Optimisation. 3/10/2010 Tibor Illés Optimisation

Transcription

1 Optimisation Lectures 3 & 4: Linear Programming Problem Formulation Different forms of problems, elements of the simplex algorithm and sensitivity analysis Lecturer: Tibor Illés tibor.illes@strath.ac.uk 3/10/2010 Tibor Illés Optimisation

2 Linear programming: problem description A company produces four different (A, B, C, D) products from three different resources (R₁, R₂, R₃). The resources are bounded by their capacities. The capacities of R₂ and R₃ should be used completely. From the products A and B in total we need to produces as much as from C. Resources Products Capacities A B C D R₁ R₂ R₃ Product prices From the product B we need to produce at least five unit more than from D. Maximize the profit! 3/10/2010 Tibor Illés Optimisation

3 Linear programming: problem formulation Decision variables: x₁ is the # of picies produced from product A x₂ is the # of picies produced from product B x₃ is the # of picies produced from product C x₄ is the # of picies produced from product D The capacities of R₂ and R₃ should be used completely. Constraints: Products A B C D R₁ x₁ + 2 x₃ + x₄ 280 R₂ R₃ x₁ + x₂ = 140 x₂ + x₃ + x₄ = 120 3/10/2010 Tibor Illés Optimisation

4 Constraints and the objective function Constraints: From the products A and B in total we need to produces as much as from C. From the product B we need to produce at least five unit more than from D. x₁ + x₂ = x₃ or x₁ + x₂ - x₃ = 0 x₂ - x₄ 5 Objective function max 4 x₁ + 5 x₂ + 6 x₃ + 8 x₄ 3/10/2010 Tibor Illés Optimisation

5 Linear programming problem Objective function Constraints max 4 x₁ + 5 x₂ + 6 x₃ + 8 x₄ x₁ + 2 x₃ + x₄ x₁ + x₂ = 140 x₂ + x₃ + x₄ = 120 x₁ + x₂ - x₃ = 0 x₂ - x₄ 5 Sign constraints x₁, x₂, x₃, x₄ 0 3/10/2010 Tibor Illés Optimisation

6 Linear programming: problem description In a factory using three machines (M₁, M₂, M₃) five different products (A₁, A₂, A₃, A₄, A₅) can be produced. All products are manufactured using all three machines. The processing time of different products are different on given machines. Specific processing times for different machines, theircapacities in working hours are shown on the following tables. The precies of different products are 2, 3, 2, 4, 2, respectively. Maximize the profit under the following constraints: Machines Products Capacities A₁ A₂ A₃ A₄ A₅ M₁ M₂ M₃ /10/2010 Tibor Illés Optimisation

7 Linear programming: problem formulation a) The capacities of the machines can not be exceeded. b) From the product A₁ at least twice as much should be produced as from A₅. c) Fr om the products A₂ and A₃ together can not be made more than 120. Decision variables: x₁, x₂, x₃, x₄, x₅ 0, are the decision variables [sign constraint on decision variables] assigned to the products A₁, A₂, A₃, A₄, A₅, respectively. Capacity constraints: [The information from the table and the verbal constraint a) is used.] x₁ + 2x₂ + 4x₃ + 3x₄ + 2x₅ 480 3x₁ + x₂ + x₃ + 5x₄ + 3x₅ 460 2x₁ + 3x₂ + x₃ + x₄ + 5x₅ 450 3/10/2010 Tibor Illés Optimisation

8 Linear programming: problem formulation Constraints: [Based on the verbal constraints b) and c)] From the product A₁ at least twice as much should be produced as from A₅. Fr om the products A₂ and A₃ together can not be made more than 120. x₁ 2x₅ in other form -x₁ + 2x₅ 0 x₂ + x₃ 120 Objective function max 2 x₁ + 3 x₂ + 2 x₃ + 4 x₄ + 2 x₅ 3/10/2010 Tibor Illés Optimisation

9 Linear programming problem Objective function max 2 x₁ + 3 x₂ + 2 x₃ + 4 x₄ + 2 x₅ Constraints Standard linear programming problem x₁ + 2x₂ + 4x₃ + 3x₄ + 2x₅ 480 3x₁ + x₂ + x₃ + 5x₄ + 3x₅ 460 2x₁ + 3x₂ + x₃ + x₄ + 5x₅ 450 -x₁ + 2x₅ 0 x₂ + x₃ 120 Sign constraints x₁, x₂, x₃, x₄, x₅ 0 3/10/2010 Tibor Illés Optimisation

10 Transportation problem/linear programming problem min 18 x x x x x x x x x x x x34 Objective function Constraints x11 + x12 + x13 + x14 = 370 x21 + x22 + x23 + x24 = 620 x31 + x32 + x33 + x34 = 440 x11 + x21 + x31 = 220 x12 + x22 + x32 = 380 x13 + x23 + x33 = 410 x14 + x24 + x34 = 420 Sign constraints x11, x12, x13, x14, x21, x22, x23, x24, x31, x32, x33, x34 0 3/10/2010 Tibor Illés Optimisation

11 Transportation problem/linear programming problem min 18 x x x x x x x x x x x x34 Objective function Priority routes: Case 3 upper bounds x₁₂ 110 x₃₃ x₂₄ 320 x11 + x12 + x13 + x14 = 370 x21 + x22 + x23 + x24 = 620 x31 + x32 + x33 + x34 = 440 Sign constraints Constraints x11 + x21 + x31 = 220 x12 + x22 + x32 = 380 x13 + x23 + x33 = 410 x14 + x24 + x34 = 420 x11, x12, x13, x14, x21, x22, x23, x24, x31, x32, x33, x34 0 3/10/2010 Tibor Illés Optimisation

12 Transportation problem: mathematical formulation Simplex tableau Linear programming formulation: min c x A x = b x 0 3/10/2010 Tibor Illés Optimisation

13 Standard linear programming problem max 2 x₁ + 3 x₂ + 2 x₃ + 4 x₄ + 2 x₅ 2x₁ + 3x₂ + 2x₃ + 4x₄ + 2x₅ - z = 0 x₁ + 2x₂ + 4x₃ + 3x₄ + 2x₅ + s₁ = 480 x₁ + 2x₂ + 4x₃ + 3x₄ + 2x₅ 480 3x₁ + x₂ + x₃ + 5x₄ + 3x₅ 460 2x₁ + 3x₂ + x₃ + x₄ + 5x₅ 450 -x₁ + 2x₅ 0 x₂ + x₃ 120 3x₁ + x₂ + x₃ + 5x₄ + 3x₅ + s₂ = 460 2x₁ + 3x₂ + x₃ + x₄ + 5x₅ + s₃ = 450 -x₁ + 2x₅ + s₄ = 0 x₂ + x₃ + s₅ = 120 x₁, x₂, x₃, x₄, x₅, s₁, s₂, s₃, s₄, s₅ 0 max z x₁, x₂, x₃, x₄, x₅ 0 x₁ x₂ x₃ x₄ x₅ b s₁ s₂ s₃ s₄ s₅ z x₁ x₂ x₃ x₄ x₅ s₁ s₂ s₃ s₄ s₅ z b s₁ s₂ s₃ s₄ s₅ z /10/2010 Tibor Illés Optimisation

14 Primal simplex method: optimal tableau... after the second iteration we get... x₁ x₂ x₃ x₄ x₅ s₁ s₂ s₃ s₄ s₅ z b s₁ x₄ s₃ s₄ x₂ z /10/2010 Tibor Illés Optimisation

15 Linear programming problem: Excel solver Initial values The matrix of the problem Objective function coefficients Variables x₁ x₂ x₃ x₄ x₅ Righthand side values Constraint formulation Relations between lefthand side and righthand side values 3/10/2010 Tibor Illés Optimisation

16 Linear programming problem: Excel solver Options Linear constraints Sign constraints on variables 3/10/2010 Tibor Illés Optimisation

17 Excel solver: answer report Microsoft Excel 12.0 Answer Report Worksheet: [tutorial-examples.xlsx]sheet2 Report Created: 30/10/ :30:52... after the second iteration we get... Target Cell (Max) Adjustable Cells Cell Name Original Value Final Value $K$14 objfun Cell Name Original Value Final Value $E$6 x₁ 0 0 $F$6 x₂ $G$6 x₃ 0 0 $H$6 x₄ $I$6 x₅ 0 0 x₁ x₂ x₃ x₄ x₅ s₁ s₂ s₃ s₄ s₅ z b s₁ x₄ s₃ s₄ x₂ z Constraints Cell Name Cell Value Formula Status Slack $K$8 const1 444 $K$8<=$L$8 Not Binding 36 $K$9 const2 460 $K$9<=$L$9 Binding 0 $K$10 const3 428 $K$10<=$L$10 Not Binding 22 $K$11 const4 0 $K$11<=$L$11 Binding 0 $K$12 const5 120 $K$12<=$L$12 Binding 0 3/10/2010 Tibor Illés Optimisation

18 Negative reduced cost have 5 variables: x₁, x₃, x₅, s₂ & s₅. s₅ has reduced cost -2.2 and since s₅ = 0, the resource 5 has been fully used. If we force s₅ to enter the basis then the objective function value will decrease (bad pivot column). Any unused unit of resource 5 will decrease the OF by 2.2. Primal simplex method: optimal tableau... after the second iteration we get... x₁ x₂ x₃ x₄ x₅ s₁ s₂ s₃ s₄ s₅ z b s₁ x₄ s₃ s₄ x₂ z Opportunity cost: the increase in the value of the OF that would occur if we could obtain an extra unit of this scare resource over and above the current supply limit 3/10/2010 Tibor Illés Optimisation

19 Linear programming problem: Excel solver Target Cell (Max) Cell Name Original Value Final Value $K$14 objfun Cell Name Original Value Final Value $K$14 objfun Adjustable Cells Constraints Cell Name Original Value Final Value $E$6 x₁ 0 0 $F$6 x₂ $G$6 x₃ 0 0 $H$6 x₄ $I$6 x₅ 0 0 Cell Name Cell Value Formula Status Slack $K$8 const $K$8<=$L$8 Not Binding 34.6 $K$9 const2 460 $K$9<=$L$9 Binding 0 Cell Name Original Value Final Value $E$6 x₁ 0 0 $F$6 x₂ $G$6 x₃ 0 0 $H$6 x₄ $I$6 x₅ 0 0 $K$10 const $K$10<=$L$10 Not Binding 19.2 $K$11 const4 0 $K$11<=$L$11 Binding 0 $K$12 const5 121 $K$12<=$L$12 Binding 0 Why? 3/10/2010 Tibor Illés Optimisation

20 Opportunity cost: how much more of this resource to use? Increasing by 1 unit the 5th resource, it will force the following changes: s₁ = = 34.6 x₄ = = 67.8 s₃ = = 19.2 s₄ = = 0 x₂ = = 121 OF = = Fully coincides with the Excel solution! x₁ x₂ x₃ x₄ x₅ s₁ s₂ s₃ s₄ s₅ z b s₁ x₄ s₃ s₄ x₂ z max 2 x₁ + 3 x₂ + 2 x₃ + 4 x₄ + 2 x₅ 4*(-0.2) + 3 *1 = 2.2 min {36/1.4, 68/0.2, 22/2.8} = = min { 25.71, 340, 7.86} = /10/2010 Tibor Illés Optimisation

21 Changing resource 5 to 127 and 128 For computation use Excel solver and read Wisniewski & Dacre: Optimization models for business and management decision making, McGraw-Hill Book Co., 1990, chapter 8 on sensitivity analysis 3/10/2010 Tibor Illés Optimisation

22 Changing resource 5 to 127 and 128 Target Cell (Max) Adjustable Cells Cell Name Original Value Final Value $K$14 objfun Cell Name Original Value Final Value $E$6 x₁ 0 0 $F$6 x₂ $G$6 x₃ 0 0 $H$6 x₄ $I$6 x₅ 0 0 Target Cell (Max) Adjustable Cells Cell Name Original Value Final Value $K$14 objfun Cell Name Original Value Final Value $E$6 x₁ 0 0 $F$6 x₂ $G$6 x₃ $H$6 x₄ $I$6 x₅ 0 0 Constraints Constraints Cell Name Cell Value Formula Status Slack $K$8 const $K$8<=$L$8 Not Binding 26.2 $K$9 const2 460 $K$9<=$L$9 Binding 0 $K$10 const $K$10<=$L$10 Not Binding 2.4 $K$11 const4 0 $K$11<=$L$11 Binding 0 $K$12 const5 127 $K$12<=$L$12 Binding 0 Cell Name Cell Value Formula Status Slack $K$8 const $K$8<=$L$8 Not Binding 24.4 $K$9 const2 460 $K$9<=$L$9 Binding 0 $K$10 const3 450 $K$10<=$L$10 Binding 0 $K$11 const4 0 $K$11<=$L$11 Binding 0 $K$12 const5 128 $K$12<=$L$12 Binding 0 3/10/2010 Tibor Illés Optimisation

23 Excel solver: sensitivity report Microsoft Excel 12.0 Sensitivity Report Worksheet: [tutorial-examples.xlsx]sheet2 Report Created: 30/10/ :30:52 Adjustable Cells If the values of c i and b j are changing in the given intervals Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $E$6 x₁ E+30 $F$6 x₂ E+30 1 $G$6 x₃ E+30 $H$6 x₄ $I$6 x₅ E+30 c₁ < c₂ c₃ < c₄ 15 c₅ < Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $K$8 const E $K$9 const $K$10 const E $K$11 const E+30 0 $K$12 const b₁ 120 b₂ b₃ 0 b₄ 0 b₅ < then the basis is unchanged, thus only x₂ and x₄ are in the basis. 3/10/2010 Tibor Illés Optimisation

24 Excel solver: sensitivity analysis test problem x₁ x₂ x₃ x₄ x₅ Same solution structure. const const const const const objfun c₁ < c₂ c₃ < c₄ 15 c₅ < Target Cell (Max) 444 Cell Name b₁ Original Value Final Value $K$14 objfun b₂ b₃ 0 b₄ Adjustable Cells Cell Name Original Value Final Value $E$6 0 x₁ b₅ < $F$6 x₂ $G$6 x₃ 0 0 $H$6 x₄ $I$6 x₅ 0 0 Constraints Cell Name Cell Value Formula Status Slack $K$8 const1 260 $K$8<=$L$8 Not Binding 240 $K$9 const2 200 $K$9<=$L$9 Binding 0 $K$10 const3 320 $K$10<=$L$10 Not Binding 180 $K$11 const4 0 $K$11<=$L$11 Not Binding 100 $K$12 const5 100 $K$12<=$L$12 Binding 0 3/10/2010 Tibor Illés Optimisation

25 Linear programming problem: Excel solver Microsoft Excel 12.0 Limits Report Worksheet: [tutorial-examples.xlsx]limits Report 1 Report Created: 30/10/ :30:53 Target Cell Name Value $K$14 objfun 632 Adjustable Lower Target Upper Target Cell Name Value Limit Result Limit Result $E$6 x₁ E $F$6 x₂ $G$6 x₃ $H$6 x₄ $I$6 x₅ /10/2010 Tibor Illés Optimisation

26 (Primal) simplex algorithm: standard problem Input: standard LP problem initial basic feasible solution is given Apply the ratio test, determine the pivot position, compute the new solution. Is current solution optimal? (optimality criterion) no Find pivot column (positive reduced cost value). yes Stop Unbounded LP no yes Is current solution unbounded? (unboundedness criterion) Stop Optimal solution 3/10/2010 Tibor Illés Optimisation

27 General LP problem and simplex algorithm: summary Linear programming problem: standard form (maximization problem, less-than-equal type constraints, sign restricted variables, non-negative right hand side vector) initial feasible basis (initial basic tableau, basic & non-basic variables, basic feasible solution) slack and artificial variables (transforming general LP problems using slack and artificial variables into the 1st phase problem that has initial feasible basis) sensitivity analysis of linear programming problems Simplex algorithm (Dantzig, 1947): elementary row operations pivoting (pivot row and pivot column, [bad and good] pivot position) optimality and unboundedness criteria solving standard linear programming problems using simplex algorithm solving general linear programming problems (two phase simplex algorithm) Excel LP solver 3/10/2010 Tibor Illés Optimisation

28 3/10/2010 Tibor Illés Optimisation

29 Standard linear programming problem max 2 x₁ + 3 x₂ + 2 x₃ + 4 x₄ + 2 x₅ 2x₁ + 3x₂ + 2x₃ + 4x₄ + 2x₅ - z = 0 x₁ + 2x₂ + 4x₃ + 3x₄ + 2x₅ + s₁ = 480 x₁ + 2x₂ + 4x₃ + 3x₄ + 2x₅ 480 3x₁ + x₂ + x₃ + 5x₄ + 3x₅ 460 2x₁ + 3x₂ + x₃ + x₄ + 5x₅ 450 -x₁ + 2x₅ 0 x₂ + x₃ 120 3x₁ + x₂ + x₃ + 5x₄ + 3x₅ + s₂ = 460 2x₁ + 3x₂ + x₃ + x₄ + 5x₅ + s₃ = 450 -x₁ + 2x₅ + s₄ = 0 x₂ + x₃ + s₅ = 120 x₁, x₂, x₃, x₄, x₅, s₁, s₂, s₃, s₄, s₅ 0 max z x₁, x₂, x₃, x₄, x₅ 0 x₁ x₂ x₃ x₄ x₅ b s₁ s₂ s₃ s₄ s₅ z x₁ x₂ x₃ x₄ x₅ s₁ s₂ s₃ s₄ s₅ z b s₁ s₂ s₃ s₄ s₅ z /10/2010 Tibor Illés Optimisation

30 LP problem: general form max 4 x₁ + 5 x₂ + 6 x₃ + 8 x₄ x₁ + 2 x₃ + x₄ x₁ + x₂ = 140 x₂ + x₃ + x₄ = 120 x₁ + x₂ - x₃ = 0 x₂ - x₄ 5 x₁, x₂, x₃, x₄ 0 Slack variables: s₁, s₅ 1st phase problem Artificial variables: w₂, w₃, w₄, w₅ 2nd phase problem Decision variables: x₁, x₂, x₃, x₄, x₅ max z 4 x₁ + 5 x₂ + 6 x₃ + 8 x₄ - z =0 x₁ + 2 x₃ + x₄ + s₁ =280 2 x₁ + x₂ + w₂ = 140 x₂ + x₃ + x₄ + w₃ = 120 x₁ + x₂ - x₃ + w₄ = 0 x₂ - x₄ - s₅ + w₅ = 5 x₁, x₂, x₃, x₄, x₅, s₁, s₅, w₂, w₃, w₄, w₅ 0 min w₂ + w₃ + w₄ + w₅ max - w₂ - w₃ - w₄ - w₅ 3/10/2010 Tibor Illés Optimisation

31 General LP problem: bounded variables max -3 x₁ - x₂ - x₃ + 2 x₄ - x₅ + x₆ + x₇ - 4 x₈ x₁ + 3 x₃ + x₄ - 5 x₅ - 2 x₆ + 4 x₇ - 6 x₈ = 7 x₂ - 2 x₃ - x₄ + 4 x₅ + x₆ - 3 x₇ + 5 x₈ = -3 0 x₁ 8, 0 x₂ 6, 0 x₃ 4, 0 x₄ 15 0 x₅ 2, 0 x₆ 10, 0 x₇ 10, 0 x₈ 3 Simplex algorithm can be modified to deal with bounded variables. For the Excel solver you should enter these bounds as constraints and simply solve the problem. 3/10/2010 Tibor Illés Optimisation

32 General LP problem: Excel solver data x₁ x₂ x₃ x₄ x₅ x₆ x₇ x₈ constraint constraint bound bound bound bound bound bound bound bound obj fun Why? 3/10/2010 Tibor Illés Optimisation

33 General LP problem: How to use Excel solver? x₁ x₂ x₃ x₄ x₅ x₆ x₇ x₈ constraint constraint bound bound bound bound bound bound Constraints Cell Name Cell Value Formula Status Slack $L$6 constraint1 7 $L$6=$M$6 Not Binding 0 $L$7 constraint2-3 $L$7=$M$7 Not Binding 0 Even if we add the bound on x₇ the same solution remains feasible and optimal, too. Thus the bound on x₈ made the $L$8 bound1 0 $L$8<=$M$8 Not Binding 8 $L$9 bound2 6 $L$9<=$M$9 Binding 0 $L$10 bound3 0 $L$10<=$M$10 Not Binding 4 $L$11 bound4 15 $L$11<=$M$11 Binding 0 $L$12 bound5 2 $L$12<=$M$12 Binding 0 $L$13 bound6 1 $L$13<=$M$13 Not Binding 9 obj fun Adjustable Cells Final Reduced Objective Allowable Allowable problem 24 infeasible. Cell Name Value Cost Coefficient Increase Decrease $C$4 x₁ E+30 $D$4 x₂ E+30 2 $E$4 x₃ E+30 $F$4 x₄ E+30 1 Optimal solution! $G$4 x₅ E+30 1 $H$4 x₆ $I$4 x₇ $J$4 x₈ E+30 3/10/2010 Tibor Illés Optimisation

34 General LP problem: unbounded or free variables max x₁ + x₇ x₁ + x₂ + x₃ + x₄ + x₅ + x₆ - x₇ = 10 x₁ + 2x₂ + x₇ = 2 x₃ + 2x₄ + x₇ = 5 x₅ + 2x₆ + x₇ = 8 x₁, x₂, x₃, x₄, x₅, x₆ 0 and x₇ is an unbounded (free) variable Express x₇ from any of the given equations and substitute back into other equations and into the objective function as well. x₇ = 8 - x₅ - 2x₆ 3/10/2010 Tibor Illés Optimisation

35 General LP problem: unbounded or free variables max x₁ - x₅ - 2x₆ + 8 x₁ + x₂ + x₃ + x₄ + 2x₅ + 3x₆ = 18 x₁ + 2x₂ - x₅ - 2x₆ = -6 x₃ + 2x₄ - x₅ - 2x₆ = -3 x₁, x₂, x₃, x₄, x₅, x₆ 0 # of variables and constrains are reduced. Multiply the 2nd and the 3rd equations by -1 x₁ + x₂ + x₃ + x₄ + 2x₅ + 3x₆ = 18 - x₁ - 2x₂ + x₅ + 2x₆ = 6 - x₃ - 2x₄ + x₅ + 2x₆ = 3 3/10/2010 Tibor Illés Optimisation

36 Solving general LP problem by simplex algorithm Linear programming problem: standard form (maximization problem, less-than-equal type constraints, sign restricted variables, non-negative right hand side vector) Simplex algorithm solves only standard LP problems. This problem has initial feasible solution: s=b, x=0 and the objective function value is 0. max cx Ax b (b 0) x 0 Add slack variables max cx Ax + s = b (b 0) x, s 0 The goal is to transform any LP problem into such that has an initial feasible basis. (Two phase.) 3/10/2010 Tibor Illés Optimisation

37 Solving general LP problem by simplex algorithm General LP problem. Using slack variables transform the constraints into equations. If we have bounded or unbounded variables then (i) eliminate unbounded variables [in this way the number of variables and equations General will be LP reduced], problems (ii) transform the the bounded variables into such state-of-the-art new variables that has 0 as lower bound. linear programming solvers Multiply any equation by -1 if it is necessary to get non-negative right hand side. (CPLEX, XPRESS-MP, Excel) If the transformed problem without has initial any feasible transformation basis then we can solve it by using the simplex algorithm can [or be modified fed. SA that handles bounded variables], otherwise we should introduce artificial variables to obtain the first phase LP problem. 3/10/2010 Tibor Illés Optimisation

38 General LP problem and simplex algorithm: summary Linear programming problem: standard form (maximization problem, less-than-equal type constraints, sign restricted variables, non-negative right hand side vector) initial feasible basis (initial basic tableau, basic & non-basic variables, basic feasible solution) slack and artificial variables (transforming general LP problems using slack and artificial variables into the 1st phase problem that has initial feasible basis) sensitivity analysis of linear programming problems Simplex algorithm (Dantzig, 1947): elementary row operations pivoting (pivot row and pivot column, [bad and good] pivot position) optimality and unboundedness criteria solving standard linear programming problems using simplex algorithm solving general linear programming problems (two phase simplex algorithm) Excel LP solver 3/10/2010 Tibor Illés Optimisation

39 LP problem: general form max 4 x₁ + 5 x₂ + 6 x₃ + 8 x₄ x₁ + 2 x₃ + x₄ x₁ + x₂ = 140 x₂ + x₃ + x₄ = 120 x₁ + x₂ - x₃ = 0 x₂ - x₄ 5 x₁, x₂, x₃, x₄ 0 x₁ x₂ x₃ x₄ s₁ s₅ w₂ w₃ w₄ w₅ z v b s₁ w₂ w₃ w₄ w₅ z v After six iterations the first phase problem has been solved: x₁ = , x₂ = , x₃ = , x₄ = 0, s₁ = , s₅ = 28,3333 It is easy to check the feasibility. 3/10/2010 Tibor Illés Optimisation

40 LP problem: general form x₁ x₂ x₃ x₄ s₁ s₅ w₂ w₃ w₄ w₅ z b s₁ x₁ s₅ x₂ x₃ z The second phase problem is not optimal. Why? Pivot position: (3,4). Variable x₄ enters and s₅ leaves the basis. 3/10/2010 Tibor Illés Optimisation

41 LP problem: general form x₁ x₂ x₃ x₄ s₁ s₅ w₂ w₃ w₄ w₅ z b s₁ x₁ x₄ x₂ x₃ z Is this basic tableau optimal? Yes, it is optimal! Why? Binding constraints are the 2nd 5th. Optimal solution is: x₁ = 59, x₂ = 22, x₃ = 81, x₄ = 17, s₁ = 42, s₂ = 0 Resource 1 is not fully used, because s₁ = 42, however there is no surplus production from product x₃, because s₅ = 0. The objective function row shows that s₅ has coefficient -2.4, therefore if we relax this constraint to 4 then the objective function value will be 970.4, however if we increase it to 6, then the objective function will decrease to /10/2010 Tibor Illés Optimisation

42 LP problem: general form x₁ x₂ x₃ x₄ s₁ s₅ w₂ w₃ w₄ w₅ z b s₁ x₁ x₄ x₂ x₃ z x₁ = = 59.2 x₂ = = 21.6 x₃ = = 80.8 x₄ = = 17.6 s₁ = = 41.6 s₅ = = 0 Objective function value: 4* * * *17.6 = /10/2010 Tibor Illés Optimisation

43 LP problem: general form x₁ x₂ x₃ x₄ s₁ s₅ w₂ w₃ w₄ w₅ z b s₁ x₁ x₄ x₂ x₃ z Increase by 1 unit of the 2nd resource will result the following solution with objective function value of How to analyze the case of the 3rd or 4th row? What does it mean the column of w₄ or w₅? Take into consideration an equality constraint for instance the 2nd. Objective function row in column w₂ contains here as well the opportunity cost. x₁ = = 59.6 x₂ = = 21.8 x₃ = = 81.4 x₄ = = 16.8 s₁ = = 40.8 s₅ = = 0 3/10/2010 Tibor Illés Optimisation

44 General LP problem: sensitivity analysis Maximization problems: interpretation of the effects of a 1 unit increase in the RHS of a binding constraint Constraint type The relevant columni is that of the associated slack variable. The opportunity cost coefficient will be negative. The coefficient indicates the increase in the objective function. The column coefficients indicates the change in the values of the basic variables. The maximum change is determined by the smallest ratio of values to negative coefficients in the slack variable column. 3/10/2010 Tibor Illés Optimisation

45 General LP problem: sensitivity analysis Maximization problems: interpretation of the effects of a 1 unit increase in the RHS of a binding constraint Constraint type The relevant columni is that of the associated surplus (slack) variable. (Surplus variable has -1 coefficient.) The opportunity cost coefficient will be negative. The coefficient indicates the decrease in the objective function. The column coefficients must have their signs reversed (positive) to indicates the change in the values of the basic variables. The maximum change is determined by the smallest ratio of values to positive coefficients in the slack variable column. 3/10/2010 Tibor Illés Optimisation

46 General LP problem: sensitivity analysis Maximization problems: interpretation of the effects of a 1 unit increase in the RHS of a binding constraint Constraint type = The relevant columni is that of the associated artificial variable. The opportunity cost coefficient will be negative. The coefficient indicates the increase in the objective function. The column coefficients indicates the change in the values of the basic variables. The maximum change is determined by the smallest ratio of values to negative coefficients in the relevant column. 3/10/2010 Tibor Illés Optimisation

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48 An oil refinery produces four types of raw gasoline: alkylate, catalytic-cracked, straight-run and isopentane. Two important characteristics of each gasoline are its performance number PN (indicating antiknock properties) and its vapor pressure RVP (indicating volatility). These two characteristics, together with the production levels in barrels per day, are as follows: PN RVP Barrels produced Alkylate Catalytic-cracked Straight-run Isopentane A simplified blending problem These gasolines can be sold either raw at per barrel, or blended into aviation gasoline (Avgas A and/or Avgas B). 3/10/2010 Tibor Illés Optimisation

49 A simplified blending problem: continue Quality standards impose certain requirements on the aviation gasolines: these requirements, together with the selling prieces, are as follows: PN RVP Price per barrel Avgas A at least 100 at most Avgas B at least 91 at most The refinery aims for the production plan that yields the largest possible profit. Formulate as a linear programming problem (in standard form). Remark: The PN and RVP of each mixture are simply weighted averages of the PNs and RVPs of its constituents. 3/10/2010 Tibor Illés Optimisation

50 Model for the blending problem x₁ Alkylate x₁₂ x₁₁ a Avgas A x₂ Catalytic-cracked x₂₁ x₂₂ x₄₁ x₄₂ Isopentane Avgas B b x₄ x₃₂ x₃₁ x₃ Straight-run 3/10/2010 Tibor Illés Optimisation

51 Decision variables Linear programming model for blending Alkylate Catalytic-cracked Straight-run Isopentane Avgas A Avgas B x₁ x₂ x₃ x₄ a b Avgas A Avgas B Alkylate x₁₁ x₁₂ Catalytic-cracked x₂₁ x₂₂ Straight-run x₃₁ x₃₂ Isopentane x₄₁ x₄₂ The decision variables are either sign restricted or bounded variables. 0 x₁ 3814, 0 x₂ 2666, 0 x₃ 4016, 0 x₄ /10/2010 Tibor Illés Optimisation

52 Bounds, flow and qualitative constraints x₁₁ + x₁₂ + s₁ = x₁ x₂₁ + x₂₂ + s₂ = x₂ x₃₁ + x₃₂ + s₃ = x₃ x₄₁ + x₄₂ + s₄ = x₄ x₁₁ + x₂₁ + x₃₁ + x₄₁ = a x₁₂ + x₂₂ + x₃₂ + x₄₂ = b 100 a 107 x₁₁ + 93 x₂₁ + 87 x₃₁ x₄₁ 91 b 107 x₁₂ + 93 x₂₂ + 87 x₃₂ x₄₂ 5 x₁₁ + 8 x₂₁ + 4 x₃₁ + 21 x₄₁ 7 a 5 x₁₂ + 8 x₂₂ + 4 x₃₂ + 21 x₄₂ 7 b max a b (s₁ + s₂ + s₃ + s₄) 3/10/2010 Tibor Illés Optimisation

53 Linear programming problem: Excel form a b x₁ x₂ x₃ x₄ x₁₁ x₂₁ x₃₁ x₄₁ x₁₂ x₂₂ x₃₂ x₄₂ s₁ s₂ s₃ s₄ RHS The LP problem is primal and dual degenerate! /10/2010 Tibor Illés Optimisation

54 a b x₁ x₂ x₃ x₄ x₁₁ x₂₁ x₃₁ x₄₁ x₁₂ x₂₂ x₃₂ x₄₂ s₁ s₂ s₃ s₄ RHS E E E E E E E E /10/2010 Tibor Illés Optimisation

55 Optimal solution of the blending problem Alkylate Catalytic-cracked Avgas A Avgas B Isopentane Straight-run /10/2010 Tibor Illés Optimisation

56 New qualitative variables: bilinear terms 100 a 107 x₁₁ + 93 x₂₁ + 87 x₃₁ x₄₁ 91 b 107 x₁₂ + 93 x₂₂ + 87 x₃₂ x₄₂ 5 x₁₁ + 8 x₂₁ + 4 x₃₁ + 21 x₄₁ 7 a 5 x₁₂ + 8 x₂₂ + 4 x₃₂ + 21 x₄₂ 7 b PNa a = 107 x₁₁ + 93 x₂₁ + 87 x₃₁ x₄₁ PNb b = 107 x₁₂ + 93 x₂₂ + 87 x₃₂ x₄₂ 5 x₁₁ + 8 x₂₁ + 4 x₃₁ + 21 x₄₁ = RVPa a 5 x₁₂ + 8 x₂₂ + 4 x₃₂ + 21 x₄₂ = RVPb b 99.8 PNa 102, 90.8 PNb 93, 6.5 RVPa, RVPb 7.5 3/10/2010 Tibor Illés Optimisation

57 Resource allocation problem Ajax Ltd. manufactures and sells three types of computer: α-pc, β-nb and γ-ws. For the moment we assume that all production during the week can and will be sold immediately. α-pc β-nb γ-ws Net profit in s Labour in hours Net profit equals the sales price of each computer minus the direct cost of purchasing components, producing computer cases, and assembling and testing computer. This week, 120 hours are available on the A-line test equipment where assembled α-pcs and β-nbs are tested, and 48 hours are available on the C-line test equipment where assembled γ-wss are tested. The testing of each computer takes 1 hour. In addition, production is constrained by the availability of 2000 labor hours for product assembly. 3/10/2010 Tibor Illés Optimisation

58 Resource allocation model Decision variables a the number of α-pcs b the number of β-nbs c the number of γ-wss max 160 a b c 10 a + 15 b + 20 c 2000 a + b 120 c 48 a, b, c 0 Final Reduced Objective Allowable Allowable Cell Name Original Value Final Value $C$9 a $D$9 b 0 0 $E$9 c 0 40 Cell Name Cell Value Formula Status Slack $G$11 production 2000 $G$11<=$H$11 Binding 0 $G$12 test1 120 $G$12<=$H$12 Binding 0 $G$13 test2 40 $G$13<=$H$13 Not Binding 8 Cell Name Original Value Final Value $G$15 objective Cell Name Value Cost Coefficient Increase Decrease $C$9 a E+30 5 $D$9 b E+30 $E$9 c Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $G$11 production $G$12 test $G$13 test E /10/2010 Tibor Illés Optimisation

59 Shadow price A fundamental ingredient in the economic analysis of an LP model is the shadow price associated with each constraint, which is defined as the change in the optimal value of the objective function if the right-hand side of the constraint is increased by one unit. For reasons of symmetry, it is also defined as the change in the optimal value of the objective function if the right-hand side of the constraint is decreased by one unit. 3/10/2010 Tibor Illés Optimisation

60 Shadow price: practical application Rounding the solution to the nearest integer? a b c Non integer optimal solution! 1 unit decrease in the right-hand side production test test objective Shadow price of the 1st constraint is 15.5 Change in the objective function value is = /10/2010 Tibor Illés Optimisation

61 Rounding to the nearest integer solution a b c solution solution solution solution 1 solution 2 solution a b c production test test objective solution 1 optimal solution of the linear programming problem with real variables solution 2 integer solution, but infeasible solution 3 it is a feasible, integer solution of the LP problem. Optimality of the integer solution? 3/10/2010 Tibor Illés Optimisation

62 Resource allocation problem: dual version The Immense Computer Co. is experiencing a rapid growth in sales. As a result, Immense has insufficient capacity for testing and assembling their computers, and the director of purchasing is looking to rent capacity from other smaller, companies. α-pc She is considering β-nb approaching γ-ws Ajax to offer Capacity to rent its capacity on a weekly basis. Net profit in s In particular, she wishes to determine nonnegative prices per hour of A-line Labour test capacity, 10 nonnegative 15 prices 20 per in hours hour of 2000 C-line test capacity A-line and nonnegative 1 prices 1 per hour 0 of in labour hours capacity 120 to offer Ajax that C-line will induce 0 Ajax into 0 agreeing 1 to rent in hours its resources 48 rather than to use them in the manufacture of its own products. At the same time she wishes to pay the least amount for these resources. A number of large manufacturing firms are concered with the economic viability of their small suppliers. 3/10/2010 Tibor Illés Optimisation

63 Resource allocation model: dual problem Decision variables u price per hour of A-line test v price per hour of C-line test z price per hour of labour min 120 u + 48 v z u + 10 z 160 u + 15 z 210 v + 20 z 310 u, v, z 0 Cell Name Original Value Final Value $C$19 u α-pc β-nb γ-ws Capacity Net profit in s $D$19 v 0 5 $E$19 z 0 0 Labour in hours 2000 Cell Name Cell Value Formula Status Slack A-line in hours 120 $G$ $G$21>=$H$21 Binding 0 $G$ $G$22>=$H$22 Not Binding 27.5 C-line in hours 48 $G$ $G$23>=$H$23 Binding 0 Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $C$19 u $D$19 v $E$19 z E+30 8 Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $G$ E+30 5 $G$ E+30 $G$ /10/2010 Tibor Illés Optimisation

64 Integerality of the solution(s) of LP problems max 160 a b c 10 a + 15 b + 20 c 2000 a + b 120 c 48 a, b, c 0 min 120 u + 48 v z u + 10 z 160 u + 15 z 210 v + 20 z 310 u, v, z 0 Name Final Value a 120 b 0 c 40 Name Final Value u 15.5 v 5 z 0 solution a b c Why these solutions are not integer? Explanation lies in deeper understanding of mathematics behind these models! 3/10/2010 Tibor Illés Optimisation

65 Shadow price & reduced cost Standard primal dual linear programming problem primal objective coefficient = dual constraint R.H. side dual objective coefficient = primal constraint R.H. side primal solution = dual shadow price dual solution = primal shadow price primal slack dual slack = dual reduced cost = - primal reduced cost Mathematics! 3/10/2010 Tibor Illés Optimisation

66 Resource allocation model: modification Since the full capacity of C-line test equipment were not used in the optimal production plan, the engineers made an invention and as a consequence of this α-pcs can be tested on C-line as well, but such a test takes 1.5 hours. Modify the model for this case. max 160 a b c a a₂ a₁ = 0 10 a + 15 b + 20 c 2000 a₁ + b a₂ + c 48 a, a₁, a₂, b, c 0 3/10/2010 Tibor Illés Optimisation

67 Solution of the modified model a1 a2 a b c a-pcs production test test objective The optimal solution of the modified resource allocation problem is optimal and all resources are fully used. Furthermore the solution is integer! 3/10/2010 Tibor Illés Optimisation

68 Resource allocation: outsourcing For the shake of illustration, suppose Ajax can outsource A-line testing equipment at 20/hour in unlimited quantities and can hire additional labour at 15/hour in unlimited quantities. Let a b c A L L quantity of rented labour hours A quantity of outsourced A-line test hours max 160 a b c 20 A 15 L 10 a + 15 b + 20 c L a + b A c 48 L, A, a, b, c 0 production test test objective /10/2010 Tibor Illés Optimisation

69 Multiperiod Resource Allocation Problem After reviewing the optimal assembly plan given by the optimal solution (120, 0, 40) of the previous LP model the Ajax marketing manager doubts that his organization can sell 120 units of α-pcs next week. Moreover, he is concerned that the plan does not call for the assembly of any β-nbs. Even if the profit margin from β-nbs is smaller relative to the other two pruducts, the marketing manager feels strongly that it must be included in Ajax product line. Thus he requests instead that the production manager develop a 4-week production strategy based on the sales forecasts shown Because Ajax has capital tied up in products, carrying costs must be charged for items held in inventory. These carrying costs 4 per week for Week 1 Week 2 Week 3 Week 4 α-pc [20,60] [20,80] [20,120] [20,140] β-nb [20,40] [20,40] [20,40] [20,40] γ-ws [20,50] [20,40] [20,30] [20,70] each α-pc, 5 per week for each β-nb, and 9 per week for each γ-ws. Initial inventory at this week equals 22 α-pcs, 42 β-nbs, and 36 γ-wss. For each product, these balance equations are of the form I(t) = I(t-1) + P(t) S(t) 3/10/2010 Tibor Illés Optimisation

70 Multiperiod Resource Allocation Model 10 a j + 15 b j + 20 c j 2000 a j + b j 120 c j 48 a j, b j, c j 0 (j = 1, 2, 3, 4) Production constraints Ia j = Ia j-1 + a j Sa j, La j Sa j Ua j Ib j = Ib j-1 + b j Sb j, Lb j Sb j Ub j Ic j = Ic j-1 + c j Sc j, Lc j Sc j Uc j Ia j, Ib j, Ic j, Sa j, Sb j, Sc j 0, (j = 1, 2, 3, 4) Inventory constraints 160 Sa j Sb j Sc j - 4 Ia j - 5 Ib j - 9 Ic j = z j, z j 0, (j = 1, 2, 3, 4) max z₁ + z₂ + z₃ + z₄ Profit constraints 3/10/2010 Tibor Illés Optimisation

71 Sp₁ Production constraints Week 1 p₁ - Sp₁ Profit constraints Week 1 Multiperiod Resource Allocation Model Ip₁ Inventory constraints Week 1 Production constraints Week 2 Sp₂ p₂ - Sp₂ Profit constraints Week 2 Inventory constraints Week 2 Production constraints Week 3 Profit constraints Week 3 3/10/2010 Tibor Illés Optimisation Ip2 Sp₃

72 Structure of the multi-period model a₁ b₁ c₁ a₂ b₂ c₂ a₃ b₃ c₃ a₄ b₄ c₄ Ia₁ Ib₁ Ic₁ Ia₂ Ib₂ Ic₂ Ia₃ Ib₃ Ic₃ Ia₄ Ib₄ Ic₄ Sa₁ Sb₁ Sc₁ Sa₂ Sb₂ Sc₂ Sa₃ Sb₃ Sc₃ Sa₄ Sb₄ Sc₄ z₁ z₂ z₃ z₄ production₁ test1₁ test2₁ production₂ test1₂ test2₂ production₃ test1₃ test2₃ production₄ test1₄ test2₄ inventory-a₁ equal 22 inventory-b₁ equal 42 inventory-c₁ equal 36 inventory-a₂ equal 0 inventory-b₂ equal 0 inventory-c₂ equal 0 inventory-a₃ equal 0 inventory-b₃ equal 0 inventory-c₃ equal 0 inventory-a₄ equal 0 inventory-b₄ equal 0 inventory-c₄ equal 0 obj-week₁ equal 622 obj-week₂ equal 0 obj-week₃ equal 0 obj-week₄ equal 0 lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds obj function /10/2010 Tibor Illés Optimisation

73 Solution of the multi-period model a₁ b₁ c₁ a₂ b₂ c₂ a₃ b₃ c₃ a₄ b₄ c₄ Ia₁ Ib₁ Ic₁ Ia₂ Ib₂ Ic₂ Ia₃ Ib₃ Ic₃ Ia₄ Ib₄ Ic₄ Sa₁ Sb₁ Sc₁ Sa₂ Sb₂ Sc₂ Sa₃ Sb₃ Sc₃ Sa₄ Sb₄ Sc₄ z₁ z₂ z₃ z₄ production₁ test1₁ test2₁ production₂ test1₂ test2₂ production₃ test1₃ test2₃ production₄ test1₄ test2₄ inventory-a₁ equal 22 inventory-b₁ equal 42 inventory-c₁ equal 36 inventory-a₂ E-14 equal 0 inventory-b₂ E-13 equal 0 inventory-c₂ E-13 equal 0 inventory-a₃ E-14 equal 0 inventory-b₃ E-13 equal 0 inventory-c₃ E-14 equal 0 inventory-a₄ E-13 equal 0 inventory-b₄ E-14 equal 0 inventory-c₄ E-13 equal 0 obj-week₁ equal 622 obj-week₂ E-11 equal 0 obj-week₃ E-11 equal 0 obj-week₄ E-12 equal 0 lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds lower bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds upper bounds obj function /10/2010 Tibor Illés Optimisation