Mathematical Programming: Modelling and Applications September 2009 - - PowerPoint PPT Presentation

Sonia Cafieri

LIX, École Polytechnique cafieri@lix.polytechnique.fr

Mathematical Programming: Modelling and Applications

September 2009

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Outline

Some basic AMPL useful operations & commands A modelling problem Formulation of the mathematical model The AMPL model Solution of the problem

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AMPL basic set operations

These operations become useful or necessary in many problems

• Let A, B, and U be sets
• AMPL allows simple set operations:

Union:

U = A union B elements of A or B

Intersection:

U = A inter B elements of A and B

Difference:

U = A diff B elements of A and not of B

Union: Example

ampl: set MONTHS := {1,2,3,4}; ampl: set MONTHS0 := {0} union MONTHS; ampl: display MONTHS0; set MONTHS0 := 0 1 2 3 4;

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AMPL commands

AMPL recognizes a lot of commands.

• Some commands you already know are, e.g.:
• model:

switch to model mode

• data:

switch to data mode

• exit:

exit AMPL

• display:

print model entities and expressions

• let:

change data values

• ……
• Other useful commands:
• fix:

freeze a variable at its current value

• unfix:

undo a fix command

• delete:

delete model entities

• purge:

delete model entities and their dependents

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AMPL commands

Note that fix, unfix, delete, purge,.. are used for changing a model : it is possible in AMPL modifying models and data.

Fix: Example

fix varname

• This command instructs AMPL to treat the indicated variable as though fixed

at its current value (e.g. in solve command): we have a constant.

• If varname is the name of an indexed collection of variables, fix (and unfix)

affects all members of the collection.

• Fixing a variable we have a further constraint in the problem.

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AMPL commands: fix

ampl: var y{1..4} >=1; ampl: let y[3] := 10; ampl: fix y[3]; ampl: display y[3]; y[3] = 10 ampl: minimize somma: sum{i in 1..4} y[i]; ampl: option solver cplex; ampl: solve; ILOG CPLEX 10.100, licensed to "ecolepolytechnique-palaiseau",

• ptions: e m b q use=8

CPLEX 10.1.0: optimal solution; objective 13 0 dual simplex iterations (0 in phase I) ampl: display y; y [*] := 1 1 2 1 3 10 4 1 ;

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Production planning problem

A firm is planning the production of 3 products A1, A2, A3 over a time horizon of 4 months (January to April). We know:

• The demand for the products over the 4 months;
• Prices, production costs, production quotas, activation costs and

minimum batches for each product; Furthermore:

• There is a different number of productive days over the 4 months;
• The activation status of a production line can be changed every month.
• Minimum batches are monthly.
• Each product needs to be stored. There are different monthly rates for

renting the storage space for each product.

• Each product takes the same amount of storage space.

The total available volume is given.

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Production planning problem: data

Demand for the products over the months:

Demand January February March April A1 5300 1200 7400 5300 A2 4500 5400 6500 7200 A3 4400 6700 12500 13200

Prices, production costs, production quotas, activation costs and minimum batches:

Product A1 A2 A3 Selling prices $124 $109 $115 Activation costs $150000 $150000 $100000 Production costs $73.30 $52.90 $65.40 Production quotas 500 450 550 Minimum batches 20 20 16

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Production planning problem: data

Write a mathematical program to maximize the income, and solve it with AMPL. Number of productive days over the months:

January 23 February 20 March 23 April 22

Monthly rates for storage space:

A1 $3.50 A2 $4.00 A3 $3.00

Total available volume: 800 units.

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Writing the mathematical model

Preliminary observations:

• There are some quantities (parameters) defined

for each product : selling price, production cost, production quota, activation cost, minimum batch, storage cost for each month: number of production days for each product and each month: maximum demand for each product in each month independently on product/month: storage capacity.

• Furthermore:

For each product, there is a different quantity produced, sold, stocked during each month and the activation status of each production line is also dependent on the months.

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Writing the mathematical model

Sets and indices:

• 3 products use an index
• 4 months use an index

define parameters / variables indexed on

I i ∈ J j ∈ j i,

What decisions should we take?

quantity produced product , month quantity sold quantity stocked also take into account the activation status

i j

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Mathematical model

• Parameters

Pj : number of production days in month j; dij : maximum demand for product i in month j; vi : selling price for product i; ci: production cost of product i; qi : maximum production quota of product i; ai : activation cost for production i; bi : minimum batch for production i; si : storage cost for product i; C: storage capacity in number of units.

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Mathematical model

• Variables

xij : quantity of product i produced during month j; wij : quantity of product i sold during month j; zij : quantity of product i stocked during month j; yij : activation status for production i during month j; yij binary

⎩ ⎨ ⎧ =

• therwise

month during active is product if 1 j i yij

All variables are non negative

• Objective function

∑ ∑ ∑ ∑ ∑

∈ ∈ ∈ ∈ ∈

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − −

I i J j J j J j J j ij i ij i ij i ij i

y a z s x c w v max

maximize the total income total income: 3 products – sum income obtained during each month

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Mathematical model

• Constraints
• demand:
• production:
• balance:
• capacity:
• activation:
• minimum batch:
• december:

ij ij

d w J j I i ≤ ∈ ∈ ∀ ,

j I i i ij

P q x J j ≤ ∈ ∀

∑

∈ ij ij ij j i

w z x z J j I i + = + ∈ ∈ ∀

−1 ,

, C z J j

I i ij ≤

∈ ∀

∑

∈ ij i j ij

y q P x J j I i ≤ ∈ ∈ ∀ ,

ij i ij

y b x J j I i ≥ ∈ ∈ ∀ ,

0 =

∈ ∀

i

z I i

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AMPL model – production planning

Starting from the mathematical formulation, try to code the AMPL model We have 2 indices for many variables/constraints:

i (products): A1, A2, A3 j (months): 1,2,3,4 It is useful to define 2 sets: set PRODUCTS; param Months; set MONTHS := 1..Months

We also have zi0 -- index 0

set MONTHS0 := MONTHS union {0}

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AMPL model – production planning

Now we can define the parameters:

all parameters are non negative

param days{MONTHS} >= 0; param demand { PRODUCTS, MONTHS } >= 0; param price { PRODUCTS } >= 0; param cost { PRODUCTS } >= 0; param quota { PRODUCTS } >= 0; param activation { PRODUCTS } >= 0; param batch { PRODUCTS } >= 0; param storage { PRODUCTS } >= 0; param capacity >= 0;

Variables:

var x { PRODUCTS, MONTHS } >= 0; var w { PRODUCTS, MONTHS } >= 0; var z { PRODUCTS, MONTHS0 } >= 0; var y { PRODUCTS, MONTHS } >= 0, binary;

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maximize revenue: sum {i in PRODUCTS} (price[i] * sum {j in MONTHS} w[i,j] - cost[i] * sum {j in MONTHS} x[i,j]

• storage[i] * sum {j in MONTHS} z[i,j]
• activation[i] * sum {j in MONTHS} y[i,j]) ;

Constraints:

subject to requirement {i in PRODUCTS, j in MONTHS}: w[i,j] <= demand[i,j]; subject to production {j in MONTHS}: sum {i in PRODUCTS} (x[i,j] / quota[i]) <= days[j]; subject to balance {i in PRODUCTS, j in MONTHS}: z[i,j-1] + x[i,j] = z[i,j] + w[i,j]; subject to capacitymag {j in MONTHS}: sum {i in PRODUCTS} z[i,j] <= capacity; subject to active {i in PRODUCTS, j in MONTHS}: x[i,j] <= days[j]*quota[i]*y[i,j]; subject to minbatch {i in PRODUCTS, j in MONTHS}: x[i,j] >= batch[i]*y[i,j];

AMPL model – production planning

Objective function:

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AMPL dat – production planning

set PRODUCTS := A1 A2 A3 ; param Months := 4 ; param days := 1 23 2 20 3 23 4 22 ; param demand: 1 2 3 4 := A1 5300 1200 7400 5300 A2 4500 5400 6500 7200 A3 4400 6700 12500 13200 ; param : price cost quota activation batch storage := A1 124 73.30 500 150000 20 3.5 A2 109 52.90 450 150000 20 4 A3 115 65.40 550 100000 16 3 ; param capacity := 800 ; let {i in PRODUCTS} z[i,0] := 0; fix {i in PRODUCTS} z[i,0];

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AMPL run – production planning

model productionplanning.mod; data productionplanning.dat;

• ption solver cplex;

solve;

• ption display_round 4;

display revenue; display x; display y;

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Solution – production planning

ILOG AMPL 10.100, licensed to "ecolepolytechnique-palaiseau". AMPL Version 20060626 (Linux 2.6.9-5.ELsmp) ILOG CPLEX 10.100, licensed to "ecolepolytechnique-palaiseau", options: e m b q use=8 CPLEX 10.1.0: optimal integer solution; objective 1581550 33 MIP simplex iterations 0 branch-and-bound nodes revenue = 1581550.0000 x := A1 1 6100.0000 A1 2 0.0000 A1 3 0.0000 A1 4 0.0000 A2 1 0.0000 A2 2 3518.1818 A2 3 0.0000 A2 4 0.0000 A3 1 4400.0000 A3 2 6700.0000 A3 3 12650.0000 A3 4 12100.0000 ; y := A1 1 1.0000 A1 2 0.0000 A1 3 0.0000 A1 4 0.0000 A2 1 0.0000 A2 2 1.0000 A2 3 0.0000 A2 4 0.0000 A3 1 1.0000 A3 2 1.0000 A3 3 1.0000 A3 4 1.0000 ;