N1.04: Section 2 Part 2
Making the logistic formula more versatile, stage one—shifting right or left
While it is useful to be able to control the rate of transition by using the rate parameter, more adjustability is needed to make this formula a good match to typical transition situations whose graphs have shapes similar to the basic logistic. We can handle this by adding more parameters to control the size and position of the curve. So in addition to the rate/slope parameter, we will add a center parameter that tells the x value where the midpoint occurs, enabling us to shift the curve to the right (with positive center values) or to the left (with negative center values). Since the logistic formula equals ½ when the exponent is equal to zero, the way to put the center parameter into the formula is to subtract center from the x value before multiplying by the rate. This leads to the following version of the logistic formula, in which both slope and left-right position can be controlled: Simple logistic formula (with center adjustment): [latex]y=\frac{1}{1+{{0.018316}^{rate\cdot(x-center)}}}[/latex] The advantage of having the center parameter is that we can use it for data where we can see that there is a logistic transition from zero to one, but we don’t know exactly when that transition occurred. By fitting a simple logistic model to the data, we can have the data tell us both when the transition occurred (the best-fit value for the center parameter) and what its maximum rate of change was (the best-fit value for the rate parameter). Example 2: Wine fermentation time As wine matures, a small amount of yeast grows and converts the sugar in the grapes to alcohol. At first growth is rapid because there is plenty of sugar and no alcohol; as more alcohol accumulates, the rate slows down as the increasing amount of alcohol and decreasing amount of sugar interferes with yeast metabolism, finally stopping when the maximum alcohol level for that type of wine is reached.. For the data shown below, [a] When was half the alcohol produced? [b] How fast was the percentage of alcohol compared to the saturation alcohol level increasing at that time? Solution approach: The data shows a transition from zero to one (since 100% = 1), so it can be modeled with the simple logistic model constructed from the General Model template. The appropriate spreadsheet formula to put in cell C3 is =1/(1+0.018316^($G$3*(A3-$G$4))) if we use cell G3 for the rate parameter and cell G4 for the center parameter. Answers: [a] Half the alcohol was produced by 8.68 days of fermentation (from the best-fit value of the center parameter). [b] At the midpoint of fermentation, the alcohol level was growing at the rate of 16.5% of the saturation level per day (from the best-fit value of the rate parameter).Days of fermentation | Saturation level % |
0 | 0% |
3 | 2% |
6 | 15% |
9 | 55% |
12 | 90% |
15 | 99% |
18 | 100% |
21 | 100% |
A | B | C | D | E | F | G | H | |
X | y data | y model | Residual | Squared | y = 1/(1+0.018316^($G$3*(x-$G$4))) | |||
1 | Days | Saturation | prediction | deviation | deviation | Simple-logistic parameters: | ||
2 | 0 | 0% | 0.003294 | -0.003294 | 1.09E-05 | 0.164513 | Rate | |
3 | 3 | 2% | 0.023243 | -0.003243 | 1.05E-05 | 8.680789 | Center | |
4 | 6 | 15% | 0.146279 | 0.003721 | 1.38E-05 | |||
5 | 9 | 55% | 0.552322 | -0.002322 | 5.39E-06 | Model and data value counts | ||
6 | 12 | 90% | 0.898822 | 0.001178 | 1.39E-06 | 2 | Number of parameters | |
7 | 15 | 99% | 0.984607 | 0.005393 | 2.91E-05 | 8 | Number of deviations averaged | |
8 | 18 | 100% | 0.997834 | 0.002166 | 4.69E-06 | |||
9 | 21 | 100% | 0.999699 | 0.000301 | 9.09E-08 | Goodness of fit of this model | ||
10 | 7.59E-05 | Sum of squared deviations | ||||||
11 | 0.003556 | Standard deviation |
Licenses & Attributions
CC licensed content, Shared previously
- Mathematics for Modeling. Authored by: Mary Parker and Hunter Ellinger. License: CC BY: Attribution.