Mathematics for economists: An introductory textbook, fifth edition

By Malcolm Pemberton and Nicholas Rau

This book is a self-contained treatment of all the mathematics needed by undergraduate and masters-level students of economics, econometrics and finance. Building up gently from a very low level, the authors provide a clear, systematic coverage of calculus and matrix algebra. The second half of the book gives a thorough account of probability, dynamics and static and dynamic optimisation. The last four chapters are an accessible introduction to the rigorous mathematical analysis used in graduate-level economics. The emphasis throughout is on intuitive argument and problem-solving. All methods are illustrated by examples, exercises and problems selected from central areas of modern economic analysis. The book's careful arrangement in short chapters enables it to be used in a variety of course formats for students with or without prior knowledge of calculus, for reference and for self-study.

The preface to the new edition and full table of contents are available from https://www.manchesterhive.com/page/mathematics-for-economists-supplementary-materials

• Language: English
• Publisher: Manchester University Press
• Release date: Nov 10, 2023
• ISBN: 9781526173522

Malcolm Pemberton

Malcolm Pemberton is Senior Lecturer in Economics at University College London

Book preview

Chapter 1

LINEAR EQUATIONS

In this chapter we introduce two of the main themes of the book. The first is the fact that a relationship between two quantities — for example price and output, or income and consumption — can often be expressed either as an equation or by means of a diagram. Building on this fact will lead us later on into curve-sketching and calculus. Here we start with the simplest case, where the relevant ‘curves’ are straight lines.

The second topic of Chapter 1 is the solution of systems of linear equations. This is also something we shall build on later in the book, using the powerful techniques of matrix algebra. But, as we shall see in this chapter, the basic method of solution can be explained very simply, as can its applications to the economics of market equilibrium and input-output analysis.

When you have studied this chapter you will be able to:

•sketch graphs of linear relations;

•solve systems of two linear equations in two unknowns, and of three linear equations in three unknowns;

•calculate price and quantity in market equilibrium, given linear supply and demand schedules;

•solve simple problems in input-output analysis.

1.1 Straight line graphs

You are probably familiar with the use of graphs in summarising data. The use of graphs in mathematics is quite similar but is usually not tied to a particular application. To cope with this greater degree of abstraction we shall need some technical terms, so we start with a few definitions.

Suppose we have two lines at right-angles to each other, as shown in Figure 1.1. The horizontal line is called the x–axis, the vertical line the y–axis and the plane containing the two lines the xy–plane. The position of a point is then specified by its distances from the two axes. The horizontal distance from the point to the y–axis is known as the x–coordinate of the point, and the vertical distance from the x–axis is known as its y–coordinate. For example, the x–coordinate of the point P in Figure 1.1 is 3 and the y–coordinate is 4; we can therefore refer to P as ‘the point (3,4)’.

When finding the coordinates of a point, we treat distances to the left of the y–axis as negative; similarly, distances below the x–axis are negative. Thus Q, R and S in Figure 1.1 are the points (−1,2), (4,−1.5) and (−2,−3.5) respectively.

Figure 1.1: Axes and coordinates

The point of intersection of the axes is known as the origin and has coordinates (0,0). More generally, all points on the x–axis have y–coordinate equal to zero and all points on the y–axis have x–coordinate zero. Thus the point (3,0) is on the x–axis, 3 units to the right of the origin, and the point (0,−4) is on the y–axis, 4 units below the origin.

The axes divide the xy–plane into four regions known as quadrants. The four points P, Q, R and S of Figure 1.1 are all in different quadrants. The part of the xy–plane consisting of points such as P whose coordinates are both non-negative is called the positive quadrant.

Sketching a straight line

We now explain how to draw a picture of all the points whose x and y coordinates satisfy a relation of the form

y=ax+b,

where a and b are given numbers. This procedure is referred to as sketching the graph of the relation. The relation under consideration here is said to be linear because its graph is a straight line.

For example, the graph of the linear relation y=2x+6 is the upward-sloping line in Figure 1.2. We could convince ourselves of this by taking a series of values of x, say 0,1,−1,2,−2 and so on, and then calculating the corresponding values of y. In this way, we see that the points (0,6),(1,8),(−1,4), (2,10),(−2,2), and so on, all lie on the graph. If we then plot these on graph paper, we would see that they lie on a straight line. But in fact this procedure would be a waste of time: to sketch the graph of a linear relation, it suffices to find two points on the graph and draw the line that passes through them.

Figure 1.2: Straight lines in the xy–plane

Example 1 Sketch the straight line y=2x+6.

As we noted above, it is enough to find two points on the line. It is convenient to choose our two points to be those where the graph crosses the axes: in other words, the point on the line where y = 0 and the point where x = 0. If 2x+6=0 then x=−3, so the line crosses the x–axis at the point (−3,0). Also, 2×0+6=6, so the line crosses the y–axis at the point (0,6). We therefore sketch the graph by marking the two points (−3,0) and (0,6), and drawing the line that passes through both of them. This gives us the upward-sloping line in Figure 1.2.

The method of Example 1 may be used to sketch the graph of y=ax+b whenever neither a nor b is zero. If a = 0 we have y=b, which is the equation of the horizontal line through the point (0,b). For example, the horizontal line y = 3 is also sketched in Figure 1.2.

If b = 0 our relation becomes y=ax, which is obviously satisfied if x=y=0. Thus in this case our line passes through the origin and we have only to find one other point on it. For example, Figure 1.2 sketches the line y=−x as the downward-sloping line through the origin and the point (−3,3).

Since a straight line is the graph of a linear relation, the points which are not on the line have coordinates that do not satisfy the relation. For example, none of the equations

y=2x+6,  y=3,  y=−x

is satisfied if x = 2 and y = 6. This means that the point (2,6) does not lie on any of the three graphs that we have sketched in Figure 1.2. The point is labelled Q in the diagram.

Slope and intercept

Returning to the linear relation y=ax+b, we call the number a the slope of the relation; b is called the intercept of the relation.

The intercept b is the value taken by y when x = 0; in geometrical terms, the line cuts the y–axis at the point (0,b). The slope a measures the steepness of the straight line graph. The line is upward-sloping if a is positive, downward-sloping if a is negative and horizontal if a = 0. A straight line with slope 8 slopes upward more steeply than one with slope 2, and a line with slope −100 slopes downward more steeply than one with slope −10. Another name for slope is gradient.

Figure 1.3 depicts three linear relations, all with intercept −4 but with different gradients:

(A) y=2x−4;  (B) y=4x−4;  (C) y=−2x−4.

In Figure 1.4 we sketche three straight lines, each with gradient 2 but with different intercepts:

(L) y=2x−4;  (M) y=2x;  (N) y=2x+2.

The six straight lines of Figures 1.3 and 1.4 may all be drawn using the same methods that we used for Figure 1.2. Thus the linear relation (C) is illustrated in Figure 1.3 by the straight line C which cuts the y–axis at (0,−4) and the x–axis where 2x=−4, namely at the point (−2,0). The linear relation (M) is illustrated in Figure 1.4 by the straight line M ihat passes through the origin and the point (1,2).

Figure 1.3: Lines with intercept −4

Figure 1.4: Lines with slope 2

More on the slope

We stated above that the slope a of the line y=ax+b measures the steepness of the line. To be more precise about this, a is the amount by which y increases when we move along the line, increasing x by one unit. Suppose for example that x increases from u to u+1. The old value of y is au+b and the new value is

a(u+1)+b=au+b+a,

a units greater than the old value.

Similar reasoning applies if ‘1’ is replaced by some other number, say h: if we move along the straight line y=ax+b, increasing x by h units, then y increases by ah units. In short,

 slope=  change in y change in x.

We may rephrase this equation in a way that may look rather intimidating but will turn out to be very useful. Suppose (x0,y0) and (x1,y1) are points on the straight line y=ax+b; then

a= y1− y0x1− x0.(1.1)

For example, the straight line y=2x−4, drawn as L in Figure 1.4, passes through the points (0,−4) and (2,0); in this case the right-hand side of (1.1) is

0+42−0=2,

which is of course the slope of the line.

The general linear equation

There is one class of straight lines in the xy–plane whose equations cannot be put in the form y=ax+b. This consists of the vertical lines: the y–axis and lines parallel to it. Such lines have equations of the form x=c, where c is a constant (in other words, a given number). For example, the straight line x = 10 is the vertical line through the point (10,0); it is parallel to the y–axis and 10 units to its right.

The general equation for a straight line in the xy–plane is

ux+vy=w,(1.2)

where u,v,w are constants such that at least one of u and v is not zero. This equation covers both the vertical and the non-vertical cases. If v = 0 we have the vertical line x=w/u. If v≠0,¹ we can rearrange (1.2) into an equation of the form y=ax+b and read off the slope and the intercept from this equation.

Figure 1.5: The straight lines of Example 2

Example 2 We sketch the straight lines 3x=12 and 3x+4y=12.

The first equation may be written x = 4, so we have the vertical line x = 4 in Figure 1.5. The second equation may be written as 4y=12−3x, or as

y=−34x+3.

This is the straight line with slope −34 and intercept 3. It is the downward-sloping line through (4,0) and (0,3) and is also shown in Figure 1.5.

There is a slightly easier way of sketching the graph of the second straight line, Instead of rearranging the equation 3x+4y=12 in the form y=ax+b, we use the equation directly to find the points where the line crosses the axes. If y = 0 then 3x=12, so x = 4: the line crosses the x–axis at the point (4,0). If x = 0 then 4y=12, so y = 3: the line crosses the y–axis at the point (0,3).

Finding the equation of a line

As we explained above, one can sketch the graph of a linear relation by finding two points on the graph and drawing the straight line through them. Sometimes we want to go in the other direction: we wish to find the equation of the straight line through two given points, say (x0,y0) and (x1,y1).

If x0=x1 we have the vertical line x=x0. Now suppose that x0≠x1. Then the equation we want is of the form y=ax+b, and the slope a is given by (1.1):

a=y1−y0x1−x0.

Having found a, we can find b simply by using the fact that our line goes through the point (x0,y0): since y0=ax0+b, b=y0−ax0.

Example 3 Find the equation of the straight line that passes through the points (−2,2) and (3,17).

The slope is

17−23+2=155=3.

Hence the equation of the line is y=3x+b, for some b. To find the intercept b, we use the fact that our line goes through the point (−2,2):

b=2−3×(−2)=2+6=8.

The equation of the line is therefore y=3x+8.

Exercises

1.1.1 Show the points ( 2 , 3 ) , (−2,3) and (2,−3) in the xy–plane. These points are three vertices of a rectangle; what are the coordinates of the fourth vertex?

1.1.2 Sketch in the same diagram the graphs of the following linear relations:

 (a) y=x+ 1,  (b) y=x− 3, (c) y=x+ 8.

What do you notice?

Find the equation of the line of slope 1 which passes through the point (−1,5).

1.1.3. Sketch the graphs of the following linear relations:

 (a) y= 2x+ 3,  (b) y=−x+ 3, (c) y=−8x+ 3.

What do you notice?

Find the equation of the line of intercept 3 which passes through the point (1,0).

1.1.4. Sketch in the same diagram the graphs of the following linear relations:

 (a) y= 2  (b) x=−3  (c) y= 4x− 1  (d) y= 23x (e) x− 3y= 5 (f) 10x+ 6y= 0

1.1.5. Find the equations of the straight lines through the following pairs of points:

 (a) (−1,3) and (1,−5) (b) (1,2) and (2,9)  (c) (−1,4) and (2,−8)  (d) (7,0) and (7,−3)

1.2 An economic application: supply and demand

As an application of straight line graphs, we consider a simple example of supply and demand in a market, say for honey. Suppose the relation between quantity of honey demanded, say q, and the price of honey p is given by

q=7−p.

This relation is called the demand schedule for the good. In our example, the graph of the demand schedule in the pq–plane is a straight line of slope −1 and intercept 7, meeting the p–axis where p = 7. The negative slope reflects the ‘law of demand’ whereby consumers demand less honey the higher its price.

Suppose also that the supply schedule for honey is given by

q=2p−2,

with the interpretation that bee-keepers wish to produce 2p−2 units of honey if the price of honey is p. This is represented in the pq–plane by a straight line of slope 2 and intercept −2, meeting the p–axis where p = 1. The line slopes upward because a higher price of honey induces people to devote more time and effort to keeping bees.

The straight lines representing the demand and supply schedules in the pq–plane are shown in Figure 1.6. The equilibrium, where quantities demanded and supplied are the same, is at the point E where the two lines intersect. At this point the two equations

q=7−p,q=2p−2

must be satisfied simultaneously. Thus at E,

7−p=2p−2.

Rearranging, we have 9=3p, whence p = 3; substituting this into either the supply or the demand equation gives q = 4. Thus in this case the equilibrium (market-clearing) price and quantity are

p=3,q=4.

Figure 1.6: Supply, demand and equilibrium

A more familiar diagram

Figure 1.6 differs in two ways from the supply-and-demand diagrams that appear in most textbooks on economics. First, for historical reasons, economists tend to draw supply and demand curves with quantity q on the horizontal axis and price p on the vertical axis. Secondly, negative price and quantities are usually of no economic relevance. For this reason, the standard supply-and-demand diagram shows not the entire qp–plane but only the positive quadrant.

To draw a standard supply-and-demand diagram for the same schedules as those of Figure 1.6, we begin by rearranging the demand and supply equations to bring p on to the left-hand side. The demand and supply schedules are respectively

p=7−q.  p=1+q2,

with the interpretation that 7−q is the largest price at which q units of honey would find buyers and 1+12q is the smallest price at which bee-keepers would be prepared to supply q units of honey.

The supply and demand schedules are redrawn in the positive quadrant of the qp–plane in Figure 1.7. The market clears if

p=7−q=1+q2

so that 3q/2=6; hence q = 4 and p=7−4=3. These are of course the same values of p and q that we obtained from our earlier calculations.

Figure 1.7: Supply and demand — the standard diagram

Exercises

1.2.1 Suppose the demand and supply schedules for milk are

q=15−3p,q=2p−5

respectively. Sketch these schedules in a standard supply-and-demand diagram with q on the horizontal axis and p on the vertical. Find the equilibrium price and quantity of milk.

1.2.2. Suppose the demand and supply schedules for wine are

q=k−5p,q=5p−4

respectively, where k is a constant. Find the equilibrium price and quantity in terms of k.

Sketch the demand and supply schedules in the same diagram for each of the cases k = 16, k = 12 and k = 6. For each case, write down the equilibrium price and quantity.

What would you expect to happen if k = 2?

1.3 Simultaneous equations

In our example of the supply and demand for honey, we found the equilibrium price and quantity by solving a system of two linear equations in two unknowns. We now explain a systematic procedure for such exercises. Later in the section we generalise the method to more complicated systems of linear equations.

Consider the pair of equations

3x+2y=8,2x+5y=9.

Each of these equations is an example of the general linear relation (1.2) and can therefore be be represented by a straight line in the xy–plane. The line representing the first equation crosses the x–axis at the point (8/3,0) and the y–axis at the point (0,4). Similarly, the line representing the first equation passes through the points (9/2,0) and (0,9/5). Both graphs are sketched in Figure 1.8.

Figure 1.8: Two equations, two unknowns, unique solution

Solving the system of two equations means finding values of x and y that satisfy both equations simultaneously. What this means geometrically is finding the intersection of the two straight lines of Figure 1.8. We can see in the figure that the point of intersection has coordinates (2,1), so that the solution to our system of equations is

x=2, y=1.

We now explain why this is so.

Our problem is to solve the system of equations

3x+2y=8,2x+5y=9.

To do this we start by eliminating one unknown from one equation, leaving one equation in one unknown. This is achieved by the following step:

(S) Leave the first equation as it is and eliminate x from the second equation by subtracting a suitable multiple of the first equation.

If we subtract t times the first equation from the second, we get an equation in which the coefficient of x is 2−3t. To eliminate x we should make this coefficient zero by choosing t to be 2/3.

Therefore, carrying out step (S) leads to the new system

3x+2y=8,(5−43)y=9−163.

The second equation of the new system says that 113y=113, so that y = 1. We substitute this into the first equation and see that

x=13(8−2y)=13(8−2)=2.

The solution to our system is indeed

x=2, y=1.

Complications

We have just solved a system of two linear equations in two unknowns which had a unique solution: the corresponding straight lines, sketched in Figure 1.8, had a unique point of intersection. To see that this does not always happen, consider the system

3x+2y=8,6x+4y=9.

In this case, carrying out step (S) means subtracting twice the first equation from the second. This gives a new system of which the second equation is the absurd statement

0=−7.

This is a signal that the system of equations we started with was inconsistent and therefore has no solution. The geometry of this is panel (A) of Figure 1.9: the two lines are parallel and therefore have no point of intersection.

The final kind of outcome of solving two linear equations in two unknowns is shown by the following:

3x+2y=8,6x+4y=16.

Here, carrying out step (S) by subtracting twice the first equation from the second gives the new system

3x+2y=8,0=0.

The second equation is true in all circumstances, and therefore uninteresting. We are left with only one equation: any values of x and y which satisfy that equation solve the system. If we assign an arbitrary value p to y we can solve the first equation for x, giving x=13(8−2p). We write the solution

x=23(4−p), y=p.

This is the complete solution in the sense that we get all the different solutions of the system by assigning all possible numerical values to p. It follows that the system has infinitely many solutions. Geometrically, the two equations of the system represent the same line and any one of the infinite number of points on that line is a solution: see Figure 1.9, panel (B).

Figure 1.9: Two equation, two unknowns — the awkward cases

Two equations, two unknowns: summary

Given a system of two linear equations in two unknowns, there are three possibilities:

In each case, the system may be solved by starting with step (S): ‘solving’ the system means finding the unique solution in the first case, tracking down an inconsistency in the second case, and describing the complete solution in the third. You do not have to know before you start which case you are in, since this is revealed by the solution procedure, as the examples above demonstrated.

The three cases can be thought of geometrically as follows. Each of the two equations represents a straight line. If the lines are different and intersect, they intersect in exactly one point: the system has a unique solution. If the equations represent parallel lines there is no solution. The remaining case is where the two equations represent the same line: here we have an infinite number of solutions.

Three equations in three unknowns

The method we used to solve a system of two linear equations in two unknowns x,y may be extended to solve systems of three linear equations in three unknowns x,y,z.

We start by solving the following system:

2x+7y+z=23y−2z=74z=4

This is an example of a ‘triangular’ system, so called because of the pattern made by the coefficients on the left hand side: x does not appear in the second equation, and neither x nor y appear in the third. Solving the third equation, z = 1; substituting this into the second equation gives y=(7+2)/3=3; and substituting our values for y and z into the first equation gives

x=2−21−12=−10.

The solution is

x=−10, y=3, z=1.

The process of repeated substitution into the equation above is called back-subst-itution. The method yields a unique solution for any triangular system of three equations with the property that x has a non-zero coefficient in the first equation, y in the second and z in the third. A triangular system with this property is said to be regular.

For systems which are not triangular, we eliminate unknowns from equations until, if possible, we arrive at a regular triangular system. This is most easily achieved by a generalisation of step (S) above.

To see how the procedure works, consider the system

2x+4y+z=5,x+y+z=6,2x+3y+2z=6

To solve this, we begin by carrying out the following generalisation of (S):

(S1) Leave the first equation alone and eliminate x from the second and third equations by subtracting from those equations suitable multiples of the first equation.

In our example this means subtracting 1/2 times the first equation from the second equation and (since 2/2=1) subtracting the first equation from the third equation. This gives the new system

2x+4y+z=5,−y+12z=72,−y+z=1

The system is not yet triangular because of the presence of y in the third equation. The next stage is to perform the following step:

(S2) Leave the first and second equations alone and eliminate y from the third equation by subtracting a suitable multiple of the second.

In this case the ‘suitable multiple’ is (−1)/(−1)=1: we subtract the second equation from the third, obtaining the regular triangular system

2x+4y+z= 5−y+12z= 7212z=−52

We can now apply back-substitution:

z=−5,y=12(z−7)=12×(−12)=−6,x=12(5−z)−2y=12×10+12=17.

The solution to the system is

x=17, y=−6, z=−5.

Gaussian elimination

In addition to the elimination steps (S1) and (S2), it is sometimes necessary to change the order of the equations. Thus, in the system

−3y+4z=−2,x+5y+2z= 9,x+y+z= 6

we cannot eliminate x from the second and third equations by subtracting multiples of the first equation. We therefore begin by interchanging the first two equations and then proceeding as above.

In general, then, the elimination procedure can be carried out using a sequence of elementary operations. These are of two kinds:

(EO1) writing the equations in a different order;

(EO2) subtracting a multiple of one equation from another equation.

The particular procedure we use to simplify a system of equations by successive elementary operations is called Gaussian elimination:² (EO1) operations, if required, alternate with elimination steps such as (S1) and (S2) in which multiples of an equation are subtracted from each of the equations below it.

The following three properties of the procedure are extremely important.

1. Elementary operations are reversible . If we exchange two equations and then exchange again, we are back where we started. And if we perform the (EO2) of subtracting 7 times the second equation from the third equation, and then perform the further (EO2) of adding 7 times the second equation to the new third equation, we are again back where we started. The fact that elementary operations are reversible is important for the following reason: when we simplify a system of equations using these operations, the simplified system is logically equivalent to the one we started off with, and therefore has the same solution. This is the reason why Gaussian elimination always works.

2. Any system of three linear equations in three unknowns which has a unique solution can be reduced by Gaussian elimination to a regular triangular system. The solution is then found by back-substitution. And the same is true of systems of n linear equations in n unknowns, where n can be 2, 6, 27 or whatever.

3. It is possible for a system of n linear equations in n unknowns not to have a unique solution. We have already seen this for n = 2 (recall Figure 1.9 ) and the same is true for larger values of n . In such cases reduction to a regular triangular system is not possible, but Gaussian elimination is still helpful. If the elimination procedure gives rise to an absurd equation such as 0=3, the system has no solution. And if the procedure leads to a vanishing equation (0=0) and there are no absurdities, there are infinitely many solutions. In such cases the complete solution can be found, as in the two-equation case. We shall explain all this in full detail in Chapter 12.

Exercises

1.3.1. Solve the following equations simultaneously:

x+2y=32x−3y=13

1.3.2. Show that the following system of equations has no solution unless c=12:

2x−5y=c4x−10y=1

If c=12, find the complete solution.

1.3.3. Solve the system of equations

x + 2 y + 2 z = 1 2 x – 2 y + z = 2 x – y + 3 z = 3

1.3.4. Solve the system of equations

y+2z= 22x+z=−1x+2y=−1

1.4 Input-output analysis

One application in economics of systems of simultaneous linear equations is the input-output model, which has proved helpful in forecasting and planning. We illustrate it with a simple example.

Suppose an economy produces three goods X, Y and Z. There may also be non-produced goods such as labour, land, imported raw materials and so on. We assume no joint production: this means that we can think of ‘industry X’ producing positive quantities only of good X, using goods Y and Z, and possibly X itself as inputs. The other two industries produce Y and Z.

We define the gross output of good X to be the total amount produced, including that fed back into the system as industrial input. This is to be contrasted with the net output of X, which is the amount of X produced and not fed back into the system, being therefore available for consumption, accumulation and export. The relation between net and gross output of good X is

net output of X=gross output of X−quantity of X required as input in industry X−quantity of X required as input in industry Y−quantity of X required as input in industry Z.

Similar relations hold for goods Y and Z.

Input requirements per unit of gross output of each produced good are given in the following table, known as an input-output table. This says for example that each unit of gross output of good Y requires the input of 0.2 units of good X, 0.5 units of good Y and 0.3 units of good Z.

Our final assumption is constant returns to scale, which in this context means that the figures given in the table for input requirements per unit of gross output are independent of the levels of gross output. Thus, if gross outputs of Y and Z are y and z, 0.1z units of Y are required as input in the Z industry and 0.3y units of Z are required as input in the Y industry, however large or small y and z happen to be.

Let the gross outputs of X, Y, Z be x,y,z and the net outputs x∗,y∗,z∗. Then the relation between net and gross output of good X may be written

x * = x – 0.2 x – 0.2 y – 0.2 z

The equivalent relations for Y and Z are:

y * = y – 0.4 x – 0.5 y – 0.1 z z * = z – 0.4 x – 0.3 y – 0.3 z

Simplifying, we have the equations

x * = 0.8 x – 0.2 y – 0.2 z y * = – 0.4 x + 0.5 y – 0.1 z z * = – 0.4 x – 0.3 y + 0.7 z

As a matter of simple arithmetic, these equations give the net outputs, given the gross outputs. More interestingly, one can also use the equations to find out how much gross output of each good must be produced if a given list of net outputs is required.

Suppose for example that the required net outputs of X, Y and Z are 10, 15 and 7 respectively. To find the gross outputs, we must solve the above system of linear equations for x, y and z, given that x∗=10, y∗=15 and z∗=7. This is done in the usual way by Gaussian elimination. Our first step is to leave the first equation alone and eliminate x from the second and third equations by adding 0.5 times the first equation to each of these equations. We obtain the system

0.8x−0.2y−0.2z=100.4y−0.2z=20−0.4y+0.6z=12

Leaving the first two equations alone and adding the second equation to the third, we obtain the triangular system

0.8x−0.2y−0.2z=100.4y−0.2z=200.4z=32

By back-substitution, z=80y=12(100+z)=90 and x=14(50+y+z)=55. The required gross outputs are 55 units of X, 90 units of Y and 80 units of Z.

Input-output analysis is much more general and practical than this simple example may suggest. As early as the 1950s the economist Wassily Leontief and his associates were constructing 80–sector input-output tables for the United States. Since then input-output analysis has been greatly extended to bring in time-lags in production, choice of technique, natural resources and many other complications.

Exercises

1.4.1 For the example of the input-output model given in the text, find the gross outputs corresponding to net outputs of 12, 10, 10 of X, Y, Z respectively.

1.4.2. Suppose an economy produces two goods X and Y under conditions of no joint production and constant returns to scale. Input requirements per unit of gross output are given by the following input-output table.

Find the gross outputs when the required net outputs of X and Y are 30 and 10 respectively.

Problems on Chapter 1

1–1. Suppose the demand and supply schedules for a good are linear.

  (i) Given that 31 units of the good are demanded if p = 4 and 11 if p = 8, find the demand schedule.

 (ii) Given that 3 units of the good are supplied if p = 8 and 15 if p = 12, find the supply schedule.

(iii) Under the assumptions of parts (i) and (ii), find the equilibrium price and quantity.

1–2. Show that the following system of equations has no solution if k=−4:

x+3y−2z=22x−5y+z=03x−2y−z=k

If k = 2, show that the system reduces to two equations in three unknowns and hence find the complete solution.

1–3. Solve the following system of equations and sketch the corresponding straight lines in an xy –plane:

2x+3y=10x+2y=6

If k is a constant, the equation

(2+k)x+(3+2k)y=10+6k

represents a straight line through the point of intersection of the first two lines. Explain why. What can you say about the third line if (i) k is large and positive, (ii) k is large and negative?

1–4. In the following macroeconomic model, the unknowns are Y (national income), C (consumption) and T (tax collection):

Y=C+I+G,C=2+0.8(Y−T),T=1+0.2Y.

I (investment) and G (government expenditure) are assumed to be known. Find Y, C and T in terms of I and G. What happens to Y, C and T when G increases by x units?

1–5. Consider again the input-output table of Exercise 1.4.2 . Find the gross outputs corresponding to net output levels a and b of X and Y respectively. How do you know that the gross outputs are positive?

¹The symbol ≠ means, and is read, ‘is not equal to’.

²After the great German mathematician and scientist Carl Friedrich Gauss (1777–1855).

Chapter 2

LINEAR INEQUALITIES

Not all economic statements can be expressed as equations. The statements that India’s GDP is greater this year than last, and that at most 10% of the British labour force is employed in manufacturing, are examples of inequalities. In this chapter we show how inequalities may be treated in a similar way to equations in Chapter 1: they may be manipulated algebraically and sketched in the xy–plane. As before, we deal with straight lines rather than curves: hence the word ‘linear’ in the chapter’s title.

Section 2.3 introduces optimisation subject to constraints, one of the main themes of this book. The exercises in that section are good training in combining algebraic and diagrammatic reasoning.

When you have studied this chapter you will be able to:

•rearrange inequalities using the three main algebraic rules;

•depict linear inequalities as regions in a plane;

•apply linear inequality diagrams in economic examples;

•solve simple problems in linear programming.

2.1 Inequalities

If u and v are numbers, then ‘uinequality and the symbol < is called an inequality sign. If u and v are represented as points on the horizontal axis, the relation u

The other inequality signs are defined as follows:

•  u>v means that u is greater than v (u is to the right of v);

•  u≤v means that u is less than or equal to v;

•  u≥v means that u is greater than or equal to v.

Figure 2.1: Points on the horizontal axis

Figure 2.1 illustrates, among other things, the facts that − 2<−1, − 1< 12 and 3> 1. In general, u>v means the same as v are called strict inequalities, while those involving ≤ and ≥ are called weak inequalities.

For the rest of this book we shall use the term positive number to mean a number x that satisfies the strict inequality x> 0. Similarly, a number x is said to be negative if x< 0, non-negative if x≥ 0 and non-positive if x≤ 0.

Rules for manipulating inequalities

Let u and v be two numbers such that uby the same amount, say 2 units, then the red button will still be to the left of the blue button: in other words, u+ 2

Rule 1 If the same number is added to both sides of an inequality, the inequality is preserved.

Notice that if we had moved our buttons to the left rather than the right, say by 3 units, the red button would again be to the left of the blue one: u− 3

Again let u and v be numbers such that u¹ For suppose u is Pemberton’s net wealth and v is Rau’s net wealth. If u and v are positive numbers, then u 0, then Pemberton is in debt while Rau has positive net wealth, and this remains true when Pemberton’s debts and Rau’s wealth are doubled.

All of this is equally valid when doubling is replaced by multiplication by any positive number, so we have:

Rule 2 If both sides of an inequality are multiplied by the same positive number, the inequality is preserved.

The third and final rule for dealing with inequalities is more subtle:

Rule 3 If both sides of an inequality are multiplied by the same negative number, the inequality is reversed.

For example, if u−v (in this case the negative multiplier is − 1) and − 5u>−5v.

The reason why Rule 3 is correct is that it follows from the other two rules, as the following argument demonstrates. Suppose u 0; we wish to show that (−a)u>(−a)v. Subtracting u+v from both sides of the inequality u−av, which is what we wanted.

Points to notice

1. The rules apply to weak as well as strict inequalities: if a ≥ b then a + 2 ≥ b + 2 , 3 a ≥ 3 b and − 4 a ≤ − 4 b .

2. A consequence of Rule 1 that will be used repeatedly throughout this book is the following: u < v if, and only if, v − u is a positive number. Similarly, u ≤ v if and only if v − u is non-negative.

3. The three rules may be combined. If u < v then 4 + 5 u < 4 + 5 v by Rules 1 and 2, and 6 − 7 u > 6 − 7 v by Rules 1 and 3. Similarly, if a ≥ b then 8 a − 3 ≥ 8 b − 3 by Rules 1 and 2.

Inequalities in the xy–plane

We now show how inequalities can be depicted in the xy–plane. Recall from Section 1.1 that the equation x=−3 represents a line parallel to the y–axis, and the equation y= 4 a line parallel to the x–axis. These lines are drawn in Figure 2.2. The dotted region consists of points above or on the line y= 4; these are the points whose y–coordinate is greater than or equal to 4. Thus the dotted region depicts the weak inequality y≥ 4. Similarly, the shaded region, consisting of points on or to the left of the line x=−3, depicts the inequality x≤−3. The region which is both dotted and shaded consists of points satisfying the two inequalities x≤−3,y≥ 4.

Figure 2.2 is simple because it depicts regions whose boundaries are parallel to the axes. We now give some more complicated examples.

Example 1 Depict the inequality 3x+ 4y≤ 6.

We start by drawing the straight line 3x+ 4y= 6 in the usual way, as the line through the point on the x–axis where x= 2 and the point on the y–axis where y= 3∕2. This line is labelled L in Figure 2.3. The inequality is satisfied by all points on or on one side of L.

Figure 2.2: Regions x≤−3, y≥ 4

Figure 2.3: The region 3x+ 4y≤ 6

To decide which side, we choose a point not on L and see whether its coordinates satisfy the inequality. The simplest point to test is the origin: if x=y= 0, 3x+ 4y= 0< 6, and the inequality is satisfied. Thus the inequality 3x+ 4y≤ 6 corresponds to the set of points which are either on the line L through the points (2,0) and (0,3∕2) or on the same side of that line as the origin. This set of points is shaded in Figure 2.3. Note that the points in the shaded region but not on the line L are those whose coordinates satisfy the strict inequality 3x+ 4y< 6.

Figure 2.4: The regions 3x+ 4y≤ 6 and x− 2y≤ 0

Example 2 Determine the region in the xy–plane that satisfies simultaneously the inequalities

3x+ 4y≤ 6, x− 2y≤ 0.

The set of points satisfying the first inequality is shaded in Figure 2.3, and again in Figure 2.4, where the line 3x+ 4y= 6 again has the label L. Now consider the second inequality. The line x− 2y= 0, labelled M in Figure 2.4, passes through the origin and has slope 12. The points (x,y) for which x− 2y≤ 0 lie either on the line M or on one side of it. To decide which side, we cannot use the origin as test point since it is on the line. Instead, we note that x− 2y≤ 0 if y= 0 and x< 0; the set of points whose coordinates satisfy the inequality therefore contains the negative part of the x–axis and consists of all points on or to the left of M. The region x− 2y≤ 0 is dotted in Figure 2.4.

It follows that the set of points (x,y) satisfying the pair of inequalities 3x+ 4y≤ 6 and x− 2y≤ 0 is the region of the xy–plane that is dotted and shaded in Figure 2.4. The boundary lines of the region intersect at the point (6∕5,3∕5): you should check this algebraically.

Exercises

2.1.1 Use the rules for manipulating inequalities to determine the values of x which satisfy the following:

(a) 2x+ 1> 0 (b)8− 3x≤ 0 (c)2x+ 7≥x− 5(d)1− 4x< 3+x

2.1.2 Shade the region in the x y –plane which satisfies

x+ 2y≤ 3, 2x− 3y≥ 13.

[HINT: if you have done Exercise 1.3.1, you will already have found the point of intersection of the relevant straight lines.]

2.2 Economic applications

As we said at the beginning of this chapter, constraints in economics often take the form of inequalities. We now show how such constraints can be represented graphically, using the methods of the last section.

The budget set and the budget line

Suppose that there are only two goods labelled 1 and 2, and that Ian consumes quantities x1 of good 1 and x2 of good 2. If the prices of the goods are p1 and p2, Ian spends p1x1 on good 1 and p2x2 on good 2; his total expenditure is then p1x1+ p2x2. If Ian’s income is m, then the statement that he cannot consume more than his income may be written

p1x1+ p2x2≤m.

This inequality is called the budget constraint. Since goods can be consumed only in non-negative quantities, x1≥ 0 and x2≥ 0. The set of points in the x1x2–plane satisfying these two inequalities and the budget constraint is called the budget set. The budget line is the straight line with equation

p1x1+ p2x2=m.

Points on the budget line whose coordinates are both non-negative represent consumption choices such that Ian spends all his income.

To sketch the budget set in the positive quadrant of the x1x2–plane, we begin by drawing the part of budget line in that quadrant. The budget line meets the horizontal axis where x1=m∕p1 and the vertical axis where x2=m∕p2. By using the origin as test point in the usual way, we see that the set of all points satisfying the budget constraint consists of the budget line and those points on the same side of it as the origin. The budget set is drawn shaded in Figure 2.5.

Figure 2.5: The budget line and the budget set

Sketching production possibilities

Suppose a firm manufactures two products X and Y. Let the production process involve three departments A, B and C, with time (in minutes) required in each department per unit output of each product given by the table below.

Let x and y be the amounts of X and Y produced each day by the firm. Then the table above states that the amount of time required each day in department A is 20x minutes for the production of X and 40y minutes for the production of Y, in all 20x+ 40y minutes. Similarly the total amount of time per day required in department B is 30x+ 30y minutes, and the amount of time per day required in department C is 45x+ 30y minutes.

Suppose that departments A and B are each available for 8 hours per day, and that department C is available for 11 hours per day. Then each department imposes a constraint on x and y of the form

 time required ≤ time available.

Using the above data and the fact that there are 60 minutes in each hour, we see that the time constraint for department A is

20x+ 40y ≤ 480.

The corresponding constraints for B and C are

30x+ 30y ≤ 480, 45x+ 30y ≤ 660.

Dividing the first of these three inequalities by 20, the second by 30 and the third by 15, we have

x+ 2y ≤ 24(constraint A) x+ y ≤ 16(constraint B) 3x+ 2y ≤ 44(constraint C)

Output levels x and y must of course be non-negative, which gives us the additional inequalities

x≥ 0, y≥ 0.

The set of points in the xy–plane satisfying these five inequalities is called the feasible set for the firm’s production plan.

To sketch the feasible set, we begin by drawing three lines corresponding to the time constraints in the three departments:

x+ 2y = 24(line A) x+ y = 16(line B) 3x+ 2y = 44(line C)

Figure 2.6: The feasible set for a two-product firm

Using the origin as test point in the usual way, we see that the set of points satisfying constraint A consist of all points on or to the left of line A. Similar results hold for B and C. Since the feasible set consists of all points which satisfy the three time constraints and have both coordinates non-negative, it is the shaded region ODEFG in Figure 2.6.

The coordinates of the corners of the feasible set are easily found using the methods of Chapter 1. D is the point where line A crosses the y–axis and is therefore the point (0,12). E is the point of intersection of lines A and B; its coordinates are found by solving simultaneously the equations

x+ 2y= 24, x+y= 16

and turn out to be (8,8). Similarly, since F is the point of intersection of lines B and C, its coordinates are (12,4). G is the point where line C crosses the x–axis and is therefore the point (44∕3,0). The entire path DEFG, that part of the boundary of the feasible set which does not coincide with either axis, is called the constraint boundary.

Notice finally that the point H where A crosses C is outside the feasible set. The interpretation of this is that the firm cannot use departments A and C to full capacity: if it did, it would be producing too much X and Y per day for department B to handle in eight hours.

Exercises

2.2.1 Judy has an income of 10 which she spends on fish and chips, the prices of which are 2 and 3 respectively. Sketch the budget set.

Also sketch the budget set if the prices of the two goods are reversed.

2.2.2 Henry has an income of 1 8 which he can spend on two goods labelled 1 and 2, with prices 1 and 3 respectively. Sketch the budget set.

Sketch also the budget set in each of the following cases:

(a) income 3 6 , prices 2 and 6 ;

(b) income 9 0 , prices 5 and 1 5 ;

(c) income 9 , prices 0 . 5 and 1 . 5 .

What do you notice? Can you formulate a general result?

2.2.3 A firm produces two products X and Y, using a production process involving two departments A and B. The time in minutes required in each department per unit output of each product is given by the following table.

 Department  A  B Product X 16 10  Product Y 8 20

Department A is available for 4 hours per day and department B is available for 5 hours per day. Sketch the feasible set.

2.2.4 Suppose the situation is as in Exercise 2.2.3 , but with the additional information that production per unit of X and Y causes the emission of 2 and 3 units of carbon respectively. Sketch the feasible set if total carbon emissions are to be restricted to 4 8 units per day.

Sketch also the feasible set if total carbon emissions per day are to be restricted to (a) 60 units, (b) 24 units.

2.3 Linear programming

Two of the most important concepts in economic theory are equilibrium and optimisation. Possibly the simplest example of economic equilibrium, clearing of a single market, was discussed in Section 1.2. Optimisation means doing as well as one can subject to given constraints.

To illustrate this, we continue with the example of the two-product firm of the last section, and assume in addition that the firm attempts to maximise profit. Suppose products X and Y yield profits of £30 and £40 per unit respectively. If the firm produces per day x units of X and y of Y, its profit is 30x+ 40y pounds per day. The firm’s problem is to choose the output combination (x,y) which makes profit as large as possible, subject to the constraint that (x,y) be feasible. Recalling from Section 2.2 the inequalities that define our firm’s feasible set of output combinations, we may write the profit-maximisation problem as follows:

 maximise30x+40y  subject to x+ 2y≤24 x+ y≤16 3x+ 2y≤44 x≥ 0,y≥ 0

This particular kind of optimisation problem, in which a linear expression is being maximised subject to linear inequalities, is called a linear maximisation programme. The expression to be maximised, in this case representing profit, is called the objective function.

To solve our maximisation programme we must search over the feasible set to locate the point of maximal profit. We start by considering the different output combinations which yield a given level of profit, say z. If (x,y) is such a combination,

30x+ 40y=z;

the point (x,y) therefore lies on the straight line which cuts the x–axis at (z∕30,0) and the y–axis at (0,z∕40). Such a straight line is called an isoprofit line.

The isoprofit line just discussed has slope − 3∕4 and intercept z∕40. Changing the value of profit z gives another isoprofit line with the same slope but a different intercept: note that a higher intercept means a higher level of profit. As we allow z to vary, we obtain what is known as a family of isoprofit lines; the lines are parallel to each other, each corresponding to a particular level of profit and with higher lines corresponding to higher profits. Four members of the family are sketched in Figure 2.7, which also reproduces the feasible set from Figure 2.6. Notice that all but one of the four lines meet the feasible set; the exception is  IP4, which corresponds to the highest of four levels of profit.

It is now easy to solve our linear programme. Moving outward from the origin to isoprofit lines corresponding to increasingly high profits, we reach the highest one that meets the feasible set. This is the line  IP3 which meets the feasible set at just one point, namely E. This point gives the profit-maximising output combination. Recalling from Section 2.2 that the coordinates of E are (8,8), we see that profit is maximised if x=y= 8: maximal profit is

30×8+ 40×8= 560.

The solution method

Trial-and-error with isoprofit lines is not an efficient method of solving linear programmes, and one feature of Figure 2.7 suggests a better way. We are referring to the fact that at the optimum point E, the slope of the isoprofit line  IP3 lies between the slopes of DE and EF, the two parts of the constraint boundary which meet at E.

Figure 2.7: Profit maximisation for a two-product firm

Why is this helpful? If you recall how we constructed the feasible set in Section 2.2, you will notice that the slope of each part of the constraint boundary is easily read off from the data of the problem. The part of the boundary from D to E is a segment of what we called line A and therefore has slope − 1∕2; the line-segment EF is part of line B and therefore has slope − 1; similarly the slope of FG is − 3∕2. The feasible set is drawn yet again in Figure 2.8; lines are labelled with their slopes, enclosed in square brackets. We know from the data on profits that the slope of each isoprofit line is − 3∕4. To solve the problem, we simply note that − 3∕4 is between − 1∕2 and − 1, which locates the optimum point at E.

To summarise, we have solved our profit-maximisation programme by the following method, which is the one we recommend for problems of this type:

1. Sketch the feasible set, calculating the coordinates of its corners and the slopes of its edges (the line-segments that make up the constraint boundary).

2. Calculate the common slope s of the isoprofit lines. It is not necessary to sketch these lines.

3. If s lies between the slopes of two adjacent edges, profits are maximised at the corner where these edges meet.

We now use the recommended method to solve some variants of the problem we have just solved, corresponding to different assumptions about profits; these solutions are also depicted in Figure 2.8. Suppose the unit profits for X and Y are not 30 and 40 as above, but zX and zY respectively. The common slope s of the isoprofit lines is then − zX∕zY . For example, if zX= 60 and zY = 50, the slope of each isoprofit line is − 6∕5, which lies between the slopes of EF and FG: hence the optimum is at F. Since F has coordinates (12,4), maximal profit is 920.

A case of some interest is that where zX= 16 and zY = 54; here the isoprofit lines are even flatter than DE and the optimum is at the point D on the y–axis; 0 units of X and 12 units of Y are produced, giving a maximal profit of 768. Notice that, in this as in all the other cases, it is the profit ratio zX∕zY that determines the optimal level of output of each product; the absolute levels of zX and zY serve only to determine the maximal amount of profit.

Another interesting case occurs where zX= zY = 25. Here zX∕zY = 1, so the slope of each isoprofit line is equal to the slope of EF. This implies that the profit-maximisation problem has multiple solutions: any (x,y) combination on the edge EF maximises profits, with a profit of 400.

Figure 2.8: The two-product problem under alternative assumptions

Generalities and complications

Having analysed the example of a two-product firm in some detail, we are now in a position to make some general points about linear programming. Recall that a linear maximisation programme is an optimisation problem in which a linear expression, called the objective function, is maximised subject to constraints taking the form of linear inequalities. The set of points in the xy–plane which satisfy the constraints is called the feasible set.

A programme is said to be feasible if it is possible to satisfy all constraints simultaneously. The profit-maximisation problem discussed above was clearly feasible, as were all its variants, each of which had the shaded region of Figure 2.5 as feasible set. On the other hand, it is possible to think of linear programmes for which no points satisfy all the constraints; in other words, the constraints are inconsistent with each other. A programme with this property is said to be infeasible. An example of an infeasible programme is

 maximise 2x+ 3y subject to x+y≤ 2 and x+y≥ 3.

In some feasible linear maximisation programmes, the objective function may be made arbitrarily large without violating the constraints. Thus in the linear programme

 maximise x+y subject to x≥ 2y and y≥ 0,

the point (4M,M) is in the feasible set for any positive number M; the objective function then takes the value 5M, which may be made as large as we like by choosing M large enough. This programme is illustrated in Figure 2.9: the feasible region is shaded and upward movement along the line y=x∕4 represents the method just described of increasing the value of the objective function indefinitely. In such cases the programme is said to be unbounded; otherwise the programme is said to be bounded.

Figure 2.9: An unbounded linear programme

The production problem considered earlier in this section is feasible and bounded in all its variants. A linear programme which is feasible and bounded always has at least one solution. If there is a unique solution, it occurs at a corner of the feasible set. If there are multiple solutions, as in our production problem when zX= zY = 25, then a solution can still be found at a corner; thus either E or F in Figure 2.8 will do).

The observant reader will have noticed that we have considered only weak inequality constraints in this section. In more general linear programming problems, equation constraints may occur, but strict inequality constraints are not allowed. To see the reason for this, consider the innocent-looking problem

 maximise x subject to x< 1.

This problem has no solution: x= 0.99 does better than x= 0.9, x= 0.999 does better still, and so on, but x= 1 violates the constraint. To avoid technical difficulties of this kind, linear programming restricts attention to constraints which are equations or weak inequalities. In many problems in economics this is the realistic way to proceed; thus in our production problem it makes sense to assume that the firm can use any of its three departments to maximum availability if it wishes.

Extensions

When there are more than two variables, the graphical approach is no longer possible. It is however possible to generalise the method. The reader should be able to visualise a ‘corner’ in three dimensions as a point where three planes meet. In dimensions higher than three, no pictorial representation is available but the concept of a corner can still be defined algebraically. The significance of corners is that, if a solution exists, there is a solution at a corner of the feasible set. Thus solution of a linear programme reduces to a search over corners. The most popular way of doing this is