Quantitative Methods for Business chapter 3 ppt

This is because the equation is linear; the slope is constant so the rate at which the value of y changes when x is changed is the same however big or small the value of x.. The equation

Dealing with curves without going round

the bend

3

Chapter objectives

This chapter will help you to:

■ deal with types of non-linear equations

■ interpret and analyse non-linear business models

■ apply differential calculus to non-linear business models

■ use the Economic Order Quantity (EOQ) model for stockcontrol

■ become acquainted with business uses of EOQ model

In the last chapter we looked at how linear equations, equations thatrepresent straight lines, can be used to model business situations andsolve business problems Although important, their limitation is thatthe relationships they are used to model need to be linear for them to beappropriate There are circumstances when this is not the case, forinstance the sort of model economists use to represent the connectionbetween volume of production and cost per unit In such a modeleconomies of scale mean that the average cost of production per unitgets lower as output increases The equation representing the situationwould be non-linear and we might be interested in analysing it to findthe least cost level of output

Example 3.1

A train operating company sets ticket prices using the equation:

y  0.8  0.2x where y is the ticket price in £ and x is the number of miles travelled.

To use this equation to work out the cost of a ticket for a 4-mile journey we simply

substitute the 4 for x:

y 0.8  0.2 * 4  0.8  0.8  £1.60The cost of a 5-mile journey will be:

y 0.8  0.2 * 5  0.8  1.0  £1.80The cost of a 10-mile journey will be:

y 0.8  0.2 * 10  0.8  2.0  £2.80

In this chapter we will look at the features of basic non-linear tions and use them to analyse business operations Following this wewill consider how to find optimal points in non-linear business modelsusing simple calculus Later in the chapter you will meet the EconomicOrder Quantity model, a non-linear business model that organizationscan use in determining their best stock ordering policy

equa-3.1 Simple forms of non-linear equations

One thing you might have noticed about the linear equations that tured in Chapter 2 was the absence of powers We met terms like 60Q and

fea-0.08x but not 3Q2or 3/x The presence of powers (or for that matter

other non-linear forms like sines and cosines, although we will not beconcerned with them here) distinguishes a non-linear equation In order

to appreciate why, consider two possible relationships between x and y:

If an equation is not linear the size of the change in y that comes about when x changes does depend on how big x is With a non-linear equation

a one-unit increase in x when x is small may result in a modest change

in y whereas a one-unit change in x when x is large may cause a much larger change in y.

The cost of an 11-mile journey will be:

y 0.8  0.2 * 11  0.8  2.2  £3.00Notice that the difference an extra mile makes to the cost, £0.20, is the same whetherthe difference is between 4 and 5 miles or between 10 and 11 miles This is because the

equation is linear; the slope is constant so the rate at which the value of y changes when

x is changed is the same however big or small the value of x The equation is plotted in

Figure 3.1

0 0.5 1 1.5 2 2.5 3 3.5

An equation that includes x to the power two is called a quadratic equation, derived from the Latin word quadrare, which means to square Similarly an equation that includes x to the power three is known as

cubic You may also meet reciprocal or hyperbolic equations These include

x to a negative power, for instance:

y  x1 1/x

To plot a linear equation you only need two points since the line is straight To plot anon-linear equation we need a series of points that track the path of the curve that rep-

resents it This entails calculating y values using the range of x values in which we are

interested, in this case from 0 (product launch) to 9

These points are plotted in Figure 3.2

Example 3.3

An economist studying the market for a certain type of digital camera concludes thatthe relationship between demand for the camera and its price can be represented bythe equation:

y  800/x where y is the demand in thousands of units and x is the price in £.

To plot this equation we need a series of points such as the following:

The equation is plotted in Figure 3.3

x (price in £)

Figure 3.3

The demand equation in Example 3.3

Some curves feature peaks and troughs, known as maximum andminimum points respectively These sorts of points are often of particu-lar interest as they may represent a maximum revenue or a minimumcost, indeed later in the chapter we will be looking at how such points

can be identified exactly using calculus.

Figure 3.4 show the sort of curve that economists might use to sent economies of scale The minimum point represents the minimumcost per unit and the point below it on the horizontal axis the level ofoutput that should be produced if the firm wants to produce at thatcost Other models economists use include maximum points.

repre-Example 3.4

The project manager of the new Machinar car plant suggests to the board of directorsthat the production costs per car will depend on the number of cars produced accord-ing to the equation:

y  x2 6x  11 where y is the cost per car in thousands of pounds and x is the number of cars produced

in millions

The equation is plotted in Figure 3.4

0 2 4 6 8 10 12

rela-y  x2 4x where y is the total revenue in millions of pounds and x is the price per holiday in

thousands of pounds

The equation is plotted in Figure 3.5.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

x (price in £000)

Figure 3.5

The curve representing the equation in Example 3.5

The maximum point in Figure 3.5 represents the maximum revenuethe firm could earn and the point below it on the horizontal axis theprice the firm should set in order to maximize its revenue

3.2 Basic differential calculus

We can find the approximate location of the maximum and minimumpoints in Examples 3.4 and 3.5 by studying the graphs carefully, marking

the maximum or minimum point then identifying the values of x and y

at which the point is located using the scales along the axes At best this

would give us an idea of where the point lies in relation to x and y, but

it is almost impossible to pinpoint it accurately by inspecting the graph

To find the precise location of maximum and minimum points wecan use techniques from the branch of mathematics known as calculus.The word calculus conveys an impression of mystery to some people

In fact it, like the word calculate, is derived from the Latin word calculare

which means to reckon with little stones, a reflection of the method of

counting in ancient times Calculare is related to the Latin word calx

which means a stone, the source of the word calcium

Calculus has two branches, differential calculus and integral calculus The former, which involves the process of differentiation, is about finding

how curves change, whereas the latter is about finding areas underneath

curves Our concern in this section is with differentiation If you wouldlike to find out about integration you may find Croft and Davison (2003)helpful.

Differentiation is concerned with slopes, and slopes in equations

reflect the way that the x variable changes the y variable In a simple

equation such as:

y  5  3x

the slope,3, tells us that an increase of one in the value of x will result

in the value of y increasing by three The slope is the rate of change in y that is brought about by a unit change in x The other number in the equation, 5, is a constant or fixed component; whatever the value of x, the amount added to three lots of x to get y is always 5.

The slopes of more elaborate equations are not so straightforward

If you look carefully at Figure 3.5 you will see that the slope changes asthe line moves from left to right across the graph It begins by climbingupward then reaches a maximum before altering course to descend

To start with it has a positive slope then at the maximum it has a zeroslope, it ‘stands still’, and finally it has a negative slope The nature ofthe slope therefore depends on where we are along the horizontal

scale, in other words, the value of x For the lower values of x the slope

is positive, for the higher ones it is negative

Figure 3.4 shows a similar pattern To begin with the slope is wards, or negative, then it ‘bottoms out’ at a minimum and finallybecomes upwards, or positive In this case the minimum is the pointwhere the slope is momentarily zero, it is a point of transition betweenthe negative-sloping and positive-sloping parts of the curve The max-

down-imum point in Figure 3.5 is a similar point of transition, or turning point

forming the coefficient, and the power reduced from 2 to 1:

x2 becomes 2x21 or simply 2x

This means that for the equation:

y  x2

the slope, or the rate at which y changes in response to a unit change in

x, is 2x This is the difference to y that a change in x makes, the

differen-tial of the equation, which is represented as dy/dx:

You may find it helpful to associate the process of differentiation asfinding such a difference A more exact definition is that it is the marginal

change in y resulting from an infinitely small marginal change in x.

Because the differential is derived from the original equation it is also

known as the derivative.

If the expression already includes a coefficient for x, like 2x2, we ply the power by the existing coefficient to give us the new coefficient:

multi-In this case the differential includes x, so the slope of the equation depends on the size of x; whatever x is, the rate of change in y arising from a change in x is four times the value of x If we apply the same pro- cedure to 3x, the result is a constant, 3 In carrying out the differentia- tion of 3x remember that 3x is 3 times x to the power one so the power

reduces to zero, and that any quantity raised to the power zero is one:

In this case the slope is simply 3, and does not depend on the value of

x: whether x is small or large the slope will always be 3.

The other type of differentiation result that you should know

con-cerns cases where x is raised to a negative power, such as the reciprocal

of x, 1/x, which can be written as x1 The process is the same, take theoriginal power and multiply it by the existing coefficient then reducethe power by one:

When you differentiate expressions like this the key is to remember

that a constant divided by x raised to a positive power is simply the stant times x raised to the negative power When x is taken above the

con-line the power becomes negative

Some of the equations you may have to differentiate will consist of

several different parts, not necessarily all involving x For the types of

equation we shall look at you need only deal with the parts one at atime to reach the differential

Note that in Example 3.6 the constant of 11 is not represented in the

derivative The derivative tells us how y changes with respect to x When

x changes so will x2and 6x but the constant remains 11.

At this point you may find it useful to try Review Question 3.1 at the

end of the chapter

When an equation has a turning point, a maximum or a minimum,

we can use the differential to find the exact position of the turningpoint At the turning point the slope is zero, in other words the differ-

ential, the rate of change in y with respect to x, is zero Once we know the differential we can find the value of x at which the slope is zero sim-

ply by equating it to zero and solving the resulting equation

Example 3.6

The production cost equation for the car plant in Example 3.4 was:

y  x2 6x  11 where y is the cost per car in thousands of pounds and x is the number of cars produced

From Example 3.6 we know that the derivative is:

The value of x at which this is equal to zero is the position of the turning point along

the horizontal axis:

2x  6  0 so 2x  6 and x  3

The turning point is located above 3 on the horizontal axis If you look back to Figure 3.4you can see that the plotted curve reaches its minimum at that point We can conclude thatthe minimum production cost per car will be achieved when 3 million cars are produced

d

y

x  x

In Example 3.7 we were able to refer back to Figure 3.4 and see fromthe graph that the turning point was a minimum But what if the costequation had not been plotted, how could we tell that it was a min-imum and not a maximum? We take the derivative and differentiate

again to produce a second derivative If the second derivative is positive

the turning point is a minimum, if it is negative the turning point is amaximum It may help to think that after a minimum the only way to

go is up, so the second derivative is positive whereas after a maximumthe only way to go is down, so the second derivative is negative

The cost per car at that level of production is something we can establish by insertingthe value of x at the turning point into the original production cost equation:

y  x2 6x  11

Minimum cost 32 6 * 3  11  9  18  11  2

If 3 million cars are produced the cost per car will be £2000

Example 3.8

Find the second derivative of the production cost equation from Example 3.4

The first derivative of the cost equation was:

If we differentiate this again we get 2, which is of course positive, confirming that theturning point is a minimum

nota-The inclusion of the 2s on the left hand side signifies that the process

of differentiation has been applied twice in reaching the result

d

2 2

y

Example 3.9

The total revenue for the firm in Example 3.5 was:

y  x2 4x where y is the total revenue in millions of pounds and x is the price per holiday in

thousands of pounds

Find the first order derivative and use it to locate the turning point Confirm that theturning point is a maximum by finding the second derivative, and work out the max-imum revenue

The first derivative is:

Turning point location:

Revenue will be maximized when the price is £2000

The second derivative is:

Since the second derivative is negative, the turning point is a maximum

The revenue when the price is £2000:

y (2)2 4 * 2  4  8  4The maximum revenue is £4 million

d

2 2

At this point you may find it useful to try Review Questions 3.2 to

3.11at the end of the chapter

3.3 The Economic Order Quantity model

An important application of the calculus we looked at in the previoussection occurs in a technique companies use to manage inventories,the Economic Order Quantity (EOQ) model This model was developed

to help companies manage stocks of materials and components, fically by enabling managers to work out the quantity to order eachtime if they want to minimize costs

speci-Before we look at the model in detail it is worth reflecting on the sons companies keep stocks or inventories Most companies keep somestock of the materials they need for their operations; a bus companywill probably have a stock of diesel fuel, a furniture maker will probablyhave a stock of wood Such stocks are important because being withoutthe material would disadvantage the company On the other hand, if acompany keeps a large amount of stock the costs of holding it are likely

rea-to be very high The srea-tock would have rea-to be srea-tored somewhere, perhapsunder certain temperature or security constraints, and these facilitieswill have a cost

For sound business reasons then a company will not want to run out

of stock, yet will not want to hold too much To resolve this, a companymight consider placing small, regular orders with their supplier Whilstthis would mean that stock would never be very high, and hence stock-holding costs would be modest, it will probably cost money every time

an order is made; a requisition may have to be processed, a deliverycharge met, a payment authorized They would find that the more ordersthey make, the higher the cost of making them

The dilemma they face is to decide how much material should beordered each time they place an order, the order quantity, so that the

combined costs of making the orders and holding the stock, the total

stock cost, is at a minimum To see how this can be resolved we need to

start by examining how the total stock-holding costs and the total ordercosts vary in relation to the quantity ordered

We shall concentrate on the simplest Economic Order Quantitymodel, in which we assume that the rate at which the material is used

is constant The amount used in one week is the same as the amountused in any other week We also assume that once an order is placedthe material will be delivered right away, there is no time lag for delivery.This latter assumption means that the company can wait until theirstock runs out before placing an order for its replenishment

Taking these assumptions together we can conclude that the highestamount that could be in stock would be the order quantity, which we

will refer to as Q, the amount that is ordered every time an order is

placed The lowest level of stock will be zero, since they don’t need toorder more until they run out The rate of use is constant, so the fluctu-ations in the stock level will follow the pattern shown in Figure 3.6.The repeating saw-tooth pattern that you can see in Figure 3.6 consists

of a series of vertical lines each of which represents a delivery of the

order quantity, Q The stock level peaks at Q at the point of each delivery

and then declines at a constant rate as the material is taken out of thestore and used until the stock level is zero and another delivery comes in