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Chapter 7: Problem 3

Use the curve-fitting criterion to minimize the sum of the absolute deviations for the following models and data set: a. \(y=a x\) b. \(y=a x^{2}\) c. \(y=a x^{3}\) $$ \begin{array}{l|llllll} x & 7 & 14 & 21 & 28 & 35 & 42 \\ \hline y & 8 & 41 & 133 & 250 & 280 & 297 \end{array} $$

Short Answer

Expert verified
Cubic model minimizes deviations best.

Step by step solution

01

Understand the Problem

We are given a dataset with values of \(x\) and \(y\), and we need to find the best-fitting model among the options \(y=ax\), \(y=ax^2\), and \(y=ax^3\) to minimize the sum of the absolute deviations.
02

Define Criterion for Curve Fitting

The curve-fitting criterion given is to minimize the sum of absolute deviations, i.e., \( \ \sum_{i=1}^{n} |y_i - \hat{y_i}| \), where \( \hat{y_i} \) is the predicted value using the model and \( y_i \) is the actual value.
03

Compute for Model \(y=ax\)

1. Set up the objective function for absolute deviations: \( \sum_{i=1}^{n} |y_i - ax_i| \).2. To find \(a\) for each model, compute \(a = \frac{y_i}{x_i}\) for each data point and then choose an \(a\) that minimizes the sum of the deviations.3. Calculate sum of deviations for potential values of \(a\) using trial and error, or estimate values of \(a\) using median of ratios.
04

Compute for Model \(y=ax^2\)

1. Similarly, set up the equation for absolute deviation as \( \sum_{i=1}^{n} |y_i - ax_i^2| \).2. Compute potential \(a\) using \(a = \frac{y_i}{x_i^2}\) for each point and track which \(a\) leads to minimal deviation.3. Test calculated \(a\) using cumulative absolute deviations.
05

Compute for Model \(y=ax^3\)

1. Formulate absolute deviations for the cubic model as \( \sum_{i=1}^{n} |y_i - ax_i^3| \).2. Determine potential \(a\) values via \(a = \frac{y_i}{x_i^3}\) for each \(x_i, y_i\) pair and compute their respective deviations.3. Compare cumulative absolute deviations for these \(a\) values.
06

Compare and Choose the Best Model

Compare the sum of absolute deviations obtained for each model (linear, quadratic, and cubic). The model with the smallest sum is considered the best fit for the provided dataset.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Curve Fitting
When we're talking about curve fitting, we're looking for a way to draw a curve that most closely matches a set of data points. Imagine having a scatter plot of dots, each representing a data pair from a table. Curve fitting is like finding the perfect curve that "hugs" these dots as closely as possible. The aim of doing this is to make predictions and understand the data better by identifying trends.
In mathematical modeling, we use equations to represent these curves. The exercises given include simple models like linear (\( y = ax \)), quadratic (\( y = ax^2 \)), and cubic (\( y = ax^3 \)) equations. Each gives a different shape of line to fit the data.
To choose the best fit among these, we need to analyze how "off" each model is from the actual data points, which directly brings us to the next topic.
Absolute Deviations
Absolute deviation is a way of measuring how far off a prediction is from an actual observed value in data. Essentially, it's looking at the distance between what's expected (the curve) and the reality (the data points), without worrying if it's above or below the curve.
For instance, if our model predicts a weight of 100 pounds but the real weight is 110 pounds, the absolute deviation is 10 pounds. To fit a curve using absolute deviations, sum up all these little deviations (or errors) across all data points. The goal is to make this total sum as small as possible.
In modeling terms, minimizing these deviations means tweaking the curve's parameters (like "\( a \)") until the curve is fitting through the data points in a way that these absolute deviations add up to the smallest possible number.
  • This method is especially useful when outliers or extreme values might otherwise skew the results.
  • Unlike squaring deviations (as in least squares), absolute deviation focuses purely on distance, providing a linear measure.
Objective Function
An objective function is like the "scorecard" used in optimization problems. When we're performing curve fitting, the objective function tells us how well our current solution is working. It's the guide to finding the best possible model fit.
For absolute deviations, the objective function is the sum of absolute deviations across all points. \[ \text{Objective Function:} \ \\sum_{i=1}^{n} |y_i - \hat{y_i}| \]The "\( y_i \)" are the actual observed values, and "\( \hat{y_i} \)" are the values predicted by our model.
By focusing on minimizing this sum, we aim to adjust our model's parameters (like the value of "\( a \)") until we get the smallest possible deviation sum. Here are steps involved in working with the objective function:
  • Define the function that measures errors (in this case, absolute deviations).
  • Calculate this error function for all model equations proposed.
  • Compare results and select the model that results in the lowest score.
This process is straightforward but powerful for systematically determining the best fitting curve.

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Most popular questions from this chapter

Nutritional Requirements \(-\) A rancher has determined that the minimum weekly nutritional requirements for an average-sized horse include \(40 \mathrm{lb}\) of protein, \(20 \mathrm{lb}\) of carbohydrates, and \(45 \mathrm{lb}\) of roughage. These are obtained from the following sources in varying amounts at the prices indicated: $$ \begin{array}{l|cccc} \hline & \begin{array}{c} \text { Protein } \\ \text { (lb) } \end{array} & \begin{array}{c} \text { Carbohydrates } \\ \text { (lb) } \end{array} & \begin{array}{c} \text { Roughage } \\ \text { (lb) } \end{array} & \text { Cost } \\ \hline \text { Hay } & 0.5 & 2.0 & 5.0 & \$ 1.80 \\ \text { (per bale) } & & & & \\ \begin{array}{c} \text { Oats } \\ \text { (per sack) } \end{array} & 1.0 & 4.0 & 2.0 & 3.50 \\ \begin{array}{l} \text { Feeding blocks } \\ \text { (per block) } \end{array} & 2.0 & 0.5 & 1.0 & 0.40 \\ \begin{array}{l} \text { High-protein } \\ \text { concentrate } \\ \text { (per sack) } \end{array} & 6.0 & 1.0 & 2.5 & 1.00 \\ \begin{array}{l} \text { Requirements } \\ \text { per horse } \\ \text { (per week) } \end{array} & 40.0 & 20.0 & 45.0 & \\ \hline \end{array} $$ Formulate a mathematical model to determine how to meet the minimum nutritional requirements at minimum cost.

A Farming Problem - A farm family owns 100 acres of land and has $$ 25,000\( in funds available for investment. Its members can produce a total of 3500 work- hours worth of labor during the winter months (mid-September to mid-May) and 4000 work-hours during the summer. If any of these work-hours are not needed, younger members of the family will use them to work on a neighboring farm for $$ 4.80\) per hour during the winter and $$ 5.10\( per hour during the summer. Cash income may be obtained from three crops (soybeans, corn, and oats) and two types of livestock (dairy cows and laying hens). No investment funds are needed for the crops. However, each cow requires an initial investment outlay of $$ 400\), and each hen requires $$ 3 .\( Each cow requires \)1.5\( acres of land and 100 work-hours of work during the winter months and another 50 work- hours during the summer. Each cow produces a net annual cash income of $$ 450\) for the family. The corresponding figures for the hen are as follows: no acreage, \(0.6\) work-hour during the winter, \(0.3\) more work-hour in the summer, and an annual net cash income of $$ 3.50 .$ The chicken house accommodates a maximum of 3000 hens, and the size of the barn limits the cow herd to a maximum of 32 head.

Scheduling Production - A manufacturer of an industrial product has to meet the following shipping schedule: $$ \begin{array}{lc} \hline \text { Month } & \text { Required shipment (units) } \\ \hline \text { January } & 10,000 \\ \text { February } & 40,000 \\ \text { March } & 20,000 \\ \hline \end{array} $$ The monthly production capacity is 30,000 units and the production cost per unit is \(\$ 10 .\) Because the company does not warehouse, the service of a storage company is utilized whenever needed. The storage company determines its monthly bill by multiplying the number of units in storage on the last day of the month by \(\$ 3 .\) On the first day of January the company does not have any beginning inventory, and it does not want to have any ending inventory at the end of March. Formulate a mathematical model to assist in minimizing the sum of the production and storage costs for the 3 -month period. How does the formulation change if the production cost is \(10 x+10\) dollars, where \(x\) is the number of items produced?

Purchasing Various Trucks - A truck company has allocated \(\$ 800,000\) for the purcha: of new vehicles and is considering three types. Vehicle A has a 10 -ton payload capaci and is expected to average \(45 \mathrm{mph}\); it costs \(\$ 26,000\). Vehicle \(\mathrm{B}\) has a 20 -ton payloz capacity and is expected to average \(40 \mathrm{mph}\); it costs \(\$ 36,000\). Vehicle \(\mathrm{C}\) is a modifie form of B and carries sleeping quarters for one driver. This modification reduces th capacity to an 18 -ton payload and raises the cost to \(\$ 42,000\), but its operating speed still expected to average \(40 \mathrm{mph}\). Vehicle A requires a crew of one driver and, if driven on three shifts per day, coul be operated for an average of \(18 \mathrm{hr}\) per day. Vehicles \(\mathrm{B}\) and \(\mathrm{C}\) must have crews of tw drivers each to meet local legal requirements. Vehicle B could be driven an average \(18 \mathrm{hr}\) per day with three shifts, and vehicle \(\mathrm{C}\) could average \(21 \mathrm{hr}\) per day with three shift The company has 150 drivers available each day to make up crews and will not be ab. to hire additional trained crews in the near future. The local labor union prohibits an driver from working more than one shift per day. Also, maintenance facilities are suc that the total number of vehicles must not exceed 30 . Formulate a mathematical mod to help determine the number of each type of vehicle the company should purchase maximize its shipping capacity in ton-miles per day.

Consider a company that carves wooden soldiers. The company specializes in two main types: Confederate and Union soldiers. The profit for each is \(\$ 28\) and \(\$ 30\), respectively. It requires 2 units of lumber, \(4 \mathrm{hr}\) of carpentry, and \(2 \mathrm{hr}\) of finishing to complete a Confederate soldier. It requires 3 units of lumber, \(3.5 \mathrm{hr}\) of carpentry, and \(3 \mathrm{hr}\) of finishing to complete a Union soldier. Each week the company has 100 units of lumber delivered. There are \(120 \mathrm{hr}\) of carpenter machine time available and \(90 \mathrm{hr}\) of finishing time available. Determine the number of each wooden soldier to produce to maximize weekly profits.
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