Chapter 3: Discrete Optimization Integer Programming

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1 Chapter 3: Discrete Optimization Integer Programming Edoardo Amaldi DEIB Politecnico di Milano Website: Academic year Edoardo Amaldi (PoliMI) Optimization Academic year / 26

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 c(x) x X where X is a discrete set and c : X R is the objective function. Example: X { 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) Optimization Academic year / 26

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 (1) x 0 integer. If in (1) all the variables x i {0, 1}, we have a Binary Linear Programming (0-1-ILP) problem. Without loss of generality: only inequalities and all coefficients are integer. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

4 Observation: The integrality restriction on x i is a nonlinear constraint since it can be formulated as sin(πx i ) = 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) Optimization Academic year / 26

5 Modeling techniques and examples Binary variables allow to model a choice between two (several) alternatives or 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 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) Optimization Academic year / 26

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) Optimization Academic year / 26

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 while completing all the projects. Assumption: every person can perform any project, and each person (project) must be assigned to a single project (person). Number of feasible solutions: n! ILP formulation 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) Optimization Academic year / 26

8 Example 3: Set Covering/Packing/Partitioning problems Given finite groundset M = {1, 2,..., m} with 1 i m, collection {M 1,..., M n} of n subsets of M indexed by N = {1,..., n} (namely M j M for j N), a cost c j for each subset M j with j N, we say that a subset of indices F N defines a cover of M if j F M j = M (i.e., each groundset element i is covered at least once), a packing of M if M j1 M j2 = for all j 1, j 2 F with j 1 j 2 (i.e., each groundset element i is covered at most once), a partition of M if it is both a cover and a packing of M (i.e., each groundset element i is covered exactly once). The total cost/weight of a subset indexed by F N is defined as j F c j. Examples: Edoardo Amaldi (PoliMI) Optimization Academic year / 26

9 Set Covering problem: Given a finite groundset M = {1, 2,..., m}, a collection {M 1,..., M n} of n subsets of M indexed by N = {1,..., n}, and a cost c j of M j for each j N, find a cover of M of minimum total cost. ILP formulation Variables: x j = 1 if subset M j is selected and x j = 0 otherwise, with j N min s.t. n j=1 c jx j j:i M j x j 1 i (2) x j {0, 1} j where the inequalities (2) are the so-called covering constraints. Proposition: Set Covering is NP-hard. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

10 Matrix notation: { 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 } N = { candidate sites } M j = { areas reachable in at most τ = 10 minutes from candidate site j } Decide where to locate ambulances so as to minimize the total cost, while guaranteeing that the next call is served in at most τ minutes. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

11 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: Combinatorial auctions Determine the winner of each item so as to maximize the total revenue (see introduction): max S M b(s)x S s.t. S M : i S x S 1 i M x S {0, 1} S M, Proposition: Set Packing is NP-hard. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

12 Set Partitioning problem: { n } min or max c j x j : Ax = 1, x {0, 1} n where the weights c j may represent costs or profits Example: Airline crew scheduling Consider a predefined planning horizon. j=1 M = { flight legs } where a flight leg consists of a single takeoff-landing phase to be carried out within a predefined time window. M j = { feasible subsets of flight legs } where a subset of flight legs is feasible if it can be carried out by a same crew while respecting different constraints (e.g., compatible flights, rest periods, total flight time,...). Assign the crews to the flight legs so as to minimize total cost. Other example: distribution planning (assign customers to routes) Proposition: Set Partitioning is NP-hard. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

13 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, number of Hamiltonian circuits : (n 1)! Proposition: ATSP is NP-hard. Variety of applications: logistics, microchip manufacturing, scheduling, (DNA) sequencing,... Edoardo Amaldi (PoliMI) Optimization Academic year / 26

14 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 VRP) -... Edoardo Amaldi (PoliMI) Optimization Academic year / 26

15 An ILP formulation Variables: x ij = 1 if arc (i, j) is included in the Hamiltonian circuit and x ij = 0 otherwise, with (i, j) A min s.t. (i,j) A c ijx ij j V :j i x ij = 1 i (3) i V :i j x ij = 1 j (4) (i,j) δ + (S) x ij 1 S V, S (5) x ij {0, 1} (i, j) A where equations (3) and (4) are the assignment constraints, δ + (S) = {(i, j) A : i S, j V \ S}, and constraints (5) are the so-called cut-set inequalities. Observation: The number of constraints (5) grows exponentially with the number of nodes n. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

16 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 (6) (i,j) E(S) where E(S) = {(i, j) A : i S, j S} for S V. Example: The number of constraints (6) is still exponential w.r.t. the number of nodes. n. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

17 MILP models 1) Pairs of continuous and binary variables allow to model objective functions with fixed costs. 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) Optimization Academic year / 26

18 MILP formulation 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 (7) y j {0, 1} j N 0 x ij 1 i M, j N where the n constraints (7) 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) Optimization Academic year / 26

19 Example 6: Uncapacitated Lot-Sizing (ULS) Plan the production of a single type of product for the next n periods. Assumption: the stock is empty at the beginning and it must be empty 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 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) Optimization Academic year / 26

20 MILP formulation 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 = 0 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 sizes? t t t t Edoardo Amaldi (PoliMI) Optimization Academic year / 26

21 2) Binary variables also allow to impose disjunctive constraints such as: 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. Illustration of feasible region: Introduce a binary variable y i for each original constraint a i x b i, with 1 i 2, and consider the following constraints: where M max 1 i 2 {a i x b i a i x b i M(1 y i ) for i = 1, 2 y 1 + y 2 = 1 y i {0, 1} for i = 1, 2 0 x u, : 0 x u}. Clearly, if y 1 = 1 then x satisfies a 1 x b 1 while a 2 x b 2 is inactive, and conversely if y 2 = 1. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

22 Example 8: Scheduling problem Given m machines and n products for each product j, 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: products cannot be processed simultaneously on the same machine whenever started, the execution of a product on a machine cannot be interrupted (non-preemptive scheduling). For simplicity s sake, also assume: each product must be processed on all the machines according to the order of the machine indices. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

23 MILP formulation Variables: t jk = time when product j is started on machine k, with 1 j n and 1 k m t time 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 (8) t jk + p jk t ik + My ijk k, i, j with i < j (9) t 0, t jk 0, y ijk {0, 1} i, j, k, where M is a large enough parameter (e.g., M = n j=1 d j). Constraints (8) and (9) ensure that no pair i, j of products are processed simultaneously on the same machine (either i preceds j or j preceds i). Extension: each product must be processed on subset of machines in arbitrary order. Edoardo Amaldi (PoliMI) Optimization Academic year / 26

24 3) An appropriate combination of continuous and binary variables allows to model arbitrary (nonconvex) 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) Optimization Academic year / 26

25 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 ) k i=1 λ i = 1 k 1 i=1 y i = 1 λ 1 y 1, λ k y k 1 λ i y i 1 + y i i = 2,..., k 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) Optimization Academic year / 26

26 Linearization of products of variables - Linearizing the product of two (several) binary variables: The product z = y 1 y 2, with y i {0, 1} for i = 1, 2 and z {0, 1}, can be replaced by z y 1 z y 2 z y 1 + y 2 1 Readily extendable to the product of three or more binary variables. - Linearizing the product of a binary variable and bounded continuous variable: The product z = x y, with x [0, u], y {0, 1} and z [0, u], can be replaced by 0 z uy z x z x (1 y)u As we shall see, the product of two continuous variables cannot be linearized exactly. Edoardo Amaldi (PoliMI) Optimization Academic year / 26