MATH 18
LOGISTICS MODEL
WORKSHEET
Logistic Model
A logistic model has an equation of the form
.
The logistic function f increases if B is positive and decreases if B is negative. The graph of the logistic function is bounded by the horizontal lines y = D and y = A + D. We call A + D the limiting value of the function.
Example: Alpha Company invented a new toy that was an immediate fad. It kept track of its revenues during the first eight months of the year and recorded them in the following table.
|
Month |
January |
February |
March |
April |
May |
June |
July |
August |
|
Revenue (thousand dollars) |
12.3 |
14.8 |
19.0 |
26.5 |
41.5 |
56.1 |
62.3 |
66.1 |
A scatter plot of the data looks like the following:
The “S” shape is typical of what is called a logistic
curve. Initially, it is nearly level,
then it becomes steeper, and finally levels off. We can fit a model to this curve using the
logistic function. The basic form of the
equation is 
. This model has a
“limiting value” which is y = a + d.
Thus, for the first eight months of the year, the revenue for Alpha Toy Company would be
thousand dollars
where m = 1 in January, m = 2 in February, etc.
The limiting value would be 55.103 + 12.386 = 67.489. This means that the revenue would level off at approximately $67,500.
The graph of the model with the scatter plot is shown below. Next to it is a graph with the line y = 67.5 introduced. This shows the leveling of the graph.
Practice Problems:
1. The table gives the total number of states associated with the national P.T.A. organization from 1895 through 1931.
|
Year |
1895 |
1899 |
1903 |
1907 |
1911 |
1915 |
1919 |
1923 |
1927 |
1931 |
|
Total number of states |
1 |
3 |
7 |
15 |
23 |
30 |
38 |
43 |
47 |
48 |
a. Find a logistic model for the data.
b. What is the limiting value of the model?
Answers:
a. For the years 1895 through 1913, the number of states is given by
where y is the year since 1895.
b. The limiting value is 53.719 – 2.996 = 50.723. Thus, the model tells us that the number of states will level off at about 51 states. This sounds fairly good until we remember that there were only 48 states at the time.
2. In 1949 the
|
Month |
Jan. |
Feb. |
March |
April |
May |
June |
July |
Aug. |
Sept. |
Oct. |
Nov. |
Dec. |
|
Total number of polio cases |
494 |
759 |
1016 |
1215 |
1619 |
2964 |
8489 |
22,377 |
32,618 |
38,153 |
41,462 |
42,375 |
a. How many new cases were diagnosed in June?
b. Find a logistic model for the total number of cases diagnosed.
c. Examine the scatter plot and graph. Determine when the spread of polio was most rapid.
Answer:
a. The total number of cases diagnosed at the end of May was 1619; at the end of June it was 2964; thus, during the month of June, 2964 – 1619 = 1345 cases were diagnosed.
b. The total number of polio cases diagnosed is given by
where m = 1 in January, m = 2 in February, …, m = 12 in December.
c. The
slope of the graph is very steep in the months of June through September. Polio was a disease that occurred mostly in
the summer.
3. A chemical reaction begins when a certain mixture of
chemicals reaches
. The reaction
activity is measured in Units (U) per 100 microliters 
of the mixture. Measurements during the first 18 minutes
after the mixture reaches are listed in the
table.
|
Time (minutes) |
0 |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
18 |
|
Activity
|
0.10 |
0.10 |
0.25 |
0.60 |
1.00 |
1.40 |
1.55 |
1.75 |
1.90 |
1.95 |
a. Examine a scatter plot of the data. Estimate the limiting value.
b. Find a logistic model for the data. What is the limiting value for this logistic function?
c. Use the model to estimate by how much the reaction activity increased between 7 and 11 minutes.
d. When was
the reaction rate 1.25 
?
Answers:
a.
It appears to be leveling off at about 2.
b. The
chemical reaction is 
t minutes after the temperature reaches . The model gives the
limiting value as 1.982 – 0.0334 = 1.9486.
c. R(7) =
0.774 and R(11) = 1.503. There was an
increase of 1.503 – 0.774 = 0.729 .
d. 9.44 minutes after the reaction began
4. The table records the volume of sales (in thousands) of a popular movie for the first 18 months after it was released on videocassette.
|
Months after release |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
Number of cassettes sold each month (thousands) |
580 |
565 |
527 |
467 |
321 |
291 |
204 |
131 |
79 |
|
Months after release |
10 |
11 |
12 |
13 |
14 |
18 |
16 |
17 |
18 |
|
Number of cassettes sold each month (thousands) |
61 |
31 |
17 |
9 |
4 |
3 |
3 |
2 |
2 |
a. Examine a scatter plot and describe its curvature.
b. Explain why this type of data could reasonably be modeled by a logistic equation.
c. Find a logistic model for the data.
Answers:
a.
b. Initially there are many copies sold; then the number starts to decease until the demand is nearly 0. This should be the limiting value.
c. There
are 
x months after it is first released onto videocassettes.
5. The table shows the number of investment clubs in existence during the 1990s.
|
Year |
1990 |
1991 |
1992 |
1993 |
1994 |
1995 |
1996 |
1997 |
1998 |
|
Number of clubs |
7085 |
7360 |
8267 |
10,033 |
12,429 |
16,054 |
25,409 |
31,828 |
36,500 |
a. Examine a scatter plot of the data and determine an appropriate model for the data.
b. Find a logistic model for the data.
c. According to the model, how many investment clubs were there in 1995? How well does the model do in predicting the accuracy?
d. According to the model, how many investment clubs are there in 2003? How comfortable are you with this prediction?
Answers:
a.
This is typical of a logistic curve. It starts flat, becomes steep, and then levels off.
b. The number of investment clubs in the 1990s is given by
clubs
t years after 1990.
c. The model predicts 17,219 clubs. This compares to the actual number of 16,054. So the model over predicted the number by 1165 clubs. While this may seem like a like it is only 7.3% of the total.
d. The
model predicts 40,531 for 2003. This
assumes that nothing has changed in the years since 1998. It is always risky to project a model too far
beyond the data.