If I place other numbers from the data I can identify a more accurate response from what the equation would be.
By placing 10 in t and 657.5 in y the calculations would be the following:
657.5 = 554.8*r 10
1.186 = r 10
R = 10√1.186
R = 1.017
MODEL: y = 554.8*1.017 t
I introduced the data given to me in excel in the following way:
I introduced this model to see how accurate it would be in reference to the original data.
This model was introduced in the first cell and dragged down so that all cells followed the same rule.
Then, I created a graph to compare my model in relation to the original graph of the original data:
Here we can see that my model has a more precise relation with the original in the first years, but later the results start to become more separate and create another geometric line which doesn’t relate with the original.
As my model presented different results seen in the graph, I will now try to change mu constants to have a more precise result in relation to the original.
I thought that when placing another constant in “to the power of” so that my results could be more accurate, meaning that my formula would now be using a system similar to the y = a℮ (kt) formula by having two upper constants, although the ℮ in this case would be another constant implemented by me.
By using trial and error I started to place different numbers into the constants of mi formula so that they would then be as accurate as possible being now:
These constants refer to my column of Model 2.
In the graph we can see that my model 2 in relation to the first model is more near to what the original data is, although it isn’t as accurate as it should.
Now I will use the formula of y=mx+c which concentrates in linear graphs. I will use this formula using P= mt+c, where:
P= Population in millions
C= 554.8 (where c is the initial population at time 0)
M= constant (gradient of line)
T= time (years)
The final model would be: Popn = 14.5t+554.8
I tried this formula and by trial and error, finally achieved the best model which fits the original one, which almost fits the original line.
2. A research suggests that the population, P at time t can be modeled by: P (t) = K /1+Le -Mt where K, L and M are parameters.
Firstly, to find what K is I will place 0 into t (time).
Through mathematical knowledge we may say that e –Mt will equal to 1, because when t=0 M will also be 0 when multiplied together, and any number to the power of 0 will always equal to 1.
When t=0, the population is 554.8, meaning the equation would be:
554.8 = K / 1+ L (*1)
If we say L = 1, then K must equal 2*554.8, which is 1109.6
Being the final equation: P (t) = 1109.6 / 1+ L
By using Microsoft Excel I tried estimating the possible values of K, L and M by entering the data and plotting graphs.
Being:
The graph shows that the research model line does not fit the actual data; therefore I will try other numbers for K, L and M so that it becomes more accurate.
I changed my constants using trial and error and finally got a model which fits better:
The model would be: 1600/1+e-0.04t.
In China, families by law are allowed to have a maximum of one child per family only; therefore this could explain why this population trend tends to be almost linear.
- Another table was given to me with the population in China in millions from year 1983 to 2008:
The research model increases too quickly and doesn’t fit with the IMF data, whereas my model 3 fits better in relation to this line, but it also increases slightly in the end.
I finally modified my final model and the research model so that they could fit the IMF data as accurately as possible, which applies to the given data from 1950 to 2008.
The graph shown more underneath, demostrates how my final model and the research model have been modified so that they can slightly fit the IMF data. As you can see, I finally obtained a very precise data which fits the actual line thorugh trial and error.
Below is also shown how I modified my models, where you can see the formula used for each one in relation to all the IMF data.