Solution

Question: What will be the total maintenance cost for a truck?

List the variables that affect the behavior you have identified.

The variables that affect the behavior from the given sentence are:

[MAINTENANCE_COST, YEAR_BUILD, TRUCK_MILAGE]

Also, we could consider some other variables that could affect the maintenance costs, such as:

[DRIVER_EXPERTISE, TRAFFIC_CONDITIONS, ROAD_CONDITIONS, ROUTE_HAS_STEEP_HILLS, TRUCK_TYPE, TRUCK_WEIGHT, NUMBER_OF_GEARS, NUMBER_OF_AXES, OIL_CHANGE_FREQUENCY, BREAKING_FREQUENCY, TIRE_CONDITION, N_DAYS_FROM_LAST_MAINTENANCE, MAINTENANCE_PARTS_COST, MAINTENANCE_JOB_DESCRIPTION, MAINTENANCE_LABOR_COST, N_HOURS_MAINTENANCE_LABOR]

Which variables will be neglected completely?

I consider that the following variables will be neglected.

[DRIVER_GENDER, DRIVER_AGE, MAX_LOAD]

Which might be considered as constants initially?

The variables that I would consider as constants in an initial form will be:

[DRIVER_EXPERTISE, TRAFFIC_CONDITIONS, ROAD_CONDITIONS, OIL_CHANGE_FREQUENCY, MAINTENANCE_LABOR_COST, N_HOURS_MAINTENANCE_LABOR]

Can you identify any submodels you would want to study in detail?

I believe that we could construct some sub models as follows:

BREAKING_FREQUENCY = f(ROAD_CONDITIONS, ROUTE_HAS_STEEP_HILLS, NUMBER_OF_GEARS)

OIL_CHANGE_FREQUENCY = f(YEAR_BUILD, TRUCK_MILAGE, NUMBER_OF_GEARS)

TIRE_CONDITION = f(DRIVER_EXPERTISE, BREAKING_FREQUENCY, ROAD_CONDITIONS, ROUTE_HAS_STEEP_HILLS, NUMBER_OF_AXES)

MAINTENANCE_LABOR_COST = f(N_HOURS_MAINTENANCE_LABOR)

Identify any data you would want collected.

I would like to collect:

[YEAR_BUILD, TRUCK_MILAGE, DRIVER_EXPERTISE, TRAFFIC_CONDITIONS, ROAD_CONDITIONS, ROUTE_HAS_STEEP_HILLS, TRUCK_TYPE, TRUCK_WEIGHT, NUMBER_OF_GEARS, NUMBER_OF_AXES, N_DAYS_FROM_LAST_MAINTENANCE, MAINTENANCE_PARTS_COST, MAINTENANCE_JOB_DESCRIPTION, MAINTENANCE_LABOR_COST, N_HOURS_MAINTENANCE_LABOR]

Solution

In order to support a proportionality model; it must satisfy as follows:

\(y = kx^3\) for a constant \(k\).

Let’s plot our given data as it is:

Now, let’s calculate \(x^3\).

y	x	x^3
0	1	1
1	2	8
2	3	27
6	4	64
14	5	125
24	6	216
37	7	343
58	8	512
82	9	729
114	10	1000

Now, let’s plot again and see it’s behavior.

As we can see, there now seems to be some sort of linearity; hence we could calculate our slope as follows:

\(k = \frac{114 - 0}{1000 - 1} = \frac{114}{999}\)

From here, we could deduce that our model will be something similar to:

\(y = \frac{114}{999} x^3\)

Let’s plot our linear model and see if it actually follows the points.

From the above visual it seems that our linear model follows all points very accurate.

Now, I will find the predicted \(y\) value rounded to two decimals and compare them to our given data in order to obtain an Error difference.

Error shows no decimals, but calculations made to two decimals.

y	x	x^3	predicted y	Error
0	1	1	0	0
1	2	8	1	-0
2	3	27	3	1
6	4	64	7	1
14	5	125	14	0
24	6	216	25	1
37	7	343	39	2
58	8	512	58	0
82	9	729	83	1
114	10	1000	114	0

From the above graph, we can find out that our maximum error difference is about 2.14.