Linear Equations in Data Analysis Assignment/Examples and Questions
Central to algebra is the use of letters, most frequently x and y, to represent numbers. Doing this allows us to deal systematically with quantities that are unknown yet of importance in an analysis. These unknown quantities are often referred to as variables, literally things that vary over a range of numbers or values. Sometimes the point is to express a quantitative procedure in a succinct way, and the use of letters merely constitutes convenient shorthand.
A sales agent is paid a basic wage of £200 per week plus 10% commission on sales. The procedure for working out his/her total wage could be written as: 4
Total wage = 200 + 10% of sales
It is often more useful to abbreviate this by using letters. If y is used to represent the total wage and x to represent sales we can express the procedure as:
y = 200 + 0.1x
Using this we can find the total wage for a week when sales were £1200:
y = 200 + 0.1 * 1200 = 200 + 120 = £320
Linear Equations in Data Analysis Example 2
The sales agent in Example1, currently receiving a wage of £200 plus 10% commission on sales, is offered the alternative of receiving 20% commission on sales with no basic wage. What is the minimum level of sales the agent would have to reach to make the alternative commission-only wage attractive?
The existing arrangement can be represented as:
y = 200 + 0.1x
where y represents the wage and x the sales.
The alternative can be expressed as:
y = 0 + 0.2x
Both equations are plotted in Figure 1
Figure 1: The lines of the equations in Example 2
In Figure 1 the lines cross at the point representing sales of £2000 and a wage of £400. The line representing the current method of determining the wage is the higher line when sales are below £2000, indicating that it would be the better arrangement for the agent when sales are less than £2000. The line representing the alternative arrangement is the higher line when sales are greater than £2000, indicating that it would be the better arrangement when sales exceed £2000. The minimum level of sales the agent would have to reach to make the alternative commission-only wage attractive is therefore £2000.
You can solve simultaneous equations without plotting their lines on a graph using a method known as elimination. As the name implies this involves removing or eliminating one of the unknown quantities with the intention of leaving a numerical value for the other.
In order to end up with a clear result conventionally simultaneous equations are arranged so that the unknown quantities and the factors applied to them, their coefficients, are located on the left hand side of the equals sign and the intercept, or constant appears to its right. This may involve rearranging the equations. In doing this we need to ensure that any manipulation preserves the equality inherent in the equation by balancing every operation performed on one side with the exact same operation on the other side.
Following this we proceed to the removal of one of the unknown quantities, either x or y. This is straightforward if the number of x’s or y’s in both equations is the same, in which case if you subtract one equation from the other you will be left with an expression which has just one unknown on the left of the equals sign and a number on its right. Using suitable multiplication or division you can then establish the value of this remaining unknown, one of the pair of x and y values that fits both equations.
Having found one of the pair of values mutually compatible with both equations you can substitute it into one of the original equations and find the value of the other unknown that, in combination with the value of the first unknown, satisfies the original equations jointly, or simultaneously.
Linear Equations in Data Analysis Example 3
In Example 1 the sales agent is presented with two possible wage arrangements represented as:
y = 200 + 0.1x (the original arrangement)
y = 0 + 0.2x (the alternative arrangement)
where y represents the wage and x the sales.
We will start by rearranging both equations so the components, or terms, involving x and y are on the left of the equals sign and the ‘stand-alone’ numbers are on the right. In the case of the first equation, representing the original arrangement, this entails moving the 0.1x from the right of the equals sign over to the left. In doing this we have to reverse the sign in front of it, as strictly speaking we are subtracting 0.1x from both sides of the equation, and hence preserving the balance:
y = 200 + 0.1x
Subtract 0.1x from both sides:
y – 0.1x = 200 + 0.1x – 0.1x
y – 0.1x = 200
For the second equation:
y = 0 + 0.2x
Subtract 0.2x from both sides:
y – 0.2x = 0 + 0.2x – 0.2x
y – 0.2x = 0
We can now set these rearranged equations alongside each another and subtract one from the other to eliminate y:
Y – 0.1x = 200
Y – 0.2x = 0
+0.1x = 200
This tells us that one-tenth of x is 200.
If we multiply both sides of this by ten we find that a ‘whole’ x is worth 2000:
0.1x * 10 = 200 * 10 so 1x = 2000
In other words both wage determination models produce the same wage when sales are £2000.
But what will the wage be? To find this put the sales figure of 2000 into the equation representing the original arrangement:
y = 200 + 0.1 * 2000 = 200 + 200 = 400
The original approach to establishing the sales agent’s wage will produce a wage of £400 when sales are £2000. The alternative, commission-only formulation will of course yield the same wage when sales are £2000:
y = 0 + 0.2 * 2000 = 400
The values of 2000 and 400 for sales and wages respectively therefore satisfy both wage determination equations simultaneously. Applying elimination in Example 3 was made easier because in both equations there was only one ‘y’, that is the coefficient on y in each equation was one. If the coefficients on an unknown are different you have to apply multiplication or division to one or both equations to make the coefficients on the unknown you wish to eliminate equal before you can use subtraction to remove it.
Linear Equations in Data Analysis Example 4
Find the level of wages at which the two procedures for determining the sales agent’s wage in Example 2 result in the same wage by eliminating x, the level of sales. The equations representing the procedure, as rearranged in Example 3, are:
y = 200 + 0.1x (the original arrangement)
y = 0 + 0.2x (the alternative arrangement)
If we multiply the first equation by two we get:
2y – 0.2x = 400
Subtracting the second equation from this:
2Y – 0.2x = 400
Y – 0.2x = 0
Y = 400
Again we find that the wage level at which the two wage determination models produce the same result is £400. If we substitute this value of y into the equation representing the original arrangement we can find the level of sales that will yield a wage of £400:
400 – 0.1x = 200
Subtract 400 from both sides:
400 – 400 – 0.1x = 200 – 400
-0.1x = -200
Multiply both sides by minus one:
(-1) *-0.1x = -200 * (-1)
0.1x = 200
Multiply both sides by ten:
x = 2000
The level of sales at which both approaches to wage determination will produce a wage of £400 is therefore £2000.
Not all pairs of equations can be solved simultaneously. These are either cases where one equation is a multiple of another, such as: 3x + 2y = 10 and 6x + 4y = 20 or cases where one equation is inconsistent with the other, such as: 2x + y = 14 and 2x + y = 20 In the first case the equations are the same; if you try to plot them you will find they produce the same line. In the second case plotting them produces lines that are parallel and therefore do not cross.
The type of linear model that we have looked at in the previous section can be used to analyze
In setting up a break-even analysis we need to make several definitions and assumptions. First we assume that there are two types of cost, fixed and variable. Fixed costs, as the name implies, are those costs that are constant whatever the level of production. These might be the costs of setting up the operation such as the purchase of machinery as well as expenses, such as business rates, that do not vary with the level of output. Variable costs on the other hand are costs that change in relation to the amount produced, such as the costs of raw materials and labor. We can define the total costs (TC) as the sum of the total fixed costs (TFC) and the total variable costs (TVC):
TC = TFC + TVC
The total variable costs depend on the quantity of output. We will assume that the variable cost of producing an extra unit is the same however many units we produce; in other words, it is linear or varies in a straight line with the amount produced. We can therefore express the total variable cost as the variable cost per unit produced, known as the average variable cost (AVC) multiplied by the quantity produced (Q), so the total cost is:
TC = TFC + AVC * Q
The total revenue (TR) is the price per unit (P) at which the output is sold multiplied by the quantity of output (Q):
TR = P * Q
Once we have defined the total cost and total revenue equations we can plot them on a graph and look at exactly how total revenue compares to total cost. This is a key comparison as the total revenue minus the total cost is the amount of profit made:
Profit = TR – TC
The point at which the lines representing the two equations cross is the point at which total cost is precisely equal to total revenue, the break-even point.
Linear Equations in Data Analysis Example 5
The Ackrana Security Company intends to manufacture video security cameras. The costs of acquiring the necessary plant and machinery and meeting other fixed costs are put at £4.5 million. The average variable cost of producing one of their cameras is estimated to be £60 and the company plans to sell them at £150 each. How many will they need to produce and sell in order to break even?
Total cost, TC = 4,500,000 + 60Q
Total Revenue. TR = 150Q
These equations are plotted in Figure 2. Conventionally the money amounts, cost and revenue are plotted on the vertical or y axis and the output is plotted on the horizontal or x axis. This arrangement reflects the assumption that the money amounts depend on the output and makes it easier to interpret the diagram.
Linear Equations in Data Analysis Figure 2
Total cost and total revenue lines in Example 5
In Figure 5 the steeper line that starts from the origin represents the total revenue equation and the other line represents the total cost equation. You can see that the lines cross when output is about 50,000 units. At this level of production both the total cost and total revenue are equal, at about £7.5 million. This is the break-even point, at which costs precisely match revenues.
We can verify the break-even point by solving the total cost and total revenue equations simultaneously:
Total cost, TC = 4,500,000 + 60Q so TC-60Q = 4,500,000
Total Revenue. TR = 150Q so TR – 150Q = 0
When total cost and total revenue are equal, subtracting one from the other will leave us with an expression in which the only unknown is the level of output, Q:
TC- 60Q = 4,500,000
TR – 150Q = 0
+90Q = 4,500,000
Dividing both sides by 90 means that the level of output at which total cost and total revenue are equal is 50,000:
4500000/90 = 50000
The total cost and total revenue when 50000 units are produced will be:
TC = 4,500,000 + 60*50,000 = 4,500,000 + 3,000,000 = 7,500,000
TR = 150 * 50,000 = 7,500,000
Break-even analysis can be extended to illustrate the levels of output that will yield a loss and those that will yield a profit. A level of output less than the break-even level, and hence to the left of the position of the break-even point along the horizontal axis of the graph, will result in a loss. A level of output higher than the break-even level, to the right of the break-even point on the horizontal axis, will yield a profit.
At any point to the left of the break-even point the total cost line is the higher line indicating that total cost is higher than total revenue; the greater the difference between the two lines, the larger the loss. At any point to the right of the break-even point the total revenue is the higher line, which means that the total revenue is higher than the total cost; the bigger the difference between the two lines, the larger the profit. The areas representing loss and profit are shown in Figure 3.
Linear Equations in Data Analysis Figure 3
Break-even graph for Example 5 with areas representing profit and loss
Using Figure 3 you can establish how much profit or loss will be achieved at a particular level of production. If for instance production were 30,000 units the graph suggests that the total cost would be £6.3 million and the total revenue would be £4.5 million resulting in a loss of £1.8 million.
We expect that a company would seek to operate at a level of production at which they would make a profit. The difference between the output they intend to produce, their budgeted output, and the break-even level of output is their safety margin. This can be expressed as a percentage of the budgeted output to give a measure of the extent to which they can fall short of their budgeted output before making a loss.
Linear Equations in Data Analysis Example 6
If the Ackrana Security Company in Example 5 aims to produce 80,000 cameras what profit should they expect and what is their safety margin?
TR = 150 * 80,000 = 12,000,000
TC = 4,500,000 + 60*80,000 = 9,300,000
Profit = 12,000,000 – 9,300,000 = 2,700,000
In Example 5 we found that their break-even point was 50,000 cameras so their safety margin is:
Budgeted output – break-even output * 100 = 80000-50000*100 = 37.5%
Budgeted output 80000
The break-even analysis we have considered is the simplest case, where both costs and revenue are assumed to be linear, that is to form straight lines when plotted graphically. In practice companies might find that with greater levels of production come economies of scale that mean their variable cost per unit is not constant for every unit produced but falls as output increases. Furthermore, they may have to reduce their price if they want to sell more products so that their total revenue would not have a linear relationship to their output level.
Despite these shortcomings the basic model can be a useful guide to the consequences of relatively modest changes in output as well as a framework for considering different levels of initial investment, pricing strategies and alternative sources of raw materials.
Linear Equations in Data Analysis Questions:
- The current system for allocating budgets for the local depots of a national office-cleaning company gives each depot a fixed sum of £35,000 plus an extra £500 for each corporate client in the area the depot covers.
(a) Express the budget allocation model as an equation.
(b) Use your equation from (a) to work out the budget allocations for the following depots:
(i) Ashford, which has 43 corporate clients
(ii) Byfleet, which has 29 corporate clients
(iii) Croydon, which has 66 corporate clients
(c) A new accountant at the company head office wants to alter the budget allocation model by reducing the fixed sum to £20,000 and increasing to £800 the extra for each corporate client. What changes will these alterations mean for the depots in (b)?
- A domestic consumer is offered two alternative methods of paying for her annual water bill. The first option is to make 8 payments of £32 at certain points over the forthcoming year. The second is to have her water metered and pay £0.08 for each cubic metre of water used plus an annual standing charge of £20. What is the maximum amount of water she would have to use over the year to make it economical for her to choose the second payment method?
- Following a crash, the owner of the Skorrost Courier Service has to purchase a new van. The make and model have already been decided, but there is a choice between petrol and diesel versions. The petrol version costs £10,000 to buy and will incur fuel costs of £0.12 per mile. The diesel version costs £11,000 and will incur fuel costs of £0.08 per mile. What is the minimum number of miles that the vehicle must travel to make purchasing the diesel version the more economical choice?
4 The Pasuda Porcelain Company is about to launch a new luxury tableware range. The selling price for a set will be £90. To make the range the company has invested £319,000 in new equipment. Variable production costs will be £35 per set.
(a) What number of sets must they sell to break even?
(b) What profit will they make if they sell 6000 sets, and what is their margin of safety?
- Samocat Automotive want to move the production of their motor scooters to a purpose-built new plant that will cost £24 m to construct. The scooters sell for £1850 and variable costs of production amount to £1100 per scooter.
(a) What is the break-even point for scooter production at the new plant?
(b) How will the break-even point change if the costs of the new plant rise to £30 m?
- A plant hire company purchases a mobile crane for £120,000. It plans to charge £190 a day for hiring out the crane. For each day on which it is hired out the company expects to incur variable costs of £40.
(a) Work out how many days the company needs to hire out the crane in order to break even.
(b) If the variable costs are actually £55 and the hire charge has to be reduced to £180 per day, what is the new breakeven point?
- Holly buys a hot dog stall for £360. In addition she has to pay a fixed charge of £200 to the local council to secure her pitch. She plans to sell her hot dogs at £1 each and the cost of preparing each one is £0.60.
(a) How many hot dogs will she need to sell to break even?
(b) She finds a supplier that can provide cheaper hot dog sausages, enabling her to save 10 p per hot dog on the preparation cost. What effect does this have on the break-even point?
- Volna Appliances want to take over a competitor that makes microwave ovens. They have bid £48 m for the company and expect their offer to be accepted. The microwave ovens are sold for £85 and variable costs amount to £25 per unit.
(a) Work out the break-even point for microwave production.
(b) Volna plan to introduce a rationalization plan that will reduce variable costs by £6 and a price reduction of £16 to increase market share. What is the new break-even point?
Content represents a modification of Chapter 2, sections 2.2 and 2.3 of Buglear, John. Quantitative Methods for Business (pp. 74-76). Taylor and Francis.