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

Transcription

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

2 Contents 1 Introduction Phases of solution to decision problems Illustrative example Mathematical Principles of linear programming The system of linear equations The system of linear inequalities Formulation of linear programming problems The production planning problem Mixing problem Partition problem Graphical method for solving linear programming problems 21 5 Simplex method Finding the initial solution Test optimality of solution Improvement of solution Simplex method examples Method of articifial variables Illustrative example Method of artificial variables examples Duality in linear programming The concept of duality and the construction of a dual problem The economic interpretation of the dual problem The dual simplex method Duality in linear programming exercise Problems on the construction of the dual problems and the determining of type of dual coupled problems Tasks to use the dual simplex method The testing tasks The transportation problem Balanced transportation problem Vogel s approximation method Optimality test The improvement of solution

3 10 Balanced transportation problem exercise Finding the initial solution of transportation problem by VAM Test optimality of solution of transportation problem The improvement of solution of transportation problem Unbalanced transportation problem and properties of solutions Degenerate solution of transportation problem Alternativ solution of transportation problem Nonbasic optimal solution Unbalanced transportation problem Unbalanced transportation problem exercise Degenerate and alternative solution of transportation problems examples Unbalanced transportation problem examples The assignment problem The assignment problem of minimization type The assignment problem of maximization type The assignment problem exercise The assignment problems of minimization type examples The assignment problems of maximization type examples The assignment problems of both types examples The multicriteria decision making problem Introduction to the multicriteria decision making problem Graphical method Weighted sum method for solving MDM Multicriteria decision making problem exercise Solved examples and the solving procedure Examples entering and results Multicriteria programming problem The lexicographic method The method of aggregation of objective functions The method of goal programming The lexicographical method examples The method of aggregation of objective functions examples The method of goal programming examples Game Theory The game in normal form Classification of decision situations

4 18.3 The antagonistic conflicts Matrix games Matrix games solved in the field of pure strategies Matrix games solved in the field of mixed strategies The matrix games of type The principle of dominating The matrix games of type 2 n and m 2. Graphical method Matrix games solved by converting to the problem of linear programming Matrix games solved by converting to the problem of linear programming exercise Matrix games solved by general way Matrix games solved by general way exercise Basic terms of graph theory Critical Path Method Marking in the CPM method CPM method procedure CPM method example CPM method exercise The PERT method PERT method example PERT method exercise 156 4

5 1 Introduction With the advancement of technology is complicated by the different production processes. The need to model decision-making situations, be able to search for optimal decision solutions caused the development of the sector called OPERATIONAL RESEARCH (also OPERATIONAL ANALYSIS or ECONOMIC - MATHEMATICAL METHODS). Operational research is the summary of the methods used to solve decision problems. It solves the problems that have more possible solutions and among them are looking for the solution that best matches the target. Such a solution is called optimal solution. 1.1 Phases of solution to decision problems Each decision problem has several solutions phases: 1. The problem is analyzed from economic point of view, that constitutes the economic model. 2. It is assembled corresponding mathematical model. 3. It is solved a mathematical model - is looking for a specific numerical solutions. 4. Interpretation of the results of the solution. 5. Implementation of the solution. ad 1. The result of the analysis of economic reality is a qualitative and quantitative description of the situation - own verbal task assignment. Problems watching on simplified model (mean simplified image of reality, consider only those aspects of reality, which are significant). The economic model should include: OBJECTIVE (Criterion) - must be clearly defined. An example might be as profit maximization, cost minimization,... PROCESSES that take place at the decision-making situation. FACTORS affecting the process and of the relationship between processes and factors. 1.2 Illustrative example Example: The company produces three types of products V 1, V 2 and V 3. The production of one product V 1 is needed 15 kg of raw material M, 3 kwh of energy E and production takes 1 h of machine time T. The production of one product V 2 is needed 20 kg of raw material M, 4 kwh of energy E and production takes 1 h of machine time T. 2 The production of one product V 3 is needed 40 kg of raw material M, 7 kwh of energy E and production takes 3 h of machine time T. The profit from each product V 1 is 700 CZK, for the product V 2 it is 500 CZK and for the product V 3 5

6 it is CZK. The company has kg of raw material M, kwh energy E and h of machine time T. What production plan will give the company maximum profit? The economic model is already themselves verbal task assignment. Often the economic model is presented in form of a table. ECONOMIC Products Available MODEL V 1 V 2 V 3 amount material M energy E machine time T profit max ad 2. In order to solve the economic model, we must express the objective conditions of the process by mathematical means (functions, equations, inequalities,... ). This will create a corresponding mathematical model. By solution of this model we obtain the corresponding solutions. If the functions, equations and inequalities used in the mathematical model are only linear (ie 1st degree), it is problem of linear programming. Example: The the company has carried out three activities, the production of goods V 1, V 2 and V 3. Therefore, the mathematical model will contain 3 variables: x 1... number of produced products V 1, x 2... number of produced products V 2, x 3... number of produced products V 3. Mathematical expressions of constraints: i) Unable to produce a negative amount of products = x 1 0; x 2 0; x 3 0. They are called nonnegativity conditions. ii) The production of one product V 1 consumes 15 kg of raw material M. If you make x 1 these products, we use 15 x 1 kg of raw material M. Same for products V 2 and V 3. Consumption of raw material M by total is: M : 15x x x 3 kg. Total consumption must not exceed the quantity available. Thus: M : 15x x x Analogous for energy consumption and of machine time: E : 3x 1 + 4x 2 + 7x , T : x 1 + 0,5x 2 + 3x The three inequalities are called own constraints. The task is to determine the production plan, ie the vector production x = (x 1, x 2, x 3 ) of maximum profit. Value of profit for a given plan of production function is given by z = 700x x x 3. 6

7 Such a function is called objective function. The objective function in this case is of maximization type. Therefore, the mathematical model of example has the form: z = 700x x x 3 max, 15x x x , 3x 1 + 4x 2 + 7x , x 1 + 0,5x 2 + 3x , x 1 0, x 2 0, x Mathematical Principles of linear programming This chapter serves to deepen and widen those parts of the curriculum of the basic course in mathematics, which we will use in solving linear programming (LP). In particular, we will find the definition of a system of linear equations (SLE), a system of linear inequalities (SLI), the solution of these systems, their properties and basic theorems of these concepts. 2.1 The system of linear equations Definition: The system of m linear equations of n unknowns is a system of equations a 11 x 1 + a 12 x a 1n x n = b 1, a 21 x 1 + a 22 x a 2n x n = b 2, a m1 x 1 + a m2 x a mn x n = b m, (1) It can be used also equivalent matrix notation: A x T = b T, where A is matix of system, x T is vector of unknown a b T je vector of the right sides. Matix R = (A b T ) is augmented matrix of system. Comment: Matix of system (1) is A = a 11 a a 1n a 21 a a 2n... a m1 a m2... a mn. 7

8 Augmented matrix of system (1) is a 11 a a 1n b 1 a 21 a a 2n b 2 R =.... a m1 a m2... a mn b m, vector of the right sides is b = (b 1, b 2,..., b m ), and thus b T = b 1 b 2. b m, a vector of unknown x = (x 1, x 2,..., x n ), and thus x T = x 1 x 2. x n. Definition: The solution of systems of linear equations (1) we understand each n-tuple x = (x 1, x 2,..., x n ), which is fulfilled for all m equations of the system. The condition of solvability of a system of linear equations gives the so-called Frobenius theorem. Frobenius theorem: System (1) has a solution if and only if the rank of matrix A is equal to the rank of matrix R, ie when h(a) = h(r). Frobenius theorem complements the theorem about the number of solutions : Theorem: a) If h(a) = h(r) = n, the system (1) has just one solution. b) If h(a) = h(r) = h < n, the system (1) has infinitely many solutions that depend on n h parameters. c) If h(a) < h(r), the system (1) has no solution. Example: Solve the system of linear equations: x 1 2x 2 + x 3 + x 4 = 1, x 1 2x 2 + x 3 x 4 = 1, x 1 2x 2 + x 3 + 5x 4 = 5. Solution: SLE we solve by modifying the augmented matrix on a staircase shape: ( ).

9 Rank of a matrix A is equal to the rank of a matrix R and this rank is less than the number of of variables (h(a) = h(r) = 2 < 4). Thus, the system of equations has infinitely many solutions depend on n h = 4 2 = 2 parameters. Apparently x 4 = 1. The parameters s and t we can choose as follows: x 2 = s, x 3 = t, where s, t R. Then x 1 = 1 + 2x 2 x 3 x 4 = 2s t. Solution x = (2s t, s, t, 1), where s, t R, is the general solution of the considered system of linear equations. If we want to specify any particular solution, we choose s and t specific numbers. Example: Determine at least four specific solutions of SLE of previous example. Solution: For the parameters s and t is sufficient to select a specific real numbers, for example: i) s = 2, t = 1 = x = (3, 2, 1, 1), ii) s = 1, t = 0 = x = (2, 1, 0, 1), iii) s = 1, t = 1 = x = ( 3, 1, 1, 1), iv) s = 0, t = 0 = x = (0, 0, 0, 1). Definition: The solution that we obtain if each parameter is equal zero is called basic solution. Unknowns of parameters (their number is n h) are called non-basic variables. Other variables (h) are called basic variables. Example: In the example, we chose for the parameters x 2, x 3, and therefore x 2, x 3 are non-basic variables while the other variables, ie x 1, x 4, are basic. Solution iv) of system, ie, x = (0, 0, 0, 1), is the basic solution of the system, because parameters (ie, non-basic variables) are equal zeros. Definition: If in basic solution all the basic variables are nonzero, the solution is called non-degenerate basic solution. If in basic solution at least one basic variable is equal zero, the solution is called degenerate basic solution. Example: Solution iv) a system of linear equations in the example, ie x = (0, 0, 0, 1), is degenerate, because the basic variable x 1 = 0. Often we work with a system of linear equations in the canonical form. Definition: We say that a system of h linear equations is in canonical form if its matrix of system contains all h columns of the unit matrix of order h. Order of the columns is not important. If we consider for example system of equations x 1 + a 1,h+1 x h a 1n x n = b 1, x 2 + a 2,h+1 x h a 2n x n = b 2, x h + a h,h+1 x h a hn x n = b h, (2) the following facts are evident: 1. It is a system of h linear independent equations. 9

10 2. Matrix of system rank is h, the rank of the extended matrix of system is also h, the number of variables is n, and therefore there are infinitely many solutions depend on n h parameters. 3. From this shape can be directly determine the basic solution. By choice of x h+1 = x h+2 = = x n = 0 (non-basic variables, ie n h parameters), we get x 1 = b 1, x 2 = b 2,..., x h = b h (basic variables). Therefore, basic solution of the system (2) is x = (b 1, b 2,..., b h, 0,..., 0). Matrix of system (2) is A h = a 1,h+1 a 1n a 2,h+1 a 2n a h,h+1 a hn. (3) It contains currently h different basic unit column vectors (in (3) form a unit submatrix). These column vectors determine the basic variables. This principle is used for solving the system (1). The matrix A of system (1) are converted using elementary equivalent adjustments to matrix A h of system, which is in canonical form. Both systems are equivalent and solution (basic) is determined from the system in a canonical form. Example: Find all basic solutions SLE: x x 2 + 4x 4 = 30, 2x 1 + 8x 2 2x 3 + 2x 4 = 18, 6x 2 + 2x 3 + 2x 4 = 22. Solution: Solving we perform by full elimination of the table. The system has three linearly independent equations of four variables. There are four basic solutions. There are just as much basic solutions how many ways can be choose three basic variables of the four variables. Or else, there are just so many ways that can be choose a secondary variable (from four). Definition: The key element of the matrix we understand chosen nonzero matrix element, whereby appropriate modifications with line matrix change this element to one and resets all other matrix elements lying in the same column as the key element. Comment: Number of basic solutions of the system m linearly independent equations of n unknowns is at most equal to the combinatorial number ( ) ( ) n n n! = = m n m m!(n m)!. 10

11 Tasks: 1. Find all basic solutions SLE: a) x 1 + x 2 3x 3 + x 4 = 2, 4x 1 + x 2 + x 4 = 14, x 1 2x 2 + x 3 x 4 = 2. b) 2x 1 + 3x 2 + x 3 = 30, x 1 2x 2 + 3x 3 = 35. c) x 1 + 3x 2 = 18, 5x 1 + 2x 2 = 25, x 1 + x 2 = 8. d) x 1 + x 2 = 10, 3x 1 x 2 = 2, x 1 + 3x 2 = 20. e) x 1 + x 2 + x 3 + x 4 = 16, x 2 + 2x 3 + 3x 4 = Find the basic solution of a system of linear equations with a given set M of the basic variables: a) M = {x 1, x 2, x 3 }, x 1 + 3x 2 + 5x 3 + 7x 4 + 9x 5 = 40, x 1 x 2 + x 3 x 4 + x 5 = 12, 2x 1 + x 3 x 5 = 23. b) M = {x 2, x 3, x 4 }, 2x 1 6x 2 + x 3 x 4 = 7, x 1 + x 2 x 3 + x 5 = 1, x 1 + 2x 2 + x 3 + 2x 4 + x 5 = 10. c) M = {x 3, x 5 }, x 1 x 2 x 3 x 4 + x 5 = 4, x 2 2x 3 3x 4 + 4x 5 = 20. d) M = {x 1, x 2, x 3 }, 2x 1 + x 2 + 3x 3 x 4 = 16, x 1 x 2 + x 3 + x 4 = 4, x 1 + x 2 + x 3 = 6. 11

12 Solutions: 1. a) (4, 4, 0, 6); (0, 12, 4, 26); ( 13, 1, 3, 0); (3, 0, 1, 2). 4 4 b) (11, 0, 8); (0, 5, 15); ( 165, 40, 0). 7 7 c) (3, 5). d) φ. e) ( 4, 20, 0, 0); (6, 0, 10, 0); ( 28, 0, 0, 20 ); (0, 12, 4, 0); (0, 14, 0, 2); (0, 0, 28, 12) a) (9, 2, 5, 0, 0). b) (0, 1, 2, 3, 0). c) (0, 0, 2, 0, 6). d) (0, 1, 5, 0), solution is degenerate. 2.2 The system of linear inequalities Linear programming 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 i1 x 1 + a i2 x a in x n b i, i = 1, 2,..., m. (4) In doing so symbol represents one of the symbols of,, or more generally one of the symbols of, =,. Definition: Solution of systems of linear inequalities (4) 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 Then inequalities (4) of type becomes to equations x i 0, i = 1, 2,..., m. (5) a i1 x 1 + a i2 x a in x n + x i = b i, i = 1, 2,..., m, (6) 12

13 and inequalities (4) of type becomes to equations a i1 x 1 + a i2 x a in x n x i = b i, i = 1, 2,..., m. (7) We get a set of m linear equations of m+n unknowns with nonnegativity conditions (5). Each solution SLI (4) corresponds to a solution of the corresponding SLE (6) or (7), 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 (4) 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 (6) or (7) (or their combinations), if nonnegativity conditions (5) are satisfied. 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. 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: Z.p. 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). 13

14 Tasks: 1. 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. Solution: 1. 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). 3 Formulation of linear programming problems 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. 14

15 3.1 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. V 1 V 2 Available amount S 1 [kg/pc] S 2 [kg/pc] profit [CZK/pc] 18 8 max Solution: 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 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. 15

16 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. Tasks: 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. 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. 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. 16

17 Solution: 1. z = 308x x 2 max, 10x 1 + 8x , 2x 1 + x 2 240, x 1 0, x z = 80x x 2 max, 6x 1 + 4x , 4x 1 + 8x , 2x 1 + 2x , x 1 + 2x 2 0, x 1 0, x z = 10x x x 3 max, 3.2 Mixing problem x 1 + 2x , 2x 1 + 2x 2 + 3x , 0,5x 1 + 2x , x 1 0, x 2 0, x 3 0. 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. Solution: It is advantageous to transfer all the data in tabular form: K 1 K 2 The minimum amount Ca [mg/kg] M g [mg/kg] cost [CZK/kg] max 17

18 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. 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. Tasks: 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. 18

19 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. 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. Solution: 1. z = 30x x 2 min, x 1 + x 2 160, 2x 1 + 5x 2 500, x 1 0, x z = 200x x x 3 min, 10x 1 + 2x x , 6x 1 + 6x 2 + 4x , x 1 0, x 2 0, x z = 12x x 2 min, x 1 + x 2 = 6, x 1 + 1,5x 2 8, 12x x 2 60, x 1 0, x

20 3.3 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. 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. 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 and overall length of waste (in meters) is 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 z = x 1 + 4x 3 + 3x 4 + 2x 5. 6 m long rope, 20

21 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. 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. 21

22 Tasks: 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. Solution: 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

23 4 Graphical method for solving linear programming problems Linear programming 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. 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 1 x 1 + c 2 x 2 was the maximum possible. Imagine the set of all lines of type c 1 x 1 + c 2 x 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 1 x 1 + c 2 x 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 oneelement, or contains infinitely many elements. 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 1 x 1 + c 2 x 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. 23

24 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. Solution: 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 =

25 This line must go through points for example [0, 170] and [ 850, 0] (by two points is a straight line 7 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. 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, we write The maximum value of the objective function is X opt = B = [50, 100]. z max = z(b) = = 950. 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

26 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. 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) = = Comment: In the two examples is seen that the objective function has a significant effect on the optimal solution. 26

27 Tasks: 1. Solve example 1 on page 14 by graphical method. 2. Solve example 2 on page 14 by graphical method. 3. Solve example 1 on page 16 by graphical method. 4. Solve example 3 on page 16 by graphical method. Solution: 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. 27

28 5 Simplex method Simplex method is a computational procedure for determining optimal solutions of linear programming problems. The algorithm of the simplex method is schematically illustrated in figure 1. Figure 1: Scheme of simplex method algorithm During solving LP problems firstly we always obtain the initial basic feasible solution. For this we require a constraints of problem in the form of a system of linear equations in canonical form. Because of the constraints in the LP problem are usually in the form of inequalities, the first step is the transfer of this system of linear inequalities (SLI) to a system of linear equations (SLE). If this system is not in canonical form, so it will convert to canonical form. If SLE is in canonical form, we obtained initial basic feasible solution of the problem. Optimality test can determine whether the basic solution is optimal or not. In the case that the initial basic feasible solution is optimal, we found the optimal solution and the algorithm ends. If not, we can find a better basic feasible solution of LP problem, ie solutions with higher value of the objective function at maximization problem, respectively solution with lower value of the objective function at minimization problem. On the better solution is again applied the optimality test, etc. Because number of the basic solutions is a finite number, after a finite number of steps we arrive at the optimal solution. 5.1 Finding the initial solution Firstly we need to find the initial basic feasible solution. Ie., that constraints are expressed in the form of system of linear equations in canonical form with non-negative right-hand side of the equations, ie, 28

29 b i 0, i = 1, 2,..., n. How to achieve this, with regard to the specific shape of the of constraints, is contained in the following discussion: 1. If some b i < 0, i = 1, 2,..., n, the equations or inequalities we multiply by the number ( 1). So we get for all of constraints nonnegative right side, ie b i 0, i = 1, 2,..., n. 2.a) If the constraints are of the form Ax b, b i 0, i = 1, 2,..., m, then the system of linear inequalities translated using additional variables x i 0, i = 1, 2,..., n to the system of linear equations, that is already in canonical form. We obtain a 11 x 1 + a 12 x a 1n x n + x 1 = b 1, a 21 x 1 + a 22 x a 2n x n + x 2 = b 2, a m1 x 1 + a m2 x a mn x n + x m = b m. From there we get initial basic solution of the form x = (x 1, x 2,..., x n ; x 1, x 2,..., x m) = (0, 0,..., 0; b 1, b 2,..., b m ). This solution is feassible, because the b i 0, i = 1, 2, dots, m (satisfies the nonnegativity conditions). 2.b) If at least one constraint is of the form a i1 x 1 + a i2 x a in x n b i, b i 0, then the system of linear inequalities convert to system of linear equations, which is not in canonical form, for example: x 1 2x 2 5, x 1 2x 2 x 1 = 5, 2x 1 + 3x 2 2, 2x 1 + 3x 2 x 2 = 2, x 1 0, x 2 0; x 1 0, x 2 0, x 1 0, x 2 0. The equations can not multiply by number ( 1), because we would received a negative right side of the equations (appropriate basic solution would not be feasible). Initial feasible solution (system of linear equations in canonical form) can be obtained using the it method of artificial base, i.e. introduction it auxiliary variables, see below. The simplex table When constraints together with nullification objective function generates a system of linear equations in canonical form, we can realize the own simlex algorithm. It implements the so-called simplex table. 29

30 The actual simplex table we describe in the following example. Example: Consider the LP problem specified by this the mathematical model 2x 1 + 5x , 4x 1 + x , z = 3x 1 + 2x 2 max, x 1 0, x 2 0. With additional variables and the annulment of the objective function we obtain a system of three linear equations (with the five variables) that is already in canonical form. 2x 1 + 5x 2 +x 1 = 1 000, 4x 1 + x 2 +x 2 = 1 100, z 3x 1 2x 2 = 0 (max), (8) x 1 0, x 2 0, x 1 0, x 2 0. From the system (8) can be obtained initial basic feasible solution in the form x = (0, 0; 1 000, 1 100). The system (8) we rewrite to simplex table. B.v. x 1 x 2 x 1 x 2 b i p = b i a ik x x z max Table 1: Simplex table of the example Mark and name the individual parts of the table: (1) 1st column marked B.v. is a column of basic variables; ( ) (2) elements constitutes matrix of system; ( ) (3) column is column of right sides b i ;

31 (4) the last column, ie two empty fields marked with p = b i a ik, is used to determine the key row; (5) the last row, ie ( ), is the row of annulled objective function coefficients; (6) element in the column b i and row z represents the value of the objective function corresponding to basic feasible solution. Here is z = Test optimality of solution 1. MAXIMIZATION PROBLEM: If at least one coefficient c j of non-basic variable x j in row (5) of the objective function is NEGATIVE, the value z can be increase, ie. that the solution is not yet optimal. If in the row (5) is more negative coefficients, we choose the smallest possible of them. If all coefficients of non-basic variables in row (5) are non-negative, the solution is optimal. 2. MINIMIZATION PROBLEM: If at least one coefficient c j of non-basic variable x j in row (5) of the objective function is POSITIVE, the value z can be decrease, ie. that the solution is not yet optimal. If in the row (5) is more positive coefficients, we choose the biggest possible of them. If all coefficients of non-basic variables in row (5) are non-positive, the solution is optimal. This way we set the variable x j, we have to include among the basic variables, ie the so-called entering variable. This gives key column. 5.3 Improvement of solution We ll find out in which equations entering variable has positive coefficients, ie the collection of all positive coefficients in the key column. The right side of b i we divide by the coefficients a ik, a ik > 0, and write to columns (4) of simplex table. Select the minimum of all fractions p = b i a ik pro a ik > 0. Equation, in which is the smallest fraction, contains outgoing variable. Comment:Procedure for determining outgoing variable is the same at maximization and minimization problems. Exchange of entering and outgoing variables we make by full elimination. At the end of paragraphs summarize the procedure of the simplex method. 1. Determination of key column, ie entering variable, depending on which negative (max. problems), respectively positive (min. problems), of the coefficient in row (5) has the largest absolute value. 2. Determination of key row, ie outgoing variable, depending on which fraction p = b i a ik pro a ik > 0 in column (4) is smallest. 3. A key point (pivot) is the intersection of the key row and the key column. 4. Execute full eliminating. 31

32 The whole process is repeated until they are in row (5) any non-negative coefficients (for max. problem), respectively not positive (for min. problems). In this case the problem has the optimal solution. In case you can not determine the key row, ie all the coefficients in the key column are negative, ie a ik < 0, the problem has not an optimal solution. Example: Solve the LP problem with mathematical model: 7x 1 + 5x 2 850, 2x 1 + 2x 2 300, z = 7x 1 + 6x 2 max, x 1 0, x 2 0. Solution: The system of linear inequalities convert to a system of linear equations (by introduction of additional variables). At the same time nullifies the objective function (subtract its right side). 7x 1 + 5x 2 + x 1 = 850, 2x 1 + 2x 2 + x 2 = 300, z 7x 1 6x 2 = 0 (max), x 1 0, x 2 0, x 1 0, x 2 0. That is a canonical form of a system of linear equations with nonnegative right sides. We therefore proceed to the simplex table, see tab. 41. B.v. x 1 x 2 x 1 x 2 b i p = b i a ik pro a ik > 0 x 1 (7) x z max x x 2 0 ( 4 7 ) z max 1 x x z MAX 4 Table 2: Simplex table 32

33 In the last step of the table, in the row of the objective function is no longer no negative coefficient (a problem is of maximization type), the optimal solution has the shape of x opt = (50, 100; 0, 0) and the maximum value of the objective function is z max = z(x opt ) = 950. Example: Solve the LP problem with mathematical model (from the the previous example it differs only in the shape of the objective function): 7x 1 + 5x 2 850, 2x 1 + 2x 2 300, z = 2x 1 + 7x 2 max, x 1 0, x 2 0. Solution: We convert SLI to SLE. Again, we obtained the canonical form of systems of linear equations with nonnegative right sides. 7x 1 + 5x 2 +x 1 = 850, 2x 1 + 2x 2 +x 2 = 300, z 2x 1 7x 2 = 0 (max), x 1 0, x 2 0, x 1 0, x 2 0. Therefore we proceed to the simplex table and by the simplex method determine the optimal solution, Z.p. x 1 x 2 x 1 x 2 b i p = b i a ik pro a ik > 0 x x 2 2 (2) z max x x z MAX Table 3: Simplex table 33

34 see tab. 3. Optimal solution and maximum value of the objective function is x opt = (0, 150; 100, 0), z max = z(x opt ) = Tasks: 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. Task solve by the simplex method. 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. Task solve by the simplex method. 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. Task solve by the simplex method. Solution: 1. x opt = (70, 100; 0, 0), z max = z(x opt ) = Kč. 2. x opt = (400, 0; 0, 1 600, 400, 400), z max = z(x opt ) = Kč. 3. x opt = (0, 6 000, 3 000; 6 000, 0, 0), z max = z(x opt ) = Kč. 34