# How Does Fuel Consumption Scale with Speed in a Fiat Car?

• skrat
In summary, the given data of fuel consumption for a Fiat car at different velocities is used to check if the fuel consumption follows a quadratic equation. By using least-squares estimation and the data, the parameters ##\Phi_0## and ##\alpha## are found. However, it is suggested to either keep all three parameters and use a 3x3 matrix or keep only two parameters and use a 2x2 matrix to solve the linear system of equations instead of inverting the matrices directly. This is because matrix inversion can lead to errors and should be avoided whenever possible.
skrat

## Homework Statement

Five measurements of fuel consumption of a random Fiat car gives us data:v (km/h) l/100 km
70 --> 3.9+-0.1
90 --> 4.8+-0.1
120 --> 6.0+-0.1
150 --> 8.1+-0.1
160 --> 9.0+-0.1

We would like to check if for velocities greater than 70 km/h the fuel consumption obeys ##\Phi =\Phi _0 +\alpha v^2##. Find ##\Phi _0## and ##\alpha ## using least-squares estimation.

## The Attempt at a Solution

Ok, I tried to do something but I assume there exist an easier way than mine...

Firstly, the law that fuel consumption should obey is ##\Phi =\Phi _0 +\beta v +\alpha v^2##. So $$\begin{bmatrix} \Phi _0\\ \beta\\ \alpha \end{bmatrix}=\begin{bmatrix} \Phi _0\\ 0\\ \alpha \end{bmatrix}=(H^TH)^{-1}H^Tz$$ Where IF I am not mistaken $$H= \begin{bmatrix} 1 & 0& 70^2\\ 1& 0 & 90^2\\ 1& 0 & 120^2\\ 1& 0 & 150^2\\ 1& 0 & 160^2 \end{bmatrix}$$ meaning $$H^TH= \begin{bmatrix} 5 &0 & 75500\\ 0 & 0&0 \\ 75500& 0&14.6\cdot 10^8 \end{bmatrix}$$ Now here the problems start. Am I really supposed to find the inverse of this? Is there really no better/faster solution to this problem?

skrat said:

## Homework Statement

Five measurements of fuel consumption of a random Fiat car gives us data:v (km/h) l/100 km
70 --> 3.9+-0.1
90 --> 4.8+-0.1
120 --> 6.0+-0.1
150 --> 8.1+-0.1
160 --> 9.0+-0.1

We would like to check if for velocities greater than 70 km/h the fuel consumption obeys ##\Phi =\Phi _0 +\alpha v^2##. Find ##\Phi _0## and ##\alpha ## using least-squares estimation.

## The Attempt at a Solution

Ok, I tried to do something but I assume there exist an easier way than mine...

Firstly, the law that fuel consumption should obey is ##\Phi =\Phi _0 +\beta v +\alpha v^2##. So $$\begin{bmatrix} \Phi _0\\ \beta\\ \alpha \end{bmatrix}=\begin{bmatrix} \Phi _0\\ 0\\ \alpha \end{bmatrix}=(H^TH)^{-1}H^Tz$$ Where IF I am not mistaken $$H= \begin{bmatrix} 1 & 0& 70^2\\ 1& 0 & 90^2\\ 1& 0 & 120^2\\ 1& 0 & 150^2\\ 1& 0 & 160^2 \end{bmatrix}$$ meaning $$H^TH= \begin{bmatrix} 5 &0 & 75500\\ 0 & 0&0 \\ 75500& 0&14.6\cdot 10^8 \end{bmatrix}$$ Now here the problems start. Am I really supposed to find the inverse of this? Is there really no better/faster solution to this problem?

You won't be able to find an inverse to this matrix because one entire row and one entire column is nothing but zeroes. Obviously, if you calculate the determinant of the original matrix, it too will equal zero; therefore, the original matrix is not invertible.

Also, this is not technically a physics problem, let alone an advanced physics problem, so I'm moving this thread to the Pre-Calc HW forum.

skrat
Ok... than what should I do?

Yes, I apologize for misplacing the thread and thank you for moving it!

Mmmmm! Generalized least squares. Powerful with the data analysis is this. Everybody who ever does data analysis should learn this one. It should not be the last method you learn, of course. Not by a wide margin. But it will get you through many a task.

The problem statement does not have a term linear in speed. Where did your linear term come from? And why did you set it to zero?

Whenever you do generalized least-squared fit method, your H matrix should never have more than one column that is all one value. Having two columns that were both constant essentially means you are pulling out two constant (phi_0) terms. That does not make sense. Also, you should not have a zero column. That would mean you were pulling out the portion of the data that was some parameter times zero. Again, it does not make sense.

By the way, once you have got your fitting params as the question suggestions, how will you check that this is a good fit? You will get values for these params from a very wide rang of data. How will you check that the values you have obtained are a good fit? There are several ways to approach that. Presumably your text or your class has told you one or more such methods. For example, if the data were really better fitted by a constant and a linear term, how would you detect that? Or what if it is really a cubic?

skrat said:

## Homework Statement

Five measurements of fuel consumption of a random Fiat car gives us data:v (km/h) l/100 km
70 --> 3.9+-0.1
90 --> 4.8+-0.1
120 --> 6.0+-0.1
150 --> 8.1+-0.1
160 --> 9.0+-0.1

We would like to check if for velocities greater than 70 km/h the fuel consumption obeys ##\Phi =\Phi _0 +\alpha v^2##. Find ##\Phi _0## and ##\alpha ## using least-squares estimation.

## The Attempt at a Solution

Ok, I tried to do something but I assume there exist an easier way than mine...

Firstly, the law that fuel consumption should obey is ##\Phi =\Phi _0 +\beta v +\alpha v^2##. So $$\begin{bmatrix} \Phi _0\\ \beta\\ \alpha \end{bmatrix}=\begin{bmatrix} \Phi _0\\ 0\\ \alpha \end{bmatrix}=(H^TH)^{-1}H^Tz$$ Where IF I am not mistaken $$H= \begin{bmatrix} 1 & 0& 70^2\\ 1& 0 & 90^2\\ 1& 0 & 120^2\\ 1& 0 & 150^2\\ 1& 0 & 160^2 \end{bmatrix}$$ meaning $$H^TH= \begin{bmatrix} 5 &0 & 75500\\ 0 & 0&0 \\ 75500& 0&14.6\cdot 10^8 \end{bmatrix}$$ Now here the problems start. Am I really supposed to find the inverse of this? Is there really no better/faster solution to this problem?

Either keep all three parameters ##\Phi_0, \alpha, \beta## and use the 3x3 matrix, or else keep only two parameters ##\Phi_0, \alpha## and use the 2x2 matrix obtained by omitting row and column 2 in the one you have above. Then just invert the matrices---or better, still, solve the linear system of equations by straight Gaussian elimination (the method you first learned in high school). Generally, matrix inversion ought to be avoided whenever possible; matrix inversion is just about the worst way to solve equations (due to stability and roudoff error issues).

Ray Vickson said:
Either keep all three parameters ##\Phi_0, \alpha, \beta## and use the 3x3 matrix, or else keep only two parameters ##\Phi_0, \alpha## and use the 2x2 matrix obtained by omitting row and column 2 in the one you have above. Then just invert the matrices---or better, still, solve the linear system of equations by straight Gaussian elimination (the method you first learned in high school). Generally, matrix inversion ought to be avoided whenever possible; matrix inversion is just about the worst way to solve equations (due to stability and roudoff error issues).

Hmm, ok, I understand the first part, but I don't quite get the part about how to avoid the matrix inversion. Could you be more specific?

For example in my case the equation is ##\begin{bmatrix}
\Phi _0\\
\alpha
\end{bmatrix}=(H^TH)^{-1}H^Tz## and I can't see how would one solve it without matrix inversion.

Aaaa, ok, I found a formula in my notes for that inverse $$(H^TH)^{-1}=\frac{1}{n\sum t_i^2-(\sum t_i)^2 }\begin{bmatrix} \sum t_i^2 & -\sum t_i\\ -\sum t_i & n \end{bmatrix}$$ and also $$H^TZ=\begin{bmatrix} \sum t_i\\ \sum t_iz_i \end{bmatrix}$$.

Caluclating this (checked with mathematica) leaves me with some rather "weird" results, saying that $$\begin{bmatrix} \Phi _0\\ \alpha \end{bmatrix}=\begin{bmatrix} 13.539\\ -0.00047538 \end{bmatrix}$$ Weird because this means that the fuel consumption is reducing with speed. I find that a bit strange. Or?

skrat said:
Hmm, ok, I understand the first part, but I don't quite get the part about how to avoid the matrix inversion. Could you be more specific?

For example in my case the equation is ##\begin{bmatrix}
\Phi _0\\
\alpha
\end{bmatrix}=(H^TH)^{-1}H^Tz## and I can't see how would one solve it without matrix inversion.

I said: just solve the equations.

Your equations ##(H^T H) (\Phi_0,\alpha)^T = H^T z## are
$$\begin{bmatrix} 5 & 75500\\75500& 1458590000 \end{bmatrix} \begin{bmatrix}\Phi_0 \\\alpha \end{bmatrix} = \begin{bmatrix}31.8 \\557040 \end{bmatrix}$$
These say
$$5 \Phi_0 + 75500 \alpha = 31.8\\ 75500 \Phi_0 + 1458590000 \alpha = 557040$$
These are just 2 linear equations in 2 unknowns, and can be solved by high-school methods.

BTW: it is always a good idea to scale a problem before setting out trying to solve it. In this case, using ##w = v/10## instead of ##v## produces a much more well-behaved problem. The equation ##\Phi = \Phi_0 + \alpha v^2## can be re-written as ##\Phi = \Phi_0 + \gamma w^2##, where ##\gamma = 100 \alpha##. This produces
$$\begin{bmatrix} 5&755\\755 & 145859 \end{bmatrix} \begin{bmatrix} \Phi_0\\ \gamma \end{bmatrix} = \begin{bmatrix} 31.8 \\5570.4 \end{bmatrix}$$
or
$$5 \Phi_0 + 755 \gamma = 31.8\\ 755 \Phi_0 + 145869 \gamma = 6670.4$$
The solution of this system is ##\Phi_0 = 2.71654548879261348 \doteq 2.7165## and ##\gamma = 0.0241288378225654673 \doteq 0.02413##, giving ##\alpha \doteq 0.2413 \times 10^{-3}##.

BTW: rounding off the matrix entry ##1458590000## to ##14.6 \times 10^8##--as you have done--produces a fair amount of inaccuracy in the solution, perhaps even an unacceptable level of inaccuracy.

## 1. What is least-squares estimation?

Least-squares estimation is a statistical method used to find the best fit line or curve for a set of data. It is based on minimizing the sum of the squared differences between the observed data points and the predicted values from the model.

## 2. How does least-squares estimation work?

Least-squares estimation works by finding the line or curve that minimizes the sum of the squared differences between the observed data points and the predicted values from the model. This is done by calculating the error or residual for each data point and finding the line or curve that has the smallest overall error.

## 3. What are the assumptions of least-squares estimation?

There are three main assumptions of least-squares estimation: 1) the errors or residuals are normally distributed, 2) the errors have constant variance, and 3) the errors are independent of each other. Violations of these assumptions can affect the accuracy of the estimated model.

## 4. When is least-squares estimation used?

Least-squares estimation is commonly used in regression analysis, where it is used to estimate the parameters of a linear or nonlinear model. It is also used in other statistical methods, such as ANOVA and time series analysis, to find the best fit for a set of data.

## 5. What are the advantages of least-squares estimation?

The main advantage of least-squares estimation is that it provides a simple and efficient way to estimate the parameters of a model. It also produces unbiased and efficient estimates, meaning that they are close to the true values and have a low variance. Additionally, the method is widely used and understood, making it easy to apply to various types of data and problems.

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