Suppose we have an island of which live 1000 immortal people.

• Suppose furthermore that 720 of these people are sane while 280 are demented.
• Suppose finally that at the end of every year 20 percent of those who were sane at the beginning of the year have become demented, while 10 percent of those who were demented at the beginning of the year will have become sane.

The figure below depicts the population transitions of the sane and demented people.

• Of the 720 sane people, 20 percent or 144 will become demented in the first year, while of the 280 demented 10 percent or 28 will become sane.
• As a result there will be a loss of 144 sane individuals and a gain of 28 sane, for a net loss of 116 sane.
• The number of sane will therefore decrease to 604.
• There will also be a loss of 28 demented individuals and a gain of 144 demented, for a net gain of 116 demented.
• The number of demented will therefore increase to 396.

Suppose the we wish to find the number of sane and demented individuals after 100 years.

• We could do the calculations year by year, ultimately performing 100 sets of calculations.
• This would be easy enough but very tedious, and if we did the calculations by hand we would have a high probability of making an error.

Video File #01

If we let S stand for the number of sane people and D for the number of demented people at the beginning of a year, then the transitions are as indicated in the diagram below.

• If the numbers of sane and demented individuals were unchanging, what would be the situation?
• One response for the class was that "the population flows would be equal". This is an excellent way of looking at the situation.
• The "flows" are .20 S (the number of sane becoming demented) and .10 D (the number of demented becoming sane).

Combining the steady-state condition with the fact that the total number of individuals is always 1000 (there are no new people, just people changing categories), we obtain the system of simultaneous equations shown below.

• We solve the system by solving first equation for D, obtaining D = 2 S.
• This means that when the population is stable, we have twice as many demented as sane people. Since the percent of sane becoming demented is twice the number of demented becoming sane, this is clearly what is required for the transition numbers to be equal.
• Substituting 2 S for D in the first equation we obtain S + 2S = 1000, which we easily solve to find S = 333.3... .
• If we wish to find D, we substitute S=333.33... into the first equation and solve for D.   We find here that D = 666.66... .

More generally if P_SD stands for the proportion of sane becoming demented and P_DS for the number of demented becoming sane, the "flows" are as indicated in the figure below.

• The condition for steady-state thus becomes P_DS * D = P_SD * S, and we obtain the system of simultaneous equations shown below.
• The steady-state condition is easily solve for D, obtaining D = P_SD / P_DS * S.

Video File #02

We now look at the original situation symbolically.

• We see that the number of sane individuals after a transition will be S - .20 S + .10 D, while the number of demented will be D - .10 D + .20 S.
• These expressions simplify to .80 S + .10 D and .90 D + .20 S.

We see that in a transition we retain 80% of the originally sane people and 90% of the original demented in their original categories, with the remaining 20% of the sane and 10% of the demented switching categories.

If we look at the second transition we see that, in terms of the original S and D, the transitions will be .10 ( .90 D + .20 S) and .20 ( .80 S + .10 D).

• We will obtain some interesting results if we analyze the sequences we obtain in this manner.
• Sequences are the subject of Chapter 10.

The topic of Chapter 9 is matrices. We will therefore develop the matrix model of this situation.

• We write a row vector [S D]` as indicated. The ` symbol indicates that while our word processor must place the symbols in a row, they really should be in a column.

• The ` symbol is called the transpose operator, and indicates that rows are to be switched to columns and columns to rows.
• DERIVE uses this symbol for the transpose operator, though it doesn't really work as it should (DERIVE still shows the column vectors as rows).

• To the left of the column vector we write a row vector with the .80 (the proportion of sane remaining sane) and .10 (the number of demented becoming sane).
• We define the product of this row and column as .80 S + .10 D, multiplying the first number in the row by the first number in the column and the second number in the row by the second number in the column, then adding the results.
• The result represents the number of sane individuals after the transition.

In the figure below we multiply the column vector [ S D ]` by a matrix consisting of two row vectors [ .80 .10 ] and [.20 .90 ].

• We could denote this matrix by [ [.80, .10] , [ .20, .90] ]. This is the DERIVE notation for the matrix, and you should go to DERIVE and see that it works (e.g.., author and simplify the products in the figure below).
• The first row vector in the matrix is multiplied by the column vector to yield .80 S + .10 D, which is placed in the first row of the column matrix.
• The second row vector is multiplied by the column vector to yield .20 S + .90 D, which is placed in the second row of the product matrix.

The result of this second product is shown below.

• We note for later reference that the population of sane individuals after two transitions is (.80^2 + .10 * .20) S + .10 ( .80 + .90) D, while the number of demented is (.90^2 + .20 * .10 ) D + .20 (.90 + .80) S.
• You should note the symmetry in these two expressions.