Quantitative Methods for Business chapter 2 doc

CHAPTER
Getting linear models
straight
2
Chapter objectives
This chapter will help you to:
■ plot and solve linear equations
■ apply basic break-even analysis
■ interpret inequalities
■ undertake simple linear programming using graphs
■ use the technology: Solver in EXCEL
■ become acquainted with business uses of linear programming
This chapter is intended to introduce you to the use of algebra in solv-
ing business problems. For some people the very word algebra conjures
up impressions of abstract and impenetrable jumbles of letters and
numbers that are the preserve of mathematical boffins. Certainly parts
of the subject of algebra are complex, but our concern here is with alge-
braic techniques that help to represent or model business situations.
In doing this we are following in the footsteps of the ‘father of
algebra’, Mohammed ibn-Musa al-Khwarizmi. In the ninth century
al-Khwarizmi wrote Al-jabr wa’l-muqabala, which might be translated as
‘Calculation Using Balancing and Completion’. The first part of the
Arabic title gives us the word algebra. Although al-Khwarizmi was a
scholar working at the House of Wisdom in Baghdad, he saw his task in
very practical terms, namely to focus on
… what is easiest and most useful in arithmetic, such as men constantly
require in cases of inheritance, legacies, partitions, law-suits, and trade,
Chapter 2 Getting linear models straight 33
and in all their dealings with one another… (cited in Boyer,
1968, p. 252)
In the course of this chapter we will confine our attention to simple

algebra and how you can use it to solve certain types of business prob-
lem. We will focus on linear equations, which are those that are straight
lines when they are plotted graphically. They form the basis of linear
models that assist business problem-solving.
2.1 Linear equations
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.
These types of expression are called equations because of the equals
sign, ‘ϭ’, which symbolizes equality between the quantity to its left and
the quantity to its right. An equation is literally a state of equating or
being equal.
An equation that involves just two unknown quantities can be drawn
as a line or a curve on a graph. Each point along it represents a com-
bination of x and y values that satisfies, or fits the equation.
To plot an equation start by setting out a scale of possible values of one
unknown along one axis, or dimension, and a scale of the possible values
Example 2.1
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:
Total wage ϭ 200 ϩ 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 ϭ 200 ϩ 0.1x
Using this we can find the total wage for a week when sales were £1200:
y ϭ 200 ϩ 0.1 * 1200 ϭ 200 ϩ 120 ϭ £320

of the other unknown along the other axis. Ensure the scales cover the
range of plausible values, and start them at zero unless interest in the line
is limited to part of it well away from zero. Plot the x values along the hor-
izontal axis, known as the x axis, and the y values along the vertical axis,
known as the y axis. This conveys that y depends on x.
Once each axis has been prepared, portraying an equation in its
graphical form involves finding two points that lay along the line that
will represent the equation. This means you have to identify two pairs of
x and y values both of which satisfy the equation. The way to do this is to
specify an x value and use the equation to work out what value y would
have to take in order to satisfy the equation, then repeat the process for
another x value. To ensure that your line is accurate it is important to
take one x value from the far left hand side of the horizontal axis and
the other from the far right hand side.
34 Quantitative methods for business Chapter 2
Example 2.2
Plot the equation that represents the procedure for working out the weekly wage for
the sales agent in Example 2.1.
The equation is:
y ϭ 200 ϩ 0.1x
where y represents the wage and x the sales.
To help us design the x axis let us suppose that the maximum sales the agent could
achieve in a week is £5000. Using the equation we can use this to find the maximum wage:
y ϭ 200 ϩ 0.1(5000) ϭ 700
This means the highest value we need to include in the scale on the y axis is £700.
We are now in a position to construct the framework for our graph, which might look
like Figure 2.1.
To plot the line that represents the equation we need to find two points that lie on
the line. One of these should be on the left hand side. The lowest number on the left
of the horizontal axis is zero, so we could use the equation to work out the wage when

sales are zero:
y ϭ 200 ϩ 0.1(0) ϭ 200
When sales are £0 the wage is £200, this pair of values gives us the position, or coord-
inates of one point on the line. The sales value, 0, positions the point along the hori-
zontal axis and the wage value, 200, positions the point along the vertical axis.
To get a second set of coordinates we should take a sales figure from the right hand
side of the horizontal axis, say the maximum figure of 5000, and work out the wage
when sales are £5000, again using the equation:
y ϭ 200 ϩ 0.1(5000) ϭ 700
Chapter 2 Getting linear models straight 35
When sales are £5000 the wage is £700. The point we plot to represent this pair of val-
ues will be positioned at 5000 along the horizontal axis and at 700 along the vertical axis.
We can now plot both points as in Figure 2.2:
If plotting an equation is new to you, or just something you haven’t done for a while,
it is a good idea to plot a third point between the first two. A third point should lie in
line with the first two so it is a good way of checking that you have plotted the other
points correctly. A suitable position for our third point in this case might be when sales
0 1000 2000 3000 4000 5000
0
100
200
300
400
500
600
700
x (sales in £)
y (wage in £)
500040003000200010000
700

600
500
400
300
200
100
0
x (sales in £)
y (wage in £)
Figure 2.1
Framework for plotting the equation in Example 2.1
Figure 2.2
Points on the line of the equation in Example 2.1
36 Quantitative methods for business Chapter 2
are £2000 and the wage will be:
y ϭ 200 ϩ 0.1(2000) ϭ 400
The point that represents these coordinates, sales of £2000 and a wage of £400, has
been plotted in Figure 2.3.
The final stage in plotting the equation is to draw a straight line linking the plotted
points. This is shown in Figure 2.4:
0 1000 2000 3000 4000 5000
0
100
200
300
400
500
600
700
x (sales in £)

y (wage in £)
Figure 2.3
Three points on the line of the equation in Example 2.1
0
1000 2000 3000 4000 5000
0
100
200
300
400
500
600
700
x (sales in £)
y (wage in £)
Figure 2.4
The line of the equation in Example 2.1
Lines that represent simple linear equations, such as the one plotted
in Figure 2.4, have two defining characteristics: a starting point, or inter-
cept, and a direction, or slope. We can think of the intercept as specifying
the point the line begins, and the slope as specifying the way in which
the line travels. In Figure 2.4 the line begins at 200, when sales are zero,
and travels upwards at a rate of 0.1 for every one-unit increase in sales,
reflecting the fact that the sales agent receives an extra £0.10 for every
additional £1 of sales.
Different lines will have different intercepts and slopes. It will help
you interpret results of this type of analysis if you can associate basic
types of intercept and slope in linear equations with their plotted forms.
To illustrate this we can extend the sales agent example to include
contrasting approaches to wage determination.

Chapter 2 Getting linear models straight 37
Example 2.3
Suppose the basic wage of the sales agent in Example 2.1 is increased to £300 and the
commission on sales remains 10%. Express the procedure for determining the wage as
an equation and plot it.
The total wage (y) in terms of sales (x) is now:
y ϭ 300 ϩ 0.1x
The line representing this has an intercept of 300 and a slope of 0.1. It is the upper
line in Figure 2.5.
0
200
400
600
800
1000
0 2000 4000 6000
x (sales in £)
y (wage in £)
Figure 2.5
The lines of the equations in Examples 2.1 and 2.3
You can see two lines plotted in Figure 2.5. The lower is the line plot-
ted in Figure 2.4, the original formulation for finding the wage. The
upper represents the equation from Example 2.3, where the basic
wage is increased to £300. It is higher because the intercept is 300 com-
pared to the 200 in the original equation but note that the two lines
are parallel since they have exactly the same slope. Lines that have the
same slope will be parallel whatever their intercept.
The bottom line in Figure 2.6 represents the equation from Example
2.4. It starts from the point where both wage and sales are zero, known
as the origin, since the intercept of the line is zero. It is parallel to the

lines above it because it has the same slope as them.
38 Quantitative methods for business Chapter 2
Example 2.4
Identify the equation that would express the calculation of the wages of the
sales agent in Example 2.1 if there were no basic wage and the commission rate
remained 10%.
The total wage would be:
y ϭ 0 ϩ 0.1x
This is plotted in Figure 2.6 together with the equations from Examples 2.1 and 2.3.
0
100
200
300
400
500
600
700
800
900
0 2000 4000 6000
x (sales in £)
y (wage in £)
Figure 2.6
The lines of the equations in Examples 2.1, 2.3 and 2.4
The equations plotted in Figure 2.7 have the same intercept, 200, but
different slopes. This means they start at the same point on the left
hand side of the graph but their paths diverge. The upper, steeper line
represents the equation from Example 2.5. It has a slope of 0.2, twice
the slope of the line representing the equation in Example 2.1, 0.1,
reflecting the greater rate at which commission is earned, 20% rather

than 10%. The slope is twice as steep since the same sales will result in
the sales agent earning double the commission.
Chapter 2 Getting linear models straight 39
Example 2.5
The basic wage of the sales agent in Example 2.1 is to remain at £200, but the rate of
commission increases to 20%. Express the procedure for determining the wage as an
equation and plot it.
The total wage (y) in terms of sales (x) is now:
y ϭ 200 ϩ 0.2x
The line representing this has an intercept of 200 and a slope of 0.2. It is plotted in
Figure 2.7 together with the equation from Example 2.1.
0
100
200
300
400
500
600
700
800
900
0 2000 4000 6000
x (sales in £)
y (wage in £)
Figure 2.7
The lines of the equations in Examples 2.1 and 2.5
Example 2.6
Identify the equation that would express the calculation of the wages of the sales agent
in Example 2.1 if the basic wage is £200 and there is no commission.
The equation in Example 2.6 is plotted as the bottom, horizontal line

in Figure 2.8. It has a zero slope; literally it goes neither up nor down.
Whatever the level of sales the wage will be unaffected.
The slopes in the equations we have looked at so far have been
upward, or positive, and in the case of Example 2.6, zero. You will also
come across equations that have negative, or downward slopes.
40 Quantitative methods for business Chapter 2
The total wage would be:
y ϭ 200 ϩ 0x
This is plotted in Figure 2.8 together with the equation from Example 2.1.
0
200
400
600
800
0 2000 4000 6000
x (sales in £)
y (wage in £)
Figure 2.8
The lines of the equation in Examples 2.1 and 2.6
Example 2.7
The company that employs the sales agent in Example 2.1 believes that its sales vary
according to the price charged for its product. They summarize the relationship in the
form of the following equation:
y ϭ 800 Ϫ 10x
where y represents the number of units sold and x the price at which they are sold in £.
The equation is plotted in Figure 2.9.
Chapter 2 Getting linear models straight 41
0
200
400

600
800
1000
0 20 40 60 80 100
x (price in £)
y (units sold)
Figure 2.9
The line of the equation in Example 2.7
In Figure 2.9 the line slopes downwards because the slope, Ϫ10, is neg-
ative. It means that for every increase of £1 in the price of the product
the number of units sold will decrease by ten.
At this point you may find it useful to try Review Questions 2.1 and
2.2 at the end of the chapter.
2.2 Simultaneous equations
In the previous section we looked at how linear equations can be used
to show the connection between two variables. Such equations repre-
sent the relationship in general terms; they are in effect recipes or for-
mulae that specify how the value of one quantity can be established
with reference to another quantity. This is how a wage that consists of
a basic component plus sales commission can be calculated or how a
phone bill made up of a fixed charge plus a cost per unit can be
worked out. In each case a single linear equation provides a clear
numerical definition of the process involved and can be used to work
out the appropriate y value for any given x value.
Sometimes it is necessary to consider two linear equations jointly, or
simultaneously, hence the fact that such combinations of equations are
known as simultaneous equations. Typically the aim is to find a pair of
specific values of x and y that satisfy both equations. You can achieve
this by plotting both equations on the same pair of axes and identifying
the point where the lines cross.

42 Quantitative methods for business Chapter 2
Finding values that fit both of two equations is known as solving
simultaneous equations. In Example 2.8 the point where the lines
Example 2.8
The sales agent in Example 2.1, currently receiving a wage of £200 plus 10% commis-
sion 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 ϭ 200 ϩ 0.1x
where y represents the wage and x the sales.
The alternative can be expressed as:
y ϭ 0 ϩ 0.2x
Both equations are plotted in Figure 2.10.
In Figure 2.10 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 arrange-
ment 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.
0
100
200
300
400
500
600
700

800
0 2000 4000 6000
x (sales in £)
y (wage in £)
Figure 2.10
The lines of the equations in Example 2.8
Chapter 2 Getting linear models straight 43
cross meets the requirements of both equations because it is a point on
both lines. The fact that it is the only point where the lines cross tells us
that it represents the only combination of wage level and sales that fits
both equations.
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.
Example 2.9
In Example 2.8 the sales agent is presented with two possible wage arrangements rep-
resented as:
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
y ϭ 200 ϩ 0.1x (the original arrangement)
and y ϭ 0 ϩ 0.2x (the commission-only alternative)
44 Quantitative methods for business Chapter 2
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 ϭ 200 ϩ 0.1x
Subtract 0.1x from both sides:
y Ϫ 0.1x ϭ 200 ϩ 0.1x Ϫ 0.1x
to get:
y Ϫ 0.1x ϭ 200
For the second equation:
y ϭ 0 ϩ 0.2x
Subtract 0.2x from both sides:
y Ϫ 0.2x ϭ 0 ϩ 0.2x Ϫ 0.2x
to get:
y Ϫ 0.2x ϭ 0

We can now set these rearranged equations alongside each another and subtract one
from the other to eliminate y:
We can break this operation down into three parts:
giving us:
0.1x ϭ 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 ϭ 200 * 10 so 1x ϭ 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 ϭ 200 ϩ 0.1 * 2000 ϭ 200 ϩ 200 ϭ 400
y Ϫ y ϭ 0y no y
Ϫ0.1x Ϫ (Ϫ0.2x) ϭϩ0.1x
200 Ϫ 0 ϭ 200
yx
yx
0.1 200
0.2 0
0.1 200
Ϫϭ
Ϫϭ
ϩϭx
Chapter 2 Getting linear models straight 45
Applying elimination in Example 2.9 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.

Example 2.10
Find the level of wages at which the two procedures for determining the sales agent’s
wage in Example 2.8 result in the same wage by eliminating x, the level of sales.
The equations representing the procedure, as rearranged in Example 2.9, are:
y Ϫ 0.1x ϭ 200
y Ϫ 0.2x ϭ 0
If we multiply the first equation by two we get:
2y Ϫ 0.2x ϭ 400
Subtracting the second equation from this:
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 Ϫ 0.1x ϭ 200
Subtract 400 from both sides:
400 Ϫ 400 Ϫ 0.1x ϭ 200 Ϫ 400
Ϫ0.1x ϭϪ200
2 0.2 400
0.2 0
400
yx
yx
y
Ϫϭ
Ϫϭ
ϭ
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 ϭ 0 ϩ 0.2 * 2000 ϭ 400
The values of 2000 and 400 for sales and wages respectively therefore satisfy both

wage determination equations simultaneously.
46 Quantitative methods for business Chapter 2
Multiply both sides by minus one:
(Ϫ1) * (Ϫ0.1x) ϭ (Ϫ1) * (Ϫ200)
0.1x ϭ 200
Multiply both sides by ten:
x ϭ 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 ϩ 2y ϭ 10 and 6x ϩ 4y ϭ 20
or cases where one equation is inconsistent with the other, such as:
2x ϩ y ϭ 14 and 2x ϩ y ϭ 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.
At this point you may find it useful to try Review Question 2.3 at the
end of the chapter.
2.3 Break-even analysis
The type of linear model that we have looked at in the previous section
can be used to analyse the relationship between the costs and revenue
of a company. The aim in doing this is to identify the point at which
the revenue matches the costs, known as the break-even point, the out-
put level at which the company makes neither profit nor loss but
breaks even.
In setting up a break-even analysis we need to make several defin-
itions 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 out-
put. Variable costs on the other hand are costs that change in relation
to the amount produced, such as the costs of raw materials and labour.
Chapter 2 Getting linear models straight 47
We can define the total costs (TC) as the sum of the total fixed costs
(TFC) and the total variable costs (TVC):
TC ϭ TFC ϩ 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 ϭ TFC ϩ 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 ϭ 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 com-
pares to total cost. This is a key comparison as the total revenue minus
the total cost is the amount of profit made:
Profit ϭ TR Ϫ 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.
Example 2.11
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 esti-
mated 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 ϭ 4500000 ϩ 60Q
Total revenue, TR ϭ 150Q
These equations are plotted in Figure 2.11. Conventionally the money amounts, cost
and revenue are plotted on the vertical or y axis and the output is plotted on the hori-
zontal or x axis. This arrangement reflects the assumption that the money amounts
depend on the output and makes it easier to interpret the diagram.
48 Quantitative methods for business Chapter 2
In Figure 2.11 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 equa-
tions simultaneously:
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:
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 ϭ 50000
The total cost and total revenue when 50000 units are produced will be:
TC ϭ 4500000 ϩ 60 * 50000 ϭ 4500000 ϩ 3000000 ϭ £7,500,000
TR ϭ 150 * 50000 ϭ £7,500,000
TC 60Q 4500000
TR 150Q
Q 4500000
Ϫϭ
Ϫϭ

ϩϭ
0
90
TC 4500000 60Q so TC 60Q 4500000
TR 150Q so TR 150Q 0
ϭϩ Ϫϭ
ϭϪϭ
0
2
4
6
8
10
12
14
16
0 20 40 60 80 100 120
Output (000)
Cost/Revenue (£m)
Figure 2.11
Total cost and total revenue lines in Example 2.11
Chapter 2 Getting linear models straight 49
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 2.12.
Using Figure 2.12 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 pro-
duction 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 per-
centage 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.
Figure 2.12
Break-even graph
for Example 2.11
with areas
representing profit
and loss
0
2
4
6
8
10
12
14

16
02040
Loss
Profit
60 80 100 120
Output (000)
Cost/Revenue (£m)
Example 2.12
If the Ackrana Security Company in Example 2.11 aims to produce 80,000 cameras
what profit should they expect and what is their safety margin?
50 Quantitative methods for business Chapter 2
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.
At this point you may find it useful to try Review Questions 2.4 to 2.9
at the end of the chapter.
2.4 Inequalities
So far we have concentrated on the use of equations to model business
situations. Under some circumstances it is appropriate to use inequal-
ities, also known as inequations, expressions of relationships in which
one or more unknowns are not necessarily equal to a specific numer-

ical value.
The basic forms of inequalities are not equal (), less than (Ͻ), and
greater than (Ͼ):
x
x
x

Ͻ
Ͼ
10 means must take a value other than 10
10 means must be less than 10
10 means must be greater than 10
x
x
x
In Example 2.11 we found that their break-even point was 50,000 cameras so their
safety margin is:
budgeted output break-even output
budgeted output
80000 50000
80000
37.5%
Ϫ
ϭ
Ϫ
ϭ**100 100
TR 4500000 60 * 80000 9,300,000
TC 150 * 80000 12,000,000
Profit 12000000 9300000 2,700,000 that is £2.7 million
ϭϩ ϭ

ϭϭ
ϭϪϭ
Chapter 2 Getting linear models straight 51
It may help you to distinguish between Ͻ and Ͼ if you think of the
sharp end of each symbol as pointing to the lesser quantity, and the
open end to the greater. In x Ͻ 10 the sharp end points the x so it is
the smaller quantity and 10 is the larger quantity whereas in x Ͼ 10 the
sharp end points to 10 so that is the smaller quantity and x is the larger.
There are two types of composite inequality that you will meet later
in this chapter. These are less than or equal to (р) and greater than or
equal to (у). In both cases the lines beneath the Ͻ and Ͼ signs signi-
fies the possibility that the two sides in the relationship are equal is
included.
These composite inequalities are useful for representing ways in
which business operations might be limited or constrained by factors
such as the amount of raw material or time available for the produc-
tion of different products or services. In such cases all or some of the
limited resource might be used, but it is not possible to use more than
the amount on hand. The available quantity of resource is a ceiling or
constraint on the business activity. An inequality can be used to repre-
sent the relationship between the amounts of the resource required
for the products and the quantity in stock so that we can tell which
combinations of products or services can be produced with the given
materials, in other words what output levels are feasible.
Example 2.13
The Sirdaria Citrus Company produces juices using exotic fruits including two made
from hoormah, ‘Anelle’ and ‘Emir’. Fruit concentrate for these products is imported
and the supply is erratic. Sirdaria has 4000 litres of hoormah concentrate in stock.
The Anelle brand consists of 8% concentrate. Emir, the luxury product, consists of 10%
concentrate.

Represent the fruit concentrate constraint as an inequality.
Since we don’t know exactly how much of each product can be produced, indeed we
want an inequality to identify what the possibilities are, we must start by defining the key
variables: the amounts of each product produced. We will use x to represent the
amount of Anelle produced and y to represent the amount of Emir produced.
Anelle requires 0.08 litres of concentrate per litre, so if we produce x litres of Anelle
we will need 0.08x litres of concentrate. Emir requires 0.1 litres of concentrate per litre
so producing y litres of Emir will use up 0.1y litres of concentrate. Whatever the volume
of Anelle and Emir produced the amount of concentrate needed will be the amount
required for Anelle production added to the amount required for Emir production:
Concentrate required ϭ 0.08x ϩ 0.1y
52 Quantitative methods for business Chapter 2
Inequalities, like equations, represent the connection or relationship
between amounts either side of the appropriate sign. Like equations
they can also be represented graphically, but whereas the graphical
form of an equation is a line the graphical form of an inequality is an
area bounded or limited by a line. To portray an inequality graphically
you have to start by plotting the line that bounds the area. If the
inequality is a composite type the line to plot is the line for the equa-
tion that represents the limits of feasibility, output combinations that
use up all of the available resource.
The concentrate required must be balanced against the available supply of 4000
litres. To be feasible the output of products must not give rise to a demand for concen-
trate that exceeds the available supply, in other words the demand must be less than or
equal to the available supply:
0.08x ϩ 0.1y р 4000
Example 2.14
Show the inequality in Example 2.13 in graphical form.
If all the hoormah concentrate that the Sirdaria Citrus Company have in stock is used
then the amounts of concentrate required for Anelle and Emir production will equal

4000 litres:
0.08x ϩ 0.1y ϭ 4000
The best way to plot this equation is to work out how many litres of each product
could be produced using this amount of concentrate if none of the other product were
made. You can work out how many litres of Anelle could be produced from 4000 litres
of concentrate by dividing 4000 by the amount of concentrate needed for each litre of
Anelle, 0.08 litres:
4000/0.08 ϭ 50,000
So if only Anelle were made, 50,000 litres could be produced using the 4000 litres of
concentrate, but there would be no concentrate remaining for Emir production. The
point on the graph representing 50,000 litres of Anelle and zero litres of Emir is the
intercept of the line of the equation on the x axis.
If all the concentrate were committed to the manufacture of Emir the number of
litres produced would be:
4000/0.1 ϭ 40,000
Chapter 2 Getting linear models straight 53
At this level of output there would be no concentrate available for Anelle production.
The point on the graph representing 40,000 litres of Emir and zero litres of Anelle is
the intercept of the line on the y axis.
The line can be plotted using these two points.
Each point on the line plotted in Figure 2.13 represents a combination of Anelle and
Emir output that would use up the entire stock of concentrate. It is of course feasible to
produce output combinations that require a lesser amount of concentrate than 4000
litres, for instance producing 20,000 litres of each product would require only 3600
litres of concentrate. We can confirm this by putting these output levels in the expres-
sion representing the concentrate requirement in Example 2.13:
Concentrate required ϭ 0.08x ϩ 0.1y
For 20,000 litres of both products:
Concentrate required ϭ 0.08 * 20000 ϩ 01 * 20000 ϭ 1600 ϩ 2000 ϭ 3600
Look at Figure 2.14 and you can see that the point representing this combination lies

below the line.
The line represents all production combinations that use precisely 4000 litres of con-
centrate. All the points below it represent combinations that require less than 4000
litres of concentrate. All the points above it represent combinations that require more
than 4000 litres of concentrate and are therefore not feasible. An example of this is the
manufacture of 30,000 litres of each product, which would require:
0.08 * 30000 ϩ 0.1 * 30000 ϭ 2400 ϩ 3000 ϭ 5400
0
10
20
30
40
50
60
0204060
x (Anelle production in 000 litres)
y (Emir production in 000 litres)
Figure 2.13
The line of the equation in Example 2.14
54 Quantitative methods for business Chapter 2
Clearly 5400 litres of concentrate is considerably higher than the available stock of 4000
so it is simply not possible to produce these quantities.
The graphical representation of the inequality that expresses the constraint is there-
fore the line that defines the limits to the production possibilities and the area under-
neath it. In Figure 2.15 the shaded area represents the output combinations that are
feasible given the amount of concentrate available.
0
10
20
30

40
50
60
0204060
x (Anelle production in 000 litres)
y (Emir production in 000 litres)
ϩ
Figure 2.14
A feasible production mix in Example 2.14
0
10
20
30
40
50
60
0204060
x (Anelle production in 000 litres)
y (Emir production in 000 litres)
Figure 2.15
The concentrate constraint in Example 2.14
Chapter 2 Getting linear models straight 55
The constraint analysed in Examples 2.13 and 2.14 is an example of
a less than or equal to form of inequality. Other constraints might take a
greater than or equal to form of inequality.
Example 2.15
The Sirdaria Citrus Company has a contractual obligation to produce 10,000 litres of
Anelle for an important customer.
This is a constraint on production because it obliges Sirdaria to produce at least
10,000 litres of Anelle. If x represents the litres of Anelle produced then the inequality

for this constraint is that x must be greater than or equal to 10,000 litres:
x у 10000
Given this limitation any output mix is feasible if it involves producing 10,000 or
more litres of Anelle. Were they to produce only 8000 litres, for instance, they would
have insufficient to fulfil their commitment to the customer.
To represent this constraint on a graph we need to plot the limit to the constraint
(x ϭ 10000) and identify which side of it represents production combinations that are
feasible.
In Figure 2.16 the vertical line represents the lower limit on Anelle production result-
ing from the contractual commitment. Any point to its left would result in too little
0
10
20
30
40
50
60
0204060
x (Anelle production in 000 litres)
y (Emir production in 000 litres)
Figure 2.16
The contractual constraint in Example 2.15
56 Quantitative methods for business Chapter 2
Typically companies face not one single constraint on their opera-
tions but several. It can be helpful to display the constraints in a single
diagram rather than separately.
The shaded area in Figure 2.17 represents the production mixes
that meet both constraints. There are some output combinations for
which they have sufficient concentrate that do not allow them to satisfy
their contractual commitments, such as producing 40,000 litres of

Emir and no Anelle. Other combinations allow them to produce
enough to fulfil their contract but require more concentrate than they
have available, such as producing 30,000 litres of each product.
Anelle being produced to meet the commitment. Any point to its right enables them to
fulfil their obligation and is therefore feasible.
Example 2.16
Show the two constraints faced by the Sirdaria Citrus Company, the 4000 litres of con-
centrate available and their contractual commitment to produce 10,000 litres of Anelle,
in graphical form. Identify the production combinations that would be possible taking
both constraints into account.
0
10
20
30
40
50
60
0204060
x (Anelle production in 000 litres)
y (Emir production in 000 litres)
Figure 2.17
Feasible output combinations in Example 2.16