What follows is a worked example of the application of cost behaviour based on real data. This page assumes that its readers are already familiar with the basics of cost behaviour analysis.
In this case we will be exploring the cost function
y = a + bx
where y = the estimate of total cost
a = fixed cost
b = variable, or marginal, cost
x = output or production or sales
In this case, we should be aware that the value of ‘a’ is incidental: we are not as bothered about it as we would be if we were working on cost per unit of a product or service, for example.
The data given are second hand car values for two different cars; and in the suggested solution to this exercise, I discuss a variety of issues including the different methods we can use to derive the behaviour of the data as well as including some pointers to the potential pitfalls of such an analysis.
Finally, I ask you to work through the second set of data yourself.
Question
The following data are actual second hand prices of two different motor cars. The prices relate to the second hand value of a vehicle in ‘A1’, excellent, condition. From these data, estimate for each vehicle:
§ An annual value for depreciation
§ The method of depreciation you feel is demonstrated by these results
Rolls Royce Corniche | Range Rover | ||
Two door automatic | Five door automatic | ||
Age (years) | A1 Value (£) | Age (years) | A1 Value (£) |
9 | 29,400 | 7 | 5,325 |
8 | 33,150 | 6 | 7,800 |
7 | 35,775 | 5.5 | 8,850 |
6 | 40,750 | 5 | 9,525 |
5.5 | 47,775 | 4.5 | 10,750 |
5 | 56,725 | 4 | 11,400 |
4.5 | 58,400 | 3.5 | 13,525 |
4 | 62,025 | 3 | 14,175 |
3.5 | 65,025 | 2.5 | 14,925 |
3 | 71,675 | 2 | 15,750 |
2.5 | 75,650 | 1.5 | 17,125 |
2 | 85,350 | 1 | 17,925 |
1.5 | 89,200 | 0 | 19,475 |
1 | 94,600 | ||
0.5 | 97,975 | ||
0 | 110,025 | ||
Notes: values taken from Parker’s Car Price Guide
an age zero indicates a car that is second hand but less than six months old
Suggested Solution: Rolls Royce Corniche
This solution relates to the Rolls Royce car only.
Let’s look at three ways of answering this question
§ The scattergraph method
§ The high low method
§ The ordinary least squares method
The Scattergraph Method
However sophisticated the data we are working with or the techniques we might eventually have to use to analyse them, the first step in data analysis is to draw one or more scattergraphs of the data: the graph of the Rolls Royce data follows:
The graph includes a linear line of best fit (LOBF) from which we can determine the slope of the data curve. The slope of the LOBF gives us an estimate of the annual average provision for depreciation. Using this LOBF gives us an estimate of -£9,095 depreciation provision per year: notice this is a negative value, confirming that the value of the car is falling as it gets older. How do we find the value of -£9,095? We find the value of two separate points on the LOBF and found the difference between them: we then work out what that meant in terms of annual depreciation.
Keep it as simple as possible by choosing two points on the LOBF that are exactly a year apart from each other. Reading the relevant values from the LOBF, here’s how:
Year 3 A1 Value = £74,370
Year 4 A1 Value = £65,275
Difference = -£ 9,095
Notice that we represent the difference as negative: because the value of the car is falling year by year. We can see from the graph that it should be negative, too, because the LOBF is falling from left to right, as is the data curve, of course.
We can use this estimate now to find the value of the car for any age of Rolls Royce Corniche: 6 years, 8¼ years and 14 years, for example:
Age (years) | Estimated Value (£) |
6 | 47,088.00 |
8¼ | 26,624.25 |
14 | -25,672.00 |
As a matter of interest, if we’d chosen 6 and 8¼ years to estimate the annual depreciation value from the LOBF, the calculation would have been:
6 years | £47,088.00 |
8¼ years | 26,624.25 |
difference | -£20,463.75 |
Therefore, annual provision = -£20,463.75 ÷ (8¼ – 6 years) = -£9,095
By the way, if you were to draw your own LOBF on this graph, you might find that it is a bit different to the one given here: adding such lines by eye or by inspection is something of a hit and miss affair and such differences are to be expected. Anything too different, though, would be a cause for concern.
The High Low Method
The high low method is a rough and ready method that is best confined to sets of data that behave in a relatively linear manner. The Rolls Royce Corniche data are fortunately relatively linear, so we can use the high low method here.
The high low method works by finding the difference between the two extreme values in the ‘x’ data range, the high and the low values, and then dividing that by the difference between the corresponding ‘y’ values. This gives an estimate of the value of ‘b’ or the slope of the LOBF where
y = a + bx
and in this case ‘b’ represents the annual provision for depreciation.
Reminder: in cost accounting and cost behaviour generally, ‘b’ represents the variable or marginal cost per unit and ‘a’ represents the fixed cost.
Despite being a rough and ready method, the high low method is very strict in terms of how we use it: the high value is the highest ‘x’ value and the low value is the lowest ‘x’ value. Let’s work through the Rolls Royce example:
Independent and Dependent Variables
Which of Age and A1 Value is the ‘x’ variable? Age is the ‘x’, or independent, variable; and Value is the ‘y’ or dependent variable. Age has to be the independent variable because it must be the age of the car that helps to determine its value and not the other way round. We couldn’t say that because the car is worth £100,000 therefore it must be 4 months old: so, value depends on age.
Here are the calculations now
Age (years) x | A1 Value (£) y | |
High | 9 | 29,400 |
Low | 0 | 110,025 |
Difference | 9 | -80,625 |
Therefore ‘b’ = -80,625 ÷ 9 = -8,958.33
This means that the high low estimate of the annual provision for depreciation is £8,958.33.
Putting this on a graph gives us a good idea of how realistic it is to use the high low method in this case. Bearing in mind that the correlation coefficient of r = 0.9833 suggests a high degree of cause and effect between age and value, the high low estimates are not that good:
Apart from where the age of the car is 9 years and 0 years old, the high low estimates are relatively far away from the actual value: we would to see these differences being as small as possible.
Compare this method with the scattergraph method: at least there the differences between the LOBF estimates, and the actual were always relatively small.
Nevertheless, despite the apparent problems with the application of the high low method in this case, it has given us an estimate of the annual provision for depreciation, £8,958.33, and we can use that to derive the kind of table that follows:
Rolls Royce Corniche | |
Two door automatic | |
Age (years) | High Low Estimate of A1 Value (£) |
9 | 29,400 |
8 | 38,358 |
7 | 47,317 |
6 | 56,275 |
… | … |
1.5 | 96,588 |
1 | 101,067 |
0.5 | 105,546 |
0 | 110,025 |
The final method we will look at is the ordinary least squares method: this is the most accurate and reliable of the three methods we will be using in the case. It is reliable and accurate because it uses statistical methods to derive the best possible estimates of ‘b’.
The Ordinary Least Squares Method
To use the ordinary least squares (OLS) method, we input the ‘x’ and ‘y’ variables into the various formulae and it can then tell us the following:
R2 | 0.9669 | |
Coefficients | t statistic | |
‘a’ | 101,658.206 | 47.864051 |
‘b’ | -9,095.7349 | -20.21504 |
This means that y = a +bx = 101,658.206 –9,095.7349x; and for any age we can now calculate an estimate of the value of one of these cars. The t statistics provide us with an indication of the reliability of the OLS results: in this case since the t statistics are very large, we can conclude that our results are reliable.
Graphically, we have:
We can leave the high low estimate on the graph, as we have here, for the purposes of comparison.
Using the LOBF and the regression equation, we are confident, since the R2 value is high at 0.9669, of saying that we can estimate the value of the car at any age. For example:
If the car is exactly 5 years old, we would estimate its value to be
y = 101,658 - 9,095.7x = £101,658 - £9,095.7 * 5 years = £56,179.5
The actual value, from the table of data we were given in the question, is £56,725: a difference between actual and estimate of £545.5. This represents a difference of only 0.9617%
The calculations required to work through the OLS method are relatively complex and are not shown here. Moreover, to carry out all of the work we have done in this example, we can use spreadsheet software. Spreadsheets are programmed to take care of OLS and can easily allow us to prepare high low and scattergraph analyses too: we can even trick Excel into doing all of our high low calculations if we want!
In fact, using Microsoft Excel, we can carry out most of the analysis graphically. Even on a graph, we can use Excel to derive the y = a + bx functions and the R2 calculations. We only need to be able to use Excel and not the underlying functions and formulae. Excel’s Help files do fill in some of the background to its calculations, however, if you need it.
Additional Question
What would you estimate the value of an A1 condition Rolls Royce Corniche to be that is
6 years old?
8¼ years old?
14 years old?
Check your answers with the solutions to be found at the end of this page.
The Depreciation Method
By using the scattergraph, high low and OLS methods, we have assumed that the straight line method is the most appropriate method to use in this situation. Take a look at the section A Final Thought, below, for further thoughts on this, however.
Conclusions
Using three different methods, we have obtained three different estimates of the value of the annual depreciation provision for a Rolls Royce Corniche motor car: which is correct?
Statistically speaking, the OLS method is the most correct in all respects. Moreover, the OLS method, if we use it in full, will also tell us how reliable its results are: the value of the R2 must be taken into account, as must the t statistics.
Nevertheless, if we are in a situation where we have some data and don’t have time or access to a computer or calculator, we might want to plot a chart and work through the scattergraph and high low methods: they will give us a good idea, at least between them, of the results we are looking for.
The scattergraph method is an especially powerful tool since most of us could plot an LOBF by eye or by inspection that wouldn’t be too bad.
Whether we should use the high low method in any given situation really depends: if the R2 value is high, then the high low method can give us a reasonable estimate. If the R2 value is low, forget the high low method, even as a rough and ready estimate.
A Final Thought
Take the OLS results and calculate the value of the Rolls Royce Corniche at the age of 11 years and then again at the age of 12 years: interesting? The results are £1,605.3 and £(7,490.4) respectively.
This means that you should be able to buy a Rolls Royce Corniche in excellent condition that is 11 years old for just £1,600; and if you can find a 12 year old Rolls Royce Corniche in excellent condition, its owner will GIVE YOU £7,490.4 to take it away!
Not very realistic is it? Two points emerge from this
§ Relevant range
§ Reliability of the model
Relevant range: this simply tells us that providing we apply our OLS model within the age range of 0 to 9 years, we can confidently use estimates. If we try to apply the model to, say 11 years, 12 years or 14 years, the results may be problematic, as we saw with the 12 year old calculations.
Reliability of the model: statistically, our OLS model is good. However, since we might want to estimate the value of a 12 year old or 14 year old Rolls Royce Corniche, we should look for a model that will allow us to do that. Take a look at the following graph and see what happens when we use the model y = 108,348 + 563.16x2 – 13,982x.
For a 12 year old Rolls Royce Corniche, we now estimate the value to be £21,659.04 and for a 14 year old Rolls Royce Corniche, the value is estimated at £22,979.36. Of course, the new model gives us new values for ALL ages of car, not just for 12 and 14 years. For a 20 year old car, the value is estimated as £53,972.
Is the new model better? Well, yes, in the sense that it prevents the value of a car in excellent condition from being written down to and beyond zero. Moreover, since we are talking about a Rolls Royce Corniche, the chances are, as we see with this new statistical model, that after a certain age its value begins to RISE: appreciation not depreciation. Isn’t this more realistic in this case, too?
Finally, this section suggests that the straight line method of depreciation is not appropriate. Moreover, a depreciation method using a power function is suggested as being more appropriate.
Solution To The OLS Additional Question
Using the equation y = 101,658 – 9,095.7x:
Age (years) | Estimated Value (£) |
6 | 47,084 |
8¼ | 26,618 |
14 | (25,682) |
What is the meaning of a negative value for a 14 year old car?: refer to the section A Final Thought for a discussion on this point.
Suggested Solution: Range Rover
The solution to this part of the question is not given here, although to start you off, here is the first graph that you might draw when working through these data:
Notice the way the data behave more in a linear way than does the Rolls Royce data.
However, work through the data in the same way that we have done for the Rolls Royce and consider the appropriacy of a power function model for the Range Rover, too.
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