advanced features of ampl

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Advanced AMPL Features

Cormac Lucas


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Table of Contents

1. Advanced Features of AMPL..........................................................1


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1. Advanced Features of AMPL

In this workshop we illustrate some advanced features of AMPL.

Quite often we include redundant columns and rows in the model removing them by

explicitly declaring upper bounds of zero. In this section we illustrate the use of the

logical expressions to implicitly handle these conditions.

The following model, taking from Williams is used to illustrate these features.

WORKSHOP Multi-Time Period Problem

An engineering factory makes seven products (PROD 1 to PROD 7) on the

following machines: four grinders, two vertical drills, three horizontal drills, one

borer, and one planer. Each product yields a certain contribution to profit (defined

as /unit) together with the unit production times (hours) required on each process

are given below. A dash indicates that a product does not require a process.

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PROD

1

PROD

2

PROD

3

PROD

4

PROD

5

PROD

6

PROD

7

Contribution

to profit

10 6 8 4 11 9 3

Grinding 0.5 0.7 - - 0.3 0.2 0.5

Vertical

drilling

0.1 0.2 - 0.3 - 0.6 -

Horizontal

drilling

0.2 - 0.8 - - - 0.6

Boring 0.05 0.03 - 0.07 0.1 - 0.08

Planing - - 0.01 - 0.05 - 0.05

In the present month (January) and the five subsequent months certain machines will be

down for maintenance. These machines will be:

January 1 grinder

February 2 horizontal drills

March 1 borer

April 1 vertical drill

May 1 grinder and 1 vertical drill

June 1 planer and 1 horizontal drill

There are marketing limitations on each product in each month. These are:

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The corresponding mathematical model is as follows

Indices

i=1..5 denotes the machine types

j=1..6 denotes the time periods

k=1..7 denotes the products

Data

Pkunit profit contribution for product p

Uk,i hours needed on machine i for a unit of product k

D j k, monthly demand for each product

Ni available number of machine type i

Wi,j number of machine type i in maintenance time j

M maximum storage of each product for all time periods

S storage cost of holding one product in inventory for 1 month

Gkdesired closing stock of product k

A hours available each month

Variables

xj,kamount of product k produced time j

yj,kamount of product k sold time j

zj,kamount product k held in inventory at the end of time period j

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maximize = == =

6

1

7

1

6

1

7

1 k j

jk

k j

kjk SzyP

Subject to

Machine hours each month

( ) jiWNAxU ijik

jkkj ,7

1

=

Production balance

x y z z j kj k j k j k j k, , , , , + = 1

and the following bounds

z M k jj k, , < 6

z G kk k6,

=

y D j kj k j k, , ,

and non negativity of the variables

x y z j kj k j k j k, , ,, , , 0

The resulting AMPL model and data file are

set product;

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set months circular;

set machine;

param contrib{i in product};

param time {k in machine,i in product};

param number{k in machine};

param down{j in months,k in machine};

param demand{j in months,i in product};

param storecst := 0.5;

param maxstore := 100;

param closestck := 50;

param maxhours := 6*4*2*8;

var sale{i in product, j in months : demand[j,i]>0 } >=0, =0;

var store{i in product, j in months} >=0,0} (contrib[i] * sale[i,j]) -

sum{i in product, j in months} (storecst*store[i,j]);

subject to

cap{j in months, k in machine} : sum{i in product} time[k,i]*prod[i,j] 0} : store[i,j] = store[i,prev(j)]

+prod[i,j]-sale[i,j];

bal12{i in product, j in months: j"jan" and demand[j,i]=0} : store[i,j] = store[i,prev(j)]

+prod[i,j];bal2{i in product, j in months: j="jan"} : store[i,j] = prod[i,j]-sale[i,j];

str2{i in product, j in months: j="jun"} : store[i,j] = closestck;

data;

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set product := 1 2 3 4 5 6 7;

set months := jan feb mar apr may jun;

set machine := grind vert hori bore plan;

param contrib := 1 10

2 6

3 8

4 4

5 11

6 9

7 3;

param time : 1 2 3 4 5 6 7 :=

grind 0.5 0.7 0 0 0.3 0.2 0.5

vert 0.1 0.2 0 0.3 0 0.6 0

hori 0.2 0 0.08 0 0 0 0.6

bore 0.05 0.03 0 0.07 0.1 0 0.08

plan 0 0 0.01 0 0.05 0 0.05;

param number := grind 4

vert 2

hori 3

bore 1

plan 1;

param down : grind vert hori bore plan :=

jan 1 0 0 0 0

feb 0 2 0 0 0

mar 0 0 0 1 0

apr 0 1 0 0 0

may 1 1 0 0 0

jun 0 0 1 0 1;

param demand : 1 2 3 4 5 6 7 :=

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jan 500 1000 300 300 800 200 100

feb 600 500 200 0 400 300 150

mar 300 600 0 0 500 400 100

apr 200 300 400 500 200 0 100

may 0 100 500 100 1000 300 0

jun 500 500 100 300 1100 500 60;

However, we can be more efficient in the declaration of the variable as in

var store{i in product, j in months} >= (if (j="jun") then closestck else 0) ,

0) then sale[i,j] else 0);

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Running AMPL in command line

The ampl model file can be created in an editor such as notepad and saved with a

name mymodel.mod. Similarly, the data file can also be edited in notepad and

stored in a file called mydata.dat. Then by opening an MSDos window theproblem can be solved in the following manner.

Microsoft Windows 2000 [Version 5.00.2195]

(C) Copyright 1985-2000 Microsoft Corp

C:\Program Files\AMPL\Models>ampl

License expires on (y)2005:(m)06:(d)08.

OptiRisk Systems license manager: valid AMPL license found

ampl: model prodplan.mod;

ampl: data prodplan.dat;

ampl: solve;MINOS 5.5: optimal solution found.

49 iterations, objective 81756.42857

ampl: display store;

store [*,*]

: jan feb mar apr may jun :=

1 100 0 0 0 0 50

2 0 0 0 0 0 50

3 0 0 0 0 100 50

4 0 0 0 0 0 50

5 0 100 0 0 100 50

6 100 0 0 0 0 50

7 0 100 0 0 100 50;

ampl: display prod;

prod [*,*]


1 600 0 0 200 0 550

2 788.571 0 0 300 100 550

3 300 200 0 400 600 0

4 300 0 0 500 100 350

5 800 500 0 200 1100 06 300 0 400 0 300 550

7 0 250 0 100 100 0;

ampl: display sale;

sale [*,*]

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1 500 100 0 200 . 500

2 788.571 0 0 300 100 500

3 300 200 . 400 500 50

4 300 . . 500 100 300

5 800 400 100 200 1000 50

6 200 100 400 . 300 5007 0 150 100 100 . 50;

Useful Commands

ampl: show;

parameters:

closestck demand maxhours number time

contrib down maxstore storecst

sets: machine months product

variables: prod sale store

constraints: bal11 bal12 bal2 cap str2

objective: contr

ampl: show bal2;

s.t. bal2{i in product, j in months: j == 'jan'} : store[i,j] == prod[i,j] -

sale[i,j];

ampl: expand bal2;

s.t. bal2[1,'jan']:

sale[1,'jan'] - prod[1,'jan'] + store[1,'jan'] = 0;

s.t. bal2[2,'jan']:


s.t. bal2[3,'jan']:


s.t. bal2[4,'jan']:


s.t. bal2[5,'jan']:


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s.t. bal2[6,'jan']:


s.t. bal2[7,'jan']:


ampl: display store >prodplan.log;

ampl: display prod >>prodplan.log;

ampl: exit;

Estimating the Model Statistics

We wish to estimate model statistics such as the number of rows, columns, bounds

and non zeros of the problem.

By expressing them as parameter expressions we see how easy it is to represent

large problems in concise statements. For the above model we compute thestatistics as follows

Variable Group Number Lower Upper

sale[product,month]

prod[product,month]

store[product,month]

Row Group Number Variable grp Number Tnonz

cap[month,machine]

bal[product,month]

Now recompute the same information taking into account that there is no sales

variable when demand is zero.

Questions/Notes

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