AMPL

An Introduction

Outline

• AMPL - What is it (good for)?

• Basics

• Starting a Problem

• Running the Problem

• Example

Outline (Continued)

• Two File Format

• Why Use a Data File?

• All About Sets

• Example (single file)

• Example with data file

• Logical Operators

Outline (A little More)

• Set Operations

• Set Indexing

• Syntax

• Where to get AMPL

AMPL

What is it (good for)?

AMPL is:

• A Mathematical Programming Language

• Supports over 30 solvers

• Only need to know syntax for AMPL to use many different solvers

AMPL does:

• Solves LP’s

• Solves IP’s

• Solves MIP’s, and non-linear programs

Basics

• Files you will need

• How to write them

• How to save them

Writing the .mod file

Open up a non-contextual text editor (e.g. GNU Emacs, OxEdit, or textedit), type the

model, save with file extension “.mod”

Syntax of File

• AMPL is a computer language, so it doesn’t understand what you mean

• Because of this, there is a particular format that AMPL files must have to run properly

• The format is fairly straightforward

• Here’s how it goes:

Order of entry

Enter the variables first, e.g.:

var x1 >=0;

var Sally >=4.5;

var smileyface <=-1;

var happy integer >=0;

Everything has to be identified before you use it, so AMPL knows it is there to use

Order of Entry- Cont.

Next enter the objective:

maximize OBJ: 3*x1 - happy + 2*Sally;

minimize COST: 3*x3 + 2*x1 - Sally;

maximize WHOCARES: 0;

A Quick Note:

AMPL is case sensitive, so “SALLY”, “Sally”, and “sally” are three different names for variables. If something isn’t working correctly, a case error is an easy thing to identify and fix.

Order of Entry - Cont.

Next enter the constraints:

subject to WOOD: 2*x1 - 40*happy<=27;

subject to WORK: x3 + 2.5*Sally<=34;

subject to MACHINE: happy - Sally<=14;

Put it all together:

var x1 >=0;

var Sally >=4.5;

var smileyface <=-1;

var happy integer >=0;

maximize OBJ: 3*x1 - happy + 2*Sally + 4.5*smileyface;

subject to WOOD: 2*x1 - 40*happy<=27;

subject to WORK: x3 + 2.5*Sally<=34;

subject to MACHINE: happy - 3*smileyface<=14;

Running the File

Once you have the file written and saved, we need to run it to get our

solution.

How to Run AMPL

• Open up AMPL• Tell AMPL what you want solved by:• “model <filepath>\project1.mod;” • If the file is in AMPL’s search path, you may be

able to just enter “model FILENAME.mod;”• If you want to use a different solver:• “option solver <solver name>;”• Tell AMPL to solve it:• “solve;”

Subject to Semicolons

• You may have noticed all the semicolons running around the file we put together, and in the commands to run the model.

• These tell AMPL where to separate lines, and things will not work properly without them.

• For instance, if you typed “solve” rather than “solve;”, AMPL would return a prompt that looks like this “ampl?”

Getting the variables

• This will only give you the optimal value, if it exists. How do you know the variable values that give this objective?

• Tell AMPL to give them to you:

• “display x1,x3,Sally,happy,smileyface;”

• This will display the variable values

• Remember, AMPL is case sensitive.

Put it all together• ampl: model data\ampl1.mod;

ampl: option solver cplex;ampl: solve;CPLEX 8.0.0, optimal integral solution found, objective value 712.2, 0 MIP iterations, 0 branch and bound nodesdisplay x1,x3,Sally,smileyface,happy;x1 = 0x3 = -1Sally = 22.33333happy = 11smileyface = 0

The .dat File

• AMPL supports using a separate file for the particular data

• The .dat file holds the values for the parameters, and the names of the variables

Why Use 2 Files?

• You may want to do similar problems

• You may need to modify the data, but not the model

• It makes finding and changing parameters easy

• It makes adding or removing variables easy

All About Sets

• AMPL lets you define sets in the model by declaring them: set SETNAME

• Once a set is declared, you can index parameters over it: param PARAMETER {i in SETNAME}

• Alternatively: param PARAMETER {SETNAME}• You can even index variables over sets: var

VARIABLE {i in SETNAME}• Also, you can use summation notation: sum {i in

SETNAME} VARIABLE[i]*PARAMETER[i]

A (Relatively) Simple Example

• Suppose there is a steel mill that can process raw steel into two products, bands and coils.

• The manager of the mill wants to maximize his profit, given the restrictions on processing time available, and the limitations on how much he can realistically sell of each

• Bands can be processed at 200 tons per hour, at a profit of $25 per ton, and at most 6000 tons can be sold

• Coils are processed at 140 tons per hour, at a profit of $30 per ton, with a market cap of 4000 tons

• There are 40 hours available per week to process the steel

The old way

• Without using sets or a .dat file, the model would look something like this:

var Bands;

var Coils;

maximize Profit: 25 * Bands + 30 * Coils;

subject to Time: (1/200)*Bands + (1/140)*Coils <= 40;

subject to B_Limit: 0 <= Bands <= 6000;

subject to C_Limit: 0 <= Coils <= 4000;

The New Way

Alternatively, we could split this into a .mod file and a .dat file:

steel.mod

set Products;param rate {i in Products};param profit {i in Products};param market {i in Products};param hours;var X {i in Products};maximize Total_Profit: sum {i in Products} X[i] * profit[i];subject to Time: sum {i in Products} X[i]/rate[i]<=hours;subject to Market {i in Products}: 0<= X[i] <= market[i];

steel.dat

set Products := bands coils;

param: rate market profit :=

bands 200 6000 25

coils 140 4000 30 ;

param hours := 40;

WHY????

• Right now, it looks like the old way is easier, right?

• What if we want to add a new product, rods?• In the old way, we would need to find every place

that something might change and alter it individually

• In the new way, we just have to change the .dat file

The Old Way, Updated

var Bands;var Coils;var Rods;maximize Profit: 25 * Bands + 30 * Coils + 28 *

Rods;subject to Time: (1/200)*Bands + (1/140)*Coils +

(1/160)*Rods <= 40;subject to B_Limit: 0 <= Bands <= 6000;subject to C_Limit: 0 <= Coils <= 4000;subject to R_Limit: 0<= Rods <= 5000;

Or, just change steel.dat

set Products := bands coils rods;

param: rate market profit :=

bands 200 6000 25

coils 140 4000 30

rods 160 5000 28 ;

param hours := 40;

Also…

• If you are dealing with large models, the .mod file can get REALLY big without a .dat file

• Changing the problem becomes very easy with a .dat file

Logical Operators

• AMPL supports some logical operators

• Boolean variables (true/false)

• If-then

• If-then-else

How to Use if-then-else

• Suppose all variables, except one, have the same upper bound, and the other has an upper bound that is twice that of the rest

• A constraint like this will take care of all the the upper bounds:

subject to UB {i in SETNAME }:

variable[i] <= 10*(if i=1 then 2 else 1);

Set Operations

• Basic set operations:– Union: U = A union B– Intersection: U = A inter B– Cartesian Product: U = {A,B}

Advantage of Set Operations

• Suppose we have 3 sets of products,– CANDY – TOYS – GAMES

• In our model we may need to declare the following parameters for each: – Price– Supply– Demand

Naïve Setup

– Param Price_C {CANDY};

– Param Price_T {TOYS};

– Param Price_G {GAMES};

– Param Supply_C {CANDY};

– Param Supply_T {TOYS};

– Param Supply_G {GAMES};

– Param Demand_C {CANDY};

– Param Demand_T {TOYS};

– Param Demand_G {GAMES};

Using Set Operation Union

Param Price {CANDY union TOYS union GAMES};

Param Supply {CANDY union TOYS union GAMES};

Param Demand {CANDY union TOYS union GAMES};

Even Better…

Set PRODUCTS = CANDY union TOYS union GAMES;

Param Price {PRODUCTS};

Param Supply {PRODUCTS};

Param Demand {PRODUCTS};

Compound Sets

• Suppose you have a set called PRODUCTS of products to sell at a store

• Consider that instead of one store you have three stores in different parts of the country

• Each store has a different level of each paramater

One Solution

• Instead of using the set PRODUCTS, make three different sets: – PRODUCTS_1 – PRODUCTS_2– PRODUCTS_3

• Let each of the three numbers represent one of your store locations

One Solution (cont…)

• Next, define each parameter for each set of products:– Param Price {PRODUCTS_1};– Param Supply {PRODUCTS_1};– Param Demand {PRODUCTS_1};– Param Price {PRODUCTS_2};– Param Supply {PRODUCTS_2};– …

Easier Solution

For a better solution use compound sets:

– Param Price {PRODUCTS, STORES};– Param Supply {PRODUCTS, STORES};– Param Demand {PRODUCTS, STORES};

Structure of {PRODUCTS, STORES}

• Suppose – PRODUCTS := oreos jenga lazertag ;– STORES := 1 2 3 ;

• Then {PRODUCTS, STORES} is the set of all combinations of products with stores:

(oreos,1) (oreos,2) (oreos,3)(jenga,1) (jenga,2) (jenga,3)(lazertag,1) (lazertag,2) (lazertag,3)

Specifying Data

• In your .dat file, your declaration of Demand could look like this:

param Demand: 1 2 3 :=

oreos 10 20 30

jenga 30 33 42

lazertag 40 30 22 ;

Set Indexing

Indexing of Sets

• Indexing of a one dimensional set

sum {i in PRODUCTS} cost[i]*make[i];

• Indexing is similar in compound sets• One can say

sum { (i,j) in {PRODUCTS, STORES}} cost[i]*make[i,j];or

sum { i in PRODUCTS, j in STORES} cost[i]*make[i,j];

When do we need indexing?

• We may declare a parameter with or without giving index values:

param Demand { PRODUCTS, STORES };

or

param Demand { i in PRODUCTS, j in STORES };

With PRODUCTS and STORES:

“The sales at each store will not exceed the demand for any given product at that store.”

Could be expressed as the following constraint:

subject to DEMAND {(i,j) in {PRODUCTS, STORES}}: Demand [i,j] >= Sell [i,j];

With PRODUCTS and STORES:

• Suppose we have another parameter:

Param Capacity {STORES};

• Let the capacity of a store be the total number of items it can sell all together

With PRODUCTS and STORES:

“The total sales at any given store can not exceed the sales capacity of that store.”

Could be expressed as follows

Subject to CAPACITY {j in STORES}:

Sum {i in PRODUCTS} Sell [i, j] <= Capacity [j];

Advanced Syntax

Tips and Shortcuts

Transposition

How to Transpose

• Recall an earlier example:

Param Demand: 1 2 3 :=

oreos 10 20 30

jenga 30 33 42

lazertag 40 30 22 ;

How to Transpose

• Data in Transposed form

Param Demand (tr): oreos jenga lazertag :=

1 10 30 40

2 20 33 30

3 30 42 22 ;

Why Transpose

• Not Necessary, but can help with data management

• What if we had 30 stores?

Param Demand (tr): oreos jenga lazertag :=1 10 30 402 20 33 303 30 42 22 ;

Omitted Data

Omitted Data Entry

Example: Consider the following data

Param: Cost Supply Demand :=

oreos10 20 30

jenga30 33 42

lazertag 40 30 22 ;

Omitted Data Entry

Suppose in addition to the data specified in the previous table, you have an additional parameter such as:

Param Calories {FOOD};

This parameter would apply to oreos, but not to jenga or lazertag.

Example

Param: Cost Supply Demand :=

oreos10 20 30

jenga30 33 42

lazertag 40 30 22 ;

Param: Calories :=

oreos 100 ;

Example

Param: Cost Supply Demand Calories:=

oreos 10 20 30 100

jenga 30 33 42 .

lazertag 40 30 22 . ;

We can use “.” to represent omitted data

Where to get AMPL

• http://www.ampl.com– Link to the web interface– Download the student edition

• For Windows, it is an easy installation• For Mac OS, use unix install. You may need to

install additional components• Walkthrough available at:

– http://www.ce.berkeley.edu/~bayen/ce191www/labs/lab3/amplInstallingUsing.pdf