his in thee happen Grope 211 (The Reddy Milks Company) Naka Midas produive both posterior and exterior paints from two raw materials, Aft the Filming table provides the basic date of the problem The first d of interior and tim of tow material per tom of Maximnow daily demand of int An implic assume negat requirement The comp A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than I ton. Also, the maximum daily demand of interior subject to paint is 2 tons Reddy Mikks wants to determine the optimum (best) product mix of interior and extenor paints that maximizes the total daily profit. The LP model, as in any OR model. has three basic components 1. Decision variables that we seek to determine 2. Objective (goal) that we aim to optimize 3. Constraints that we need to satisfy Any values solution F The proper definition of the decision variables is an essential first step in the develop- ble becaus ment of the model. Once done, the task of constructing the objective function and the restrictions constraints is more straightforward. constraint. For the Reddy Mikks problem, we need to determine the amounts to be produced which is les of exterior and interior paints Thus the variables of the model are defined as function a x) = Tons produced daily of exterior paint (thousand From *2 = Tons produced daily of interior paint feasible s To construct the objective function, the company wants to increase its profit as straints. Er much as possible. Letting z represent the total daily profit (in thousands of dollars), the an infinite objective of the company is expressed as will locat algebraic Maximize z = 5x, + 4x2 Next, we construct the constraints that restrict raw materials usage and demand. In The raw materials restrictions are expressed verbally as Linearity Usage of a raw material ) Maximum raw material by both paints availability 1. objective From the data of the problem, we have the valu Usage of raw material MI per day = 6x, + 4x, tons and 4x2 respecti Usage of raw material M2 per day = 1x, + 2x2 tons tionality Because the daily availabilities of raw materials MI and M2 are limited to 24 and 6 when s tons, respectively, the associated restrictions are given as the proCHAPTER 2 Introduction to Linear Programming Linear programming (LP) applies to optimization models in which the objective and constraint functions are strictly linear. The technique is used in a wide range of applications, including agriculture, industry, transportation, economics, health sys- tems, behavioral and social sciences, and the military. It also boasts efficient compu- tational algorithms for problems with thousands of constraints and variables. Indeed, because of its tremendous computational efficiency, LP forms the backbone of the solution algorithms for other OR models, including integer, stochastic, and nonlinear programming. This chapter starts with the case of a two-variable model and presents its graphi- cal solution. The graphical solution provides insight into the development of the gen- eral algebraic simplex method (see Chapter 3). It also gives concrete ideas for the development and interpretation of sensitivity analysis in LP. The chapter closes with the formulation and solution interpretation of a number of realistic applications. 2.1 TWO-VARIABLE LP MODEL This section deals with the graphical solution of a two-variable LP. Though two- variable problems hardly exist in practice, the treatment provides concrete ideas for the development of the general solution algorithm presented in Chapter 3. 1114 Chapter 2 introduction to Linear Programming 2. Adduivity stipulates that the total contribution of all the variables in the 2.2.1 offjective function and their requirements in the constraints are the direct sum of the individual contribution or requirements of each variable, In the Reddy Mikks model, the total profit equals the sum of the two individual profit components. If however, the two products compete for the same market share in a manner that an increase in sales of one adversely affects the other. then the additivety property is not satisfied PROBLEM SET 2.1A 1. For the Redily Mikks model, construct each of the following constraints and express them with a constant right-hand side: (a) The daily demand for interior paint exceeds that of exterior paint by at least I ton (b) The daily usage of raw material M2 is at most 6 tons and at least 3 tons (c) The demand for interior paint cannot be less than the demand for exterior paint. (d) The minimum quantity that should be produced of both the interior and the exterior paint is 3 tons (e) The proportion of interior paint to the total production of both interior and exterior paints must not exceed S. 2. Determine the best feasible solution among the following (feasible and infeasible) solu- tions of the Reddy Mikks model: (8) - 1.1 = 4 (b) * = 2, 13 = 2 (c) * - 3, x, - 1.5 (d) ry - 2.x - 1 (e) 1 = 2. 17 = -1 3. For the feasible solution x, = 2,x, = 2 of the Reddy Mikks model, determine (a) The unused amount of raw material MI. (b) The unused amount of raw material M2. 4. Suppose that Reddy Mikks sells its exterior paint to a wholesaler at quantity discount. The profit per ton is $5000 if the contractor buys no more than 2 tons daily and $4500 otherwise. Can this situation be modeled as an LP model? GRAPHICAL LP SOLUTION The graphical procedure includes two steps: 1. Determination of the solution space that defines all feasible solutions of the model. solution space. 2. Determination of the optimum solution from among all the feasible points in the The procedure uses two examples to show how the maximization and minimiza- tion objective functions are handled.2.2 Graphical LP Solution 15 variables in the 2.2.1 Solution of a Maximization Model the direct sum of the Reddy Mikks Example 2.2-1 t components It. a manner that an This example solves the Reddy Mikks model of Section 2.1. tivity property is Step 1. Determination of the Feasible Solution Space: First, we account for the nonnegativity constraints x, 2 0 and x, 2 0. In Figure 2.1, the horizontal axis r, and the vertical axis x, represent the exterior and interior paint variables, respectively. It then follows that the nonnegativity constraints will restrict the solution space area to the first and expos them quadrant that lies above the x-axis and to the right of the x,-axis. To account for the remaining four constraints, first replace each inequal- by at hour I ton ity with an equations and then graph the resulting straight line by locating two distinct points on it. For example, after replacing ox, + 4x, $ 24 with exterior paint. the straight line ox, + 4x2 = 24, two distinct points can be determined first rand the exterior by setting x, = 0 to obtain t, = = 6 and then setting .x2 = 0 to obtain K = = 4. Thus, the line passes through the two points (0. 6) and (4. 0), as mor and exter shown by line (1) in Figure 2.1. Next, consider the effect of the inequality. All the inequality does is edcable) solo divide the (x), x,)-plane into two half-spaces, one on each side of the plotted line. Only one of these two halves satisfies the inequality. To determine the correct side, choose any reference point in the first quadrant. If it satisfies the inequality, then the side in which it lies is the feasible half-space. Else, the other side is. It is convenient computationally to select (0, 0) as the reference. FIGURE 2.1 Feasible space of the Reddy Mikks model Constraints 6x] + 412 5 24 1 x, + 2x, 5 6 2 125 2 4 (3) 205 12 0 6 2 of the model. points in the (4 E D Solution nd minimiza- pace 2 32.1 Two-Variable Le Model 12 6 + 4s, $ 24 (Raw material I) A + 21, s 6 (Raw material M2) derials A The first demand restriction says that the difference between the daily production of interior and exterior paints, , - x, does not exceed 1 ton, which translates to 4-4 1. The second demand restriction stipulates that the maximum daily demand of interior paint is limited to 2 tons, which translates to r = 2. An implicit for "understood-to be") restriction is that variables a, and x, cannot assume negative values The nonnegativity restrictions, , = 0.x 2 0, account for this requirement. The complete Reddy Mikks model is Maximize z = Sx, + 4x subject to Ox, + 40, $ 24 erion and 4+ 20 56 12 2 1.1, 20 Any values of x, and r. that satisfy all the constraints of the model constitute a feasible solution For example, the solution., = 3 tons per day and x, = 1 ton per day, is feasi- ble because it does not violate any of the constraints, including the nonnegativity restrictions To verify this result, substitute (x, = 3,x, = 1) in the left-hand side of each constraint. For example, in the first constraint, or, + 4x, - 6 X 3 + 4 x 1 = 22. which is less than the right-hand side of the constraint (= 24). The value of the objective function associated with the solution (x, = 3, x, = 1) is : = 5 X 3 + 4 x 1 = 19 (thousand dollars). From the standpoint of the entire model, we are interested in finding the optimum feasible solution that yields the maximum total profit while satisfying all the con- straints Enumeration of the feasible solutions is not acceptable because the model has an infinite number of feasible solutions. Instead, we need a systematic procedure that will locate the optimum solution efficiently. The graphical method in Section 2,3 and its algebraic generalization in Chapter 3 address this point. In the preceding example, the objective and constraints functions are all linear, Linearity implies that the LP must satisfy two properties: proportionality and additivety 1. Proportionality requires the contribution of each decision variable in the objective function and its requirements in the constraints to be directly proportional to the value of the variable. For example, in the Reddy Mikks model, the quantities 5x, and 4x, give the profits for producing x, and x, tons of exterior and interior paint, respectively, with the unit profits per ton. 5 and 4, providing the constants of propor- tionality. If. on the other hand, Reddy Mikks grants some sort of quantity discounts and 6 when sales exceed certain amounts, then the profit will no longer be proportional to the production amounts, r, and xz.2.2 Graphical LP Solution 17 The values of x, and x, associated with the optimum point Care deter- mined by solving the equations associated with lines (1) and (2)-that is, 6x, + 41 = 24 * + 212 -6 The solution is x, = 3 and x, = 1.5 with z - 5 x 3 + 4 x 1.5 - 21. This calls for a daily product mix of 3 tons of exterior paint and 1.5 tons of inte- rior paint. The associated daily profit is $21,000. It is not accidental that the optimum solution occurs at a corner point of the solution space where two lines intersect. Indeed, if we change the slope of the profit function z (by changing its coefficients), we will discover that the optimum solution always occurs at one of these corner points. This obser vation is key to the development of the general simplex algorithm presented in Chapter 3. PROBLEM SET 2.2A 1. Determine the feasible space for each of the following independent constraints, given that r, x, 2 0. (2) -3 + 12 56 (b) x - 2r, 2 5 (c) 21 - 3x, $ 12 (d) x - 1 50 (e) - x + x,20 2. Identify the direction of increase in z in each of the following cases: (a) Maximize z = 11 - X2 (b) Maximize z = -5.x - 612 (c) Maximize z = -X1 + 2x2 (d) Maximize z = -3x1 + X2 3. Determine the solution space and the optimum solution of the Reddy Mikks model for each of the following independent changes: (a) The maximum daily demand for exterior paint is at most 2.5 tons. (b) The daily demand for interior paint is at least 2 tons. (c) The daily demand for interior paint is exactly 1 ton higher than that for exterior paint. (d) The daily availability of raw material MI is at least 24 tons. (e) The daily availability of raw material MI is at least 24 tons, and the daily demand for interior paint exceeds that of exterior paint by at least 1 ton. 4. For the (original) Reddy Mikks model, identify the corner point(s) that define the opti- mum solution for each of the following objective functions: (a) z = 3x, + X2 (b) z = x1 + 3x2 (C) z = 6x1 + 4.12 How does the solution in (c) differ from those in (a) and (b)?1 Chapter 2 Introduction to Linear Programming & lock is an aspiring freshman at Uler University. He realizes that "all work and no play subject make Jack a dull boy " As a result, Jack wants to apportion his available time of about 10 hours a day between work and play He estimates that play is twice as much fun as work He also wants to study at least as much as he plays However, Jack realizes that if he is going to get all his homework assignments done, he cannot play more than 4 hours a day. How should Jack allocate his time to maximize his pleasure from both work and play? 2.2.2 Solution of a Minimization Model Fig Mikks Example 2.2-2 (Diet Problem) gin. To Ozark Farms uses at least 800 lb of special feed daily. The special feed is a mixture of obtainc com and soybean meal with the following compositions: able. F or Xy and (20 Ib per lb of feedstuff and 3 b Cost (S/b) or (0,] Protein Fiber Be 02 30 Corn need Soybean meal 90 The 21x The dietary requirements of the special feed are at least 30% protein and at most mum 5% fiber, Ozark Farms wishes to determine the daily minimum-cost feed mix. Because the feed mix consists of corn and soybean meal, the decision variables of the model are defined as 1 = lb of corn in the daily mix x2 = lb of soybean meal in the daily mix The objective function seeks to minimize the total daily cost (in dollars) of the feed mix and is thus expressed as 1500 minimize z = .3x1 + .9x2 The constraints of the model reflect the daily amount needed and the dietary requirements. Because Ozark Farms needs at least 800 lb of feed a day, the associated constraint can be expressed as x1 + x2 2 800 1000 As for the protein dietary requirement constraint, the amount of protein included in x, lb of corn and x, lb of soybean meal is (.09x, + .6x,) 1b. This quantity should equal at least 30% of the total feed mix (x, + x2) lb; that is .09.x, + . 6.x2 2 . 3(x1 + x2) In a similar manner, the fiber constraint is constructed as .02x1 + 06.x2 5 .05(x, + x2) The constraints are simplified by grouping all the terms in x, and x, and moving them to the left-hand side of each inequality, leaving only a constant on the right-hand side. The complete model thus becomes minimize z = .3x, + 9.x22.2 Graphical LP Solution 19 subject to Ny @ 800 21.x, - 305 5 0 03x, - 015, 2 0 Figure 2.3 provides the graphical solution of the model. Unlike those of the Reddy Mikks model (Example 2 2-1), the second and third constraints pass through the ori- gin. To plot the associated straight lines, we need one additional point, which can be obtained by assigning one of the variables a value and then solving for the other vari- able. For example, in the second constraint. r, = 200 will yield .21 x 200 - 3x, - 0, or x; - 140. This means that the straight line .21x, - 3x, - 0 passes through (0, 0) and (200, 140). Note also that (0, 0) cannot be used as a reference point in constraints 2 and 3 because both lines pass through the origin. Instead, any other point [e.g, (100,0) or (0, 100)] can be used for that purpose. Because the present model seeks the minimization of the objective function, we need to reduce the value of z as much as possible in the direction shown in Figure 2.3. The optimum solution is the intersection of the two lines x, + x2 - 800 and 216 - 3x; = 0, which yields x, = 470.6 lb and x, = 329.4 lb. The associated mini- mum cost of the feed mix is z = .3 x 470.6 + .9 x 329.4 = $ 437.64 per day. FIGURE 2.3 Graphical solution of the diet model 1500- Minimize z - 35, + .9X2 03x, -01, 20 1000 21x, - 3x, 50 500 Optimum: 1 = 470.6 1b *) = 329.4 1b Ifx2800 2 = $437.64 500 1000 150016 Chapter 2 Introduction to Linear Programming point, unless the line happens to pass through the origin, in which case another point must be selected, The use of the reference point (0, 0) is illustrated with the constraint Or, + 41 5 24, Because 6 2 0 + 4 X 0 - 0 is less than 24, the half-space representing the inequality includes the origin (as shown by the arrow in Figure 2.1) To demonstrate the use of other reference points, consider (6.0) Here, 6 x 6 + 4 X 0 - 36, which is larger than the right-hand side of the first constraint, meaning that the side in which (6.0) lies is not feasible for the incquality. This result is consistent with the one obtained using (0. 0) as the reference point. Application of the reference point procedure to all the constraints of the model produces the feasible space shown in Figure 2.1. Step 2. Determination of the Optimum Solution: The feasible space in Figure 2.1 is delineated by the line segments joining the corner points A. B. C. D. E. and F. Any point within or on the boundary of the space ABCDEF is feasible, in the sense that it satisfies all the constraints. Because the feasible space ABCDEF consists of an infinite number of points, the need for a systematic procedure to identify the optimum solution PROBLE is obvious The determination of the optimum solution requires identifying the 1. De direction in which the profit function z = Sy, + 4x, increases (recall that we tha are maximizing :). We can do so by assigning arbitrary increasing values to z. For example, using : = 10 and z = 15 would be equivalent to graphing the (a) two lines 5x, + 4x. = 10 and 5x, + 41. = 15. Thus, the direction of increase in z is as shown Figure 2.2. The optimum solution occurs at C, which is the point in the solution space beyond which any further increase in z will put us outside the boundaries of ABCDEF. FIGURE 2.2 2. Id Optimum solution of the (Maximize = = 5x, + 4x7) Reddy Mikka model Increasing 17 - 2 1 + 2x, =6 3. D 2 D Optimum: x = 3 tons *2 = 1.5 tons