Combinatorial Optimization

Transcription

1 Combinatorial Optimization Winter course 6/7 Paul Göpfert University of Wuppertal Business Computing and Operations Research

2 Information concerning the course Lecture: Monday, : pm - : pm in M. Thursday, : pm - : pm in M. Start: October nd, 6 Lecturer: Paul Göpfert Office: M. Office hour: Wednesday, : pm - 6: pm (only after registration per ) Secretary office: M. pgoepfert@winfor.de Business Computing and Operations Research

3 Information concerning the course Tutorial: Wednesday, : pm - : pm in M..9 Start: October 6 th, 6 (will be lecture this time) First assignment sheet is already available Supervisor: Paul Göpfert Office: M. Office hour: Wednesday, : pm - 6: pm (registration per is mandatory) pgoepfert@winfor.de Coordinates the Tutorial Business Computing and Operations Research

4 Information concerning the course Weekly assignment sheet submission Submit in Moodle as PDF or postbo in room M. Write down course name and make sure you can identify your submission on return Mo Thu Wed Lecture Assignments online in Moodle am: due date for assignment sheet submission pm: Tutorial: Assignment review Business Computing and Operations Research

5 Preliminary Agenda. Linear programming. Applications. The Simple Algorithm. Geometry of the solution space. How fast is the Simple Method?. Working with tableaus. Duality. The dual problem. The Dual Simple Algorithm. The Revised Simple Algorithm. The Hitchcock Transportation Problem. Using the standardized Simple Algorithm. The MODI Algorithm Business Computing and Operations Research

6 Preliminary Agenda. The Primal-Dual Simple Algorithm 6. Shortest Path and Ma-Flow Problems. Shortest Path Problems. Ma-Flow Problems. Min-Cut Problems. A Primal-Dual Algorithm. Ford-Fulkerson Algorithm 7. The alpha-beta algorithm. Problem definition and analysis. Analyzing the reduced primal (RP). Solving the DRP 8. Integer programming. Basics. Cutting Plane Methods algorithm of Gomory. Branch&Bound Algorithm Business Computing and Operations Research 6

7 Preliminary Agenda 9. Matri Games. Introducing eamples. Basic definitions. Games and Linear Programming Business Computing and Operations Research 7

8 Selected Literature Brucker, P.; Knust, S.: Comple Scheduling.. Aufl., Springer, Berlin, Heidelberg, New York,. Domschke, W.; Drel, A.; Klein, R.; Scholl, A.; Voß, S.: Übungen und Fallbeispiele zum Operations Research. 8. Aufl., Springer Gabler,. Domschke, W.; Drel, A.: Einführung in Operations Research. 9. Aufl., Springer Gabler,. Myerson, R.B.: Game Theory. Analysis of Conflict. Havard University Press, 997. Papadimitriou, C.H.; Steiglitz, K.: Combinatorial Optimization. Algorithms and Compleity. Prentice-Hall, 98. Suhl, L.; Mellouli, T.: Optimierungssysteme.. Aufl., Springer Gabler,. Taha, H.A.: Operations Research. An Introduction. 9 th ed, Pearson Education,. Chvátal, V.: Linear Programming. 6th print. W.H. Freeman and Company, New York,. Wolsey, L.A.: Integer Programming. John Wiley&Sons, 998. And thousands of other good books dealing with Optimization, Linear Programming, or Combinatorial Optimization Business Computing and Operations Research 8

9 Simple calculators Ecel Solver Can be activated under Etras Add-Ins ( Version), File Options Add-Ins ( Version) Subsequently, you may use the Solver by Etras Solver ( Version), Data Solver ( Version) It is not powerful but nice to play around with our simple eamples Online Simple calculator: Business Computing and Operations Research 9

10 Linear Programming We deal with a large class of problems in this first section These problems can be mapped as Linear Programs, i.e., We define continuous variables We define linear constraints to be fulfilled by the values of the variables We define an objective function that provides an evaluation of each solution found We want to find optimal solutions, i.e., solutions that fulfill all restrictions (then we denote them as feasible) and maimize or minimize the objective function value Business Computing and Operations Research

11 Linear Program Main attributes continuous decision variables linear constraints that must be fulfilled by the values of the decision variables objective function that provides an evaluation of each solution found Finding an optimal solution Solution: vector of decision variables Feasible solution: solution that fulfills all constraints Optimal solution: feasible solution with maimal or minimal objective function value Business Computing and Operations Research

12 . Linear Programming Applications We commence our work with representative applications of Linear Programming Production Program Planning Hitchcock problem, i.e., standardized balanced transportation problem Diet Problem Business Computing and Operations Research

13 Application Production Program Planning The production management of a plant of an orange juice producer plans the production program There are two types of orange juices that are pressed and mied in this plant For simplicity reasons, let us denote them as A and B Both are produced on stages in a predetermined sequence, i.e., is the production sequence for both product types This is illustrated by the following figure Business Computing and Operations Research

14 Production Program Planning - Illustration $ A Stage Ma. amount B Stage Capacity h Stage 8 $ Capacity h Capacity h Ma. amount Business Computing and Operations Research

15 and just the values All types are produced on all stages Capacity on stages and are h, Capacity on stage is h Product A Price per gallon: $ Variable costs per gallon: $ Thus, we obtain a marginal profit of $ per gallon Ma. sales volume: gallons In order to produce one gallon of A on stage, we need hours, on stage, hour, and on stage, hours Business Computing and Operations Research

16 and product B Product B Price per gallon: 8$ Variable costs per gallon: $ Thus, we obtain a marginal profit of 6$ per gallon Ma. sales volume: gallons In order to produce one gallon of B on stage, A gallon B requires hours on stage, A gallon B requires hours, and on stage, A gallon B requires hours Business Computing and Operations Research 6

17 Optimal production program Clearly, we want to maimize our profit, i.e., the maimally obtainable total marginal profit Thus, we analyze what we are able to sell maimally Both types of orange juice are worth to produce Each item of A brings us a marginal profit of $ per gallon Each item of B even 6$ Consequently, we want to produce as much as possible of both products If the maimum volumes of sale can be produced, we have found the optimal production program Business Computing and Operations Research 7

18 We calculate the maimum demand We have the following demand levels Stage :.. 6> Thus, since demand is larger than capacity, we have a bottleneck! Stage :.. > Thus, since demand is larger than capacity, we have a bottleneck! Stage :.. 8> Thus, since demand is larger than capacity, we have a bottleneck! Business Computing and Operations Research 8

19 What is to do? Since B provides 6$ instead of $, we produce as much as possible of B; and then we fill up the rest with A? Business Computing and Operations Research 9

20 If we do so, it turns out that B needs on stage h. Maimally produce min{, /} on stage h. Maimally produce min{, /} on stage h. Maimally produce min{, /} Thus, B can be produced in its maimum volume of sales A needs on stage h. Maimally produce min{, (-6)/} on stage h. Maimally produce min{, /} on stage h. Maimally produce min{, 8/} Thus, items of A can be produced additionally Business Computing and Operations Research

21 Results in a total profit of. $. 6$ $ $ $ However, in order to analyze the problem more in detail, we want to formalize it Business Computing and Operations Research

22 Linear Program of the production Decision variables: : Produced gallons of juice A A B : Produced gallons of juice B Objective function: Maimize z 6 Maimize the total revenue Constraints: subject to A B A B Maimium volume of sales Production capacities A A A A B B B, Non-negativity constraints B Business Computing and Operations Research

23 A Graphical solution Restriction Optimal solution 7 Restriction Z Restriction 7 Business Computing and Operations Research B

24 Business Computing and Operations Research Optimal solution Obviously, the optimal solution is located at the point of intersection of restriction and Thus, we have to solve ( ) 6,6 6,6 8 / / 8 / 6 / / 8 / B A B A B A B B A B B A B B B A B B B A B A B A B A B A

25 a total (optimal) profit of Results in $ 6.6. $.$ 9.9$ 9.$ Business Computing and Operations Research

26 Consequence OK! I admit that I was wrong! However, what can we do if there are more than variables?? Then, we will formulate a general version of that LP and use the SIMPLEX ALGORITHM Business Computing and Operations Research 6

27 i, j Production Program Planning [ ] [ ] [ ] p : Marginal profit per item of product type j,..., n MU/PU j c : Production coefficient of product type j,..., n on machine i,..., m CU/PU C : Maimum capacity of machine i,..., m CU X i j : Maimum numberof items of product type j j { } [ ] the planning horizon PU i, j [ ],..., n salable in : Number of items of product type j,..., n to be produced j Maimize in the planning horizon PU n p j subject to i,..., m : c j { } j j,..., n : X n j j C j i Business Computing and Operations Research 7

28 Using Ecel The program Ecel comprises a standard solver for Linear Programs It is neither really high-performance nor convenient to use but available and sufficient for our eemplary problem constellations Activate the Solver by Etras Add-Ins ( Version), File Options Add-Ins ( Version) Subsequently, you may use the Solver by Etras Solver ( Version), Data Solver ( Version) Let us try it out Microsoft Ecel-Arbeitsblatt Business Computing and Operations Research 8

29 Application The Hitchcock Problem A service agent has three sales offices (C X, C Y, and C Z ) in Wuppertal These offices are supplied by three local printing plants (P A, P B, and P C ) In order to satisfy the numerous soccer fans in Wuppertal, the service agent has an eclusive license of sale for the famous football/soccer club Borussia Mönchengladbach While the tickets are printed in the printing plants at equal costs, the transport to the offices causes individual costs Additional information Each printing plant has an individual inventory for the net day. Note that this inventory is etremely perishable Each sales office has an individual maimum amount of sales Business Computing and Operations Research 9

30 The Hitchcock Problem Illustration Sales Office X Sales Office Y Sales Office Z Demand: pallets Demand: pallets Demand: 6 pallets Printing Plant A pallets Printing Plant B 7 pallets Printing Plant C pallets Business Computing and Operations Research

31 The Matri Producers A B C Supply 7 Consumers X Y Z Demand 6 Business Computing and Operations Research

32 Transportation Distances Distance X Y Z A B 6 8 C 7 Business Computing and Operations Research

33 What is the objective? Obviously, we have to decide about the quantities to be transported along each relation between a printing plant and a ticket office Specifically, we determine the precise number of product units that are transported along each relation Since quality is assumed to be negligible, a transportation cost minimization is appropriate to compare generated assignments Thus, a solution is solely rated by the incurred transportation costs Business Computing and Operations Research

34 Let s solve the problem When you try something out, you usually provide a heuristic solution Heuristic solutions do not always guarantee a certain quality Usually, their performance is empirically validated or roughly anticipated for worst case scenarios In the following, we want to obtain some insights into the problem structure by applying some well-known simple heuristics Business Computing and Operations Research

35 Minimum Method Basic idea: Find the least possible combination of costs and use it ehaustively. Afterwards, proceed with the second lowest one, Let us do so Business Computing and Operations Research

36 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research 6

37 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research 7

38 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research 8

39 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research 9

40 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research

41 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research

42 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:7 () (6) (8) C S: () () (7) Business Computing and Operations Research

43 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:6 () (6) (8) C S: () () (7) Business Computing and Operations Research

44 Minimum Method Distance X D: Y D: Z D:6 A S: () () () B S:6 () (6) (8) C S: () () (7) Business Computing and Operations Research

45 Minimum Method Total costs 68 Distance X D: Y D: Z D: A S: () () () B S: () (6) (8) 6 C S: () () (7) Business Computing and Operations Research

46 Vogel s Approimation Method Basic Idea: Avoid larger deteriorations by identifying critical relations Specifically, calculate the differences between the best and the second best relation for all producers and all consumers Select the best relation for the one with the largest difference Proceed until a complete solution is found Business Computing and Operations Research 6

47 Vogel s Approimation Method Distance X D: Diff: Y D: Diff: Z D:6 Diff: A S: Diff: () () () B S:7 Diff: () (6) (8) C S: Diff: () () (7) Business Computing and Operations Research 7

48 Vogel s Approimation Method Distance X D: Diff: Y D: Diff: Z D: Diff: A S: Diff: () () () B S:7 Diff: () (6) (8) C S: Diff: () () (7) Business Computing and Operations Research 8

49 Vogel s Approimation Method Distance X D: Diff: Y D: Diff: Z D: Diff: A S: Diff: () () () B S:7 Diff: () (6) (8) C S: Diff: () () (7) Business Computing and Operations Research 9

50 Vogel s Approimation Method Distance X D: Diff: Y D: Diff: Z D: Diff: A S: Diff: () () () B S: Diff: () (6) (8) C S: Diff: () () (7) Business Computing and Operations Research

51 Vogel s Approimation Method Distance X D: Diff: Y D: Diff: Z D: Diff: A S: Diff: () () () B S: Diff: () (6) (8) C S: Diff: () () (7) Business Computing and Operations Research

52 Vogel s Approimation Method Total costs 6 Distance X D: Diff: Y D: Diff: Z D: Diff: A S: Diff: () () () B S: Diff: () (6) (8) C S: Diff: () () (7) Business Computing and Operations Research

53 Local Improvement Operations We may improve an eisting solution by applying specific transformation moves, i.e., we slightly modify a current solution in a way that feasibility is maintained, and solution quality is improved A simple eample is the pairwise shift Specifically, we select two consumer-producer relations (P &C, P &C ) and ask for the change in costs by transporting one unit from P to C rather than from P to C For P, we simultaneously consider the same Note that feasibility is ensured by the simultaneous consideration of both constellations Business Computing and Operations Research

54 Pairwise shift Total Costs 68 Distance X D: Y D: Z D: A S: () () () B S: () (6) (8) 6 C S: () () (7) Business Computing and Operations Research

55 Pairwise shift Total Costs are 68-6 Distance X D: Y D: Z D: A S: () () () B S: () (6) (8) C S: () () (7) Business Computing and Operations Research

56 Pairwise shift Total Costs 6 Distance X D: Y D: Z D: A S: () () () B S: () (6) (8) C S: () () (7) Business Computing and Operations Research 6

57 Pairwise shift Total Costs are 6-6 Distance X D: Y D: Z D: A S: () () () B S: () (6) (8) C S: () () (7) Business Computing and Operations Research 7

58 The (balanced) Transportation Problem c i, j i, j : Delivery costs for each product unit that is transported from [ ] [ ] [ ] supplier i,..., m to customer j,..., n CU/PU a : Total supply of i,..., m PU i b : Total demand of j,..., n PU j [ ] : Quantity that supplier i,..., m delivers to the customer j,..., n PU Minimize m n i j c i, j i, j { } subject to i,..., m : a { } j i i, j j,..., n : b n m i, j { } { } i,..., m : j,..., n : j i i, j Business Computing and Operations Research 8

59 The balanced Transportation Problem ( ci, j) ( a i) ( bj) Sales Office X b[]: pallets Sales Office Y b[y]: pallets Sales Office Z b[z]: 6 pallets Printing Plant A a[a]: pallets Printing Plant B a[b]: 7 pallets Printing Plant C a[c]: pallets Business Computing and Operations Research 9

60 The (balanced) Transportation Problem With the previously defined parameters our problem is as follows: 6 8 ; 7 ; 7 6 ( ci, j) ( ai) ( bj) Minimize subject to,,,,,,,,,,,,,,,,,, 7,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, 6,,,,,,,,,,,,,,,,,,,,,,,,,, Business Computing and Operations Research 6

61 Alternative depiction,,, 6 8 ; 7 ; 7 6 ( ci, j) ( ai) ( bj) Minimize subject to,,,,,,,,,,,,,,,,,,,,,, 7,, 6,,,,,,,,,,,,,,,,, Business Computing and Operations Research 6

62 The (balanced) Transportation Problem We identify the characteristic structure of the problem. ( ) T P Minimize c subject to T n a T... n T a n m E b n En En En E n (,, n,, n m, m, n),...,,,...,,...,,..., T Business Computing and Operations Research 6

63 Application The Diet Problem Susan wonders how much money she has to spend on food in order to get the energy that brings her through the day Now, she thinks it is time to analyze Altogether, she chooses si foods that seem to be cheap sources of the nutrients her body needs Food Size per serving Energy (kcal) Protein (g) Calcium (mg) Price ($ Cents) Oatmeal 8 g Chicken g Eggs large 6 Whole milk 7 cc Cherry pie 7 g Pork with beans 6 g Business Computing and Operations Research 6

64 Additional information Susan needs per day, kcal g protein 8 mg calcium Iron and vitamins are satisfied by pills Consequently, servings of pork and beans are sufficient per day Imagine, times pork and beans per day This is disgusting Ok We need to impose servings-per-day limits Oatmeal: at most servings per day Chicken: at most servings per day Eggs: at most servings per day Milk: at most 8 servings per day Cherry pie: at most servings per day Pork with beans: at most servings per day Business Computing and Operations Research 6

65 The Diet Problem,,,,, servings per day of the respective food 6 Minimize 9 9 subject to Business Computing and Operations Research 6

66 and in general ai, j: Amount of ith nutrient in a serving of the jth food, i,...,m, j,...,n. r: Daily requirement for the ith nutrient, i,...,m. i c : Cost per serving for the jth food, j,...,n. j X j : Maimum number on servings of the jth food, j,...,n. Decision variables: : Daily consumption of the jth food, j,...,n. A diet j n is denoted by a choice of a vector, IR. Objective function: Minimize { } j subject to i,..., m : a r { } n j c j j i, j j i j,..., n : X n j j Business Computing and Operations Research 66

67 Consequence All applications are completely different, but their mathematical definitions are somehow strongly related All LPs have in common that the variables are continuous the objective function is linear the restrictions are linear the objective function is either a maimization or minimization restrictions require the fulfillment of a minimum or maimum bound Business Computing and Operations Research 67

68 LP in general In what follows, we introduce general forms in order to define what a Linear Program (LP) is In Literature, different forms of LPs are distinguished. Specifically, it can be found for instance LP in general form LP in canonical form LP in standard form The Reader should be warned that this classification is far away from being unambiguous Moreover, what we will denote as a Linear Program in standard form is frequently introduced as the LP in canonical form Business Computing and Operations Research 68

69 General Form a T m n n m i { } T a m Let A IR with A... a IR, i,...,m, and b IR. Furthermore, let M be the set of row indices corresponding to equality constraints, and let M be the set of row indices corresponding to inequality constraints. Additionally, let N be the set of column indices corresponding to constrained variables, and let N be the set of column indices corresponding to unrestricted variables. Then, the feasible solution space P is n T T { j i i i i} P IR j N : i M : a b i M : a b. n T Furthermore, for c IR, we pursue the maimization of z c. { } Note that M and M form a partition of,...,m. Moreover, N and N are a {,...,n} partition of. ( ) Business Computing and Operations Research 69

70 Canonical Form m n m Let A IR and b IR : Then, the set of feasible solutions is defined as follows: n { and } P IR A b Solutions that belong to P are denoted as feasible. In order to evaluate a solution that is found, we introduce an additional ( ) n T vector. Hence, let c IR : z c. In the following, we pursue the maimization of z under the constraints and A b. Business Computing and Operations Research 7

71 Standard Form m n m Let A IR and b IR : Then, the set of feasible solutions is defined as follows: n { and } P IR A b Solutions that belong to P are denoted as feasible. In order to evaluate solution that is found, we ( ) n T vector. Hence, let c IR : z c. introduce an additional In the following, we pursue the minimization or maimization of z under the constraints and A b. Business Computing and Operations Research 7

72 Problem transformations In order to prove that it is sufficient to consider LPs in standard form only, we have to think about problem transformations Obviously, in particular, the diet problem does not correspond to our class In addition, what about equalities? And what about unrestricted variables, i.e., variables that may become negative? This is briefly considered in the following Business Computing and Operations Research 7

73 Business Computing and Operations Research 7 Transformation Equality I ( ) ( ) ( ) i n j j i,j n j i j i,j n j i j i,j b a b a b a inequalities by two Replace

74 Transformation Equality II n Replace a b by introducing a j i, j j i new (positive) variable that is called a slack variable: n j a y b y i, j j i i i We call all variables that belong to the original problem structure variables to distinguish them from the slack variables. Business Computing and Operations Research 7

75 Transformation Objective function Just replace the original objective function Minimize n j by the modified equivalent objective function n j c j Maimize c ( ) j j j Business Computing and Operations Research 7

76 Transformation Free variables Create two additional variables, for each unrestricted variable IR, and substitute. j j j j j j T This leads to a doubling of the corresponding jth column in c and in A while the added column is multiplied with -: j ( c,..., cj,..., cn) j ( c,..., cj, cj,..., cn) and A ( a,..., aj,..., an) j j n n n... j m ( a,..., aj, aj,..., an),with: i {,..., n} : ci IR, i {,..., n} : ai IR j... n Business Computing and Operations Research 76

77 Conclusion Since all forms of LPs are equivalent, we switch between them arbitrarily I.e., we always use the form that seems to be most useful Business Computing and Operations Research 77

78 . The Simple Method Coming back to the Production Program Planning, we consider now the following problem constellation This time, a producer has to decide about product types (,, and ) to be produced on stages Product Marginal Profit: Production Coefficients:,, and Product Marginal Profit: Production Coefficients:,, and Product Marginal Profit: Production Coefficients:,, and Business Computing and Operations Research 78

79 Business Computing and Operations Research 79 Altogether, we get,, 8 s.t. Ma z

80 Getting equalities by slack variables Ma z s.t. 8,, Ma z s.t. 8 6,,,,, 6 The variables in the original LP are denoted as structure variables In contrast to this, the variables that are additionally introduced in the second LP (with equalities) are denoted as slack variables Business Computing and Operations Research 8

81 Getting a first solution Ma z s.t. 8 6,,,,, 6 This is a dictionary. We call the n m variables that appear on the right-hand side non - basic variables and set their value to zero. The m variables on the left-hand side (i.e., the basic variables) form a basic solution: 8 z. 6 We call the objective function now reduced costs. Business Computing and Operations Research 8

82 Quality of the solution found is obviously not really convincing (z) How can we improve it? The value is not surprising Only slack variables are unequal to zero Slack variables which are unequal to zero do not provide any benefit according to the objective function value Let us consider the set of variables For this purpose, we consider the objective function coefficients belonging to the current solution Owing to positive coefficients, an increase of,, or will raise z. Since has the largest positive coefficient, we first try it out. We call this pivoting strategy the largest coefficient rule How much can we increase? Ma z Business Computing and Operations Research 8

83 How much can we increase? Ma z s.t { } 8 min,, Business Computing and Operations Research 8

84 And now? BUT, how can we keep the structure? We want to introduce on the left-hand side. This is only possible for the first restriction.! Business Computing and Operations Research 8

85 Business Computing and Operations Research 8 Transforming the dictionary ( ) ( ) ( ),,,,, s.t. 7 Ma,,,,, 8 s.t. Ma z z z

86 Consider the solution Ma 7 z The objective function reveals us that we can improve the solution further. This is possible by increasing. Again, we ask for bounds limiting the increase of this variable. For this purpose, we have to consider the dictionary. Business Computing and Operations Research 86

87 Business Computing and Operations Research 87 How much can we increase? { }, min s.t. 7 Ma 6 z

88 Business Computing and Operations Research 88 Transforming the dictionary ( ) ( ),,,,, s.t. Ma s.t. 7 Ma z z z

89 And now? Since there remains no variable promising to increase z, we have found an optimal solution. Consider the objective function z 6! Business Computing and Operations Research 89

90 Calculation with dictionaries Left-hand side Variables that are allowed to be unequal to zero Here, we have altogether m variables We call these variables basic variables Right-hand side Variables that are equal to zero Here, we have altogether at least n-m variables We call these variables non-basic variables Objective function Positive coefficients increase the objective function value and vice versa. Later on, coefficients that belong to structure variables are denoted as reduced costs We eecute a swap of a basic and a non-basic variable in each step. By doing so, we try to improve the solution quality. Moreover, we jump along the edge of the solution space. More specifically, from corner point to corner point Business Computing and Operations Research 9

91 The solution space and the Simple Algorithm A Iteration Optimal solution Iteration A corner point Feasible solution space Iteration B Business Computing and Operations Research 9

92 The Primal Simple Algorithm with Dictionaries. Transform the problem into a canonical form and generate equations.. Initialization with a feasible basic solution.. Are there strictly positive reduced cost coefficients in the current solution? Yes : Iteration: - Largest coefficient rule: Choose a variable B that has the largest positive reduced cost coefficient. - Determine a positive upper bound on B. - If there eists an upper bound on feasible values for B, set B to the minimal upper bound that is given by equation i ; otherwise terminate since the solution space is unbounded and no optimal solution eists. - Transform equation i such that B appears on the left-hand side. - Substitute B in all other equations as well as in the objective function with the obtained equation i. - Rearrange the equation system. - Go to step. No : Termination. An optimal basic solution is found. Business Computing and Operations Research 9

93 Pitfalls and how to avoid them The presented calculation went pretty smoothly The danger that may occur was not pointed out Three kinds of pitfalls have to be considered Initialization Obviously, we need an initial solution Are there constellations thinkable where this is not possible? Iteration Is there a danger of getting stuck throughout the calculation? Is it always possible to swap from one basic solution to the net one? Termination Is the calculation always finite? Are cyclical computations possible? Business Computing and Operations Research 9

94 Initialization For what follows, we need at first a feasible solution to the LP. Fortunately, this is quite simple to provide If b is positive, we may just make use of the introduction of slack variables; i.e., all structure variables are set to zero and slack variables equal the right-hand side b Otherwise, we apply the simple procedure that is depicted on the following slides Business Computing and Operations Research 9

95 Pitfall: Initialization with a feasible solution If b, take the trivial solution: all structure variables are set to zero and all slack variables equal the right-hand side b If there is an i with b i <, apply the Two-Phase Method. Auiliary LP: Maimize z subject to Am b, ini ini ini n i i { m} ( n ). Initial solution to the auiliary LP:,...,, ini ini ini ini { { }} with... min b b < i,..., m ini Since i,..., eists with b <, is feasible *. Solve the auiliary LP and get its optimal solution. If z>, terminate (this procedure) because the original LP is not solvable * * *. Initial feasible solution to the original LP:,..., n i Business Computing and Operations Research 9 T ( ) T

96 Two-Phase Method Conclusions I.. Observation: Since the objective function value is lower bounded by zero, the auiliary LP is solvable.. Lemma: If and only if the optimal solution to the auiliary problem has the objective function value zero, the original LP is solvable Proof: : Since z holds follows. The optimal auiliary LP solution yields a feasible solution to the original LP : If the original problem is solvable, we have and, therefore, z Business Computing and Operations Research 96

97 Two-Phase Method Conclusions II Optimal auiliary LP solution: > is basic: The original LP is not solvable because at least one constraint is violated is non-basic or is basic: Erase and switch to the original LP with the feasible solution just generated Special case here: is basic: Consider the net-to-last step where the objective function becomes zero. Here, was decreased to zero. Consequently, in this step was a candidate for being erased. Hence, we can adjust this step accordingly Business Computing and Operations Research 97

98 Initialization Eample I Ma z s.t.,, Ma z s.t. min 6,,,,,, 6 Business Computing and Operations Research 98

99 Business Computing and Operations Research 99 Initialization Eample II ( ) ( ) ( ),,,,,, 9 s.t. Ma,,,,,, s.t. Ma z z

100 Business Computing and Operations Research Initialization Eample III,,,,,, s.t. Ma Introduce,,,,,, 9 s.t. Ma z z

101 Business Computing and Operations Research Initialization Eample IV ( ) ( ) ( ) Introduce,,,,,, 7 s.t. Ma,,,,,, 9 s.t. Ma z z

102 Business Computing and Operations Research Initialization Eample V feasible solution is a 8,,,,,,,, 8 s.t. Ma,,,,,, s.t. Ma z z

103 Initialization Eample VI Consequently, we resume with the Simple applied to the following dictionary in order to solve the original problem ( ) ( 8 ) Ma Ma Ma 6 z 6 6 s.t ,,,,, 6 6 Business Computing and Operations Research

104 Cases to be distinguished Altogether, we have to deal with the following cases after solving the auiliary problem. is non-basic, i.e., we have the simple case where we can directly switch to the original problem with the feasible solution justly generated. is erased. > is basic, i.e., the original problem is not solvable at all because at least one constraint is violated. is basic, i.e., this variable can be erased from the basis without affecting the solution quality. In order to make this obvious, consider the net-to-last step where the objective function becomes zero. Here, was decreased to zero. Consequently, in this step was a candidate for being erased. Hence, we can adjust this step accordingly Business Computing and Operations Research

105 Pitfall: Iteration and Termination In each iteration, we erase one variable from the basis and replace it by another variable with a positive contribution to the objective function value However, this choice is ambiguous There may be more than one non-basic candidate for entering the basis Thus, we may choose the one with the largest improvement factor If there is no candidate at all, the current solution is optimal (this point will be addressed thoroughly in Section.) In addition, the choice of the leaving variable is ambiguous as well If there is no candidate, the solution is unbounded, i.e., we can improve the solution arbitrarily Otherwise, if there are several equal bounds, we have alternative choices. But, here we obtain a degenerate solution Business Computing and Operations Research

106 Primal Degeneration Eample I Consider the following dictionary Ma 8 z s.t. 6, with,,,, 6 6 We introduce into the basis Ma 8 z s.t. 6 6, with,,,, 6 Business Computing and Operations Research 6

107 Business Computing and Operations Research 7 Primal Degeneration Eample II,,,, with, s.t. Ma 6 6 z We observe that we have two basic variables (, 6 ) with value zero Although this is not harmful in its own right, it may have annoying side effects Specifically, primal degeneracy may cause cyclical calculations, i.e., in this case it prevents termination

108 Termination Termination may be prevented by cyclical calculations Note that cycling is only a rare phenomenon. Specifically, such kind of instances are hard to generate But, how does cycling become possible? Primal degeneration may cause non-improving moves Specifically, a basic variable with value zero leaves the basis and is replaced by a non-basic one Note that a calculation only comprising improving moves cannot cycle Business Computing and Operations Research 8

109 Smallest subscript rule Pivoting strategy (smallest subscript rule): Choose the nonbasic variable with the smallest inde that has positive reduced costs to become a basic variable Choose the basic variable with the smallest inde to become a non-basic variable from all equations that provides the minimal upper bound on the new basic variable.. Theorem: The Simple Method terminates as long as the entering and leaving variables are selected by the smallest subscript rule Proof by contradiction: Assume that the opposite holds (i.e., there is a cycle with the smallest subscript rule applied) and show that this leads to a logical contradiction Business Computing and Operations Research 9

110 Proof of Theorem.. Basics Let us assume that we have a cycle of dictionaries D -D - -D k, with D D k A variable is denoted as volatile if this variable is basic as well as non-basic throughout these dictionaries Let t be the volatile variable with the largest subscript D is the dictionary where t is basic and becomes nonbasic in the net dictionary s is non-basic in D and becomes basic in the net dictionary Further along in the sequence, there is a dictionary D* where t becomes basic again Business Computing and Operations Research

111 Proof of Theorem.. Illustration t leaves the basis t enters the basis t is basic t is non-basic t is about to become basic again D D D* s is non-basic s is basic s enters the basis Business Computing and Operations Research

112 Proof of Theorem.. Dictionaries D, D* i i i, j j j B j B j j ( ) b a i B z v c * Consider the calculation from D to D. Since we have a cycle, all these dictionaries are degenerate and the objective function value * is kept unchanged. Hence, we obtain for the dictionary D the objective function z v cj j, with B as the basis of D. j B Business Computing and Operations Research

113 Dictionaries D and D*... z v cj j, with c j, j B j This dictionary D is generated by algebraic manipulations out of D. Therefore, each feasible solution of D is feasible for D and thus for each feasible solution of D it holds: j j z v c j Business Computing and Operations Research

114 We know that enters basis B in dictionary D' (net to D) i i i, s s i s By introducing y as the value for solution which is obviously feasible for D ( ) b a y i B Proof of Theorem.. A new solution y, i B i s ( ) s, we generate the following We substitute all new basic variables of this solution in the objective function of the dictionaries D and D* and obtain: z v cs y cj j v cs y cj j v csy j s j B j B j s: j Bc : v c y c v c y c j s j B s j j s j j j B j s: j j j Business Computing and Operations Research

115 Proof of Theorem.. Transforming Therefore, we obtain: v c y v c y c, with c, j B s s j j j j B c y c y c s s j j j B ( ) ( ) c y c y c b a y s s j j j, s j B c c y c b c a y s s j j j j, s j B j B cs cs cj aj, s y cj bj j B j B Is a constant independent of y Business Computing and Operations Research

116 Proof of Theorem.. Conclusion cs cs cj aj, s y cj bj j B j B Is a constant independent of y Consequently, both sides have the value zero. Thus, c c c a s s j j, s j B Since enters the basis in D, it holds c >. Thus, s is volatile and therefore s< t. Since not enters the s basis in D, it holds c. s s s Business Computing and Operations Research 6

117 Proof of Theorem.. Conclusion Since c c c a c > c s s j j, s s s j B c c > r B:c a < c s s r r, s r Since r B, is basic in D. Since c, we know that r B is volatile. Note that r t. Since t enters in D, we have c t r r >. In addition, t is leaving in D and thus, we conduct the following transformation bt a t t, j b a, a, j B j s ats, ats, j B j sats, t t t s s t j j s j a > c a < c a t r ts, r rs, t ts, r Business Computing and Operations Research 7

118 Proof of Theorem.. Conclusion Consequently, r < t, but has not entered in D. Although it is not basic, has entered in D. r t c, actually, c < since c a < a > r r r r, s r, s Since all solutions between D and D are degenerate, and volatile, we have in all solutions. and are b b in dictionary D and both were candidates for leaving r the basis B. t r But, we choose although t > r. This violates the Smallest Subscript rule and is therefore a contradiction. t t r t This completes the proof. Business Computing and Operations Research 8

119 What we know so far We have learned how to solve general LPs by applying the Simple procedure that eplores a sequence of basic solutions We have seen that under certain circumstances (i.e., if we make use of a specific subscript rule) this algorithm always terminates We have learned to deal with problems where an initial solution is not directly available In order to do this, we have generated the Two-Phase Method It terminates either with an initial solution or with the cognition that the problem is not solvable at all In what follows, we will show why it is sufficient to concentrate the search process to basic solutions Business Computing and Operations Research 9

120 . The Geometry of the solution space In what follows, we have to do a little bit mathematics By doing so, we get (hopefully!) some insights into the problem structure First of all, we focus on conveity Then, we learn about the solution space that it is sufficient to focus our search on the corner points Therefore, let the solution space P be given as defined above in the standard form Conveity is a very convenient attribute of solution spaces. Note that it causes among other advantages that each local optimum is also a global optimum Business Computing and Operations Research

121 Minimization problem In what follows, we consider minimization problems I.e., unless it is indicated differently, we consider minimization problems of the following structure Minimize and P z ( ) c T, with c IR { n IR and A b} n, P, Business Computing and Operations Research

122 .. Definition (Conve combination) Conve combinations k n k i Let a,..., a IR and α,..., αk IR, α i. Then, α i ia is denoted as a non-negative linear combination and as a i i { } conve combination if additionally α. If i,...,n : α >, k i then α a is a strict conve combination. i k i i Business Computing and Operations Research

123 Illustration Conve combinations conve combination strict conve combination Business Computing and Operations Research

124 .. Definition C if ( k a,..., a ), ( k C a,a ) Conve sets denotes the set of all conve combinations of a denotes the direct connection between a,..., a and a k... Definition A set S IR of pairs of n is conve if points,y S, i.e., C it contains all conve combinations (,y) S,,y S. Business Computing and Operations Research

125 Eamples Who is conve, who not? Business Computing and Operations Research

126 Solution space of LPs is conve ( ) { n } Consider two elements y,z P IR A b. ( ) Then, consider λ with λ and λ y λ z. Obviously,. ( ) ( ) ( ) ( ) A A λ y λ z λ A y λ Az ( ) λ b λ b b { n } P IR A b. Additionally, since λ and y,z, ( ) it holds λ y λ z. Business Computing and Operations Research 6

127 Intersection.. Lemma The intersection of any number of conve sets S i is conve. Proof: Let us consider two elements of the set S i. Then, each conve combination belongs to every set S i. Consequently, it also belongs to the intersection of all sets S i. This completes the proof. Business Computing and Operations Research 7

128 Linear combinations in conve sets.. Lemma ( k) Let S be a conve set and a,..., a k S, then C a,..., a S. We provide this proof by induction Beginning of induction: Show that the lemma holds for k. This is obviously trivial. Induction step: k> The proposition is held for all values up to k- Let us now consider k Business Computing and Operations Research 8

129 Proof of Lemma.. k k i i i i k k k i αi a, αi αk αi i i i Let us now consider: Thus, by assumption of induction: αi i α a a a S α α k ( ) k Consequently, it holds: α a α a S k We get: ( αk) a αk a ( αk) k k i k i αi a αk a αia i i k k k k i α i α k i k a αk a Business Computing and Operations Research 9

130 Hyperplanes { } { } n n T Let a IR \ and α IR. Then, H IR a α is denoted as a hyperplane. Hyperplanes are obviously conve. This can be easily shown: Let,., H λ Let us now consider: ( ( ) ) λ λ λ ( λ) ( ). T T T a a a λα λ α α Business Computing and Operations Research

131 Half spaces { } { } n n T Let a IR \ and α IR. Then, H IR a α is denoted as a half space. Half spaces are obviously conve. This can be easily shown as follows: Let, H, λ. Let us consider: ( ( ) ) T ( ) ( ) T a λ λ λ a λα λ α α. T λ a Business Computing and Operations Research

132 Observation A hyperplane in an n-dimensional space has the dimension n- A hyperplane defines two separated half spaces, i.e., it divides the space into two parts In the H n { n T IR, the hyperplane H IR a α} H { n T spaces IR a α} and { n T IR a α} the two half determines Business Computing and Operations Research

133 Conve hull..6 Definition,..., k,..., k CH M C a a a a M, k IN { } ( ) ( ) n is denoted as the conve hull to M IR. The set CH(M) is conve since a conve combination of two conve combinations of elements of set M is again a conve combination of elements of set M..7 Observation It holds: CH ( M ) { K M K K conve }, i.e., CH M is the smallest conve set that contains M. ( ) Business Computing and Operations Research

134 Proof of Observation..7 ( ) ( ) CH ( M ) SCH ( M ) ( ) ( ) Let SCH M be the smallest conve set that contains M. SCH M : This is correct since is the smallest conve set that contains M, CH M is conve, and it holds that M CH M. ( ) SCH ( M ). CH M : ( ) ( ) ( ) k i k αi a, a,...,a M i Consider CH M. Then, we know with and M SCH M. ( ) By applying Lemma.. and the conveity of set SCH M, we obtain SCH M. Business Computing and Operations Research

135 Illustration conve hull Business Computing and Operations Research

136 ..8 Definition ( ) Etreme points n A point K,K IR conve is denoted as an etreme point of K if it is not defineable by a strict conve combination. Let ε K be the set of all etreme points of K. Business Computing and Operations Research 6

137 Important attributes of etreme points..9 Observation The following propositions are equivalent. is an etreme point. a,b K, C a,b ab a b { } {} ( ). \ : n y IR y K y K. K\ is conve Business Computing and Operations Research 7

138 Proof of Observation..9 : ( ) trivial, since if, i.e., if,,, we can conclude that this conve combination is not strict. Thus, a b. : n ( ) y K y K y IR { } Let us assume holds and, \. Consider λ ( λ ) C a,b a a b b a b IR ( y) ( λ) ( y) y y y y y Let λ... ( ), λ λ λ λ λ λ y. a y b y a b This contradicts Business Computing and Operations Research 8

139 : Proof of Observation..9 ab K { } λ a ( λ) b λ ( ) ( ) { } Consider, \ and, with. We have to distinguish two cases:. λ a λ b, then we know λ a λ b K \ due to the conveity of K a λ a ( λ) b y b., then we set We consider: y y ( λ ) if λ,. In both cases we have y. otherwise ( ) ( ) λ ( λ) ( λ) λ ( λ) a if λ, a λ a λ b if λ, b otherwise b a b otherwise Since all weights are positive and, we conclude that { } ( ) λ ( λ) { } y K y K.This contradicts. Hence, we obtain a b K \ : Let K and K\ be conve. We assume that is not an etreme point, i.e., a,b K : a, b IR : a a b b But then it holds: { } { } K\ { } a b a b a,b K\ a a b b K\ a b a, b This contradicts the conveity of Business Computing and Operations Research 9

140 . ε. ε. ε. ε Eamples of etreme points ({ n IR } ) {} ( H ) ({ n }) { n IR IR } ({ n IR < } ) Business Computing and Operations Research

141 .. Definition Conve polyhedron A conve polyhedron is an intersection of a finite number of half spaces, i.e., m i { } P H A b If P P < (limited), then P is a conve polytope. { } T n { α } T n { α } A hyperplane H a a IR is significant if H P and P H a a IR for one half space. If H P, then is denoted as a corner point. If H P ab, then ab is denoted as an edge. Business Computing and Operations Research

142 Conve polyhedron - Eamples Consider a two-dimensional solution space P unbounded empty polytope, which is polyhedron polyhedron a bounded and non-empty polyhedron Business Computing and Operations Research

143 Conclusion.. Theorem Let P be a conve polyhedron. Then, all corner points are etreme points. Business Computing and Operations Research

144 Proof of Theorem.. Let be a corner point of P. In addition, let { n T } { α } H IR a a significant hyperplane with P H. We now make use of Observation..9 and consider y IR with y P y P y H y H Thus, it holds: ( ) α ( ) T T T T T T a y a a y a y a a y α and since T T α a a y Therefore, it holds: { } P H ( ) T T T a y a a y α { } y H y P y P H y Business Computing and Operations Research n

145 Basic solutions In what follows, we are able to bridge the gap to basic solutions, i.e., it is now possible to provide a definition of basic solutions Recall that these are just the solutions the Simple Algorithm is focusing on We feel that basic solutions are just corner points or etreme points of the solution space Business Computing and Operations Research

146 Basic solutions and corner points.. Definition Let B : and {,,..., m} {,..., n}, N :{,..., n m} {,..., n} B( {,..., m} ) N( {,..., n m} ) {,..., n} injective with A B invertible matri. ( B() B( m) ) m m B() B( m) a,...,a IR, a,...,a IR m an B Let B AB b and N. Then, is denoted as a N basic feasible solution (bfs) of A b, if. Business Computing and Operations Research 6

147 Business Computing and Operations Research 7 Observations ( ) ( ) ( ) () ( ) { } m m B B B N N B B N N B B N N B B N B N B IR a a A b b A b A A A b A A A A A A A base of is a,..., is invertible.,. Obviously, it holds :

148 .. Theorem Conclusions m n m Let A IR with rank A m n and let b IR. ( ) { n } Furthermore, let P IR and A b, for P. The following propositions are equivalent:. is an etreme point of { j a } j >. are linearly independent. is a basic feasible solution (bfs). is a corner point of P P Business Computing and Operations Research 8

149 () ( ) : Proof of Theorem.. Let us consider.... We sort the entries so that it holds: n,..., > and,..., r r n Case : r { j } j >. Then, a. By definition, this set is a set of linearly independent vectors. Business Computing and Operations Research 9

150 Case : r > Proof of Theorem.. i We introduce α,...,α IR : α a ( ) i Now, consider y α,...,α,,..., IR ( ) We compute i.e., y, y, ± y Consider r r A ± y A ± A y b b r i T n Thus, it holds: y, y P Business Computing and Operations Research

151 Proof of Theorem.. Thus, it holds: y, y P Since is an etreme point and we are making use of Observation..9, we can conclude that y r are linearly independent. a,...,a Business Computing and Operations Research

152 Proof of Theorem.. ( ) ( ) r We assume that are linearly independent. r a,...,a ( ) a,...,a ( ) ( ) If r < m since rank A m, we altogether have m linearly independent columns in A. Let li A be this set. Then, w.l.o.g., we can assume li A ( ) { r a,...,a } li A A B IR m m. We define B and N accordingly. If r m, we define. Then, it holds: is invertible and it holds: A b A A A A A B B B B B N N B Since P, we know and, therefore, is t feasible solution. he basic Business Computing and Operations Research

153 Proof of Theorem.. ( ) ( ) B We assume that is the basic feasible solution. N Let a,...,,,...,. Furthermore, let m elements nm elements { n T } H IR a. We can conclude the following: T T T T. Let P: a a a a since H P H B B N N N N Business Computing and Operations Research

154 Proof of Theorem... Consider a a a H Thus, H P T T T B B N N yb T. We consider y H P an yn. Since yn y, it holds y N Additionally, it holds: b A y A y A y A y y A b B B N N B B B B B { } y H P Consequently, is a corner point Business Computing and Operations Research

155 Proof of Theorem.. ( ) ( ) Assuming is a corner point. A b { n } n P IR A b IR A b E Obviously, P is a conve polyhedron (Definition..). P Since is the corner point and all corner points in P are etremal points, is an etremal point. Business Computing and Operations Research

156 .. Observation. P ε ( ) ( P) ( ) { } Conclusions This is true due to the fact that if P of Theorem.. is a corner point. ε P > m i i fulfills Restriction In order to conceive this proposition, we again make use of { } j Theorem... Owing to the fact that a j > are ( ) linearly independent and rank A m, the proposition immediately follows n n!. ε ( P) Binomial coefficient m m! ( m n)! Business Computing and Operations Research 6

157 .. Definition { } Degeneration ( P) A basic feasible solution ε is denoted as degenerated if > < m. i i..6 Observation The respective set of base vectors is unambiguously defined ( P) for each non-degenerate basic feasible solution ε. Business Computing and Operations Research 7

158 The structure of the solution space In what follows, we are able to provide a very compact and fundamental definition for the solution space P For this purpose, however, we have to distinguish if the solution space is bounded or unbounded Consequently, if the latter case applies, an infinite number of new elements of P can be generated iteratively Business Computing and Operations Research 8

159 Preliminary Definitions In what follows, we consider an LP with a solution space { n } P IR A b..7 Definition ( ) { n λ λ } Let D P y IR P: > : y P, for P...8 Lemma D ( ) { n P y IR y A y } Business Computing and Operations Research 9

160 Proof of Lemma..8 ( ) { n }. D P y IR y A y ( ) ( ) { n } Let y D P y IR P: λ > : λ y P, for P. Then, it holds: P: λ > : λ y P P: λ > : A λ y A λ A y b λ A y b λ A y A y. In addition, we know that P: λ > : λ y { } n y y y IR y A y ( ) { n } D P y IR y A y Business Computing and Operations Research 6

161 Proof of Lemma..8 { n y IR y A y } D( P). { n } Let y y IR y A y. Consider λ y, P λ >. Then, it holds: A λ y A λ A y b λ b ( ) λ y since λ y y D P. ( ) Business Computing and Operations Research 6

162 Final result The solution space..9 Minkowski s Theorem ( ε ( )) ( ) { n } P IR y z y C P z D P A non - empty polyhedron P is represented by a conve combination yof its ( ) etreme (corner) points ε and by z D P. If z, then zis called a ray. ( P) Business Computing and Operations Research 6

163 Minkowski s Theorem Eample I Let P. It is ε ( P), and D( P) λ λ λ, λ that is an infinite set. λ with λ. λ with λ. Business Computing and Operations Research 6

164 Minkowski s Theorem Eample II ( ) { n } Since A bin P, Lemma..8 becomes D P y IR y Ay. Note that this is satisfied by all y D( P). Theorem..9 states that P is equivalently written as n P IR α ( α) λ λ λ, λ Note that D( P) is omitted in the representation of Pbecause it is neutral to. Business Computing and Operations Research 6

165 Proof of Theorem..9 ( ε ( )) ( ) { n }. IR y z y C P z D P P Let ( ( )) ( ) ( ( )) ( ) { n } IR y λ z y C ε P z D P y z y C ε P z D P K A ( y z) A y A z Aζ k yk Az, k ( ) with: y,..., y ε P ζ K k K K ζk A y k A z ζk b A z b b k b k Additionally, it holds: y z K k Business Computing and Operations Research 6

166 Business Computing and Operations Research 66 Proof of Theorem..9 ( ) ( ) ( ) { } { } ( ) ( ) ( ) P D y P ε,..., λ y λ IR:,...,λ λ : n P:l IN: l j n P D z P C y z y IR P k k i i k i i i k j n > ε λ We show : is conducted by induction by The proof.

167 Proof of Theorem..9 We commence with n ε P ( ) { n ( ε ( )) ( )} IR y z y C P z D P Now, we assume that the proposition holds for all with n < l. Consider P with n l. Obviously, if ε P ( ), the proposition immediately follows. P Business Computing and Operations Research 67

168 Proof of Theorem..9 ( ) ( ) ( ) ( ) Consequently, we assume n Since ε P, it holds: y IR : y P y P Therefore, we can assume y, j We compute ( ) A λ y A λ A y b b, λ IR Thus, it holds: ε P. A y A y AA A y A b b λ y P λ y j j Business Computing and Operations Research 68

169 Business Computing and Operations Research 69 Proof of Theorem..9 [ ] ( ) ( ) ( ) Calculate ma, ˆ... Furthermore, let:... ˆ Obviously, it holds:, j j j j j j j j n Induction y y y y y y y n n v w v C P w D P λ λ ε > < Case :

170 ( ) Proof of Theorem..9 ( ε( )) ( ) Obviously, it holds: n < n v w, v C P w D P v wλ y, v C P ε ( ) Note that λ < λ y< wλ y and Induction λ y v w λ y ( ) λ λ λ ( ) A w y Aw A y w y D P ( ε ( )) ( ) { n } IR y z y C P z D P Business Computing and Operations Research 7

171 Proof of Theorem..9 Case : y j j Calculate λ min y j < [, ] yj y j ˆ... j Furthermore, let: λ y y y j j... ˆ n ε ( ) ( ) ( ) Obviously, it holds: n < n v w, v C P w D P Induction Business Computing and Operations Research 7

172 Proof of Theorem..9 Obviously, it holds:, ε ( ) Induction n ( ) ( ) v w v C P w D P < ( ) ( ( )) λ y v w λ y v wλ y, v C ε P Note that λ > wλ y since y and n ( ) λ λ λ ( ) A w y A w A y w y D P ( ε ( )) ( ) { n } IR y z y C P z D P Business Computing and Operations Research 7

173 Business Computing and Operations Research 7 Proof of Theorem..9 [ ] [ ] : Calculate min, Calculate ma, ˆ... Furthermore, let:... ˆ j i j j j j j j j j j j j j j n j i y y y y y y y y y y y λ λ λ < > < > Case :

174 Proof of Theorem..9 ε ( ) ˆ... j Furthermore, let: λ y y y j j... ˆ n since the considered entries of y are < ( ) ( ) ( ) Obviously, it holds: n < n a b, a C P b D P Obviously, it holds: n Induction < n a b, a C ε ( P) b D P Induction ( ) ( ) Business Computing and Operations Research 7

175 Proof of Theorem..9 Consider the following conve combination (note that λ < λ λ > ) λ λ λ y λ λ λ λ λ λ λ ( λ y) λ λ λ λ λ y λ λ λ y λ λ λ λ ( λ ) ( λ λ ) λ λ λ λ ( λ λ ) y y λ λ λ λ. Business Computing and Operations Research 7

176 Thus, we know: Proof of Theorem..9 λ λ λ λ a b a b λ λ λ λ λ λ λ λ ( ) ( ) λ λ λ λ a a b b λ λ λ λ λ λ λ λ C P A b b Ab Ab λ λ λ λ λ λ λ λ λ λ b b D( P) λ λ λ λ ( ε ( )) λ λ λ λ ( ε ( )) ( ) ( ε ( )) ( ) { n } IR y z y C P z D P { n } P IR y z y C P z D P Business Computing and Operations Research 76

177 Now, we are almost done with it We have shown that each solution in the solution space is a conve combination of the corner points and this enables the main cognition if there is an optimal solution then there is a corner point with an identical objective function value! Business Computing and Operations Research 77