Mixed-Integer Linear Programming. Mixed-Integer Linear Programming (MILP): Model Formulation. Mixed-Integer Linear Programming

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1 Mixed-Integer Linear Programg (MILP): Model Formulation Benoît Chachuat McMaster University Department of Chemical Engineering ChE 4G03: Optimization in Chemical Engineering Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 1 / 26 Mixed-Integer Linear Programg Integer Programs (IP) An optimization model is an Integer Program if any of its decision variables is discrete If all variables are discrete, the model is a pure integer program Otherwise, the model is a mixed-integer program Integer variables appear in many problems: Trays in a distillation column Number of employees (1000 s) Number of parallel chemical reactors Whether or not to operate boiler#2 on Monday Scheduling people and equipment to tasks over time Can be solved continuous, then rounded to nearest integer Not appropriate to solve continuous and round after Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 2 / 26 Mixed-Integer Linear Programg Class Exercise: Give more examples of integer decisions in the field of Chemical Engineering: 1 Fluid flow: 2 Heat transfer: 3 Mass transfer: 4 Reactor design: Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 3 / 26 Mixed-Integer Linear Programg Linear vs. Nonlinear Integer Programs An IP model is an integer linear program (ILP) if its (single) obective function and all its constraints are linear Otherwise, it is an integer nonlinear program (INLP) Standard Mixed-Integer Linear Programg (MILP) Formulation: x,y z = c T x + d T y s.t. Ax + Ey = b x x x max, y {0,1} ny Mixed-Integer Nonlinear Programs (MINLPs) are very difficult to solve This is a great motivation for the continued use of LP methods! Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 4 / 26

2 Outline Modeling MILP requires knowledge and ingenuity We will learn some typical MILP formulations here: 1 Native MILP Formulations Lumpy Linear Programs Knapsack Models Assignment and Matching Models Scheduling Models 2 Approximations of Nonlinear Models as MILPs Separable Programg For additional examples, see Rardin (1998), Chapte1 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 5 / 26 Lumpy Linear Programs One broad class of lumpy LPs is when either/or decisions are added to what is otherwise an LP model Expressing All-or-Nothing Requirements All-or-nothing variable requirements of the form x = 0 x = u can be modeled by substituting x u y, s.t. y {0,1} Example: In a blending model, use none or all of a given ingredient x 1,x 2,x 3 18x 1 + 3x 2 + 9x 3 s.t. 2x 1 + x 2 + 7x x x 2 30 x 3 = 0 x 3 = 20 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 6 / 26 Lumpy Linear Programs (cont d) Another source of lumpy LP models is when the obective involves fixed charges (i.e., start-up cost) Θ(x ) Expressing Fixed-Charge Requirements { Fixed-cost requirements of the form: Θ(x ) = f + c x, if x > 0 with nonnegative fixed charge f > 0 and variable x u, can be modeled by substituting Θ(x ) f y + c x, s.t. x u y and y {0,1} f c u x Formulating Models for Fixed-Charged Obectives Class Exercise: Consider a fixed-charge obective function where and x 1, x 2 satisfy x 1,x 2 Θ 1 (x 1 ) + Θ 2 (x 2 ) { Θ 1 (x 1 ) = x1, if x 1 > 0 { Θ 2 (x 2 ) = x2, if x 2 > 0 x 1 + x x x 2 8. Formulate a corresponding MILP model. Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 7 / 26 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 8 / 26

3 Lumpy Linear Programs (cont d) Yet another source of lumpy LP models is when a decision variable is semi-continuous c x Expressing Semi-Continuous Requirements Semi-continuous variables of the form: x = 0 l x u, with l > 0 can be modeled by introducing 2 new continuous variables x l,xu 0, 1 new binary variable y {0,1}, and 2 additional constraints: l c u x Formulating Models for Semi-Continuous Variables Class Exercise: Consider the following blending problem, where the ingredient x 3 is a semi-continuous variable: x 1,x 2,x 3 18x 1 + 3x 2 + 9x 3 s.t. 2x 1 + x 2 + 7x x x 2 30 Formulate a corresponding MILP model. x 3 = 0 10 x 3 20 x = l x l + u x u 1 = y + x l + xu Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 9 / 26 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 10 / 26 Knapsack Models The problem is to select a maximum value collection of items subect to limitations on resources consumed Knapsack models are the simplest of all (pure) integer linear programs (ILPs) Value 000 I I I 3 I 4 Each element is either all in or all out of the selection: y = { 1, if item selected Capacity max. Knapsack Models (cont d) 1 Dependence of choice on choice i can be enforced on corresponding binary variables by adding constraints of the form: 2 Mutually exclusiveness conditions allowing at most one of a set of choices J can be enforced by adding constraints of the form: Similarly, at most p of a set of choices J can be enforced as: Example: Selection of proects subect to limitations on budgets Class Exercise: Readily available Canadian coins are 1, 5, 10 and 25. Formulate an ILP model to imize the number of coins needed to provide change amount for q. Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 11 / 26 3 Constraints can also enforce selection of exactly p or at least p of a set of choices J Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 12 / 26

4 Formulating ILP Models Class Exercise: The Department of Chemical Engineering at McMaster is acquiring optimization software for use in ChE 4G03. The 4 codes available and the types of algorithms they provide are indicated by s in the following table: Algorithm Code, Type LP IP NLP Cost Formulate a MILP model to acquire a imum cost collection of codes providing LP, IP and NLP capability 2 Formulate a MILP model to acquire a imum cost collection of codes with exactly one providing LP, one providing IP, and one providing NLP capability Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 13 / 26 Assignment Models The issue here is optimal matching or pairing of obects of two distinct types It is standard to model all assignment forms with the decision variables: Set I Set J y i, = { 1, if i of the 1st set is matched with of the 2nd Examples: Assign sales personnel to customers, obs to machine, etc. Assignment constraints forcing every i and/or every to be assigned are of the form: Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 14 / 26 Assignment Models (cont d) Generalized assignment constraints encompass cases where allocation of i to requires fixed space s i, within capacity b : Linear assignment models imize/maximize a linear obective function of the form: c i, y i, i I J where c i, is the cost of assigning i to Class Exercise: Items i = 1,...,100 of volume c i are being stored in an automated warehouse. Storage location = 1,...,20 are at a distance d from the input/output station, and all have capacity b. Formulate a MILP model to store all items at imum total travel distance. Matching Models Unlike assignment models, matching models eliate the distinction between the sets Decision variables of matching models typically are: { 1, if i is paired with i y i,i = where i > i, by convention, to avoid double counting Set I Matching constraints forcing some i I to pair with and only with some other i I are of the form: Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 15 / 26 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 16 / 26

5 Matching Models (cont d) Class Exercise: The instructor of ChE 4G03 wants to assign his students to 2-person teams for a graded tutorial. each student s has scored his/her preference p s,s for working with each other student s. 1 Formulate a MILP model to form teams in a way that maximizes total preference. Process Scheduling Models Fee Scheduling is the allocation of resources over time Single-process scheduling problems seek an optimal sequence in which to complete a given collection of tasks on a single process that can accommodate only one ob at a time Typical Data: p = estimated process time for task p 2 r = release time at which task becomes available for processing d = due date by which task should be completed Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 17 / 26 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 18 / 26 Fee Decision Variables and Related Constraints: x = start time for task resource availability: x r, Handling of Due Dates: p 2 Due dates are usually handled as goals to be reflected in the obective function rather than explicit constraints This avoids the situation where there is no feasible schedule that meets all due dates Dates that must be met are termed deadlines to distinguish and can be easily enforced as: x + p d, Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 19 / 26 Fee p 2 Obective Functions Often imizes one of the following: Max. completion time (makespan) max{x + p : = 1,..., n} Mean completion time 1 nx (x + p ) n Max. lateness max{x + p d : = 1,..., n} Mean lateness 1 nx (x + p d ) n Max. tardiness Mean tardiness =1 =1 max{0, max{x + p d : = 1,..., n}} 1 nx max{0, x + p d } n Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 20 / 26 =1

6 Class Exercise: The following table shows the process time, release times, due dates, and scheduled starts for three obs. Compute the corresponding value of each of the six obective functions listed previously. Job 1 Job 2 Job 3 Process time, p i Release time, r i Due Date, d i Scheduled start, x i Completion times (x i + p i ) are: = 24, = 30, = 9 Maximum completion time is max{24, 30, 9} = 30 Mean completion time is 1 3 ( ) = 21 2 Lateness of the three obs (x + p d ) is: = 4, = 5, = 27 Maximum lateness is max{4, 5, 27} = 5 Mean completion time is 1 3 ( ) = 6 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 21 / 26 Fee Conflict Constraints: p 2 x + p x x + p x,, = 1,...,n, Reformulation Using Disunctive Variables: { 1, if binding comes before y, = 0, if binding comes before Illustrative Example: y 1,2 = y 1,3 = 1, y 2,3 = 0 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 22 / 26 Fee Big-M Method 01 p 2 Conflicts between tasks and can be prevented with disunctive constraint pairs: x + p x + M(1 y, ) x + p x + My, where M is a large positive constant (e.g., the max. makespan) Class Exercise: Formulate integer linear constraints for feasible schedules on a single process with 3 tasks having process times 14, 3, and 7. Separable Programg Consider the following mathematical program: x z = s.t. f (x ) = f 1 (x 1 ) + + f n (x n ) =1 a i, x b i, i = 1,...,m =1 x 0, = 1,...,n The obective consists of n nonlinear, separable terms f (x ), each a function of a single variable only The m constraints are linear When each f is convex or concave on the feasible region, the separable program can be approximated with an LP Otherwise, the separable program can be approximated with an MILP Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 23 / 26 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 24 / 26

7 Approximating General Separable Programs as MILPs Approximating General Separable Programs as MILPs Piecewise affine approximation: (N intervals) f (1) f (3) f (2) f (0) f (x ) c 2 = f (2) f (1) x (1) 0 ω,2 x (1) x (0) x (1) x (3) x Define: c (k) = f (k) f (k 1) 0 ω,k Substitute each variable x and function f by: x x (0) f (x ) f (0) + + N k=1 N k=1 ω,k c (k) ω,k Piecewise affine approximation: (N intervals) f (1) f (3) f (2) f (0) f (x ) c 2 = f (2) f (1) x (1) 0 ω,2 x (1) x (0) x (1) x (3) x Disunctive Variables: y,k = { 0, if ω,k = 0 1, if ω,k > 0 Disunctive Constraints: [ ] ω,k y,k, ω,k [ k = 1,...,N ] y,k+1, k = 1,...,N 1 Do you see why this works? Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 25 / 26 Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 25 / 26 Approximating Separable Programs as MILPs (cont d) Approximate Mixed-Integer Linear Program (on N Intervals): ω,y s.t. =1 f (0) + N c (k) ω k =1 k=1 N a i, ω,k b i =1 k=1 [ ω,k ω,k [ ] =1 a i, x (0), i = 1,...,m y,k, k = 1,...,N ] y,k+1, k = 1,...,N 1 y,k {0,1}, = 1,...,n, k = 1,...,N The solution can be made as accurate as desired by using enough intervals One pays the price in terms of increased problem size! Benoît Chachuat (McMaster University) MILP: Model Formulation 4G03 26 / 26