- Level: International Baccalaureate
- Subject: Maths
- Word count: 2103
Population trends in China
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Introduction
Claudia Nevado 003656-008
MATHS COURSEWORK
POPULATION TRENDS IN CHINA
Claudia Nevado
INDEX
1. Finding my own model(s)
2. Research model using my own method and find K, L and M
3. Input new data using previous model and research model.
- The aim in this coursework is to investigate the different functions that best model the population of China from 1950 to 1995.
The following table shows the population of China between these years:
Year | 1950 | 1955 | 1960 | 1965 | 1970 | 1975 | 1980 | 1985 | 1990 | 1995 |
Population in Millions | 554.8 | 609.0 | 657.5 | 729.2 | 830.7 | 927.8 | 998.9 | 1070.0 | 1155.3 | 1220.5 |
The relevant variables in this investigation are the population in millions in different years. The parameter is the initial population growth.
The data points from the table above are shown in the graph below, using Microsoft Excel showing the population in China from 1950 to 1995 (presenting the years from 0 being 1950, to 45 being 1995):
What I can observe according to the graph beside is that as the years pass, the population of China (in millions) increases gradually.
The functions which could model the behaviour of the graph can be any of the following:
· y = mx+c
The graph appears linear therefore we could use the
Middle
14.5
Years | Year | Population in Millions | Model 3 |
1950 | 0 | 554.8 | 554.8 |
1955 | 5 | 609 | 627.3 |
1960 | 10 | 657.5 | 699.8 |
1965 | 15 | 729.2 | 772.3 |
1970 | 20 | 830.7 | 844.8 |
1975 | 25 | 927.8 | 917.3 |
1980 | 30 | 998.9 | 989.8 |
1985 | 35 | 1070 | 1062.3 |
1990 | 40 | 1155.3 | 1134.8 |
1995 | 45 | 1220.5 | 1207.3 |
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:
K | L | M |
1109,6 | 1 | 0,06 |
Years | Year | Population in Millions | Research Model |
1950 | 0 | 554,8 | 554,8 |
1955 | 5 | 609 | 637,4 |
1960 | 10 | 657,5 | 716,4 |
1965 | 15 | 729,2 | 788,9 |
1970 | 20 | 830,7 | 852,8 |
1975 | 25 | 927,8 | 907,2 |
1980 | 30 | 998,9 | 952,2 |
1985 | 35 | 1070 | 988,5 |
1990 | 40 | 1155,3 | 1017,3 |
1995 | 45 | 1220,5 | 1039,7 |
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:
K | L | M |
1600 | 2 | 0,04 |
Years | Year | Population in Millions | Research Model |
1950 | 0 | 554,8 | 533,3 |
1955 | 5 | 609 | 606,6 |
1960 | 10 | 657,5 | 683,6 |
1965 | 15 | 729,2 | 762,8 |
1970 | 20 | 830,7 | 842,7 |
1975 | 25 | 927,8 | 921,8 |
1980 | 30 | 998,9 | 998,5 |
1985 | 35 | 1070 | 1071,5 |
1990 | 40 | 1155,3 | 1139,8 |
1995 | 45 | 1220,5 | 1202,5 |
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:
Year | 1983 | 1992 | 1997 | 2000 | 2003 | 2005 | 2008 |
Population in Millions | 1030.1 | 1171.7 | 1236.3 | 1267.4 | 1292.3 | 1307.6 | 1327.7 |
Conclusion
1171,7
1160,6
1252,4
1997
14
1236,3
1233,1
1400,3
2000
17
1267,4
1276,6
1490,1
2003
20
1292,3
1320,1
1580,1
2005
22
1307,6
1349,1
1639,7
2008
25
1327,7
1392,6
1728,4
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.
Years | Years | Population in Millions | My model 3 | Research Model |
1950 | 0 | 554.8 | 554.8 | 533.3 |
1955 | 5 | 609 | 627.3 | 606.6 |
1960 | 10 | 657.5 | 699.8 | 683.6 |
1965 | 15 | 729.2 | 772.3 | 762.8 |
1970 | 20 | 830.7 | 844.8 | 842.7 |
1975 | 25 | 927.8 | 917.3 | 921.8 |
1980 | 30 | 998.9 | 989.8 | 998.5 |
1983 | 33 | 1030.1 | 1033.3 | 1042.8 |
1985 | 35 | 1070 | 1062.3 | 1071.5 |
1990 | 40 | 1155.3 | 1134.8 | 1139.8 |
1992 | 42 | 1171.7 | 1163.8 | 1165.5 |
1995 | 45 | 1220.5 | 1207.3 | 1202.5 |
1997 | 47 | 1236.3 | 1236.3 | 1225.9 |
2000 | 50 | 1267.4 | 1279.8 | 1259.2 |
2003 | 53 | 1292.3 | 1323.3 | 1290.3 |
2005 | 55 | 1307.6 | 1352.3 | 1309.8 |
2008 | 58 | 1327.7 | 1395.8 | 1337.2 |
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.
Years | Years | Population in Millions | My model 3 | Research Model |
1950 | 0 | 554.8 | =14.5*B2+$C$2 | =1600/(1+2*2.71828^(-0.04*B2)) |
1955 | 5 | 609 | =14.5*B3+$C$2 | =1600/(1+2*2.71828^(-0.04*B3)) |
1960 | 10 | 657.5 | =14.5*B4+$C$2 | =1600/(1+2*2.71828^(-0.04*B4)) |
1965 | 15 | 729.2 | =14.5*B5+$C$2 | =1600/(1+2*2.71828^(-0.04*B5)) |
1970 | 20 | 830.7 | =14.5*B6+$C$2 | =1600/(1+2*2.71828^(-0.04*B6)) |
1975 | 25 | 927.8 | =14.5*B7+$C$2 | =1600/(1+2*2.71828^(-0.04*B7)) |
1980 | 30 | 998.9 | =14.5*B8+$C$2 | =1600/(1+2*2.71828^(-0.04*B8)) |
1983 | 33 | 1030.1 | =14.5*B9+$C$2 | =1600/(1+2*2.71828^(-0.04*B9)) |
1985 | 35 | 1070 | =14.5*B10+$C$2 | =1600/(1+2*2.71828^(-0.04*B10)) |
1990 | 40 | 1155.3 | =14.5*B11+$C$2 | =1600/(1+2*2.71828^(-0.04*B11)) |
1992 | 42 | 1171.7 | =14.5*B12+$C$2 | =1600/(1+2*2.71828^(-0.04*B12)) |
1995 | 45 | 1220.5 | =14.5*B13+$C$2 | =1600/(1+2*2.71828^(-0.04*B13)) |
1997 | 47 | 1236.3 | =14.5*B14+$C$2 | =1600/(1+2*2.71828^(-0.04*B14)) |
2000 | 50 | 1267.4 | =14.5*B15+$C$2 | =1600/(1+2*2.71828^(-0.04*B15)) |
2003 | 53 | 1292.3 | =14.5*B16+$C$2 | =1600/(1+2*2.71828^(-0.04*B16)) |
2005 | 55 | 1307.6 | =14.5*B17+$C$2 | =1600/(1+2*2.71828^(-0.04*B17)) |
2008 | 58 | 1327.7 | =14.5*B18+$C$2 | =1600/(1+2*2.71828^(-0.04*B18)) |
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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