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. [...]... 1000, and should be large Chapter 2 Getting linear models straight 67 y (Soodnas chartered) 120 100 80 Cargo 60 Passenger 40 20 0 0 20 40 60 x (Lotkas chartered) 80 100 80 100 Figure 2. 22 The feasible region in Example 2. 23 y (Soodnas chartered) 120 100 80 60 40 A 20 0 0 20 40 60 x (Lotkas chartered) Figure 2. 23 The feasible region and iso-cost line for £80,000 in Example 2. 23 enough to position the... their budgeted output before making a loss 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 TR ϭ 4500000 ϩ 60 * 80000 ϭ 9,300,000 ϭ 12, 000,000 TC ϭ 150 * 80000 Profit ϭ 120 00000 Ϫ 9300000 ϭ 2, 700,000 that is 2. 7 million In Example 2. 11 we found that their... 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: y Ϫ 0.1x ϭ 20 0 y Ϫ 0.2x ϭ 0 ϩ0.1x ϭ 20 0 We can break this operation down into three parts: y Ϫ y ϭ 0y no y Ϫ0.1x Ϫ (Ϫ0.2x) ϭ ϩ0.1x 20 0 Ϫ 0 ϭ 20 0 giving us: 0.1x ϭ 20 0 This tells us that one-tenth of x is 20 0... function are plotted in Figure 2. 19 As you can see they are parallel to each other and the higher the profit, the higher the line representing it  62 Quantitative methods for business Chapter 2 y (Emir production in 000 litres) 70 Figure 2. 19 Values of the Sirdaria Citrus Company objective function 60 50 Profit ϭ £18,000 Profit ϭ £15,000 Profit ϭ £ 12, 000 40 30 20 10 0 0 20 40 60 80 x (Anelle production... 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 24 00 ϩ 3000 ϭ 5400 54 Quantitative methods for business Chapter 2 y (Emir production in 000 litres) 60 50 40 30 ϩ 20 10 0 0 20 40 x (Anelle production in 000 litres) 60 Figure 2. 14 A feasible production mix in Example 2. 14 Clearly 5400 litres... 42 Quantitative methods for business Chapter 2 Example 2. 8 The sales agent in Example 2. 1, currently receiving a wage of 20 0 plus 10% commission on sales, is offered the alternative of receiving 20 % commission on sales with no basic wage What is the minimum level of sales the agent would have to reach... of this by ten we find that a ‘whole’ x is worth 20 00: 0.1x * 10 ϭ 20 0 * 10 so 1x ϭ 20 00 In other words both wage determination models produce the same wage when sales are 20 00 But what will the wage be? To find this put the sales figure of 20 00 into the equation representing the original arrangement: y ϭ 20 0 ϩ 0.1 * 20 00 ϭ 20 0 ϩ 20 0 ϭ 400 Chapter 2 Getting linear models straight 45 The original approach... production in 000 litres) 60 50 40 30 20 10 0 0 20 40 x (Anelle production in 000 litres) 60 Figure 2. 16 The contractual constraint in Example 2. 15 In Figure 2. 16 the vertical line represents the lower limit on Anelle production resulting from the contractual commitment Any point to its left would result in too little 56 Quantitative methods for business Chapter 2 Anelle being produced to meet the commitment... Quantitative methods for business Chapter 2 Multiply both sides by minus one: (Ϫ * (Ϫ0.1x) ϭ (Ϫ1) * ( 20 0) 1) 0.1x ϭ 20 0 Multiply both sides by ten: x ϭ 20 00 The level of sales at which both approaches to wage determination will produce a wage of £400 is therefore 20 00 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... this constraint is: 0.008x ϩ 0.015y ϭ 480 This is represented by the bold line in Figure 2. 21 y (Emir production in 000 litres) 60 50 40 30 20 D 10 0 0 20 40 60 x (Anelle production in 000 litres) 80 Figure 2. 21 The feasible region in Example 2. 22 The feasible region is now rather smaller than it was in Figure 2. 20 There are production combinations that previously were feasible but are now not possible . 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 20 00 and the. 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. 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