DATA 609: Week 2 Homework
Keith Folsom
February 7, 2017
Page 69: #12
A company with a fleet of trucks faces increasing maintenance costs as the age and mileage of the trucks increases.
Identify a problem you would like to study. List the variables affect the behavior you have identified.
- A problem to study would be the total cost of ownership of a truck. This would help a company better evaluate and determine the size of the fleet and develop a more accurate way to determine when to rotate trucks out of operation.
Variables to consider:
- Type of truck: make and model, mpg (highway and local), hauling capacity, age, weight, maintenance costs, routine maintenance schedule, labor costs, insurance, average usage per month in miles/hours
Which variables would be neglected completely?
- Hauling capacity and weight might be two variables that would be neglected completely. Although useful, these would be reflected in the mpg variable.
Which might be considered as constants initially?
- Routine maintenance might be considered a constant initially. The assumption here is that all trucks would need standard maintenance such as oil changes, tire rotations, tire replacement.
Can you identify any submodels you would want to study in detail?
- One submodel would be to model the ROI of a single truck where ROI = Revenue $ - TCO
- Another submodel would be to predict when the incurred costs of an aging truck no longer make financial sense to continue using the truck. This would be when the cost of the truck exceeds the revenue generated by the truck.
Identify any data you would want collected.
- I would want data relating to the highway and city MPG for the given truck, yearly maintenance costs, cost of insurance by year, associated labor costs, and major service milestones by mpg and cost.
Page 79: # 11
Determine whether the data set supports the stated proportionality model.
\[ y \propto x^3 \]
\(y \propto x^3\) if and only if \(y = kx^3\) for a constant k. Determine whether there is a positive constant \(k\) satisfying \(y = kx^3\).
We can calculate the ratio of \(y^{1/3}/x\) or \(y/x^3\) as both should be true if the data set is proportional.
y <- c(0, 1, 2, 6, 14, 24, 37, 58, 82, 114)
x <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
data <- data.frame(y, x)
data$ratio <- data$y/data$x^3
ggplot(data, aes(x=x, y=ratio)) + geom_point() + ggtitle('Ratio of y/x^3')
# calculate the average ratio of y/x^3
k <- (mean(data$ratio))\[y = 0.0963571x^2\]
data$predicted <- k*data$x^3
df <- rbind(data.frame(value = 'Actual', x = data$x, y = data$y),
data.frame(value = 'Predicted', x = data$x, y = data$predicted))
ggplot(df, aes(x=x, y=y, color=value)) + geom_line()
We see some an increasing difference between the actual versus predicted values starting at \(x = 5\)
Recalculate \(k\) this time with the average ratio starting from the x >= 5
# calculate the average ratio of y/x^3 starting with x = 5
k <- mean(subset(data, x > 4, select=ratio)$ratio)\[y = 0.1117912x^3\]
data$predicted2 <- k*data$x^3
rbind(data.frame(value = 'Actual', x = data$x, y = data$y),
data.frame(value = 'Predicted', x = data$x, y = data$predicted2)) %>%
ggplot(aes(x=x, y=y, color=value)) + geom_line()
With the second approach, we see that the data set does support proportionality with the constant \(k = 0.1117912\) producing a close fit. The second \(k\) value produces a much closer fit. Worth noting is that the line generated by this data set does not pass through the origin, although it is very close.
Page 94: #4
Lumber Cutters–Lumber cutters wish to use readily available measurements to estimate the number of board feet of lumber in a tree. Assume they measure the daimeter of the tree in inches at waist height. Develop a model the predicts board feet as a function of diameter in inches.
x <- c(17, 19, 20, 23, 25, 28, 32, 38, 39, 41)
y <- c(19, 25, 32, 57, 71, 113, 123, 252, 259, 294)
data <- data.frame(x, y); data## x y
## 1 17 19
## 2 19 25
## 3 20 32
## 4 23 57
## 5 25 71
## 6 28 113
## 7 32 123
## 8 38 252
## 9 39 259
## 10 41 294The variable \(x\) is the diameter of a ponderosa pine in inches, and \(y\) is the number of board feet divided by 10.
Consider two separate assumptions, allowing each to lead to a model. Completely analyze each model.
- Assume that all trees are right-circular cylinders and are approximately the same height.
The characteristic dimension of a cylinder is the diameter \(d\). The number of board feet is a function of the volume which is given by \(V = \pi r^2h\). Assuming that the radius is proportional to the diameter, we derive:
\[ V \propto r^2 \propto d^2 \]
where \(h\) is constant since all trees are the same height.
For this model we assume that all trees are of the same height so the volume is a function of the diameter squared which is the \(x\) variable.
\[ V \propto d^2 \] or \[ y \propto x^2\]
where \(y = board feet\) and \(x = diameter\)
Using this, we calculate proportionality using:
\(y \propto x^2\) if and only if y = kx^2 for a constant \(k\) which leads to \(y \propto x^2\)
Calculate the ratio of \(y/x^2\)
data$ratio <- data$y/data$x^2
k <- (mean(data$ratio))\[y = 0.122029 * x^2 \]
data$predicted <- k*data$x^2
df <- rbind(data.frame(value = 'Actual', x = data$x, y = data$y),
data.frame(value = 'Predicted', x = data$x, y = data$predicted))
ggplot(df, aes(x=x, y=y, color=value)) + geom_line()
- Assume that all trees are right-circular cylinders and that the height of the tree is proportional to the diameter.
For this model we assume that the height of the tree is proportional to the diameter. From above, we have volume is proportional to diameter squared (\(V \propto d^2\)). In this problem, we are to assume that \(h\) is proportional to the diameter.
\[ V \propto d^2 * h \propto d^3 \]
Therefore the volume of each tree is a proportional to the diameter cubed or \(V \propto d^3\).
\[ y \propto x^3 \]
where \(y = board feet\) and \(x = diameter\)
x <- c(17, 19, 20, 23, 25, 28, 32, 38, 39, 41)
y <- c(19, 25, 32, 57, 71, 113, 123, 252, 259, 294)
data <- data.frame(x, y); data## x y
## 1 17 19
## 2 19 25
## 3 20 32
## 4 23 57
## 5 25 71
## 6 28 113
## 7 32 123
## 8 38 252
## 9 39 259
## 10 41 294\(y/x^3 = k\)
data$ratio <- data$y/data$x^3
k <- mean(data$ratio)
data$predicted <- k*data$x^3\[y = 0.004286668 * x^3 \]
df <- rbind(data.frame(value = 'Actual', x = data$x, y = data$y),
data.frame(value = 'Predicted', x = data$x, y = data$predicted))
ggplot(df, aes(x=x, y=y, color=value)) + geom_line()
- The second model appears to be better. The fit of the model appears to be more accurate than compared to the first.