications of the mean and the standard deviation). You will see that the

curve changes each time you press F9. Use a relatively small growth ratio (for

instance 0.05) and a standard deviation that is about twice to three times as

small as the ratio; compare with standard deviations that are considerably

bigger).

2.5.3. We shall now investigate the effect of the randomly varying growth

ratio on the end state of our model. The end state can be found in the last step

of the model, which should be cell D101. Both Poptools and Paul’s Functions

contain a menu option that allows you to automatically recalculate a pre-

specified variable (for instance the last cell of your model) as many times as

you wish, that keeps the values of each run and gives you a summary statistics

of all the values obtained. We shall investigate the values of the randomly

varying growth rates and of the resulting end states of the model. In cell G1,

write =D101 (cell G1 now refers to the end state of the model). In cell G2,

write =Average(F2:F101). This equation calculates the average growth rate

over the 100 steps of the model. Press F9 a few times and watch the numbrs

change. Activate the menu option Functies/Monte Carlo Simulatie (see chapter

1, Step 10). The first field of the Monte Carlo window should contain a

reference to the cells that you want to recalculate with each Monte Carlo run.

These are the cells G1:G2. The second field (which eventually compares the

6

Monte Carlo run with empirically observed values) does not need to be

specified. The third field contains a reference to an output cell. Click, for

instance, on cell H1. Check the option “Keep output on a separate

Worksheet”. You will be asked to specify a name for a new worksheet that

will contain the 1000 end states and 1000 average growth rates that your

Monte Carlo run will produce. Type OP1.

Average 5020

Median 4593

Minimum 957

Maximum 17644

Perc 0.025 1813

Perc 0.975 10797

StDev 2335.761

Skewness 1.277

p-value 1.000

# of

simulations 1000

2.5.4. After finishing the Monte Carlo run, first take a look at the summary

statistics of the run (which will begin in cell H1). In order to make sense of the

numbers, recall that the model is supposed to mimic the growth of the child’s

lexicon. The table shows a list of possible results fro the first variable, namely

the end state of the model. The average number of words learned at the end is

5020. Look at the values specified by Perc 0.025 and Perc. 0.975. Those

values form the boundaries of the 95% interval (0.975 minus 0.0025 equals

0.95). That is, 95% of the values obtained in the simulations lie between 1813

words and 10,797 words. Look at the skewness of the distribution, which is

1.277. This means that the set of outcomes is strongly skewed to the right:

there’s a long tail on increasing numbers to the right of the distribution’s

center. In order to obtain a graphic representation of the distribution, select

worksheet OP1, then activate Functies/Maak frequentietabel and type 15 in the

text box that asks you for the number of categories (see Chapter 1, Step 5).

Describe the distribution of the end states. Compare this with the distribution

of the growth rates. Calculate the correlation between the end state and the

7

growth rate. What do you expect to find? Enter the following equation in an

empty cell =correl(a2:a1001,b2:b2002), which will calculate the correlation

between all growth rates and the corresponding end states. Does the outcome

confirm your expectations?

2.6. Let us conclude this section on proportional growth by trying to build a model of

lexical growth between the ages of 1 year and 18 years.

Assignment: Assume that the person learns his first word at the age of 1 year, and

knows 50,000 words at the age of 18 years. Model lexical growth in steps of one

month (every cell of your model will represent a month in the life of our subject).

What is the monthly growth rate, provided the subject must know about 50,000 words

at the age of 18? Assume that the subject’s language contains 200,000 words in total

(an imaginary number). With what growth rate can the subject have learned all the

words in his language by the age of 18? And how many words (if they were available)

would he have learned with a growth rate that were 10% bigger?

Solution of Assignment 2.6

A short historical note

The model we have just worked with has already a long history. It

was introduced in 1798 by Thomas Robert Malthus (1766-1834), an

English economist. According to Malthus, the model of exponential

growth described the growth of the human population. Malthus was

well aware of the fact that all growth must be supported by

resources, for instance, food and housing in the case of humans.

Malthus predicted that such commodities would grow in an additive

way. Such growth cannot keep up with the exponential growth of

the human population. Malthus foresaw an overpopulated earth,

with people fighting over scarce food supplies. He pleaded for an

active policy of birth control in order to avoid such disaster. The

growth rate r in the proportional growth model is still called the

Malthusean term

8

A second classical growth model: restricted growth

A striking property of the preceding model was the absence of any limit on the growth

of the variable at issue. Relatively small increases in the growth rate could cause the

process to reach unrealistically high levels.

In the real world, growth depends on resources. However abundant they may be, they

are always limited. For instance, a language contains many words, but a person cannot

learn more words than there are words in his or her language. In practice, however,

we learn considerably less words than the number of words that our mother tongue

possesses. Many words will occur so rarely that we will probably never be confronted

with them. Other words belong to highly specialized regions of activity or profession,

which we will most likely never participate in. Some people are simply more verbally

talented than others (whatever that talent may mean in practice) and will therefore

learn words more easily. Others take a high interest in language, and actively seek for

new words (for instance, people who like to do crossword puzzles). All these

properties, some of which are in the environment and some of which are in the person

himself, form the resources for word learning and will determine how many words,

approximately, a person will actually learn during the person’s lifetime.

In the first model, that of proportional growth, the notion of resource was inexistent.

The second model, however, considers the limited resource as the only factor that

determines the growth process (in addition to the growth rate, that is, but the growth

rate as such is not considered a resource factor).

A good example of the growth process that we will discuss in this section is the

learning by heart of a limited list of facts. Let us take a rather old-fashioned example,

namely the learning of 100 historical dates (or think of a taxi driver who has to learn

the names and addresses of all the hotels in town). The only things that matters to the

model is how much facts (historical dates) you have to learn by heart and how fast

you do this (the growth rate). Assume that you spend one hour a day learning the

historical dates. According to the model of restricted growth, each study hour we

9

learn a fraction or proportion of the list of historical dates, e.g. 5%. It is obvious that

this 5% applies to the number of dates you don’t know, i.e. that you still have to learn.

The first study hour, you don’t know any of the dates. You learn 5%, which leaves

you 95 dates that are still to learn. The second hour you learn 5% of the 95 remaining

dates (which is 4.75), and so forth. You may wonder what it might mean to learn

“0.75 date”. This decimal number has a natural interpretation, though: see it as the

probability that you will be able to retrieve that date if asked for it (which, in this case

would be a 75% probability).

A short historical note: Ebbinghaus

Note that this type of learning was extensively studied by one of the

founding fathers of experimental psychology, Herman Ebbinghaus

(1850-1909). In 1885, Ebbinghaus published a book - Über das

Gedächtnis – in which he reported the results of learning

experiments, in which he was himself the sole subject. For instance,

Ebbinghaus took a list of nonsense syllables, read them aloud and

then tried to recite them from memory. Every reading-aloud session

was considered a learning trial, and Ebbinghaus carefully noted how

much of the syllables he could correctly recall after an increasing

number of learning sessions.

The equation for restricted growth

The equation for restricted growth is as follows. Let G be the number of facts you

have to learn at the start, e.g. the 100 historical dates. Let L be the number of words

you already know. L/t is the increase in your knowledge of dates per unit time,

which we have defined as one hour of study. The equation for the growth of

knowledge is

L/t = r.(G – L) equation 2.3

Thus, if you know L historical dates at time t (let us say, after study hour 10), you will

know the following number of dates after time t + 1 (after study hour 11)

Lt + r.(G – Lt) equation 2.4

10

The symbol r refers, as you will recall, to the rate of growth. It will also be clear from

the equation that the only thing that matters for growth, in this model that is, is what

you still don’t know or how much you still have to learn.

Building a model of restricted growth

2.7. Open a new worksheet or rename an existing, empty worksheet. Name or rename

the worksheet as Restricted_growth (see Exceltip Move Worksheets)

Introduce the variables rate_restricted, ini_restricted and limit_restricted (see step

2.2; exceltip Introduce Named Variables). Put the names inc ells A1 to A3, which

implies that the corresponding variables will appear in cells B1:B3. Assume that, at

the beginning of the study process, the student doesn’t know any of the historical

dates he has to lean be heart. Hence, ini_restricted = 0. Remember that the list

contains 100 of such dates, thus, the limit_restricted = 100.

2.8. Use column D for the model, and E and F for the components of the model. In

column E we will put the amount of dates the student still has to learn after each study

hour. We will call this the Study Volume (e.g. if the student already knows 10 of the

100 dates, the study volume is 90 dates). In column F we will write how many of such

dates the student will have learned after a study hour. Thus, column D, the model, will

add the number of dates learned (e.g. at study hour t) to the number of dates the

student already knows at the beginning of that study hour.

2.8.1. In D1 write MODEL

2.8.2. In D2 =ini_restricted

2.8.3. In E1 STUDY VOLUME

2.8.4. In E2 = limit_restricted – D2. Note that since ini_restricted is supposed

to be 0, this equation could just as well be written as =limit_restricted.

However, the current equation allows you to experiment with a model in

which ini_restricted is bigger than 0.

2.8.5. In F1 INCREASE

2.8.6. In F2 = E2*rate_restricted

2.8.7. In D3 = D2+F2

2.8.8. In E3 =E2-F2

11

2.8.9. In F3 = E3*rate_restricted

2.8.10. Copy range D3:F3 to D4:D21. Select cells D3:F3, press the copy icon

or Ctrl-C, then Ctrl_G (GoTo instruction), fill in D4:F21, then press the paste

icon of Ctrl-V (see exceltip Quick Copy Procedure)

2.8.11. Make a line graph of the model in D1 to D21.

2.8.12. Show that the set of equations over the three columns expresses the

mathematical model given under equation 2.4

2.9. The variable rate_restricted is a mathematical variable that refers to a host of

psychological variables, components and properties, such as intelligence, effort,

motivation, … For instance, provided all other components are equal, a more

motivated student will memorize more historical dates in one study hour than a less

motivated student. Hence, the motivated student’s growth rate is higher than the less

motivated student’s rate, even if for both students all other components are exactly

equal. Assume that the student wishes to spend no more than 15 study hours and that

the teacher requires that the student knows 80% of the dates in order to pass the exam

with an A-grade

Assignment: What is the value for rate_restricted that will allow the student to

achieve at least 80% (but not much more) knowledge in 15 study hours? Imagine a

student who intends to invest only the smallest possible amount of effort over the

entire 20 study hours and who’s perfectly happy with the minimally required score,

which is 50% knowledge. What should that student’s growth rate be? See Exceltip

Mark a cell by a color

Solution of Assignment 2.9

2.10. We shall now investigate the effect of random variation of the growth rate on

the outcome of the process. Recall that the growth rate depends in factors such as

motivation and effort. It is highly likely that such factors will vary over the course of

the study process. We will follow the procedure that has been explained in step 2.5.2.

First, introduce a new variable, rate_restricted_stdev. Write the name in A4, specify

the variable in B4 ( see exceltip Introduce Named Vasriables). Go to cell G1, write

VARIABLE RATIO. In G2 to G21, write the equation for the variable growth rate,

12

=dNormalDev(rate_restricted, rate_restricted_stdev). Change the equation that

calculates the growth per study hour: In cell F2, replace the word rate_restricted by

G2, which is the cell that contains the variable ratio for that study hour. The equation

in the cell will now be =E2*G2. Copy F2:G2 to F3:G21 (see exceltip Quick Copy

Procedure). Investigate the effect of random variation on the final state of knowledge,

i.e. the number of dates known after 20 study hours.

2.10.1 Recall our lazy student, who wanted to reach just 50% knowledge after

20 hours. Now assume he wants to be on the safe side and calculate what the

student’s fixed growth ratio should be in order for him to achieve just 55%

knowledge after 20 hours. You can change the stochastic model into a

deterministic one by setting the standard deviation value (rate_restricted stdev)

to 0 (in cell B4)

2.10.2. We will now investigate the effect of random variation, with a standard

deviation that is equal to half the growth rate. Change the value of

rate_restricted_stdev to =rate_restricted/2. What we want to know is this:

what is the probability that the student fails his exam (thus has a score lower

than 50), if the average growth rate is good for 55% knowledge, but

meanwhile varies with the standard deviation as indicated?

2.10.3 In cell I1, type Level Achieved and in cell I2 =D21. In cell J1 type

Level Required, and in cell J2 type 50. Press F9 repeatedly and check

whether I2 changes (if it does not, you probably forgot to change the standard

deviation variable).

2.10.4. Activate Functions/MonteCarloSimulation (If there is no Functions

option in your Excel Menu, you probably forgot to open the file Paul’s

Functions.xla) In the first window, which asks for the cell that contains the

value(s) recalculated by the stochastic procedure, select cell I2. We will now

also use the second window. This window contains the reference to a

comparator value. You want to compare the achieved level with the required

level. Thus, 50 is the comparator value. In the second window, refer to cell J2,

which contains the value 50. In the third window, select an empty cell that will

contain the results of the Monte Carlo run.

2.10.5. In the Results table, take a special look at the p-value cell. The p-value

is based on the number of times that the Monte Carlo Procedure produced a

13

result equal to or bigger than the comparator value. Thus, with 1000 runs, a p-

value of 0.25 means that there were 250 runs in which the result was equal to

or bigger than the comparator value (which is a knowledge level of 50). The

more runs you do (1000 or more), the better the p-value approximates the true

probability of your stochastic model. Use the p-value to answer the following

question: Given, first, that the student has a growth rate that on average

produces 55% knowledge and, second, that this rate fluctuates with a standard

deviation equal to half that rate, what is the probability that the student will

fail the exam (thus, that he obtains a score less than 50)?

Solution of Assignment 2.10.5

2.11. Instead of writing the model equation over the three columns – which we do

because it makes the components of the model more clear – we could just as well

have written the whole equation in column D. Given that D2 contains a reference to

ini_restricted, in cell D3 write =D2+rate_restricted(limit_restricted-D2). First clear

the contents in columns E and F (select E1:F21, then press the delete button), then

copy D3 to D4:D21. Show that this equation is mathematically similar to equation

2.4. and thus also to the model written over three columns.

A special case: restricted exponential growth (the Gompertz model)

The model of restricted growth describes a kind of flow. In the case of the historical

dates, there was a list of dates and the dates were supposed to “get into the person’s

head” by learning them by heart. The learning represents like a flow from the list to

the learner’s head. The less there is left in the list, the slower the flow. A “true”

growth model, however, specifies growth as an increase that is proportional to the

level or size that is already attained. But we have seen that this growth has the

advantage of an exponential explosion, which will outrun the resources. A

contemporary of Malthus (see the historical note in the section on the proportional

growth model) , with the name of Benjamin Gompertz (1779-1865), solved the

problem by counterbalancing the proportional increase in the level of the grower by a

14

proportional decrease in the growth rate. The Gompertz model takes the exponential

model and adds an equation that describes the decay of the growth rate

L/t = r.L r/t = -a r equation 2.5

and thus

Lt+t = Lt + ( rt – a rt ).Lt equation 2.6

(note that the parameter a is smaller than 1 and bigger than 0).

Let us try to apply this model to lexical learning. According to this model, the rate of

lexical learning decreases over time. The reasons can be various. For instance, the

older you are, the less interested you will be in learning new words, or, to put it

differently, your motivation to learn new words gradually declines. Another

explanation could be that the longer you have spent learning new words, the more

difficult it will be to find words that you don’t know yet. Note that the decline in the

growth rate is a function of time, or, more precisely, of the number of learning steps

preceding your current state.

Let us try to model lexical growth between 1 and 18 years, as described in step 2.6.

Recall that this period covers the growth from 1 to about 50,000 words. We take a

month as our step length and define the number of steps as 18 years times 12 months,

which is 216. We know that 50,000 words is about the limit of words our imaginary

subject will ever learn. Thus, our model should level off around 50,000. We have seen

that the exponential model did not: it continued its explosive increase.

Building a Gompertz growth model

2.12. Open a new worksheet or rename an existing, empty worksheet. Name or

rename the worksheet as Gompertz_growth. (exceltip)

2.13. Introduce the variables rate_gomp, ini_gomp and decay_gomp (see step 2.2)

(exceltip). Put the names in cells A1 to A3, which implies that the corresponding

variables will appear in cells B1:B3. Assume that, at the beginning of the study

process, the learner knows one word, i.e.ini_gomp = 1.

2.14. Use column D for the model, and E and F for the components of the model. In

column E we will write the equation for the decreasing growth rate, in column F we

will write the equation for the increase.

2.14.1. In D1 write MODEL

15

2.14.2. In D2 =ini_gomp

2.14.3. In E1 GROWTH RATE

2.14.4. In E2 =rate_gomp .

2.14.5. In F1 INCREASE

2.14.6. In F2 = D2*E2 (D2 contains the current level, E2 the rate of growth)

2.14.7. In D3 = D2+F2 (the new level is the preceding level plus the increase)

2.14.8. In E3 =E2 - E2*decay_gomp

2.14.9. In F3 = D3*E3

2.14.10. Copy range D3:F3 to D4:F217 and make a line graph of the model in

D1:D217.

2.15. Manipulate the model parameters such that the growth curve levels off at around

50,000 words and that it reaches this level at about the 19th birthday (which is step

216). What is your solution? Is it possible to construct a model that complies with

these requirements?

Solution

ini_gomp 1

rate_gomp 0.3

rate_decay_gomp0.0258

A fourth classical growth model: logistic growth

The logistic model combines the principle of proportional or exponential growth with

the decelerating effect of resource limitations. As an example of a growing variable,

we will take a soccer player’s ball handling skills. The player’s ball handling skill

level at time t is represented by Lt. The player’s skill grows as an effect of practice,

training, teaching by a coach and so forth. With each training or practice event of

length t, the skill increases in accordance with the proportional growth model

L/t = r Lt

(see equation 1)

The rate of improvement, r, which is a constant in the proportional model, decreases

as the level, L, increases. There are many reasons why this could be so. For instance,

16

our soccer player becomes less motivated to spent effort in further improving his skill

as his ball handling skill improves. The reason could be that he can better compete

with the other players and feels less need to improve his ball handling skill (a).

Another reason could be that as his skill improves, his coach pays less attention to

him but prefers to focus on the players with lesser skills (b). Still another reason could

be that, as the ball handling skill improves, further improvement requires more from

the body, the reflexes, the speed of movement and so forth. That is, as the skill

improves, the player comes closer to the physical limitations of the body (c). Another

reason is that as the skill improves, more training is needed to maintain the level

already achieved, and thus less time and opportunities are left to practice new skills

(d). Notice that we gave a letter to each of the factors. This will make it easier to turn

our assumptions into a mathematical model.

Let us begin with the motivation factor, denoted by a. It says that motivation

decreases as the skill level increases and thus, that the rate of improvement, r,

diminishes by a factor that depends on a and on the level achieved, Lt

r/t = -aLt rt equation 2.7

The same applies to reduced attention from the coach, physical limitations and the

need to practice more to maintain the current level

r/t = -aLt rt - bLt rt - cLt rt - dLt rt equation 2.8

Thus, the next level of the rate of improvement will be equal to the preceding level of

the rate of improvement, minus the effect of the deceleration factors

rt+1 = rt - aLt rt - bLt rt - cLt rt - dLt rt equation 2.9

For reasons of simplicity, we combine all the decelaration factors into one factor,

which we denote by the letter d (for deceleration)

rt - aLt - bLt - cLt - dLt rt - dLt rt = rt (1 – dLt) equation 2.10

(the symbol means “is defined as”).

Equation 2.10 refers only to the decrease in the growth rate. Let us now plug this in

into the equation for the increase in skill level, which was

L/t = r Lt (see equation 1)

We replace r by the equation for the decelerating r and we obtain

L/t = r (1 – dLt) Lt equation 2.11

Thus, the next level of L, at time t+1, will be

Lt+1 = Lt + r Lt (1 – dLt) equation 2.12

17

We have seen that proportional growth will continue to increase. Thus, however small

the proportion of decelaration d may be, there will come a moment where L has

grown so big that d times Lt equals 1 (or is bigger than 1, for instance, if d is 0.01, and

Lt = 100, d times Lt is 1). If that occurs, 1 - dLt will of course be 0. Let us substitute

this 0 in equation 2.12 and we obtain

Lt+1 = Lt + r Lt 0 = Lt + 0 = Lt

Thus, any next state of L will be equal to the preceding state. This is just another way

to say that L will have reached a stable state. The stable state is the state in which the

level simply reproduces itself (instead of increasing itself). If we know d, it is easy to

calculate at which level L will reach the stable state. This is the state at which the

increase component of the equation will be equal to 0

r Lt (1 – dLt) = 0 equation 2.13

With a litlle algebraic substitution we find that

r Lt = r d Lt2 = 0 equation 2.14

We know that both r and Lt are bigger than 0 (if r were 0 then there would be no

growth; if Lt were 0 it would have been 0 from the beginning of the growth process).

Thus, we may divide both members of the equation by r Lt and we find

1 = d Lt equation 2.15

We know that this equation is true if Lt is stable. Thus, it follows that Lt is stable if it

is equal to

Lt = 1 / d equation 2.16

(Recall the example of d = 0.01, we found that if L = 100, then d times L is 1, thus 1 –

dL = 0, thus there is no longer an increase).

Thus, given all the decelerating effects that we described (and many others, probably,

that we didn’t describe), the soccer player will improve his ball handling skill up to a

level equal to 1 / d. The level at which the growth of a variable comes to a halt, due to

the decelarating factors, is called the carrying capacity of the growth system. It is a

property of the entire system. Thus, in the case of the soccer player, the system

contains all aspects related to the ball handling skill. Some of them refer to properties

of the person himself (such as the biological limitations), others refer to properties

outside the person (such as the coach’s willingness to spend time and effort coaching

the player.). The word carrying capacity implies that the system can “carry”, i.e.

sustain, only a limited level of the grower at issue. Systems with greater carrying

18

capacities are able to sustain higher levels. For instance, another soccer player who

trains in exactly the same environment as the player in our example, but whose body

is more suited for doing the rapid tricks and things with the ball, has a system with a

higher carrying capacity than the first player.

The emphasis on the decelerating aspect may make us forget that the factors in

question are those that make the (final) level possible, that they are needed to develop

and later sustain that level. As Johan Cruyf, undoubtedly the best Dutch soccer player

of all times (or, that’s what I’ve been told), used to say, “Every advantage has its

disadvantage” (but he said “Elk voordeel hep zun nadeel”). But things are the other

way down, here: What we introduced as a negative factor (the deceleration) refers in

fact to a positive factor, or collection of such factors. Aspects such as motivation,

training, help from a coach, the build of the human body and so forth, are factors that

support and sustain the ball handling skill. They are the resources that make the

growth of the skill possible. But they are limited resources: They are not

inexhaustible. It is their limited nature that is responsible for the fact that they both

support and limit the growth.

The carrying capacity is the level of a skill or grower in general that can be sustained

by the collection of limited resources in a system. This level is equal to 1 / d and is

most often referred to by the symbol K (which stands for carrying capacity). We can

change the original logistic equation

Lt+1 = Lt + r Lt (1 – dLt) equation 2.12

into a format that contains the carrying capacity or K-level.

We know that K = 1 / d and therefore, that d = 1 / K. Hence, equation 2.12 can be

changed into

Lt+1 = Lt + r Lt (1 – Lt / K) equation 2.17

In the next chapter, we will see that equation 2.12 which is based on factors that

support (and thus also limit) the growth (see equation 2.8) will allow us to easily

extend the model of the simple grower into a model of connected growers. Each

supporting factor – and every factor that will later emerge and support the grower at

issue – will be part of a network, a web of mutually supporting and competing

growers.

A short historical note

19

The word logistic growth comes from the French word ‘Logis”,

which means lodging, usually for the military troops. We still use

the word logistics for that part of an operation or business that takes

care of the food, shelter and other supplies. The equation for the

logistic growth was found by a Belgian statistician and

demographer, Pierre François Verhulst (1804-1849). Verhulst tried

to solve the Malthusian problem of the geometric, that is infinite,

increase of the population (see historical note). The decelaration

term introduces the effect of limited resources on the growth of the

population. Based on his studies of resource limitations, Verhulst

predicted that the limit of the Belgian population would be

9,400,000 (in 1994, it was 10,118,000). Verhulst was a student of

Quetelet, a Belgian social statistician and mathematician (1796-

1874). It was Quetelet who introduced the notion of the average

man, by which he meant the central value of a human variable in a

normal distribution. The notion of the average man as the essence of

what the human being would be if it were not concealed by the

accidental variations that we observe in real life, has been

abandoned. However, it still survives, tacitly, in the tendency that

we see in so much psychological research, which is focus on the

average of groups. For instance, many developmentalists measure a

variable in a group of three-year-olds, four-year-olds and five-year-

olds, take the averages over the groups and then present the line that

connects those averages as the developmental curve of the variable

in question. By doing so, they implicitly follow an idea that was in

fact already abandoned at the end of the 19th century. In dynamic

systems modeling, we conceive of development as the process that

occurs in an individual over time.

Building a model of logistic growth

2.16 Insert a new worksheet and rename it as Logistic growth (or rename an existing

empty worksheet) (exceltip). If Poptools is not open yet, open it by clicking on

20

Poptools.xla in the Poptools program directory (which is probably C:\Program Files\

Poptools)

2.17 Introduce the named variables rate_logist, ini_logist and K_logist (exceltip). See

that the values of those variables appear in cells B1, B2 and B3. As staryting values,

fill in 1, 0.1 and 100 for the three variables respectively.

2.18 We will write the actual model in column D and specify the part of the equation

that specifies the increase in column E.

2.18.1 In D1 type Model

2.18.2 In D2 =ini_logist

2.18.3 In E1 Increase

2,18.4 In E2 =D2*rate_logist*(1-D2/K_logist)

2.18.5 In D3 =D2+E2

2.18.6 Copy E2 to E3 and then copy D3 to E3 to D4 to E101 (or, to put it in Excel

format, copy D3:E3 to D4:E101)

2.18.7 Make a line graph for the entire model (D1:D101) and make a second line

graph for the first part of the model, which runs from D21 to D40). We can use this

second, shorter model to obtain a better image of growth processes that have a very

high growth rate.

2. 19 Calculate the model for a growth rate from 0.1 to 3. Let the rate_logist increase

in steps of 0.1 (thus, calculate the model with rate_logist 0.1, then 0.2, 0.3 up to 3).

This process of stepwise increase can be automatized by means of a Poptools

function. In the cell where the value of rate_logist is stored – which should be B2,

write the equation =incrementer1(0.1, 0.1,3,FALSE). Each time you press F9 the

value in the cell will begin at 0.1 (the start-value, which is the first value between the

brackets), and increase with 0.1 (the step-value, which is the second value between

the brackets. It will stop at 3 (the stop-value) and then automatically return to the

start-value. The logical term FALSE tells the function to increment its value each time

the F9 button is pressed. If the term is set to TRUE, the value remains at the start-

value.

If the model is recalculated with an increasing growth rate (up to 3), it goes through

four qualitatively distinct growth patterns (look at the graph that covers cells

D21:D40).

21

Can you see the four distinct growth patterns? What is characteristic of each

growth pattern? At which value of the growth rate does one pattern change into

another pattern?

Answer

The first type is asymptotic growth. The pattern approaches a single value. This

pattern occurs with growth rates smaller than 1. The second pattern is approximate

growth. It oscillates until it finally reaches a single end state. It occurs with growth

rates larger than 1 and smaller than 2. The third pattern is oscillatory growth. It occurs

with rates between 2 and 2.57. The pattern oscillates between 2, 4, 8, … states. For

instance, with r = 2.2 the pattern oscillates between two states and with r = 2.5 there

are four states. If r is bigger than 2.57, the pattern becomes chaotic. The pattern visits

an arbitrary number of states that are, apparently, completely randomly scattered.

A short historical note

The idea that chaotic patterns could arise out of very simple,

deterministic models become very popular in the late 1980s with the

publication of Gleick’s book Chaos: Making a new science (1987).

Chaos is an intriguing phenomenon that can be explored by means

of simple equations, such as the stepwise logistic model that we

explained in 2.19. Chaotic oscillations are very sensitive to initial

conditions (which you can check by setting up two logistic growth

models with a similar growth rate bigger than 2.57, but with a very

small difference in the initial state, for instance 0.1 and

0.1000000001). This sensitivity to small differences became popular

under the name of butterfly effect. The effect is related to a simple

model of atmospheric movements studied by Lorenz, a

meteorologist who did this work in the 1950s and 1960s. The

model, consisting of three coupled equations, demonstrated chaotic

behavior. That is, very small differences in the initial level of the

variables upon which the equations were applied led to major

differences in the final outcome, comparable to the potential effect

of a butterfly flapping its wings in Amsterdam and causing a storm

in Boston. In the early 1990s, numerous books were published

22

further popularizing the notion of chaos, which culminated in a

major role for chaos theory (more precisely, the Hollywood version

of chaos theory) in Steven Spielberg’s Jurassic Park. It’s chaos

theory that is responsible for all the problems in the park. However

interesting chaos theory may be, it is rather unlikely that it will play

a major role in the explanation of behavior and development.

Behavior and development show fluctuations that are on first view

quite similar to the fluctuations of a chaotic pattern, but that most

likely have very different causes than those of the chaotic patterns

displayed by the logistic equation, for instance.

A short technical note

The chaotic behavior of the logistic growth equation with rates bigger than 2.57 is

caused by the fact that the growth occurs in discrete steps. Every step represents a

discrete experience or event that causes a specific amount of progress (or regression)

in the growing variable. This model of discrete steps fits in very well with the notion

of actions and experiences as discrete events (although they may overlap and

eventually show smooth transitions). Historically, however, models of increasing or

decreasing variables referred to continuous events. Examples are the flow of heat

from a furnace to a container, such as in a steam engine. In such models, time cannot

be counted in discrete units but must be continuous. Continuous time is specified in

the form of differential equations, which form the subject of calculus. The differential

form of the exponential (or proportional) growth equation is

Nt = N0 eat

A simple way to model this equation in Excel goes as follows.

Specify two variables, N_0 (don’t forget the underscore) and rate. N_0 is the initial

level and rate of course the growth rate. Try values of 1 and 0.1 respectively. In

column D specify time by the series 1 to 50. Writing this series can be greatly

simplified as follows. Write 1 in D1 and 2 in D2. Select D1 and D2, place the cursor

over the small black square at the bottom right. The cursor will change into a cross.

Click the left mouse button and meanwhile drag the cursor down, for instance to D50.

Excel will automatically fill the range with increments of 1 (it will take the pattern in

23

the originally selected range, D1:D2 as an example and expand that pattern in the

range up to D50.

In E1, write =N_0*exp(rate*D1) and copy this equation to E2:E50. Make a graph of

the range E1:E50.

The differential equation for restricted growth is as follows

Nt = Nmax-(Nmax – N0) e-at

Specify an additional variable, N_max, fill range G1:G50 with a time series from 1 to

50. In cell H1, type =N_max-(N_max-N_O)*EXP(-rate*G1) and copy this equation

to H2:H50.

The differential equation for logistic growth is a little more complicated.

Nt = Nmax / ( 1 + (N_max / N_0 – 1) e-at ).

For the Excel model, write the time series 1 to 50 in J1:J50 and in K1, write

=N_max/(1+(N_max/N_O-1)*EXP(-rate*J1)) and copy to K2:K50.

Another method for calculating the growth models in the form of differential

equations is to use Poptools’ ODEIntegration function (ODE stands for Ordinary

Differential Equations). Open Poptools/Demos/ODEIntegration. This will open an

Excel file containing various classical growth models. Follow the instructions in the

file.

The computer approximates the solution of the differential equation by reducing the

step length. A step in our models has a step length of 1. We can reduce the step

length, for instance by a factor 100, by splitting our single step into 100 small steps. If

we do that, we have to reduce the growth rate by a similar factor. Thus, for each mini-

step, the growth rate becomes 0.1 / 100 = 0.0001. This method – step reduction – for

approximating the differential form of the equation goes back to Leonhard Euler

(1707-1783). A better way of approximating the differential equation uses the so-

called Runge-Kutta method, which was published in 1901.

Building a model of growth and decay

The logistic equation describes many forms of growth in many different fields,

ranging from economics to biology. It is obviously a very good model for the growth

of knowledge and skills. However, if we learn something, there is always a fair

24

chance that we forget part of what we have already learned. Everybody who has

studied a foreign language and had to learn new words knows that many of the words

once learned tend to be forgotten. Let us define forgetting as a process of proportional

reduction. That is, it is the opposite of proportional increase, which we explored in

our first model. Thus, instead of increasing the number of words by some proportional

number, for instance 5%, forgetting results in decreasing the number of words by a

proportional number. For instance, per week, we forget 5% of the foreign words we

already know.

2.20 Insert a new worksheet and rename it as Learn_and_forget (or rename an

existing empty worksheet) (exceltip). If Poptools is not open yet, open it by clicking

on Poptools.xla in the Poptools program directory (which is probably C:\Program

Files\Poptools)

2.21 Introduce the named variables rate_LF, ini_LF, K_LF and forget_LF (exceltip).

See that the values of those variables appear in cells B1, B2 and B3. As starting

values, fill in 1, 0.1, 100 and 0.05 for the four variables respectively. The value 0.05

specifies the forgetting rate

2.22 We will write the actual model in column D and specify the part of the equation

that specifies the increase in column E.

2.22.1 In D1 type Model

2.22.2 In D2 =ini_LF

2.22.3 In E1 Increase

2,22.4 In E2 =D2*rate_LF*(1-D2/K_LFt)

2.22.5 In F1 write Forgetting

2.22.7 In F2 = D2*forget_LF

2.22.8 In D3 =D2+E2-F2

2.22.9 Copy E2:F2 to E3:F3 and then copy D3:F3 to D4:F101 2.18.7 Make a line

graph for the entire model (D1:D101).

2.23 Try different values for rate_LF and forget_LF. Compare the outcome of the

models with forget_LF = 0. What is the effect of the forgetting term on the growth

process?

Solution

25

The effect of forgetting is that the upper limit of the growth process is lower than

without forgetting. If K is the upper limit without forgetting, r is the growth rate and f

the rate of forgetting, the new upper limit, K’ = K (r – f)/r . For instance, if r = 0.1, f =

0.05, and K = 1, the new K-level is (0.1-0.05)/0.1 = 0.5. You can find this result by

setting the increase term of the equation to 0. This is the level at which the growth

comes to a halt (the increase is 0). The increase term is r L (1 – L/K) – fL. If it is set

to 0, it follows that rL(1-L/K) = fL. With a little algebraic manipulation it can be

shown that L = K (r-f)/r, which gives the value of the new K (with forgetting).

2.24 Try the effect of random variation in the growth and forgetting rate. Specify two

additional variables, rate_LF_sd and forget_LF_sd. These variables will specify the

standard deviations of the variable growth and forgetting rates. Select E2, move the

cursor to the formula bar and edit the equation: Replace rate_LF by

dNormaldev(rate_LF, rate_LF_sd). That is, the fixed growth rate rate_LF is now

replaced by a randomly varying growth rate, with rate_LF as its mean and rate_LF_sd

as its standard deviation. Select F2, move to the formula bar and replace forget_LF in

the equation by dNormaldev(forget_LF, forget_LF_sd). By so doing you replace the

fixed forgetting rate by a variable one. Then copy E2:F2 to E3:F101.

The ‘hyperlogistic” model: towards a universal growth model

So far, we have discussed three growth models, the proportional (or exponential)

model, the restricted growth model and the logistic growth model. The equations

differed on essential points, such as the contribution of limited resources or the

contribution of the level already achieved.

Let us repeat the equation for logistic growth

L/t = r.L .(1 – d.L)

in which both limited resources (expressed by the d-parameter) and the level already

achieved (L) play a role.

We can rewrite this equation in the following form

L/t = r.L1 .(1 – d.L1) 1

The exponent 1 has no particular meaning here (L1 = L; the reader will probably be

more familiar with the exponent 2, of course, which means the second power, e.g. L2).

26

We can now treat the exponent as a variable parameter. If the three exponents are 1,

the equation represents the standard logistic equation, because

L/t = r.L1 .(1 – d.L1) 1 = L/t = r.L .(1 – d.L)

If the first exponent is 0

L/t = r.L0 .(1 – d.L1) 1

the equation amounts to the standard equation of restricted growth. Since L0 = 1 (any

value to the zero power equals 1)

L/t = r.L0 .(1 – d.L1) 1 = L/t = r.1 .(1 – d.L1) 1 = L/t = r.(1 – d.L)

Note that d = 1/K. Thus

r.(1 – d.L) = r. ( 1 – L/K) = r . (K – L)/K = (r /K) . (K-L)

the constant r divided by the constant K gives a new constant, which we call r’. Thus,

L/t = r’. (K-L)

and the last equation is of course easily recognised as the equation of restricted

growth. If the third exponent is 0, the equation amounts to that of exponential or

proportional growth

L/t = r.L1 .(1 – d.L1) 0 = L/t = r.L .1 = r.L

If the second exponent is 0, we obtain a variant of the exponential equation

L/t = r.L1 .(1 – d.L0) 1 = r.L.(1-d)

The constant r times the constant (1-d) gives a new constant, r’’ and thus

 L/t = r’’ .L

which is of course the exponential growth equation.

We can conclude that the three classical models are in fact special cases of a general

growth model, where each component is specified by a specific exponent. The model

is known as the hyperlogistic model (see Banks, 1992 ; see also van Geert, 1997, for

an introduction in the field of developmental and learning models).

The question is of course whether this model is more than just a common form for the

three classical models. The answer is that the hyperlogistic model does indeed add a

new dimension to the existing models. The exponents can be considered variables,

which can take any value. The general hyperlogistic model has the form

L/t = r.Lp .(1 – d.Ls) q

A simplified form is

L/t = r.Lp .(1 – d.L) q

27

Building an example of a hyperlogistic growth model

28

Preliminary technical note

The mathematical models that we shall deal with in this chapter are deterministic

models. For each step in the model, the parameters are fixed. Mathematically, the

outcome of the model is completely determined by the initial state and the parameter

values (hence deterministic models). In reality, however, the parameters represent

psychological properties and events that fluctuate (within certain limits, that is). If we

don’t know the causes of those fluctuations, we can in fact conceive of them as

accidental or random influences. In order to obtain an idea of how such random

fluctuations influence the outcomes of the models, the effect of such fluctuations

much be simulated numerically. In that case, we are working with stochastic instead

of deterministic models. After discussing the deterministic form of a model, we will

introduce a stochastic version of it and investigate its properties.

Finally, note that Excel tips and tricks appear in white font on a dark background. For

Excel novices, we will refer to earlier step-by-step explanations of the somewhat

more complicated procedures that we will use. In principle, all model-building

exercises should be doable by even inexperienced users of Excel.

29

Exceltip 1: Rename a Worksheet

To rename a worksheet, select the Worksheet Tab, double-click with the left mouse

button. The name is selected (black background, white font.). Type the new name

proportional growth, then Enter. An alternative procedure is to click on the worksheet

tab with the right mouse button, a pop-up window appears, select the option Rename,

then type the new name, then Enter

Exceltip Introduce Named Variables

To introduce named variables, proceed as follows. Select cell A1 and type Ini_prop.

Select cell A2 and type Rate_prop. These cells contain the variable names (not the

variables!). Select the range A1 to B2. In the menu, activate Insert/Name/Create/OK.

To check whether the named variables have been defined successfully, select cell B1.

The Name Box (upper left corner) should show the name Rate_prop instead of B1.

Exceltip: write equation by selection

Instead of writing this equation, you can write = in cell D2, then select cell B1 with

the cursor (which is the cell containing the variable Ini_prop, then Enter. This

alternative procedure is particularly convenient with long variable names and allows

you to avoid typing errors.

You can also compose equations in this way. For instance, write =, then select cell

D2, then write * (asterisk) , then select cell B2, then Enter. This will specify the

equation =D2*B2. The procedure is especially useful with equations that use named

variables.

Exceltip Quick Copy Procedure

A quick-copy procedure works as follows. After having copied D3 to D4:D101, select

cell E2. Around the cell you will see a black border with a small square at the bottom

right. If you place the cursor on the square, your cursor will change into a plus-sign.

With the plus-sign on, double-click with the left mouse button. The content of E2 will

be copied to E3 to E101. Excel will use the adjacent range (which is in column D to

“imitate” the length of the range to which the cell content will be pasted.

30

Exceltip Make Graphs

Graphs can be specified in various ways. The simplest way is as follows. Etc.

Exceltip Expand columns

You will notice that you can see only part of the text in cell A3. In order to expand the

width of the A column and to allow you to see the entire text, move the cursor over

the line that separates the column tabs A and B. The cursor will change into a plus-

sign with two arrow heads. Double-click with the left mouse button: the column will

automatically expand to a width which is sufficient to read the entire text.

Exceltip Move Worksheet

If you wish to replace a worksheet, i.e. change the place at which it appears in the

series of worksheets, go with the cursor to the name tab, click the left mouse button,

hold the button and meanwhile drag the name tab to the place where you want to have

it.

Exceltip Mark a cell by a color

If you experiment with values that appear somewhere in the model, e.g. on study hour

15, it might be handy to mark the cell in which that value appears. Select the cell at

issue, click on the Fill color icon (which is the icon with the little paint can), select a

marker color. If you need to replace a graph, for instance because it overlaps with the

cells of your model, move the cursor to the graph, press the left mouse button and

while holding it, drag the graph to the place you want..

31

Solutions to Assignments

32