Operations Research. Mathematical modeling exercise.

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1 Operations Research Mathematical modeling exercise.

2 The system of linear inequalities problems often leads to a system of linear inequalities (SLI). We will learn to convert the SLI to system of linear equations (SLE). Already known way we will find the SLE which is in canonical form and therefore also obtained initial basic solution. Definition: If A is a matrix system, x T the unknown vector and b T vector of right sides, we understand the system m linear inequalities of n unknowns the system A x T b T, tj. a i1x 1 + a i2x a inx n b i, i = 1, 2,..., m. (1) In doing so symbol represents one of the symbols of,, or more generally one of the symbols of, =,.

3 Definition: Solution of systems of linear inequalities (1) is each n-tuple x = (x 1, x 2,..., x n), which is satisfied for all m inequalities of the system. Inequalities we transfer to equations using the additional variables x i, i = 1, 2,..., m. Additional variable expresses the absolute value of the difference between the left and right side (L i and P i) of relevant inequality, ie x i = L i P i, i = 1, 2,..., m. Their meaning is obvious that they must satisfy the conditions of the nonnegativity x i 0, i = 1, 2,..., m. (2) Then inequalities (1) of type becomes to equations a i1x 1 + a i2x a inx n + x i = b i, i = 1, 2,..., m, (3) and inequalities (1) of type becomes to equations a i1x 1 + a i2x a inx n x i = b i, i = 1, 2,..., m. (4)

4 We get a set of m linear equations of m + n unknowns with nonnegativity conditions (2). Each solution SLI (1) corresponds to a solution of the corresponding SLE (3) or (4), and vice versa. The claim is the subject of the following sentence. Theorem: n-tuple (x 1, x 2,..., x n) is the solution a system of linear inequalities (1) if and only if (n + m)-tice (x 1, x 2,..., x n, x 1, x 2,..., x m) is the solution a system of linear equations (3) or (4) (or their combinations), if nonnegativity conditions (2) are satisfied.

5 Example: The system of linear inequalities 7x 1 + 5x 2 850, 2x 1 + 2x translate using additional variables to SLE and use complete elimination to find one of its basic solution that the position of the non-basic variables are additional variables.

6 Solution: SLI convert to SLE, ie introduce additional variables x 1 0 a x x 1 + 5x 2 + x 1 = 850, 2x 1 + 2x 2 + x 2 = 300. SLE is already in canonical form (basic variables are x 1 a x 2), proceed to table, which gradually determine the required solution: B.v. x 1 x 2 x 1 x 2 b i x x 2 2 (2) x 1 (2) x x x x 1 = x 2 = 0 x (1) = (0, 0; 850, 300) x 1 = x 2 = 0 x (2) = (0, 150; 100, 0) x 1 = x 2 = 0 x (3) = (50, 100; 0, 0) In the final step of the table, we obtain the desired solution, i.e. x = (50, 100; 0, 0).

7 Tasks: Convert SLI to SLE with additional variables and by complete elimination find one of its basic solution that the position of non-basic variables are only additional variables. a) 6x 1 + 2x 2 12, x 1 + 5x b) 6x 1 + 2x 2 12, x 1 + 5x c) 2x 1 + x 2 10, x 1 + 3x 2 9. d) x 1 x 2 3, x 1 + 2x 2 0. e) x x 2 + 4x 3 30, x 1 + 4x 2 + x 3 9, 3x 2 + x f) x 1 + x 2 6, x 1 x 2 2, x 1 + 2x 2 6.

8 Solutions: a) (1, 3; 0, 0). b) (1, 3; 0, 0). c) (3, 4; 0, 0). d) (2, 1; 0, 0). e) ( 8, 6, 7; 0, 0, 0). f) (4, 2; 0, 0, 6), (2, 4; 0, 4, 0).

9 In this section we will focus on formulating linear programming problems, we will provide a compilation of mathematical models of typical problems that are solved by methods LP. Using specific examples, we will show what we will understand of production planning problem, the mixing problem and the partitioning problem, but also how to get a mathematical model of the problem and what the elements of this model mean.

10 The production planning problem The problem of this type we have already met in introductory chapter. Let us give another example. Example: The company produces two types of products V 1, V 2. Table shows the consumption of raw materials S 1 and S 2 in kg required to produce 1 product. Profit from one product V 1 is 18 CZK, profit from one product V 2 is 8 CZK. Further, the table indicates the quantity of raw materials, which company owns. The task is to determine the optimal production plan for that the company achieve the maximum possible profit. At this point, however, will not solve the problem, but create a mathematical model of the problem.

11 Solution: V 1 V 2 Available amount S 1 [kg/pc] S 2 [kg/pc] profit [CZK/pc] 18 8 max

12 If we create a mathematical model, we have to express the problem mathematically. First, you must define variable: x i means number of products V i, i = 1, 2. From meaning of the variables is evident why the nonnegativity conditions must be satisfied: x i 0, i = 1, 2. Vector x = (x 1, x 2) we call production vector. This vector uniquely determines the production plan of company. Furthermore, we formulate an own constraints. Term 4x 1 + 2x 2 expresses the amount of raw materials S 1 (in kg), which is required for the production of x 1 of productsv 1 and x 2 of products V 2. This amount can not exceed the amount that the company has to the period available for now. Thus: 4x 1 + 2x In the same way can be obtained also the constraint on the consumption of raw material S 2: 4x 1 + x

13 The total profit attaching a given production plan is determined by the function z = z(x 1, x 2) = 18x 1 + 8x 2. The company has achieved maximum profit, thus the objective function is of maximizing type. We write z = z(x 1, x 2) = 18x 1 + 8x 2 max. We are talking about maximization problem. We got all the elements of the mathematical model, we write it summarized: z = 18x 1 + 8x 2 max, 4x 1 + 2x , 4x 1 + x , x 1 0, x 2 0.

14 Example 1: The company produces two types of products V 1 and V 2. For their production needs raw materials S and industrial equipment Z. For planning period is available kg of raw material, and 240 hours of machine time. The production of 1 pc of product V 1 needs 10 kg of raw material S and 2 hours of the machine time Z; the production of 1 pc of product V 1 needs 8 kg of raw material S and 1 hour of machine time Z. Profit from one product V 1 is 308 CZK, profit from one product V 2 is 214 CZK. The task is to determine the optimal production plan for that the company achieve the maximum possible profit. Create a mathematical model of the problem.

15 Solution 1: z = 308x x 2 max, 10x 1 + 8x , 2x 1 + x 2 240, x 1 0, x 2 0.

16 Example 2: The company produces two types of products A and B. It has kg of material raw S and hours of machine time Z on equipment and hours of available employees time F. Production 1 pc of product A needs 6 kg S, 4 h Z and 2 h F. Production 1 pc of product B needs 4 kg S, 8 h Z, 2 h F and 2 pc of products A. Product A is component for the product B. Profit of 1 pc of A is 80 CZK, profit of 1 pc of B is 200 CZK. The task is to determine the optimal production plan with maximum possible profit. Create a mathematical model of the problem.

17 Solution 2: z = 80x x 2 max, 6x 1 + 4x , 4x 1 + 8x , 2x 1 + 2x , x 1 + 2x 2 0, x 1 0, x 2 0.

18 Example 3: The company produces three types of products V 1, V 2 and V 3 from three types of materials S 1, S 2 a S 3, which is available after a series 12, 21 and 12 tons. Production 1 pc of product V 1 needs 1 kg of material S 1, 2 kg S 2 and 0,5 kg S 3; Production 1 pc of product V 2 needs 2 kg of material S 2 and 2 kg S 3; Production 1 pc of product V 3 needs 2 kg of material S 1 and 3 kg S 2. Profit of 1 pc of V 1, resp. V 2, resp. V 3 is 10, resp. 15, resp. 18 CZK. The task is to determine the optimal production plan with maximum possible profit. Create a mathematical model of the problem.

19 Solution 3: z = 10x x x 3 max, x 1 + 2x , 2x 1 + 2x 2 + 3x , 0,5x 1 + 2x , x 1 0, x 2 0, x 3 0.

20 Mixing problem The essence of mixing tasks, which is also known as nutritional problem, we will once again demonstrate with an example. Example: The company has to create the feed mixture, containing at least 308 mg of calcium and at least 214 mg of magnesium. It uses two types of feed. One kilogram of feed K 1 contains 10 mg of calcium, 8 mg of magnesium, and costs 150 CZK. One kilogram of feed K 2 contains 8 mg of calcium, 1 mg of magnesium, and costs 24 CZK. The task is to determine the composition of the resulting feed mixture, so that the cost of its have been minimized. At this point we restrict ourselves to the creation of a mathematical model.

21 Solution: It is advantageous to transfer all the data in tabular form: K 1 K 2 The minimum amount Ca [mg/kg] Mg [mg/kg] cost [CZK/kg] max We will create mathematical model of mixing problem. The variables x i denotes quantities (in kg)of feed K i in the resulting mixture, i = 1, 2. The solution is represented by a vector x = (x 1, x 2), which we will call a mixing vector. Again, must be satisfied nonnegativity conditions, x i 0, i = 1, 2, because can not be used a negative amount of feed. If an company use x 1 kg of feed K 1 a x 2 kg of feed K 2, the amount of calcium and magnesium in the resulting mixture is: Ca : 10x 1 + 8x 2, Mg : 8x 1 + x 2.

22 In connection with the minimum required amount of calcium and magnesium in the resulting mixture, own constraints have shape: Ca : 10x 1 + 8x 2 308, Mg : 8x 1 + x Cost z of creation a mixture is given by objective function z = z(x 1, x 2) = 150x x 2. The objective function has the minimization type. Overall, the mathematical model of the problem is: z = 150x x 2 min, 10x 1 + 8x 2 308, 8x 1 + x 2 214, x 1 0, x 2 0.

23 Example 1: The task of company is from materials S 1 and S 2 as cheaply as possible produce a mixture, which contains at least 160 g of vitamin A, and at least 500 g of vitamin C. 1 kg raw material S 1 contains 1 g of vitamin A and 2 g of vitamin C and costs 30 CZK, 1 kg raw material S 2 contains 1 g of vitamin A and 5 g of vitamin C and costs 50 CZK. The task is to determine the optimal mixing vector. Create a mathematical model of the problem.

24 Solution 1: z = 30x x 2 min, x 1 + x 2 160, 2x 1 + 5x 2 500, x 1 0, x 2 0.

25 Example 2: Metallurgical factory has produced at least kg of zinc Zn and at least kg of tin Sn. There are three kinds of ores. 1 ton of ore A contains 10 kg Zn, 6 kg Sn and costs 200 CZK; 1 ton of ore B contains 2 kg Zn, 6 kg Sn and costs 150 CZK; 1 ton of ore C contains 12 kg Zn, 4 kg Sn and costs 250 CZK; The task is to determine in what quantity is necessary to use each kinds of ores, the total purchase price of ores was minimal. Create a mathematical model of the problem.

26 Solution 2: z = 200x x x 3 min, 10x 1 + 2x x , 6x 1 + 6x 2 + 4x , x 1 0, x 2 0, x 3 0.

27 Example 3: The task of is from raw materials S 1 and S 2 as cheaply as possible create a mixture of the total weight of 6 kg so that the mixture contains at least 8 g of magnesium and at least 60 g of calcium. 1 kg of raw material S 1 contains 1 g of magnesium, 12 g of calcium and its price is 12 CZK, 1 kg of raw material S 2 contains 1,5 g of magnesium, 10 g of calcium and its price is 20 CZK. Create a mathematical model of the problem.

28 Solution 3: z = 12x x 2 min, x 1 + x 2 = 6, x 1 + 1,5x 2 8, 12x x 2 60, x 1 0, x 2 0.

29 Partition problem Even this type of problem and creating a mathematical model will be showen on example. Example: Assume that we have sufficient number of basic rope length of 32 m. For further use we need at least 12 pieces of 20 m rope, at least 20 pieces of 11 m rope and at least 26 of 6 m rope. The task is to determine the optimal mix of cutting plans with respect to minimal waste. Construct a mathematical model of the problem.

30 Solution: First we have each cutting plans (ie ways to cut basic ropes length of 32 m) set and determine how many basic ropes each of these ways to cut up. We prepare table of cutting plans, see the table below: 32 m Cutting plan Required I. II. III. IV. V. amount 20 m m m Waste [m] min The cutting plan I. means that the basic rope of length 32 m cut into 1 pc of 20 m long rope, 1 pc of 11 m long rope and has waste of 1 m. Meaning other of cutting plans can be interpreted analogously.

31 We will create mathematical model of the partition problem. The x i denotes the number of basic 32 m long ropes cut up by i-th cutting plan, i = 1, 2,..., 5. The solution is then represented by a vector x = (x 1, x 2, x 3, x 4, x 5), which we will call partition vector. Also, nonnegativity conditions must be satisfied, x i 0, i = 1, 2,..., 5. If cut up x 1 ropes by cutting plan I., dots, x 5 ropes by cutting plan V., cut into x 1 + x 2 pc of 20 m long rope, x 1 + 2x 3 + x 4 pc of 11 m long rope, 2x 2 + x 3 + 3x 4 + 5x 5 pc of 6 m long rope, and overall length of waste (in meters) is z = x 1 + 4x 3 + 3x 4 + 2x 5.

32 In total, the mathematical model of the problem is: z = x 1 + 4x 3 + 3x 4 + 2x 5 min, x 1 + x 2 12, x 1 + 2x 3 + x 4 20, 2x 2 + x 3 + 3x 4 + 5x 5 26, x i 0, i = 1, 2,..., 5.

33 Example: Modify the previous example as follows. All conditions and requirements remain the same, just change the optimality criterion. Let is now an objective to determine the optimal mix of cutting plans from terms of the minimum number of cut up ropes of basic length of 32 m. Construct a mathematical model of the problem LP. Solution: The mathematical model will differ from the previous one only the shape of the objective function: z = x 1 + x 2 + x 3 + x 4 + x 5 min.

34 Examples: 1 The company cut rods of basic length 80 cm to rods of length 50 cm, 40 cm and 25 cm. The minimum required amounts are 50 pc, 80 pc and 95 pc. The task is to determine the optimal mix of cutting plans with minimal waste. Create a mathematical model of the problem. 2 The company has sufficient stocks of basic ropes of 52 m. You need to cut them out of at least 60 pc of 18 m length ropes, at least 100 pc of 11 m length ropes. The task is to determine the optimal mix of cutting plans with minimize waste. Create a mathematical model of the problem. 3 There are bars base of 1,5 m. You need to cut them out of at least 120 pc of 90 cm, at least 80 pc of 60 cm, and at least 110 cm pc of 45 cm. The task is to determine the optimal mix of cutting plans with minimal waste. Create a mathematical model of the problem.

35 Solutions: 1 z = 5x x 3 + 5x 4 min, x 1 50, 2x 2 + x 3 80, x 1 + x 3 + 3x 4 95, x 1 0, x 2 0, x 3 0, x z = 5x 1 + x 2 + 8x 3 min, 2x 1 + x 2 60, x 1 + 3x 2 + 4x 3 100, x 1 0, x 2 0, x z = 15x x x 5 min, x 1 + x 2 120, x 1 + 2x 3 + x 4 80, x 2 + 2x 4 + 3x 5 110, x 1 0, x 2 0, x 3 0, x 4 0, x 5 0.

36 problems that involved only two variables can be solved graphically. Each point in the plane - has coordinates x 1, x 2 - represents only one solution. Graphical solution of linear programming problem (with two variables x 1, x 2) takes place in Cartesian coordinate system (of two dimensions), where each coordinate represents true value of each variable. The axes are labeled x 1, x 2. First we need to represent the set of all feasible solutions X P. This set is the intersection of a set of solutions various restrictive conditions (self-limiting conditions and conditions of nonnegativity). Solution set of a linear inequality is satisfactory half-plane and the set of solutions satisfying a linear equation is a line. The intersection of these sets is either bounded sets - in that case it is convex k-polygon, k m + n, where m is the number of self-limiting conditions, n is the number of variables or uncontained convex set, having k edges and k 1 peaks.

37 We consider the maximization problem. Then we look for such a feasible solution (x 1, x 2) X P, that the value the objective function z = z(x 1, x 2) = c 1x 1 + c 2x 2 was the maximum possible. Imagine the set of all lines of type c 1x 1 + c 2x 2 = s, s R. These are parallels, all with normal vector n = (c 1, c 2). Points (x 1, x 2) on each such line have the same value of the objective function z = c 1x 1 + c 2x 2 = s. Just take a line perpendicular to the normal vector n and parallel to it in the direction of the vector n move into last place of intersection with the set of feasible solutions X P. This last intersection corresponds to a set of optimal solutions X opt. It is obvious that this set can be empty, or the one-element, or contains infinitely many elements.

38 Comment: The case where the set X opt contains infinitely many elements comes in two qualitatively different cases: 1 final intersection is a line. The endpoints of the line then correspond to two basic optimal solutions of the problem. 2 final intersection is the half-line. The endpoint of the half-line then corresponds to a single basic optimal solution of the problem. Comment: The value of the objective function z = z(x 1, x 2) = c 1x 1 + c 2x 2 grows in the direction of the normal vector n = (c 1, c 2). In the case of minimization problem because final intersection perpendicular to the vector n and a set X P is looking as far as possible against the direction of the vector n. Comment: Graphical method uses direct calculations, such as coordinates of a point (ie, the values of variables in the solution) is determined by solution of the system of two equations (equations of lines intersecting at the point) of two unknowns.

39 Example: Solve the LP problem specified of mathematical model by graphical method: 7x 1 + 5x 2 850, 2x 1 + 2x 2 300, z = 7x 1 + 6x 2 max, x 1 0, x 2 0.

40 Solution:

41 First, find a set X P of all feasible solutions. A solution that satisfies constraint 7x 1 + 5x fills in the half-plane with boundary line 7x 1 + 5x 2 = 850. This line must go through points for example [0, 170] and [ 850, 0] 7 (by two points is a straight line determined uniquely). As each line divides the plane into two opposite half-planes, we have to decide which of them corresponds to the inequality 7x 1 + 5x It is sufficient to take an arbitrary point not on the line (and thus 7x 1 + 5x 2 850) and find out if inequality is satisfied. If so, the point lies in the wanted half-plane; if not, wanted half-plane is the opposite one. Preferably use origin, ie the point O = [0, 0]. Constraint 7x 1 + 5x is satisfied (0 850). Beginning therefore lies in the wanted half-plane (1). This half-plane we mark in the graph and arrow marking boundary lines based on the half-plane. Using the same procedure we can also find half-plane (2) of solutions satisfying the second condition, ie 2x 1 + 2x Similarly, we find the half-planes (3) and (4) corresponding nonnegativity conditions.

42 The solution is feasible if and only if it satisfies all its own constraints and nonnegativity conditions. Therefore, the set X P of all feasible solutions we obtain by the intersection of the four half-planes. The set X P there is a quadrangle OABC, where O = [0, 0], A = [ 850, 0], B = [50, 100] and 7 C = [0, 150]. There is a set of X P marked in gray colour in Figure. For example the coordinates of the B, we obtain by solution of the system of equations 7x 1 + 5x 2 = 850, 2x 1 + 2x 2 = 300 (point B is simultaneously at boundary straight lines semiplanes (1) and (2)). We need to find the optimal solution, ie one feasible solution that maximizes the value of the objective function. Hence mark the normal vector n = (7, 6) (70, 60), which belongs the objective function. We find perpendicular line to the vector n, which is as far as possible in the direction of the vector n, while the non-empty intersection with the set X P. This line is marked with z max in Figure. Mark the well final intersection (ie the intersection of the line z max and set X P ). It is a point B. The set of all optimal solutions is therefore one-element: X opt = B = [50, 100]. The maximum value of the objective function is z max = z(b) = = 950.

43 Example: Solve the LP problem specified of mathematical model by graphical method. 7x 1 + 5x 2 850, 2x 1 + 2x 2 300, z = 2x 1 + 7x 2 max, x 1 0, x 2 0.

44 Solution: Because the constraints and conditions of nonnegativity are the same as in the previous example, set X P of all feasible solutions of the problem is the same.

46 The objective function is different. We construct a normal vector n = (2, 7) corresponding of objective function, see the figure below. Again, we find perpendicular line to the vector n, which has final (in the direction of the vector n) nonempty intersect with set X P. Thus, we obtain: X opt = C = [0, 150], z max = z(c) = =

47 Comment: In the two examples is seen that the objective function has a significant effect on the optimal solution.

48 Examples: 1 Solve example 1 on page 14 by graphical method. 2 Solve example 2 on page 16 by graphical method. 3 Solve example 1 on page 23 by graphical method. 4 Solve example 3 on page 27 by graphical method.

49 Solutions: 1 X opt = [70, 100], z max = CZK. 2 X opt = [400, 0], z max = CZK. 3 X opt = [100, 60], z min = CZK. 4 X opt = [2, 4], z min = 104 CZK.

50 Operations Research Mathematical modeling exercise.

Study support. Course Name: Operations Research. Supervisor: doc. RNDr. Jiří Moučka, Ph.D. Author: RNDr. Michal Šmerek, Ph.D.

Study support. Course Name: Operations Research. Supervisor: doc. RNDr. Jiří Moučka, Ph.D. Author: RNDr. Michal Šmerek, Ph.D. Study support Course Name: Operations Research Supervisor: doc. RNDr. Jiří Moučka, Ph.D. Author: RNDr. Michal Šmerek, Ph.D. 2015, University of Defence 1 Contents 1 Introduction 3 1.1 Phases of solution