Linear Programming Essay Example

Linear Programming Essay Example Linear Programming Essay Linear Programming Essay Linear Programming After completing this chapter you should be able to: identify a product which maximises the contribution per unit of scarce resource when there is only one scarce resource, and determine the optimum solution. formulate an LP model to solve for the optimal product mix which maximises profits, or for cost minimisation problems to minimise costs. solve 2 variable problems graphically. use a spreadsheet to solve LP problems with any number of variables. interpret the sensitivity reports of spreadsheet solutions to LP problems to test objective function coefficient sensitivity, determine shadow prices and RHS ranging. rform throughput accounting and solve problems using the concept of the Theory of Constraints. Introduction In this chapter we continue with our profit planning, or product mix, decisions. We extend CVP analysis in the last chapter by introducing the notion of scarce resources. Although CVP analysis does not provide answers regarding optimum product mixes (mixes which maximise profits or minimise costs) one advantage of CVP analysis is that it focuses attention on products with high contribution margins. Managers and salespeople can often direct their efforts to increasing output and sales of high ontribution margin products and thereby maximise the contribution towards fixed costs and profits. Unfortunately, it is not always desirable to attempt to maximise the sales of high contribution margin products, at the expense of those with lower margins. Most firms have constraintsl (in the short run) on production constraints which limit the volume of output. In such cases it is preferable sometimes not to emphasise the sales of high contribution margin products. Rather, attention should be directed to the contribution margin per unit of scarce resources required to produce the products. Thus, sometimes products with low contribution margins also have low resource requirements, and can be highly profitable. We shall now focus on this type of analysis. Multi 9 9-1 A firms two products are both produced on a single machine. Product 1 requires 2 hours of machine time per unit of product while Product 2 requires 4 hours per unit. There is a maximum of 20 000 hours of machine time available per time period. Product 1 has a contribution margin of $4 per unit, Product 2 a margin of $5 per unit. profitable, generating $5 per unit as against Product 1 s $4 per unit. But let us check this out. If all machine time were devoted to Product 2, which requires 4 hours per unit, it would be possible to produce 20 000/4 = 5000 units per period. These 5000 units would each generate $5 contribution margin, providing a total contribution (variable profit) of 5000($5) = $25 000. If, instead, we concentrated on Product 1, which requires 2 hours of machine time per unit, it would be possible to produce 20 000/2 = 0 000 units, each earning $4 contribution margin, giving a variable profit of 10 000($4) = $40 000. This is $15 000 more than it is possible to earn from Product 2. It may seem surprising that the product with the lower contribution margin is the more profitable to produce. What is important, however, is not the contribution margin per unit of product, but the contribution margin per unit of the limiting resource. As Table 9-1 demonstrates, Product 1 earns the higher contribution margin per machine hour, $2. 00 per MH versus only $1. 25 per MH for Product 2. Contribution Margin per unit Machine Hours per unit Contribution Margin per Machine Hour Product 1 $4 2 $4/2=$2. oo Product 2 $5 4 $5/4=$1. 25 Table 9-1 : Contribution margin per unit of limiting resource Thus, with one limiting resource, we should produce only one product, the one with the highest contribution margin per unit of that limiting resource. In example 9-1 we should maximise the production of Product 1 which has the higher contribution margin per machine hour. It would not pay to produce a mix of the two, that is, some of each product. The reason is that for every unit of Product 2 which is produced a ontribution of $5 would be earned but an opportunity cost of $8 is incurred. That is, in the time that it takes to produce one unit of Product 2 (4 hours of machine time) two units of Product 1 could be made earning a contribution margin of 2x$4 = $8, leaving the firm $3 worse off. Two Limiting Resources production, because it may not be possible to obtain the same profitability rankings on different scarce resources. Consider example 9-2. Example 9 9-1, both products have to proceed through a labour-intensive finishing process. Both products require hour of finishing labour, but there is a maximum of 8000 hours of skilled finishing labour available per time period. Having learned our lesson about calculating contribution margin per unit of scarce resource, we now proceed to calculate the contribution margin per hour of finishing labour for each product, as shown in Table 9-2. Finishing Labour Hours per Finishing Hour $411-$4. 00 $511-$5. 0 Table 9-2: Contribution margin per unit of limiting resource Product 1 generates $4 per finishing hour, while Product 2 generates $5 per finishing hour. Thus Product 2 is more profitable in terms of contribution per hour of finishing abour. We are now faced with a problem of conflicting profitability rankings. There are two scarce resources, machine time and finishing time. Product 1 is more profitable in terms of machine resources consumed but Product 2 is more profitable in terms of finishing resources. Since these profitability rankings conflict, we cannot determine the optimum action using this type of analysis. We have to use a technique known as linear programming. In general, when there are multiple constraining resources there are usually conflicting profitability rankings, and linear programming must be used to determine the optimum product mix. Linear Programming achieve some objective such as profit maximisation or cost minimisation. It is assumed that there are constraining resources (such as limited production or distribution facilities) or market factors (demand etc. ) which prevent a firm from producing and selling unlimited quantities of individual products, and hence the aim is to optimise the use of these constraining resources. The word linear indicates that linearity is an assumption behind this technique. It is assumed that cost and revenue functions are linear, and that there is a linear relationship between inputs and outputs. If the output of a product is doubled, resource requirements are doubled. Care is needed when using this analysis that linearity is a reasonable assumption, and that the data are reasonably accurate. A convenient way of introducing LP is by means of the graphical method, but this restricts us to somewhat trivial examples involving two variables. For problems with more than two variables we need to use the simplex method. There are many specialised LP computer programs employing the simplex method. Also, these problems can be solved using most popular spreadsheets. Spreadsheet solutions will lso be illustrated. Maximisation Problems Example 9 9-2, where we have two products and two constraints (machine hours and finishing hours). In addition, period fixed costs are $18 000. We are required to formulate an LP model for this example and solve it to determine the product mix which will maximise the firms period net profit. A linear programming model consists of two parts, an objective function and a set of constraints. First, we formulate the objective function. Let xl represent the number of units of Product 1 to be produced, and x2 the number f units of Product 2. The objective is to maximise period net profit, which could be expressed in the form Maxtrntse P = 41+52 18 OOO (9-1) where the coefficients of xl and x2 are 4 and 5, which are the respective contribution margins of products 1 and 2, and 18 000 represents the period fixed costs. We can, however, simplify the analysis by ignoring fixed costs at this stage, and simply maximise variable profit (total contribution margin): Maximise P = 41 +52 (9-2) Fixed costs can then be subtracted from the optimum variable profit to give optimum net profit. Now we turn to the constraint set. First, there is a machine constraint: Product 1 there is a maximum of 20 000 machine hours available per period. Then, 2X1+4X2 = 20 ooo (9-3) is one of several ways of stating this information. This expression states that the machine time is to be fully used (i. e. to capacity). Graphically, Equation (9-3) would appear as shown in Figure 9-1 . Figure 9-1 The graph of Equation (9-3) is obtained by finding the two points 10 000 and 5000 on the xl and x2 axes respectively, and Joining them with a straight line. If no units of product 2 are made, then x2 = O, and we have 21 + 20000, e 21 = 20000, and hus xl = 10 000. Similarly; if no units of product 1 are made, then xl = O, and we have 2(0) + 42 = 20 000 and hence x2 = 5000. Strictly, the line should extend further at each end, but since on either of these extensions one of the variables would be negative, indicating negative production, interest is focused only in the positive quadrant. The line segment in Figure 9-1 represents combinations of xl and x2 which fully utilise the machine capacity of 20 000 hours. There are many possible solutions, some being integer values of xl and x2 while others are non-integer solutions. In eneral, LP solutions may be non-integral, and if only integers are acceptable one should use Integer Programming. Some possible solutions which represent combinations of whole units of production of xl and x2 are: (0,5000), ie xl † O, 12 (2000,4000), 2 000, (4000000), 4 000, 6 000, 2000 (8000,1000), 8 000, 1 ooo (10000,0), 10000, If the assumption of fully utilising machine capacity is relaxed, so that although 20 000 hours represents maximum machine time, less than 20 000 hours may be consumed if desirable, Equation (9-3) is converted to the inequality (which is graphed in Figure 9-2): 2X1+4X2 20 ooo Figure 9-2 Graphically, possible solutions are now extended to encompass not only points along the line segment, but also all of the area below the line, anywhere in the shaded portion of Figure 9-2. In addition to the solutions possible in (9-3), further possibilities exist in (9-4), such as Xl = 3000, = 1000, or Xl = 2000, 12 = 2000. It may be noted that in general, equations are tighter than inequalities, and restrict the number of alternative solutions. The second constraint, finishing labour, must also be considered. Given that both products require 1 hour of finishing time, and a maximum availability of 8000 hours, his constraint may be expressed in the form IXI+1X2 8000, or Simply Xl+ 12 08000 (9-5) Considered on its own, this constraint is shown in Figure 9-3. But this second constraint cannot be considered alone. It must be considered simultaneously with the first constraint. That is, the two inequalities must be considered together: Xl+X20 8000 (9-6) Both of these constraints should be drawn on one graph (superimposed on each other) to form the area of feasible solutions, illustrated as the shaded portion of Figure 9-4. Only the shaded area satisfies both constraints, being the area common to both constraints. This shaded area is a polygon containing the solution set. The set of points described by the polygon is convex. Figure 9-4 It will be noticed that once the second constraint was added the area of feasible solutions was reduced. Whenever another constraint having no new variables is added, the area of feasible solutions is less than or equal to the prior situation. We have already mentioned that we ignore negative values for xl and x2, because negative production values are not economically meaningful. Nevertheless, such a restriction is made explicit in an LP model, so that to (9-6) we should add the non- egativity constraints xl ,x200: That is, xl or x2 may be greater than zero, or equal to zero, but not less than zero (e, not negative). Although the constraints in (9-7) limit the available alternatives, they do not provide a unique answer to the decision problem of how much of each product to produce. The objective function in (9-2) which indicates the aim of maximising variable profit, is used to appraise the various possible solutions, and must be added to (9-7) to present a full statement of the problem: subject to 21+42 20 OOO so-profit Line Solution Method (machine constraint) finishing constraint) (9-8) can be added to the graph depicted in Figure 9-4. The objective function can take an infinite number of positions on the graph, each position being an so-profit line. If we examine the objective function P = 41+52 we note that it is the equation of a straight line with a negative slope of -4/5 (which can be read straight from the coefficients if xl is plotted on the horizontal axis). Formally, we can show that this is correct. The equation of a straight line is usually of the form y = mx+b where y is plotted on the vertical axis, m is the gradient and b is the y-intercept. We have placed x2 on the vertical axis, so let us convert the objective function to a form which corresponds with y = mx+b: 41+52 is converted to 52 = -4xl 12 = +P/5 So in (9-9) we verify that the slope (m) is equal to -4/5. The x2 intercept (not of great interest to us) is P/5. Thus the objective function has a slope of -4/5, and can take on many parallel positions each of which gives a different value for P. So, if one position is plotted, other positions are merely parallel shifts of this line. The further from the origin the objective function line is, the higher is the profit level. Therefore the optimum solution occurs when the objective function is as far from the origin as possible, but is Just touching the solution set. At its highest point it may touch a vertex on either axis, such as xl=8000 or x2=5000, or at the point of intersection of the two constraints, in all three cases there being a unique solution. In some cases it may lie along a constraint, indicating multiple optima. To make an initial plot of this objective function which has a slope of -4/5, we Join the points 4 on the vertical axis and 5 on the horizontal, or some multiple of them; for xample, 4000 on the vertical (x2) axis and 5000 on the horizontal (xl) axis, shown as a broken line in Figure 9-5. We then use a ruler and move it parallel to and up from the initial plot until we get a line Just touching the solution set. In this case there is a unique solution because the objective function (broken line) is Just touching a vertex at the point (6000,2000) meaning a production of 6000 units of Product 1 and 2000 units of Product 2. We either read these values directly from the graph, or else we determine them by solving simultaneously the equations of the two straight lines hich intersect at the desired point. Figure 9-5 Solving simultaneously: 8 ooo so 12=2000 and by substitution Xl = 6000 Note that the optimum solution, given two constraints, calls for production of both products. In fact, there can never be more products than the number of constraints, but there may be fewer. The solution (6000,2000) is the point of maximum profit. We calculate the variable profit by substituting 6000 for xl and 2000 for x2 into the objective function: p = 4xl +512 = $34 000. Finally, we can subtract the period fixed costs to obtain net profit: Net profit = $34 000-$18 OOO = $16 000. Relative Gradient Solution Method Instead of actually plotting the objective function to determine the optimum product mix we could simply compare the gradients of the objective function and the constraints. Reading from (9-8) we have the following gradients: Objective function -4/5 = -0. Machine constraint -2/4 = -1/2=-0. 5 Finishing constraint -1/1 † We note that the gradient of the objective function (-0. 8) lies between the gradients of the two constraints (-0. 5 and -1. 0). Therefore the optimum solution will be at the intersection of the two constraints. If the objective function were flatter than either onstraint (for example, a gradient of -0. 4) the optimum solution would be on the x2 axis at (0,5000); if it were steeper than either constraint (say -2) the optimum solution would be on the xl axis at (8000,0). Check these out by altering the angle of your ruler and pushing it away from the origin. Corner Point Solution Method A third method for determining the optimum solution is to evaluate the objective function at each corner point of the feasible set. This is the way the simplex method works, starting from the point (0,0) and working around from vertex to vertex until the largest objective function value is found. So we simply substitute the co-ordinates for each extreme vertex (corner point) in the feasible set into the objective function to find the maximum value. In Figures 9 9 9-3, plus the additional information that no ore than 3000 units of Product 2 can be sold per time period. Determine the optimum mix and period net profit. The new LP model will be simply the previous one in (9-8) augmented by the inclusion of the additional market constraint: Maximise 41+52 Subject to 21+42 0 20 000 (machine constraint) xl+ x2 0 8 000 (finishing constraint) 12 C] 3000 (market constraint) (9-10) Observation of Figure 9-6 reveals that the previous solution still holds: xl=6000, x2=2000, net profit = $16 000. Although the new constraint has reduced the size of the solution set, it has had no impact on the optimum solution. Spreadsheet