Integer Linear Programming

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Introduction Introduction to Linear Programming and Integer Linear Programming ► Outline - brief introduction to… § Modeling § Linear Programming RMIT, Melbourne, Australia § Integer Linear Programming § Solution Methods for LP and ILP Helena Ramalhinho Lourenço § AMPL

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Real Problem OR Methodology Linear Programming

Data Analysis ► Linear programming (LP) is a widely used mathematical Set of Models modeling technique designed to help managers in Optimization Model planning and decision making relative to resource maker

- Solution Methods Results & Interpretation allocation. ► Integer Linear Programming (ILP) Simulation

Decision § An integer programming model is one where one or more of the decision variables has to take on an integer or binary values in Solution & Analysis OR provides many methods the final solution. and techniques to analyze Scenarios ► Combinatorial Optimization (next) the consequences of decisions Make a decision before their introduction to reality.LP & ILP introduction 3 LP & ILP introduction 4

OR Methodology OR Methodology

► Identify a problem ► Identify the problem ► Get to know the problem context (get Data) § Describe in detail the problem, and identify all the components. § Identify the qualitative and quantitative aspects of the problem. ► Build a Model (Mathematical Model) ► Get to know the problem context ► Obtain a solution to the model (Algorithm) § Identify the relationships of the problem with the context of the ► Understand the solution the real context organization. ► Take the decision § Obtain the relevant data. § Verify the data.

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► Build a model ► Obtain a solution to the model § A Model is an abstract representation of a real world system § Which are the adequate solution methods to be applied to the § Simplification is the very essence of Modeling model? § Which components should be included in the model? * Exact Methods § Types of models: * Heuristics and Metaheuristics • Linear programming § Which is the best software to be applied? • Integer Linear Programming * Commercial software • Combinatorial Optimization * Algorithm design and code • Networks • Non-linear • Simulation • Etc.

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OR Methodology Linear Programming

► Understand the solution the real context ► Linear programming (LP) is a widely used mathematical § Does the solution obtained makes sense in the real life? modeling technique designed to help managers in § Did we have considered all relevant components? planning and decision making relative to resource ► Take the decision allocation. § Evaluate the impact of the solution § Evaluate the decision process (model + method) § Review frequently the impact of the decision.

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Linear Programming Linear Programming

► The Linear Programming Models have 4 properties in ► Formulating a linear program involves developing a common: mathematical model to represent the managerial problem. § All problems seek to maximize or minimize some quantity (the ► The steps in formulating a linear program are: objective function). § Completely understand the managerial problem being faced. § Restrictions or constraints that limit the degree to which we can § Identify the objective and the constraints. pursue our objective are present. § Define the decision variables. § There must be alternative courses of action from which to choose. § Use the decision variables to write mathematical expressions for § The objective and constraints in problems must be expressed in the objective function and the constraints. terms of linear equations or inequalities.

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► Understand well the problem (by words) ► A call center is hiring personnel since it is § What is the problem expanding to a 24h working period. § What is the decision we have to take? Can we make quantitative? § The call center works 24h a day, and needs personnel to § What are the constraints on the decisions? attend the customers every hour. § What is the objective or how we are going to evaluate the solution? § The human recourses and the operations directors have ► Define the decision variables (math) estimate the number of persons in each interval of time. ► Define the constraints on the variables (math) § There are 6 intervals of time (4 hours each). ► Define the objective function (math) § The contract of the new employees is for 8h in a row. ► Solve it!!!

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Example 2: Call Center Example 2: Call Center

► Need of additional employeesInterval for ofinterval times of time: ► Describe the problem by words. § What is the problem? Period 1 2 3 4 5 6 § What is the decision we have to take? Can we make quantitative? 0:00 – 4:00 – 8:00 – 12:00 – 16:00 – 20:00 – 4:00 h 8:00 h 12:00 h 16:00 h 20:00 h 24:00 h § What are the constraints on the decisions? Employees § What is the objective or how we are going to evaluate the needed 9 8 3 7 5 4 solution?

► Which is the minimal number of employees that should be hired?

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Example 2: Call Center Ejemplo 1: Personal laboratorio

► Define the decision variables: ► Define the constraints § Number of employees to be hired to start at the beginning of § In period 1, there is the need of 9 employees… period 1 (x1) § In period 2, there is the need of 8 employees… § Number of employees to be hired to start at the beginning of § In period 3, there is the need of 3 employees… period 2 (x2) Periods § … 1 2 3 4 5 6 § Number of employees to be hired to start at the beginning of 0:00-4:00 4:00-8:00 8:00-12:00 12:00-16:00 16:00-20:00 20:00-24:00 period 3 (x3) 0:00 X1 X1 4:00 X2 X2 § Number of employees to be hired to start at the beginning of 8:00 X3 X3 period 4 (x4) 12:00 X4 X4 16:00 X5 X5 § … 20:00 X6 X6 Personnel ≥ ≥ ≥ ≥ ≥ ≥ 9 8 3 7 5 4

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► The constraints… ► Define the constraints: § The number of persons that start at beginning of period 1 + the § X6 + X1 ≥ 9 ones that continue from period 6 must be greater or equal to 9: § X1 + X2 ≥ 8 * X6 + X1 ≥ 9 § X2 + X3 ≥ 3 § The number of persons that start at beginning of period 2 + the § X3 + X4 ≥ 7 ones that continue from period 1 must be greater or equal to 8. § X4 + X5 ≥ 5 * X1 + X2 ≥ 8 § X5 + X6 ≥ 4 § … § Xj ≥ 0, j= 1,...,6.

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Example 2: Call Center Example 2: Call Center

► Define the objective function ► The complete model ► Min Z= X1 + X2 + X3 + X4 + X5 + X6 ► Min Z= X1 + X2 + X3 + X4 + X5 + X6 § Subject to: * X6 + X1 ≥ 9 * X1 + X2 ≥ 8 * X2 + X3 ≥ 3 * X3 + X4 ≥ 7 * X4 + X5 ≥ 5 * X5 + X6 ≥ 4 * Xj ≥ 0, j= 1,...,6.

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Solution Methods Solution Methods

► Obtain the optimal solution of the LP Model ► LINEAR PROGRAMMING SOFTWARE SURVEY ► Exact Method § https://www.informs.org/ORMS-Today/Public-Articles/June- Volume-44-Number-3/Linear-Programming-Software-Survey § The PL models are “easy” to be solved § June 2017 § Simplex Method (Dantzig 1947) * https://www.informs.org/ORMS-Today/OR-MS-Today-Software- * Primal simplex Surveys/Linear-Programming-Software-Survey * Dual simplex § Vendors § Interior Point Method (Karmarkar 1984) * GLPK (GNU Linear Programming Kit), Gurobi Optimizer, IBM ILOG § ALGORITMS / SOFTWARE CPLEX Optimization Studio, LINDO & LINGO, Premium Solver Pro, * Excel Solver etc * Commercial Software § Prices * Open Software * Free to $4000/licence LP & ILP introduction 23 LP & ILP introduction 24

► Dantzig (1947) ► In Euclidean space, an object is convex if for every pair of § The set of solutions is a convex set. points within the object, every point on the straight § If there is a optimal solution, there exist an optimal solution in a line segment that joins them is also within the object. corner point (or extreme point) § An extreme point always have at least two adjacent extreme points. § If a extreme point has no adjacent extreme points with better value for the objective function, then it is the optimal solution.

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Simplex Method Simplex Method

► Basic theorem ► Example of Linear Programming Maximizar Z =60x +40x ► If exist a optimal solution there exist one in a extreme 1 2 sa point. + x1 x2 £8 ► Simplex method 2x2 £12 § Look only in the extreme point (or corner points of the polytope) 4x1+2x2 £22 § There are a finite number of extreme points x1³0, x2 ³0

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Only five extreme Infinite number of points feasible solutions

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© Helena Ramalhinho-Lourenço, (2019) 5 Simplex Method Initialization Find initial feasible extreme-point solution Look for the optimal ► An iterative procedure solution in the 5 § Move from extreme extreme/corner point to extreme point Is the current Yes points feasible extreme- Stop point solution § Optimality test in the the actual optimal? solution is Simplex Method: optimal * If a extreme-corner No solution has no adjacent solutions that Move to a better are better, then it must adjacent feasible extreme-point be an optimal solution solution

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step 1. Solution (0,0) step 2. Adjacent with z=0. solution(0,6) with z=240.

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Step 3. Adjacent solution(2,6) with Step 4. Adjacent z=360 solution (3,5) with z=380

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Step 5. Adjacent z=380 solution (5.5,0) with z=330 z=330

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Excel Solver

Objective function cell So, (3,5) is the Variables (changing cells) optimal solution with z=380!!Paso 4. Solución Constraints adyacente (0,5.5) entonces z=330

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Example 2: Call Center Example 2: Call Center

► Need of additional employeesInterval for ofinterval times of time: ► The complete model

Period 1 2 3 4 5 6 ► Min Z= X1 + X2 + X3 + X4 + X5 + X6 § Subject to: 0:00 – 4:00 – 8:00 – 12:00 – 16:00 – 20:00 – 4:00 h 8:00 h 12:00 h 16:00 h 20:00 h 24:00 h * X6 + X1 ≥ 9 Employees * X1 + X2 ≥ 8 needed 9 8 3 7 5 4 * X2 + X3 ≥ 3 * X3 + X4 ≥ 7 * X4 + X5 ≥ 5 ► Which is the minimal number of employees that should be * X5 + X6 ≥ 4 hired? * Xj ≥ 0, j= 1,...,6.

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© Helena Ramalhinho-Lourenço, (2019) 7 Excel Solver Excel Solver Objective function Z= X1 + X2 + X3 + X4 + Objective function Función X5 + X6 ► Excel file Objetivo min ∑ X ► Min Z= X1 + X2 + X3 + X4 + X5 Variablesmin 0 + X6 Función Variables:Objetivo min ∑ X X1 0 ► Specify where to X1,X2 ,X3 ,X4 , X5 ,X6 min 0 X2 0 ► X3 0 Sujeto a: X1 0 locate … X4 0 X2 0 X5 Constraints0 X6 + X1 ≥ 9 X3 0 X6 0 § parameters Parameters X4 0 X1 + X2 ≥ 8 X5 0 Restricciones Constraints X6 0 X6 + X1 ≥ 9 0 >= 9 Parameters § variables X1 + X2 ≥ 8 0 >= 8 X2 + X3 ≥ 3 Restricciones X2 + X3 ≥ 3 0 >= 3 X6 + X1 ≥ 9 0 >= 9 § constraints X3 + X4 ≥ 7 0 >= 7 X3 + X4 ≥ 7 X1 + X2 ≥ 8 0 >= 8 X4 + X5 ≥ 5 0 >= 5 X2 + X3 ≥ 3 0 >= 3 X5 + X6 ≥ 4 0 >= 4 X4 + X5 ≥ 5 X3 + X4 ≥ 7 0 >= 7 § Objective function X4 + X5 ≥ 5 0 >= 5 X5 + X6 ≥ 4 0 >= 4 X5 + X6 ≥ 4 X ≥ 0, j= 1,...,6. j =B4+B5

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Excel Solver Example 2: Call Center Objective Function min ∑ X

Función min 19 Objetivo min ∑ X min 0 ► Excel file X1 8 X2 0 X1 0 X2 0 § callcenter.xlsx X3 5 X3 0 X4 2 X4 0 X5 3 X5 0 X6 1 X6 0

Restricciones Constraints X6 + X1 ≥ 9 0 >= 9 X6 + X1 ≥ 9 9 >= 9 X1 + X2 ≥ 8 0 >= 8 X1 + X2 ≥ 8 8 >= 8 X2 + X3 ≥ 3 0 >= 3 X2 + X3 ≥ 3 5 >= 3 X3 + X4 ≥ 7 0 >= 7 Minimize X4 + X5 ≥ 5 0 >= 5 X3 + X4 ≥ 7 7 >= 7 X5 + X6 ≥ 4 0 >= 4 X4 + X5 ≥ 5 5 >= 5 X5 + X6 ≥ 4 4 >= 4

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Example 2: Call Center Integer Linear Programming

► We need to hired 19 employeesInterval of times ► Not every problem faced by businesses can easily be modeled as linear programming model… Period 1 2 3 4 5 6 ► A large number of decision problems can be solved only if 0:00 – 4:00 – 8:00 – 12:00 – 16:00 – 20:00 – 4:00 h 8:00 h 12:00 h 16:00 h 20:00 h 24:00 h variables have integer values. Employees needed ► Examples? 9 8 3 7 5 4

Empoyees working 8+1 0+8 5+0 2+5 3+2 1+3

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► An integer programming model is one where one or more ► Flipcar, a car rental company, is planning to offer of the decision variables has to take on an integer value in electrical cars to its clients. the final solution. ► The choice is limited to two models, and the manager ► An integer variable that can only take a value equal to 0 or wishes to decide on how many cars of each type to buy, if to 1 is called binary variable § There is limited parking space for charging the cars § Pure integer programming where all variables have integer values. § There is a budget limit § Mixed-integer programming where some but not all of the § The maintenance services of the company has limited capacity variables will have integer values. ► The goal is to maximize profits. § Zero-one integer programming are special cases in which all the decision variables must have integer solution values of 0 or 1.

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Integer programming Approaches for solving the problem

Max Z =X1+ 1.4X2 ► Solve the LP relaxation s.a. § Forget about the integer constraints

X1 + 0.5X2 £ 6 Max Z =X1+ 1.4X2 0,5X1 + X2 £ 5.5 s.a. X1 + 0.5X2 £ 6 X1 + X2 £ 6.8 0,5X1 + X2 £ 5.5 X1, X2 >= 0 and integer X1 + X2 £ 6.8 X1, X2 ≥ 0 Being § What is the solution? X1 = Number of electrical cars of type 1 to buy § Can you round the solution? X2 = Number of electrical cars of type 2 to buy LP & ILP introduction 51 LP & ILP introduction 52

X2 P0 Approaches for solving the problem Branch and Bound algorithm 4,2

P0 Z = 8,5; 2,6 X1 ► Solve the LP relaxation and round-off values X1 = 2,6 X2 = 4,2 § Rounding may provide with a non-optimal solution. X2 P1 X2 P2 4,5 P1 P2 3,8 P0 + X1 £ 2 P0 + X1 ³ 3 § Rounding may provide non-feasible solutions. Z1 = 8,3 Z2 = 8,3 X1 = 2 X2 = 4,5 X1 = 3 X2 = 3,8 ► Enumerate all possible solutions 2 X1 3 X1

§ Enumeration is generally not possible for large problems. X2 P11 X2 P12 X2 P21 X2 P22 P11 P12 P21 P22 5 P1 + X2 £ 4 P1 + X2 ³ 5 P2 + X2 £ 3 P1 + X2 ³ 4 4 No Z11 = 7,6 Z21 = 8,0 factible ► Use Branch-and-bound algorithm Z12 = 8,0 3 X1 = 2 X2 = 4 X1 = 3,8 X2 = 3 Discarded X1 = 1 X2 = 5 Discarded Discarded Optimal ► … Z12 < Z11 Infeasible solution 2 X1 1 X1 3,8 X1 X1

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► You’re the manager of an airplane plant and you want to ► The profit and fixed costs are given in the following table: determine the best product-mix of your six models to produce. The six models currently under production are Plane Profit Setup the Rocket, Meteor, Streak, Comet, Jet, and Biplane. Rocket 30 35 Each plane has a known profit contribution. There is also Meteor 45 20 a fixed cost associated with the production of any plane in Streak 24 60 a period. Comet 26 70 Jet 24 75 Biplane 30 30

Source: Optimization Modeling with Lingo. LP & ILP introduction 55 Source: Optimization Modeling with Lingo. LP & ILP introduction 56

Product-mix planning in an Airplane Plant Product-mix planning in an Airplane Plant

► Each plane is produced using six raw materials—steel, ► The problem is to determine the final mix of products that copper, plastic, rubber, glass, and paint. maximizes net profit (gross profit – setup costs) without ► The units of these raw materials required by the planes as exceeding the availability of any raw material. well as the total availability of the raw materials are: ► Your brand new Meteor model has the highest profit per

Rocket Meteor Streak Comet Jet Biplane Available unit of anything you’ve ever manufactured and the lowest Steel 1 4 0 4 2 1 800 setup cost. Copper 4 5 3 0 1 0 1160 ► Maybe you should build nothing but Meteors? Then again, Plastic 0 3 8 0 1 0 1780 Rubber 2 0 1 2 1 5 1050 maybe not. Glass 2 4 2 2 2 4 1360 Paint 1 4 1 4 3 4 1240

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Product-mix planning in an Airplane Plant Binary variables

Qi = Quantity to be produced of airplane i; (i =1(Rocket),...,6 (Biplane)) ► Model ïì1 if plane i is build. ► Binary variables add a new layer of flexibility in modeling Bi = í for (i =1(Rocket),...,6 (Biplane)) îï0 if plane is not build. real situations where appears selections, choices or other 6 MAX Z profit *Q setup * B = ∑( i i − i i ) conditions. i=1 s.t. ► There are several prototype problems such as: 6 ∑usageij *Qi ≤ availablej for j =1,..., 6 (raw materials) § Location problems i=1

Qi ≤ M * Bi for i =1,..., 6 (planes) § Routing problems Q ≥ 0 and integer; B binary i i Rocket Meteor Streak Comet Jet Biplane Available § Set covering problems Steel 800 200 10000 200 400 800 800 Copper 290 232 386.6667 10000 1160 10000 1160 § Scheduling problems Plastic 10000 593.3333 222.5 10000 1780 10000 1780 Rubber 525 10000 1050 525 1050 210 1050 § and many, many more … What is the value of M? Glass 680 340 680 680 680 340 1360 Paint 1240 310 1240 310 413.3333 310 1240 400 290 200 222.5 200 400 210

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► High-level computer programming languages for ► AIMMS describing and solving mathematical optimization ► AMPL problems. ► GAMS ► Advantages ► LINGO § Computer model resembles mathematical notation ► OPL ide (CPLEX) § Faster modeling cycles ► § Modeling and future maintenance becomes easier and more MOSEL-XPRESS reliable. ► CMPL COIN-OR (open source) § Direct change of solver ► … § No need to know specifics from each solver § Easy deployment for real-sized large-scale problems LP & ILP introduction 61 LP & ILP introduction 62

AMPL AMPL workflow

► www.ampl.com Symbolic model file Data file § http://ampl.com/products/ampl/ampl-for-students/ (.mod) (.dat) § Student version is limited to 300 variables and 300 constraints and objectives ► Download the student version (amplcml.zip) AMPL Solution § AMPL program (ampl.exe) (.run) Report § Solver Gurobi (gurobi.exe) LP & MIP § Solver CPLEX (cplex.exe) LP & MIP & QP § Solver MINOS (minos.exe) NLP SOLVER (Cplex, Gurobi, Minos, …)

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Example: Call Center Example: Call Center

► A call center is hiring personnel since it is ► Need of additional employeesInterval for ofinterval times of time: expanding to a 24h working period. Period 1 2 3 4 5 6 § The call center works 24h a day, and needs personnel to 0:00 – 4:00 – 8:00 – 12:00 – 16:00 – 20:00 – attend the customers every hour. 4:00 h 8:00 h 12:00 h 16:00 h 20:00 h 24:00 h § The human recourses and the operations directors have Employees needed estimate the number of persons in each interval of time. 9 8 3 7 5 4 § There are 6 intervals of time (4 hours each). § The contract of the new employees is for 8h in a row. ► Which is the minimal number of employees that should be hired?

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► The complete model ► Model callcenter.mod ► Min Z= X1 + X2 + X3 + X4 + X5 + X6 § set EMPL := 1 .. 6; § Subject to: § var X {j in EMPL} >= 0; * X6 + X1 ≥ 9 § minimize Total: sum {j in EMPL} X[j]; * X1 + X2 ≥ 8 § subject to Cap:X[6] + X[1] >= 9; * X2 + X3 ≥ 3 § subject to Cap1:X[1] + X[2] >= 8; * X3 + X4 ≥ 7 § subject to Cap2:X[2] + X[3] >= 3; * X4 + X5 ≥ 5 § subject to Cap3:X[3] + X[4] >= 7; * X5 + X6 ≥ 4 § subject to Cap4:X[4] + X[5] >= 5; * Xj ≥ 0, j= 1,...,6. § subject to Cap5:X[5] + X[6] >= 4;

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Example: Call Center High Note Sound Company

► AMPL console ► .mod file (highnote.mod) § ampl: reset; var x1 >= 0; # Number of CD players to § ampl: model callcenter.mod; produce § option solver cplex; var x2 >= 0; # Number of stereo receivers § ampl: solve; to produce § CPLEX 12.8.0.0: optimal solution; objective 19 maximize Profit: 50*x1 + 120*x2; 6 dual simplex iterations (0 in phase I) subject to Electrician: 2*x1 + 4*x2 <= 80; subject to Technician: 3*x1 + x2 <= 60;

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AMPL commands file AMPL

► .run file (highnote.run) ► AMPL console § reset; ampl: include highnote.run; § option solver cplex; CPLEX 12.6.1.0: sensitivity § option cplex_options 'sensitivity'; CPLEX 12.6.1.0: optimal solution; objective 2400 § 1 dual simplex iterations (1 in phase I) § suffix up OUT; model highnote.mod; § suffix down OUT; solve; § suffix current OUT; § x1 = 0 display x1,x2; § x2 = 20 display Profit; § Profit = 2400 § ampl: LP & ILP introduction 71 LP & ILP introduction 72

► Use meaningful names («production» not «x») ► highnote2.mod ► # Symbolic data Use comments to describe any declaration set PRODUCT; set PROCESS; ► Organize programs in a clear flow param profit{i in PRODUCT}; param hours{j in PROCESS, i in PRODUCT}; param maxHours{j in PROCESS};

# Symbolic model # Variables var production{i in PRODUCT} >= 0; # Objective maximize Profit: sum{i in PRODUCT} profit[i]*production[i];

# Constraints subject to ResourcesAvailable{j in PROCESS}: sum{i in PRODUCT} hours[j,i]*production[i] <= maxHours[j]; LP & ILP introduction 73 LP & ILP introduction 74

High Note Sound Company High Note Sound Company

► highnote2.dat ► highnote2.run data; set PRODUCT:= CDplayer stereoRecv; # High Note Sound Company - Example 2 set PROCESS:= Electrician Technician; # Commands file ------

param profit:= reset; CDplayer 50 option solver cplex; stereoRecv 120; model highnote2.mod; data highnote2.dat; param hours: expand; CDplayer stereoRecv:= Electrician 2 4 solve; Technician 3 1; display production; display Profit; param maxHours:= Electrician 80 Technician 60; LP & ILP introduction 75 LP & ILP introduction 76

Some AMPL keywords AMPL commands

► Declarations: ► display ; § is any set, parameter or variable of the model § set § Arranges the information into lists and tables. § param ► expand; § var § Displays the current instantiated model. ► Objectives: ► show; § Displays the names of the sets, parameters, variables, constraints, and § maximize objective function of current LP model. § minimize ► reset; ► Constraints: § Cleans the current working LP problem. § reset data; Keeps the models but cleans the data information. § subject to

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► Many real problems do not have this property… ► Many real problems have a large-scale dimension § Large number of variables § Large number of constraints § Need to be solved in very short time… ► Metaheuristics need to be applied! ► Examples * Location problems * Routing problems * Set covering problems * Scheduling problems LP & ILP introduction 79