Question

Civil and transportation engineers must often estimate the future traffic flow on roads and bridges to plan for maintenance or possible future expansion. The following data gives the number of vehicles (in millions) crossing a bridge each year for 10 years. Fit a cubic polynomial to the data and use the fit to estimate the flow in year 2000.
Year 1990 1991 1992 1993 1994 Vehicle flow (millions) 2.1 3.4 4.5 5.3 6.2
Year 1990 1991 1992 1993 1994 Vehicle flow (millions) 6.6 6.8 7 7.4 7.8

Answer 1

Solution:

$$f(x)=a_0+a_1\times x+a_2\times x^2+a_3\times x^3$$

$$3a_0+a_1\times\sum x_i+a_2\times\sum x_i^2+a+3\times\sum x_i^3=\sum y_i$$

$$a_0\times\sum x_i+a_1\times\sum x_i^2+a_2\times\sum x_i^3+a_3\times\sum x_i^4=\sum x_i\times y_i$$

$$a_0\times\sum x_i^2+a_1\times\sum x_i^3+a_2\times\sum x_i^4+a_3\times\sum x_i^5=\sum x_i^2\times y_i$$

For case number 1:

$$x_1=1990; x_2=1991; x_3=1992; x_4=1993; x_5=1994$$

$$y_1=2.1; y_2=3.4; y_3=4.5; y_4=5.3; y_5=6.2$$

$\sum x_i=9960; \sum x_i^2=19840330; \sum x_i^3=3.95\times 10^{10}; \sum x_i^4=7.87\times 10^{13}; \sum x_i^5=1.57\times 10^{17}; \sum x_i^6=3.12\times 10^{20}$

$$\sum y_i=21.5; \sum x_i\times y_i=42838.1; \sum x_i^2\times y_i=8.54\times 10^7; \sum x_i^3\times y_i=1.06\times 10^{31}$$

$f(x)=-181779.1+0.00019\times x^3+2579.15\times x-1.215676\times x^2$

$$f(2000)=5.8$$

For case number 2:

$$x_1=1990; x_2=1991; x_3=1992; x_4=1993; x_5=1994$$

$$y_1=6.6; y_2=6.8; y_3=7.0; y_4=7.4; y_5=7.8$$

$\sum x_i=9960; \sum x_i^2=19840330; \sum x_i^3=3.95\times 10^{10}; \sum x_i^4=7.87\times 10^{13}; \sum x_i^5=1.57\times 10^{17}; \sum x_i^6=3.12\times 10^{20}$

$\sum y_i=35.6; \sum x_i\times y_i=70918.2; \sum x_i^2\times y_i=1.41\times 10^8; \sum x_i^3\times y_i=1.76\times 10^{31}$

$$f(x)=-475741.2+801.2\times x - 0.4449\times x^2+0.000082\times x^3$$

$f(2000)=8.2$