29. Linear Programming: Graphical Methods Linear programming (LP) is a mathematical technique designed to help managers in their planning and decision making. It is usually used in an organization that is trying to make most effective use of its resources. Resources typically include machinery, manpower, money, time, warehouse space, or raw materials.A few examples of problems in which LP has been successfully applied are:Development of a production schedule that will satisfy future demands for a firm’s product and at the same time minimize total production and inventory costs.
3Establishment of an investment portfolio from a variety of stocks or bonds that will maximize a company’s return on investment.Allocation of a limited advertising budget among radio, TV, and newspaper spots in order to maximize advertising effectiveness.Determination of a distribution system that will minimize total shipping cost from several warehouses to various market locations.Selection of the product mix in a factory to make best use of machine and man hours available while maximizing the firm’s profit.
4These are but a few of the applications of linear programming that we shall discuss in the next few units of this book. All LP problems. As you can see above, seek to maximize or minimize some quantity (usually profit or cost). We refer to this property as the objective of an LP problem.The second property that LP problems have in common is the presence of restrictions, or constraints, that limit the degree to which we can pursue our objective. We want, therefore, to maximize or minimize a quantity (the objective function) subject to limited resources (the constraints).
5Formulating Linear Programming Problems A very common linear programming problem is the “product mix” problem. Two or more products are usually formed using limited resources, such as personnel and machines. The profit, which the firm seeks to maximize, is base on the profit contribution per unit of each product. The company would like to determine how many units of each product they should produce so as to maximize overall profit given their limited resources.
6EXAMPLE 9.1The Dress-Rite clothing manufacturer, which produces men’s shirts and pajamas, has two primary resources available: sewing machine time (in the sewing department) and cutting machine time (in the cutting department). Over the next month, Dress-Rite can schedule up to 280 hours of work on sewing machines and up to 450 hours of work on cutting machines.Each shirt produced requires 1 hour of sewing time and 1½ hours of cutting time. To output each pair of pajamas, 3/4 hour of sewing time and 2 hours of cutting time are needed.
7Our goal here is to develop relationships to describe these restrictions. One general relationship is that the amount of resource used is ≤ the amount of resource available.In the case of the sewing department, for example, the total time used is:(1 hour/shirt) X (number of shirts produced) + (3/4 hour/pajama) X (number of pajamas produced)To express the LP constraints for this problem mathematically, we letX1 = number of shirts producedX2 = number of pajamas produced
8Then1st constraint: 1X1 + ¾ X2 ≤ 280 (hours of sewing machine time available—our first scarce resource)2nd constraint: ½ X1 + 2 X2 ≤ 450 (hours of cutting machine time available—our second scarce resource)Note: This means that each pair of pajamas takes about 2 hours of the cutting resource).
9EXAMPLE 9.2In Example 9.1, Dress-Rite established two constraints which will keep the company from exceeding the machine time available for production. But the company wants to determine the product mix (of shirts and pajamas) that will maximize its profits. Its accounting department analyzes cost and sales figures and states that each shirt produced will yield a $4 contribution to profit and that each pair of pajamas will yield a $3 contribution to profit.
10This information can be used to create the LP objective function for this problem. Objective function: maximize total contribution to profit = $4X1 + $3X2where againX1 = number of shirts producedX2 = number of pajamas produced
11EXAMPLE 9.3Electro Corp. manufactures two electrical products, air conditioners and fans. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes 4 hours of wiring and 2 hours of drilling. Each fan must go through 2 hours of wiring and 1 hour of drilling. During the next production period, 240 hours of wiring time are available and up to 100 hours of drilling time may be used. Each air conditioner sold yields a profit of $27—each fan assembled may be sold for a $15 profit.Electro would like this production mix situation formulated as a linear programming problem.
12LetX1 = number of air conditioners to be producedX2 = number of fans to be producedObjective function: maximize profit = $27 X1 + $15 X21st constraint: 4X1 + 2X2 ≤ 240 (hours of wiring time available)2nd constraint: 2X1 + 1X2 ≤100 (hours of drilling time available)Constraints added to ensure nonnegative solutions:X1 ≥ 0X2 ≥ 0
13We see in the previous examples that each linear programming problem contains an objective function,certain constraints (resulting from limited resources) which keep the firm from producing unlimited quantities and hence making unlimited profits, andcertain interactions between variables: namely, the more units of one product produced, the less the firm can make of other products.
14It is possible for a problem to have many more variables, or products, involved as we shall see in forthcoming units. It is also possible to have more than two constraints in even a simple LP problem. Constraints may involve not only “less than or equal to (≤) inequalities,” but actual “equalities” (=) and “greater than or equal to (≥) inequalities” as well.
15EXAMPLE 9.4Referring back to the Electro Corp. in Example 9.3, we are informed that the firm’s management wishes to reformulate their LP production mix problem. In particular, management has become very sensitive to wasted, or unused, time on their expensive new drilling machine and wants all 100 hours of available time used in production. The firm also decides that to ensure an adequate supply of air conditioners for a contract, at least 25 air conditioners should be manufactured. Since it incurred an oversupply of fans the previous period, it also insists that no more than 70 fans be produced.We can use this new information to reformulate the constraints of Example 9.3
16Objective function: maximize profit = 27X1 + 15X2 (no change) Wiring-time constraint: 4 X1 + 2 X2 ≤ 240 (no change)Drilling-time constraint: 2X1 + 1X2 = 100 (now an “equality”)Air-conditioner minimum constraint: X1 ≥ 25 [ new greater than or equal to (≥) constraint ]Fan maximum constraint: X2 ≤ 70 (new fan constraint)
17GRAPHICAL REPRESENTATION OF CONSTRAINTS In order to determine the optimal solution to a linear programming problem, we must first identify a set, or region, of feasible solutions. When there are only two variables in the problem, as in the examples and problems above, this can be accomplished by plotting the constraints on a two-dimensional graph.The variable X1 is usually treated as the horizontal axis and the variable X2 as the vertical axis. Because of the nonnegativity constraints (i.e., X1 ≥ 0, X2 ≥ 0), we are always working in the first (or northeast) quadrant of the graph. The graphical method of dealing with LP problems is best demonstrated by way of an example.
18EXAMPLE 9.5The Flair Furniture Company produces inexpensive tables and chairs. it has formulated the following LP problem:Maximize profit =$7X1 + $5X2subject to: X1 + 3X2 ≤ 24 (constraint A)2 X1 + 1X2 ≤ 10 (constraint B)X1 ≥ 0, X2 ≥ 0where X1 stands for the number of tables produced and X2 stands for the number of chairs produced. We would like to graphically represent the constraints of this problem.The first step is convert the constraint inequalities into equalities (or equations):4X1 + 3X2 = 24 (A) 2X1 + 1X2 = 10 (B)
19The equation on the left (A) is plotted in Figure 9 The equation on the left (A) is plotted in Figure 9.1(A) and the one on the right in Figure 9.1(B). To plot
20The line in Figure 9.1(A), all we needed to do was find the points at which the line intersected the X1 and X2 axes. When X1 = 0 (The location where the line touches the X2 axis), it implies that 3 X2 = 24 or that X2 = 8. Likewise, when X2 = 0, we see that 4 X1 = 24 and that X1 = 6. Thus, the A constraint is bounded by the line running from (X1 = 0, X2 = 8) to (X1 = 6, X2 = 0 The shaded area represents all points that satisfy the original inequality.Constraint B is illustrated in Figure 9.1(B) in a similar fashion. When X1 = 0, then X2 = 10, and when X2 = 0, then X1 = 5. The B constraint then is bounded by the line between (X1 = 0, X2 = 10) to (X1 = 5, X2 = 0) and the shaded area represents the original inequality.
21Figure 9.2 shows both constraints together- the shaded region is the part that satisfies both restrictions.
22The shaded region in Figure 9 The shaded region in Figure 9.2 is called the area of feasible solutions or simply the feasible region. This region must satisfy all conditions specified by the program’s constraints, and is thus the region where all constraints overlap. Any point in the region would be a feasible solution to the Flair Furniture Company problem—any point outside the shaded area would represent an infeasible solution. Hence, it would be feasible to manufacture 3 tables and 2 chairs (X1 =3, X2=2), but it would violate the constraints to produce 7 tables and 4 chairs.
23THE CORNER-POINT SOLUTION METHOD There are several approaches that can be taken in solving for the optimal solution once the feasible region has been established graphically. The simplest one conceptually is called the corner-point method. The mathematical theory behind linear programming states that the optimal solution to any problem (that is, the values of the X1 variables which yield the maximum profit or minimum cost) will lie at a corner point, or extreme point, of the feasible region. Hence, it is only necessary to find the values of the variables at each corner—the maximum profit or optimal solution will lie at one of them. This concept is illustrated in Example 9.6
25We skipped corner point c momentarily because to accurately establish its coordinates, it is necessary to solve for the intersection of the two constraint lines. To do so, we apply the method of simultaneous equations to4X1 + 3X2 = 24 and 2X1 + 1X2 = 10Multiply the second equation by -2 and add it to the first equation:4X1 + 3X2 = 24-4X1 - 2X2 = [ which is –(2)(2X1+ 3X2 = 10)]1X2 = 4When X2= 4, then 4X1 + (3)(4)= 24, implying that4X1= 12 or X1= 3. Now we can evaluate point cpoint C: (X1 = 3, X2 = 4) profit = ($7)(3) + ($5)(4) =$41
26Because point c produces the highest profit of any corner point, the product mix X1 = 3 tables and X2 = 4 chairs is the optimal solution to Flair Furniture’s problem. This solution will result in a profit of $41.
27Note: Although the optimal solution to Example 9. 6 and Problems 9 Note: Although the optimal solution to Example 9.6 and Problems 9.6 and 9.7 fell at a corner point which was located at the intersection of two constraint equations, this is by no means always the case. The optimal solution could just as easily have been at a corner bordering on the X1 or X2 axis (such as points b or d in Example 9.6). It all depends upon the values of the objective function coefficients and the angle of the profit line. This topic is discussed next.
28THE ISO-PROFIT-LINE SOLUTION METHOD A second approach to graphically solving linear programming problems employs the iso-profit line. This technique is often more speedy than the corner-point method, for we do not have to evaluate the profit at every corner. Instead, we draw a series of parallel profit lines. Any point along a particular iso-profit line will have the same profit or value of the objective function. The highest profit line (or one farthest from the zero origin) which touches the feasible region pinpoints the optimal solution.
29EXAMPLE 9.7 The objective function for an LP problem is given as: maximize profit = 2X1 + 8X2The corresponding feasible region has been graphed in Figure 9.4. We wish to determine which corner point is optimal using the iso-profit-line solution method.
31We may begin by selecting any profit level, for example, P=$200, and drawing its (dashed) iso-profit line. Is there a higher possible profit line which touches a corner of the feasible region? The answer appears to be yes, and by drawing a series of parallel profit lines farther and farther away from the origin, we eventually reach the tip of the feasible region (at corner point b). The profit there will be P = $2(X1 = 0) + 8(X2 = 50) = 2(0) + 8(50) = $400. Thus, the maximum profit line is 400 = 2X1 + 8X2
32MINIMIZATION PROBLEMS IN LINEAR PROGRAMMING Many linear programming problems involve minimizing an objective such as cost, instead of maximizing a profit function. Basically the same graphical approaches may be applied to solving these types of problems.
33EXAMPLE 9.8The dean of the Western College of Business must plan his school’s course offering for the fall semester.Student demands deem it necessary to offer at least 30 undergraduate and 20 graduate courses in the term.Faculty contracts also dictate that at least 60 courses be offered in total.Each undergraduate course taught costs the college an average of $2,500 in faculty wages, while each graduate course costs $3,000.We may formulate this as a minimization LP problem as follows.LetX1= number of undergraduate business courses offered in fallX2 = number of graduate business courses
34Minimization problems can be solved graphically by first establishing the feasible region, and then by using either the corner-point method or an iso-cost-line approach (analogous to the iso-profit approach in maximization problems) to find the values of X1 and X2 which yield the minimum cost. The corner-point method is employed below to solve Example 9.8.
35EXAMPLE 9.9To solve Example 9.8 graphically, we construct the problem’s feasible region (Figure 9.5). Minimization problems are often unbounded outward (that is, on the right side and on the top), but this causes no problem in solving them. As long as they are bounded inward (on the left side and the bottom), corner points may be established. The optimal solution will lie at one of the corners.
37In the case above, there are only two corner points, a and b In the case above, there are only two corner points, a and b. It is easy to determine that at point a, X1= 40 and X2= 20 and that at point b, X1= 30 and X2= 30. The optimal solution is found at the point yielding the lowest total cost.total cost at a = $2,500 X1 + $3,000 X2= (2,500)(40) + (3,000)(20) = $160,000total cost at b = $2,500 X1 + $3,000 X2= (2,500)(30) + (3,000)(30) = $165,000The lowest cost to the college is at point a; hence, the dean should schedule 40 undergraduate courses and 20 graduate courses.
38As mentioned, the iso-cost-line approach may also be used to solve LP minimization problems. As with iso-profit lines (see Example 9.7 and Problem 9.8), we need not compute the cost at each corner point, but instead draw several parallel cost lines, The lowest cost line to touch the feasible region provides us with the optimal solution corner.