Chapter 3: Discrete Optimization Integer Programming

Transcription

1 Chapter 3: Discrete Optimization Integer Programming Edoardo Amaldi DEIB Politecnico di Milano Sito web: A.A Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

2 3.1 Integer Programming models A huge variety of practical decision-making problems arising in science, engineering and management can be formulated/approximated as linear optimization problems where (some of) the variables must take integer/discrete values. Generic discrete optimization problem: min x X c(x) where X is a discrete set and c : X R is the objective function. Example: X family of all subsets of a given finite set. A natural and systematic way to study/investigate such problems is to express them as integer (programming) optimization problems. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

3 Definitions: A Mixed Integer Linear Programming (MILP) problem is an optimization problem like min c t 1 x + ct 2 y s.t. A 1x + A 2y b x 0 integer, y 0 with matrices A 1 Z m n 1 and A 2 Z m n 2, and vectors c 1 Z n 1, c 2 Z n 2 and b Z m. If all the variables are restricted to be integer, we have an Integer Linear Programming (ILP) problem: min c t x s.t. Ax b x 0 integer. If all the variables must take binary values, we have a Binary Linear Programming (0-1-ILP) problem. Without loss of generality: only inequalities and all coefficients are integer. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

4 Observation: The integrality restriction on x is a nonlinear constraint since it can be formulated as sin(πx) = 0 Proposition: 0-1-ILP is NP-hard, and ILP/MILP are at least as difficult. Theory: No algorithm can find, for every instance of 0-1-ILP (ILP/MILP), an optimal solution in polynomial time (in the instance size), unless P=NP. Practice: Many medium-size (M)ILPs are extremely challenging! Examples of feasible regions of an ILP and a MILP: (Mixed) Integer Linear Programming is a powerful and versatile modeling framework for expressing and tackling discrete optimization problems. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

5 Modeling techniques and examples Binary variables allow to model a choice between two (several) alternatives and the association between two (several) entities. Example 1: Binary Knapsack problem Given n objects profit p i and weight a i for each object i, with 1 i n knapsack maximum total weight b (capacity) decide which objects to select so as to maximize total profit while respecting the capacity constraint. ILP formulation Decision variables: x i = 1 if the i-th object is selected and x i = 0 otherwise, 1 i n max n i=1 p ix i n i=1 a ix i b x i {0, 1} i Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

6 A number of direct applications (projects, investments,...) or indirect applications (as subproblem). Proposition: Binary Knapsack is NP-hard. Variants with additional constraints - at most one object among a given subset of objects - if i-th object selected then also j-th one - multiple resource constraints (e.g., on volume, cost,...) -... Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

7 Example 2: Assignment problem Given n projects (jobs) and n persons (machines) cost c ij for assigning project i to person j, i, j {1,..., n} decide which project to assign to each person so as to minimize the total cost to complete all the projects. Assumption: every person can perform any project, and each person (project) must be assigned to a single project (person). n! feasible solutions (permutations) ILP formulation Decision variables: x ij = 1 if i-th project is assigned to j-th person and x ij = 0 otherwise, min s.t. n i=1 n j=1 c ijx ij n i=1 x ij = 1 n j=1 x ij = 1 x ij {0, 1} j i i, j with 1 i, j n Variants with: bipartite graph representing competences, different number of projects and persons, resource constraints,... Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

8 Example 3: Set Covering problem (SC) Given finite groundset M = {1, 2,..., m} collection {M 1,..., M n} of n subsets of M ( M j M for j = 1,..., n) cost c j of M j for each j, with 1 j n decide which subsets to select so as to minimize the total cost while covering all the elements of M. Example: ILP formulation Decision variables: x j = 1 if subset M j is selected and x j = 0 otherwise, with 1 j n min s.t. n j=1 c jx j j:i M j x j 1 i (1) x j {0, 1} j where the inequalities (1) are the so-called covering constraints. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

9 Set Covering problem: { n } min c j x j : Ax 1, x {0, 1} n j=1 where A = [a ij ] with a ij = 1 if i M j and a ij = 0 otherwise, and 1 = (1, 1,..., 1) t Example: emergency service location (ambulances or fire stations) M = { areas to be covered }, M j = { areas reachable in at most 10 minutes from candidate site j } Proposition: Set Covering is NP-hard. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

10 Set Packing problem: { n } max c j x j : Ax 1, x {0, 1} n j=1 where the parameters c j represent profits Example: location of facilities with large environmental impact (rubbish dump or incinerators) M = { cities }, M j = { cities with environmental impact level for candidate site j above threshold } Proposition: Set Packing is NP-hard. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

11 Set Partitioning problem: { n } min or max c j x j : Ax = 1, x {0, 1} n where the parameters c j may represent costs or profits j=1 Example: aircrew scheduling (assign flights to pilots) Consider a predefined planning horizon M = { flights } where flight = single flight leg (takeoff-landing) to be carried out within a predefined time window M j = { feasible subsets of flights } where feasible subset of flights = a subset that can be combined and carried out by a single pilot while respecting different constraints (e.g., compatible flights, rest periods, total flight time,...) N.B.: Possibly huge number of feasible subsets of flights Other example: distribution planning (assign customers to routes) Proposition: Set Partitioning is NP-hard. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

12 Example 4: Asymmetric Traveling Salesman Problem (ATSP) Given a complete directed graph G = (V, A) with n = V nodes a cost c ij R for each arc (i, j) A (in case c ij = ) determine a Hamiltonian circuit (tour), i.e., a circuit that visits exactly once each node and comes back to the starting node, of minimum total cost. Example: Since G is complete, there are (n 1)! Hamiltonian circuits. Proposition: ATSP is NP-hard. Variety of applications: planning, logistics, microchip manufacturing, (DNA) sequencing,... Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

13 Also symmetric TSP version with undirected graph G. Website devoted to TSP: Many variants with - time windows (earliest and latest arrival time) - precedence constraints - capacity constraint - several vehicles ( Vehicle Routing Problem ) -... Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

14 An ILP formulation Decision variables: x ij = 1 if arc (i, j) is included in the Hamiltonian circuit and x ij = 0 otherwise, with (i, j) A min (i,j) A c ijx ij s.t. j V :j i x ij = 1 i (2) i V :i j x ij = 1 j (3) (i,j) δ + (S) x ij 1 S V, S (4) x ij {0, 1} (i, j) A where equations (2) and (3) are the assignment constraints, δ + (S) = {(i, j) A : i S, j V \ S}, constraints (4) are the so-called cut-set inequalities. Observation: The number of constraints (4) grows exponentially with the number of nodes n. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

15 Alternative ILP formulation Substitute the cut-set inequalities x ij 1 (i,j) δ + (S) S V, S with the so-called subtour elimination inequalities: x ij S 1 S V, 2 S n 1 (5) (i,j) E(S) where E(S) = {(i, j) A : i S, j S} for S V. Example: Observation: The number of constraints (5) grows exponentially with the number of nodes n. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

16 MILP models A pair of continuous and binary variables allow to model a typical nonlinear fixed charge cost function. Example 5: Uncapacitated Facility Location (UFL) Given M = {1, 2,..., m} set of clients N = {1, 2,..., n} set of candidate sites where a depot can be located fixed cost f j for opening a depot in candidate site j, j N c ij transportation cost if the whole demand of client i is served from depot j, i M and j N decide where to locate the depots and how to serve the clients so as to minimize the total (transportation and fixed) costs while satisfying all demands. Example: Proposition: UFL is NP-hard. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

17 MILP formulation Decision variables: x ij = fraction of demand of client i served by depot j, with 1 i m and 1 j n y j = 1 if depot j is opened and y j = 0 otherwise, with 1 j n min i M j N c ijx ij + j N f jy j s.t. j N x ij = 1 i M i M x ij my j j N (6) y j {0, 1} j N 0 x ij 1 i M, j N where the n constraints (6) link the corresponding variables x ij and y j. Capacitated FL variant: If d i is the demand of client i and k j the capacity of depot j, capacity constraints: d i x ij k j y j j N i M N.B.: The solution with, for all i and a given j, x ij = 0 and y j = 1 is feasible for MILP, but it cannot be optimal (minimization with f j > 0). Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

18 Example 6: Uncapacitated Lot-Sizing (ULS) A company has to plan the production of a single type of product for the next n periods. Suppose that the stock is empty at the beginning of the planning horizon and that 50 units must be in stock at the end. Given f t fixed cost for producing during period t p t unit production cost in period t h t unit storage cost in period t d t demand in period t determine a production plan for the next n periods that minimizes the total (production and storage) costs, while satisfying the demand in each period. MILP formulation Decision variables: x t = amount produced in period t, with 1 t n s t = amount in stock at the end of period t, with 0 t n y t = 1 if production occurs in period t and y t = 0 otherwise, with 1 t n Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

19 MILP formulation Decision variables: x t = amount produced in period t, with 1 t n s t = amount in stock at the end of period t, with 0 t n y t = 1 if production occurs in period t and y t = 0 otherwise, with 1 t n min n t=1 ptxt + n t=1 htst + n t=1 ftyt s.t. s t = s t 1 + x t d t t x t My t s 0 = 0, s n = 50 s t, x t 0 y t {0, 1} where M > 0 is large enough (upper bound on the maximum amount produced during any period). For instance: x t ( n t=1 dt + sn s0)yt N.B.: Since s t = t i=1 x i + s 0 t i=1 d i, it is possible to delete the storage variables s t How can we account for a minimum lot size? t t t t Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

20 An appropriate combination of continuous and binary variables allows to model arbitrary piecewise linear cost functions. Example 7: Minimization of piecewise linear cost functions Consider an arbitrary (not necessarily convex) piecewise linear function f (x), with f : [x 1, x k ] R. Suppose that x 1 < x 2 <... < x k and that f (x) is specified by the points (x i, f (x i )), for i = 1,..., k. Example of min x [x 1,x k ] f (x): Any x [x 1, x k ] and the corresponding value f (x) can be expressed as k k k x = λ i x i and f (x) = λ i f (x i ) with λ i = 1 and λ 1,..., λ k 0, i=1 i=1 i=1 Clearly, the choice of the coefficients λ i is not unique. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

21 It becomes unique if we require that at most two consecutive λ i can be nonzero. Any x [x i, x i+1 ] is then represented as x = λ i x i + λ i+1 x i+1 with λ i + λ i+1 = 1 and λ i 0, λ i+1 0. By defining y i = 1 if x i x x i+1 and y i = 0 otherwise, for i = 1,..., k 1 the problem min x [x 1,x k ] f (x) can be formulated as follows: min s.t. k i=1 λ if (x i ) λ 1 y 1, λ k y k 1 λ i y i 1 + y i i = 2,..., k 1 k i=1 λ i = 1 k 1 i=1 y i = 1 λ i 0, y i {0, 1} i = 1,..., k N.B.: If y j = 1 then λ i = 0 for all i, 1 i n, different from j or j + 1. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

22 Binary variables also allow to impose disjunctive constraints: either a 1 x b 1 or a 2 x b 2 with x R and 0 x u, where u is the upper bound vector. Example 8: Scheduling problem Given m machines and n products for each product j, we know the deadline d j and the time p jk needed to process product j on machine k, for k m determine an optimal schedule so as to minimize the time needed to complete all products, while satisfying all deadlines. Assumptions: - Each product must be processed on all the machines according to the order of the machine indices. - Products cannot be processed simultaneously (on the same machine) and, when processed, they cannot be stopped. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 23

23 Extension: each product must be processed on (a subset) of machines in a possibly different Edoardorder. Amaldi (PoliMI) Ottimizzazione A.A / 23 MILP formulation Decision variables: t jk = time when product j is started on machine k, with 1 j n and 1 k m t time in when all the products are completed y ijk = 1 if product i is processed before product j on machine k, and y ijk = 0 otherwise, with 1 i, j n and 1 k m min t s.t. t jm + p jm t j t jm + p jm d j t jk + p jk t j,k+1 j, k {1,..., m 1} j t ik + p ik t jk + M(1 y ijk ) k, i, j with i < j (7) t jk + p jk t ik + My ijk k, i, j with i < j (8) t 0, t jk 0 y ijk {0, 1} j, k i, j, k where M is a large enough parameter (e.g., M = n j=1 d j) Constraints (7) and (8) ensure that no pair i, j of products are processed simultaneously on the same machine (either i preceds j or j preceds i).