Optimization in Process Systems Engineering

Transcription

1 Optimization in Process Systems Engineering M.Sc. Jan Kronqvist Process Design & Systems Engineering Laboratory Faculty of Science and Engineering Åbo Akademi University

2 Most optimization problems in production planning, scheduling and design, involves some kind of discrete decision making. To model such decisions we need to use discrete variables. This session is intended as an introduction to some of the basic concepts in integer optimization

3 Background Standard problem formulation min f ( ) s.t. g ( ) 0 j 1,, m, j h ( ) 0 i 1,, m i N Conve if f and g i are conve functions and h i are all linear. Integer problems are classified as conve if an integer relaation results in a conve problem

4 Basic properties of linear programming

5 Linear programming (LP) All functions defining the objective and constraints are linear. An important type of optimization problems. Efficient algorithms and software for solving these problems. We can handle problems with millions of variables. For eample minimization of l 1 -norm

6 Linear programming (LP) Consider the following eample min st ,

7 Linear programming (LP) Consider the following eample min st , The feasible region of an LP problem is always a conve set!

8 Definition: An etreme point of a conve set is a point which cannot be obtained as a conve combination of two other points within the set. A conve combination z of the points and y is given by z = α + 1 α y, α 0,

9 Etreme points 0, 3.5 min st , 2, ,

10 Theorem 1: Any point within a conve set can be obtained as a conve combination of the etreme points C = α 1 e 1 + α 2 e 2 + α n e n, n i=1 α i = 1, α i 0, where e i are the etreme points of the conve set C

11 Theorem 2: For an LP problem, one of the etreme points corresponds to an optimal solution

12 Back to our eample problem As stated by Theorem 1, any point within the feasible set can be obtained as a conve combination of the etreme points. We can, thus, rewrite the problems as: min st ,, 0 3 min st , The etreme points are 2, 0, 0, 3.5 and 4.21,

13 By substituting 1 and 2 in the objective we obtain: min st.. 3 1, 3,, 0. 3 The optimal solution is obviously α 3 = 1. And the optimal solution to the LP problem is the etreme point 4.21,

14 Solving LP problems in Matlab Gurobi is one of most powerful solvers for LP, QP, QPQC, MILP, MIQP and MIQCQP available. Free academic licenses! Do not use Matlab s own solvers!

15 Solving LP problems in Matlab The problems must be written as min st.. T c A b lb ub

16 Mied-integer linear programming (MILP) Linear optimization problems containing some discrete variables. Non-conve problems! NP-hard. Today many MILP problems can be solved efficiently

17 MILP Consider the following optimization problem min st ,

18 MILP How can we solve this problem? min st , Rela the problem by removing the integer restrictions

19 MILP We begin by solving the relaed problem min st ,

20 Branch and Bound Divide the problem into subproblems The search space is divided into subregions to force the variables to take on integer values

21 Branch and Bound

22 Branch and Bound

23 Branch and Bound

24 Branch and Bound

25 Branch and Bound The number of nodes to eplore can grow rapidly with the number of integer variables! Worst case: With 25 binary variables we have 2 25 possible solutions, and nodes in the BB tree. How does Gurobi solve the eample problem?

26 MILP How can Gurobi solve the problem without branching???

27

28 Cutting planes We want to introduce cuts that reduce the search space without cutting of any feasible integer combination! These cuts are given by additional linear inequality constraints

29 Strengthening the relaation min st , First, let s eamine the variable bounds

30 Strengthening the relaation min st , First, let s eamine the variable bounds

31 Chvatal-Gomory Cuts A simple rounding procedure. Suppose we have a constraint of the type: a 1 + b 2 c, Where 1 and 2 are positive integer variables. We can round a and b down to take on integer values. a 1 + b 2 c Now the left hand side can only take on integer values! The Chvatal-Gomory Cut is given by: a 1 + b 2 c

32 Strengthening the relaation min st ,

33 Chvatal-Gomory Cuts included to our problem min st ,

34 Now, solving the integer relaation gives us an integer solution min st ,

35 Chvatal-Gomory Cuts Consider a pure integer problem with the constraints A b, where A is a m n matri with columns {a 1, a 2,, a n }. All valid cuts for such a problem can be obtained by n where u. n j=1 ua j j ub, All such problems can be solved as an LP problem by adding a finite number of Chvatal-Gomory Cuts! Why are integer problems difficult then? How to find the correct set of cuts? The number of cuts needed may be enormous!

36 Cuts for MILP problems Similar cuts can be made for problems containing both integer and continuous variables. Mied-integer rounding (MIR) cuts or mied-integer Gomory (MIG) cuts. Many different types of cuts! Lift and project cuts, Clique cuts, Disjunctive cuts, Cover Cuts, Network cuts,.. Still an active research area

37 Why are MILP problems an important type of optimization problems? Possible to model discrete quantities. Possible to incorporate logic decisions in the optimization problem!

38 Consider the following optimization problem: min st or , How can we formulate this as an MILP problem?

39 Let s start with a simple case: or We want to write these constraints as an MILP problem

40 Let s use the same technique here! min st or ,

41 Big-M formulation min st M (1 b ) M (1 b ) M (1 b ) M (1 b ) M (1 b ) M (1 b ) b 2 b b {0,1}, b {0,1},

42 How to choose M? M has to be big enough! Choice of M also affects the integer relaation! A too large value for M can make the problem difficult to solve! min st M (1 b ) M (1 b ) M (1 b ) M (1 b ) M (1 b ) M (1 b ) b 2 b b {0,1}, b {0,1},

43 How to choose M? M = 4 M =

44 How to choose M? M = 7 z = 11.3 M = 10 z = 12.7 M = 20 z = 18 z denotes the optimal objective value of the integer relaed problem. The optimal objective value for the problem is

45 The big-m formulation is not the only techinque by which we can epress the problem as an MILP problem. Conve hull formulation The conve hull formulation gives a tighter integer relaation, but the formulations requires more variables and constraints. The LP subproblems becomes more difficult

46 Disjunctive programming min z st.. A 0 0 Y1 Y2 A b A b Y1 Y2 T T c1 z c2 z Y Y Y Y Y Y lb ub b N {True,false} {True,false}

47 Disjunctive programming Eample: We want to produce two products p1 and p2. In order to produce these products we need to buy some new equipment, and we can choose between two machines m1 and m2. However, we can only choose one of the machines. The machines have the following restrictions: m1 is limited to 600 tones of p1 and 500 tones p2 and the total production is limited to 1000 tones. m2 is limited to 300 tones of p1 and 700 tones p2 and the total production is limited to 900 tones. Furthermore, the machines have different operating costs: The operating cost of m1 is given by: z = p p 2. The operating cost of m2 is given by: z = p p 2. Selling p1 gives a profit of 250 /ton and p2 gives a profit of 130 /ton. Determine the most profitable production strategy!

48 ma 250 p 130 p z st.. Y1 Y2 p 600 p p2 500 p2 700 p1 p p1 p2 900 z p 70 p z p 80 p Y p Y Y 1 2 Y, p {True,false} {True,false}

49 ma 250 p 130 p z st.. Y1 Y2 p1 600 p1 300 p2 500 p2 700 p1 p p1 p2 900 z p 70 p z p 80 p Y p 1 2 Y, p Y {True,false} Y {True,false} ma 250 p 130 p z st.. p p 600 (1 b ) M (1 b ) M 2 1 p p 1000 (1 b ) M 1 z p 70 p (1 b ) M p p 300 (1 b ) M 700 (1 b ) M 2 2 p p 900 (1 b ) M 2 1 z p 80 p (1 b ) M b p b b 1, p {0,1}, b {0,1}

50 A scheduling problem We are producing 3 products and we want to complete the production as fast as possible. The products are produced in 4 process steps, and the times for each step are given in the table. Processing times in minutes. Process step Product 1 Product 2 Product

51 A scheduling problem We begin by introducing variables for the starting time of each product in each production step t p,s

52 Process step 1 Only product 2. Processing times in minutes. Process step Product 1 Product 2 Product

53 Process step 2 Processing times in minutes. Process step Product 1 Product 2 Product Product 1 and 3 cannot be handled simultaneously. t 1,2 t 3,2 + 9 or t 3,2 t 1,2 + 5 (1 b 1 )M + t 1,2 t 3,2 + 9 b 1 M+ t 3,2 t 1,2 + 5 b 1 =1 Product 3 before product 1 b 1 =0 Product 1 before product

54 Process step 3 Processing times in minutes. Process step Product 1 Product 2 Product The previous processing step has to be completed before the new step starts: t 1,3 t 1,2 + 5, t 2,3 t 2,1 + 3, t 3,3 t 3,

55 Process step 3 Processing times in minutes. Process step Product 1 Product 2 Product Processing order: t 2,3 t 1,3 + 3 or t 1,3 t 2,3 + 4 t 3,3 t 1,3 + 3 or t 1,3 t 3,3 + 5 t 3,3 t 2,3 + 4 or t 2,3 t 3,

56 Process step 3 Processing times in minutes. Process step Product 1 Product 2 Product Processing order: 1 b 2 M + t 2,3 t 1,3 + 3 b 2 M + t 1,3 t 2, b 3 M + t 3,3 t 1,3 + 3 b 3 M + t 1,3 t 3, b 2 =1 Product 1 before product 2 b 2 =0 Product 2 before product 1 b 3 =1 Product 1 before product 3 b 3 =0 Product 3 before product 1 1 b 4 M + t 3,3 t 2,3 + 4 b 4 M +t 2,3 t 3,3 + 5 b 4 =1 Product 2 before product 3 b 4 =0 Product 3 before product

57 Process step 4 Processing times in minutes. Process step Product 1 Product 2 Product The previous processing step has to be completed before the new step starts: t 1,4 t 1,3 + 3, t 3,4 t 3, t 3,4 t 1,4 + 4 or t 1,4 t 3,4 + 6 (1 b 5 )M + t 3,4 t 1,4 + 4 b 5 M+ t 1,4 t 3,4 + 6 b 5 =1 Product 1 before product 3 b 5 =0 Product 3 before product

58 Processing times in minutes. Process step Product 1 Product 2 Product Finally we want to minimize the longest production time t final t 1,4 + 4 t final t 2,3 + 4 t final t 3,

59 min t final s.t. (1 b 1 )M + t 1,2 t 3,2 + 9 b 1 M+ t 3,2 t 1,2 + 5 t 1,3 t 1,2 + 5 t 2,3 t 2,1 + 3 t 3,3 t 3, b 2 M + t 2,3 t 1,3 + 3 b 2 M + t 1,3 t 2, b 3 M + t 3,3 t 1,3 + 3 b 3 M + t 1,3 t 3, b 4 M + t 3,3 t 2,3 + 4 b 4 M +t 2,3 t 3,3 + 5 t 1,4 t 1,3 + 3 t 3,4 t 3,3 + 5 (1 b 5 )M + t 3,4 t 1,4 + 4 b 5 M+ t 1,4 t 3,4 + 6 t final t 1,4 + 4, t final t 2,3 + 4, t final t 3,4 + 6 b 1,, b 5 {0,1}, t 1,2, t 3,2,, t final R

60 Other types of integer optimization problems MIQP Ok, if conve. Gurobi. MIQCQP More difficult, but tractable if conve. Gurobi. MINLP Several techniques are available for conve MINLP

61 Conve MINLP Nonlinear branch and bound. Solve a conve NLP in each node. Linear approimation techniques. Construct an equivalent MILP problem

62 Thank you! Any questions?