Now that we have a model we shall analyze it as a mathematical object. We will set aside, at least for the moment, the connection between the mathematics and the reality. Thus, for example, you should not be concerned when our calculations produce a value for S of 44,446.6 persons at a certain time--a value that will never be attained in reality. In the following analysis S is just a numerical quantity that varies with t, another numerical quantity. Using only mathematical tools we must now extract from the rate equations the information that will tell us just how S and I and R vary with t.
We already took the first steps in that direction when we used the rate
equation R' = I/14 to predict the value of R two days into the
future (see page 
). We assumed that I remained
fixed at 2100 during those two days, so the rate R' = 2100/14 = 150
was also fixed. We concluded that if R = 2500 today, it will be 2650
tomorrow and 2800 the next day.
A glance at the full S-I-R model tells us those first steps have to be modified. The assumption we made--that I remains fixed--is not justified, because I (like S and R) is continually changing. As we shall see, I actually increases over those two days. Hence, over the same two days, R' is not fixed at 150, but is continually increasing also. That means that R' becomes larger than 150 during the first day, so R will be larger than 2650 tomorrow.
The fact that the rates are continually changing complicates the mathematical work we need to do to find S, I, and R. In chapter 2 we will develop tools and concepts that will overcome this problem. For the present we'll assume that the rates S', I', and R' stay fixed for the course of an entire day. This will still allow us to produce reasonable estimates for the values of S, I, and R. With these estimates we will get our first glimpse of the predictive power of the S-I-R model. We will also use the estimates as the starting point for the work in chapter 2 that will give us precise values. Let's look at the details of a specific problem.
The Problem. Consider a measles epidemic in a school population of 50,000 children. The recovery coefficient is b = 1/14. For the transmission coefficient we choose a = .00001, a number within the range used in epidemic studies. We suppose that 2100 people are currently infected and 2500 have already recovered. Since the total population is 50,000, there must be 45,400 susceptibles. Here is a summary of the problem in mathematical terms:
Tomorrow. From our earlier discussion, R' = 2100/14 = 150 persons per day, giving us an estimated value of R = 2650 persons for tomorrow. To estimate S we use
Hence we estimate that tomorrow
Since S + I + R = 50000 and we have S + R = 47096.6 tomorrow, a final subtraction gives us I = 2903.4 persons. (Alternatively, we could have used the rate equation for I' to estimate I.)
The fractional values in the estimates for S and I remind us that the S-I-R model describes the behavior of the epidemic only approximately.
Several days hence. According to the model, we estimate that tomorrow S = 44446.6, I = 2903.4, and R = 2650. Therefore, from the new I we get a new approximation for the value of R' tomorrow; it is
Hence, two days from now we estimate that R will have the new value 2650 + 207.4 = 2857.4. Now follow this pattern to get new approximations for S' and I', and then use those to estimate the values of S and I two days from now.
The pattern of steps that just carried you from the first day to the
second will work just as well to carry you from the second to the third.
Pause now and do all these calculations yourself. See exercises 15 and
16 on page . If you round your calculated values of
S, I, and R to the nearest tenth, they should agree with those in
the following table.
Estimates for the first three days
t S I R S' I' R'
0
1 2 3
2100.0 2903.4
3986.5 5422.1
2500.0 2650.0
2857.4 3142.1
-953.4 -1290.5
-1720.4
803.4 1083.1
1435.7
150.0 207.4 284.7
Yesterday. We already pointed out, on page ,
that we can use our models to go backwards in time, too. This
is a valuable way to see how well the model fits reality, because we can
compare estimates that the model generates with health records for the
days in the recent past.
To find how S, I, and R change when we go one day into the future we multiplied the rates S', I', and R' by a time step of +1. To find how they change when we go one day into the past we do the same thing, except that we must now use a time step of -1. According to the table above, the rates at time t = 0 (i.e., today) are
Therefore we estimate that, one day ago,
Just as we would expect with a spreading infection, there are more susceptibles yesterday than today, but fewer infected or recovered. In the exercises for §3 you will have an opportunity to continue this process, tracing the epidemic many days into the past. For example, you will be able to go back and see when the infection started--that is, find the day when the value of I was only about 1.
There and back again. What happens when we start with tomorrow's values and use tomorrow's rates to go back one day--back to today? We should get S = 45400, I = 2100, and R = 2500 once again, shouldn't we? Tomorrow's values are
To go backwards one day we must use a time step of -1. The predicted values are thus
These are not the values that we had at the start, when t = 0. In fact, it's worth noting the difference between the original values and those produced by ``going there and back again.''
| original | there and | ||||||||||||||
| value | back again | difference | |||||||||||||
|
|
|
|
Do you see why there are differences? We went forward in time using the rates that were current at t = 0, but when we returned we used the rates that were current at t = 1. Because these rates were different, we didn't get back where we started. These differences do not point to a flaw in the model; the problem lies with the way we are trying to extract information from the model. As we have been making estimates, we have assumed that the rates don't change over the course of a whole day. We already know that's not true, so the values that we have been getting are not exact. What this test adds to our knowledge is a way to measure just how inexact those values are--as we do in the table above.
In chapter 2 we will solve the problem of rough estimates by recalculating all the quantities ten times a day, a hundred times a day, or even more. When we do the computations with shorter and shorter time steps we will be able to see how the estimates improve. We will even be able to see how to get values that are mathematically exact!
Delta notation. This work has given us some insights about the way our model predicts future values of S, I, and R. The basic idea is very simple: determine how S, I, and R change. Because these changes play such an important role in what we do, it is worth having a simple way to refer to them. Here is the notation that we will use:
The symbol ``
'' is the Greek capital letter delta; it
corresponds to the Roman letter ``D'' and stands for difference.
Delta notation gives us a way to refer to changes of all sorts. For
example, in the table on page , between day 1 and
day 3 the quantities t and S change by
We sometimes refer to a change as a step. For instance, in this
example we can say there is a ``t step'' of 2 days, and an ``S
step'' of -3010.9 persons. In the calculations that produced the
table on page we ``stepped into'' the future, a day
at a time. Finally, delta notation gives us a concise and vivid way to
describe the relation between rates and changes. For example, if S
changes at the constant rate S', then under a t step of 
,
the value of S changes by
Using the computer as a tool. Suppose we wanted to find out what happens to S, I, and R after a month, or even a year. We need only repeat--30 times, or 365 times--the three rounds of calculations we used to go three days into the future. The computations would take time, though. The same is true if we wanted to do ten or one hundred rounds of calculations per day--which is the approach we'll take in chapter 2 to get more accurate values. To save our time and effort we will soon begin to use a computer to do the repetitive calculations.
A computer does calculations by following a set of instructions called a program. Of course, if we had to give a million instructions to make the computer carry out a million steps, there would be no savings in labor. The trick is to write a program with just a few instructions that can be repeated over and over again to do all the calculations we want. The usual way to do this is to arrange the instructions in a loop . To give you an idea what a loop is, we'll look at the S-I-R calculations. They form a loop. We can see the loop by making a flow chart.
The flow chart. We'll start by writing down the three steps that take us from one day to the next:
, 
, and 
over the course of
a day by using the equations
, , and , we find the new values of S, I, and R
a day later by using the equations
Each step takes three numbers as input, and produces three new numbers as output. Note that the output of each step is the input of the next, so the steps flow together. The diagram below, called a flow chart, shows us how the steps are connected.
The calculations form a loop , because the output of step III is the input for step I. If we go once around the loop, the output of step III gives us the values of S, I, and R on the following day. The steps do indeed carry us into the future.
Each step involves calculating three numbers. If we count each calculation as a single instruction, then it takes nine instructions to carry the values of S, I, and R one day into the future. To go a million days into the future, we need add only one more instruction: ``Go around the loop a million times.'' In this way, a computer program with only ten instructions can carry out a million rounds of calculations!
Later in this chapter (§3) you will find a real computer program that lists these instructions (for three days instead of a million, though). Study the program to see which instructions accomplish which steps. In particular, see how it makes a loop. Then run the program to check that the computer reproduces the values you already computed by hand. Once you see how the program works, you can modify it to get further information--for example, you can find out what happens to S, I, and R thirty days into the future. You will even be able to plot the graphs of S, I, and R.
Rate equations have always been at the heart of calculus, and they have been analyzed using mechanical and electronic computers for as long as those tools have been available. Now that small powerful computers have begun to appear in the classroom, it is possible for beginning calculus students to explore interesting and complex problems that are modelled by rate equations. Computers are changing how mathematics is done and how it is learned.
Analysis without a computer. A computer is a powerful tool for exploring the S-I-R model, but there are many things we can learn about the model without using a computer. Here is an example.
According to the model, the rate at which the infected population grows is given by the equation
In our example, I' = 803.4 at the outset. This is a positive number, so I increases initially. In fact, I will continue to increase as long as I' is positive. If I' ever becomes negative, then I decreases. So let's ask the question: when is I' positive, when is it negative, and when is it zero? By factoring out I in the last equation we obtain
Consequently I' = 0 if either
The first possibility I = 0 has a simple interpretation: there is no infection within the population. The second possibility is more interesting; it says that I' will be zero when
If S is greater than 100000/14 and I is positive, then you can check that the formula
tells us I' is positive--so I is increasing. If, on the other hand, S is less than 100000/14, then I' is negative and I is decreasing. So S = 100000/14 represents a threshold. If S falls below the threshold, I decreases. If S exceeds the threshold, I increases. Finally, I reaches its peak when S equals the threshold.
The presence of a threshold value for S is purely a mathematical result. However, it has an interesting interpretation for the epidemic. As long as there are at least 7143 susceptibles, the infection will spread, in the sense that there will be more people falling ill than recovering each day. As new people fall ill, the number of susceptibles declines. Finally, when there are fewer than 7143 susceptibles, the pattern reverses: each day more people will recover than will fall ill.
If there were fewer than 7143 susceptibles in the population at the outset, then the number of infected would only decline with each passing day. The infection would simply never catch hold. The clear implication is that the noticeable surge in the number of cases that we associate with an ``epidemic'' disease is due to the presence of a large susceptible population. If the susceptible population lies below a certain threshold value, a surge just isn't possible. This is a valuable insight--and we got it with little effort. We didn't need to make lengthy calculations or call on the resources of a computer; a bit of algebra was enough.
stands for a change