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

Sonia Cafieri

LIX, École Polytechnique cafieri@lix.polytechnique.fr

Mathematical Programming: Modelling and Applications

October 2009

2

Outline

A modelling problem: Carelland Formulation of the mathematical model The AMPL model Solution Another modelling problem: Dyeing Formulation of the mathematical model The AMPL model Solution

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Carelland problem

The independent state of Carelland mainly exports four goods: steel, engines, electronic components and plastics. We want to maximize exports and minimize imports. The unit prices on the world markets for the 4 products are given. For each product it is also known:

• the number of unities of the other 3 products,
• the price of other imported goods,
• the man-months of work

needed to produce 1 unit of product. Furthermore: There is a maximum possible production for electronics and plastic materials. The total available workforce for each year is known. Steel, engines, electronics and plastics cannot be imported. Write a mathematical program that maximizes the gross internal product and solve the problem with AMPL.

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Carelland: data

Producing 1 steel unit requires: 0.02 engine units, 0.01 plastics units, 250$ in imported goods and 6 man- months of work. Producing 1 engine unit requires: 0.8 steel units, 0.15 electronics units, 0.11 plastics units, 300$ in imported goods and 1 man-year. Producing 1 electronics unit requires: 0.01 steel units, 0.01 engine units, 0.05 plastics units, 50$ in imported goods and 6 man-months. Producing 1 plastics unit requires: 0.2 steel units, 0.03 engine units, 0.05 electronics units, 300$ in imported goods and 2 man-years.

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Carelland: data

Unit prices on the world markets :

steel 500 engines 1500 electronics 300 plastics 1200

Maximum possible production: Total available workforce: 830000 man-year.

steel

• engines
• electronics

650000 plastics 60000

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Writing the mathematical model

• Objective function

What do we have to maximize? Note: the profits can be expressed in terms of exported quantities and produced quantities.

• Variables

What do we have to decide? Note : we have a certain number of goods and, for each of them, an amount of produced and exported units.

• Parameters and Sets

Consider a set containing the products. What are the parameters?

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Writing the mathematical model

• Constraints
• 1. The amount of work needed to produce all the products (man-months)

cannot exceed the total available workforce per year.

• 2. There is a maximum possible production for some of the products.

Note: this limit is not imposed on all products. How can this condition be expressed?

• 3. For each product, the produced quantity must balance the exported

quantity and the quantity needed to produce the other products. Suggestion: start writing, for each product, the relation among the total produced quantity, the exported quantity, the quantity needed for each

• ther product. Then, try to write these constraints in a more compact

form.

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Mathematical model

• Parameters and Sets

P = {a,m,e,p}: set of products (a = steel, m = engines, e = electronic, p = plastic) H: total available amount of work (man-years) Mi : maximum possible production for product i∈P hi : amount of work required to manufacture a unit of product i∈P mi : price of raw materials necessary to manufacture a unit of product i∈P pi : market price for product i∈P

• Variables

xa , xm , xe , xp : produced units of steel, engines, electronics and plastics ya , ym , ye , yp : exported units of steel, engines, electronics and plastics

All variables are non negative

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Mathematical model

• Objective function

∑ ∑

∈ ∈

−

P i i i i P i i

x m y p max

maximize the profits profits: exported quantities minus produced quantities

05 . 11 . 01 . 05 . 15 . 03 . 01 . 02 . 2 . 01 . 8 .

e m a p p p m e e p e a m m p e m a a i i P i i i

x x x y x x x y x x x x y x x x x y x M x P i H x h + + + = + + = + + + = + + + = ≤ ∈ ∀ ≤

∑

∈

• Constraints
• amount of work:
• maximum production:
• balance:

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Mathematical model

05 . 11 . 01 . 05 . 15 . 03 . 01 . 02 . 2 . 01 . 8 .

e m a p p p m e e p e a m m p e m a a

x x x y x x x y x x x x y x x x x y x + + + = + + = + + + = + + + =

can be written as:

∑

∈

+ = ∈ ∀

P j j ji i i

x a y x P i

where: Note that the following constraints

⎪ ⎩ ⎪ ⎨ ⎧ = = ∈ ∀

• therwise

product

• f

unit 1 produce to needed product

• f

units if , j i j i a P j i

ij

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AMPL model – Carelland

set PRODUCTS; param p {PRODUCTS} >= 0; param HMan >=0; param Max {PRODUCTS} >=0; param m {PRODUCTS} >= 0; param h {PRODUCTS} >= 0; param a {PRODUCTS, PRODUCTS} >=0; var x { PRODUCTS } >= 0; var y { PRODUCTS } >= 0; maximize profit: sum {i in PRODUCTS} (p[i]*y[i] - m[i]*x[i]); subject to limit{i in PRODUCTS}: x[i] <= Max[i]; subject to work: sum {i in PRODUCTS} h[i]*x[i]<= HMan; subject to balance{i in PRODUCTS} : y[i] + sum{j in PRODUCTS}(a[j,i]*x[j]) = x[i];

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AMPL dat – Carelland

set PRODUCTS := steel plastics electronics engines; param HMan:= 830000; param : p m h Max := steel 500 250 0.5 2000000 plastics 1200 300 2 60000 electronics 300 50 0.5 650000 engines 1500 300 1 2000000 ; param a: steel plastics electronics engines := steel 0 0.01 0 0.02 plastics 0.2 0 0.05 0.03 electronics 0.01 0.05 0 0.01 engines 0.8 0.11 0.15 0 ;

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Solution – Carelland

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",

• ptions: e m b q use=8

CPLEX 10.1.0: optimal solution; objective 435431250 9 dual simplex iterations (6 in phase I) x [*] := electronics 74375 engines 475833 plastics 60000 steel 393958 ; y [*] := electronics 0 engines 465410 plastics 0 steel 547.917 ;

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Dyeing problem

A fabric dyeing plant has 3 dyeing baths. Each batch of fabric must be dyed in each bath in the order: first, second, third bath. The plant must color five batches of fabric of different sizes. For each batch, the time for dyeing in each bath is known. Write a mathematical program for scheduling the dyeing operations in the baths so that the ending time of the last batch is minimized. Solve the problem with AMPL.

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Dyeing problem: data

3 dyeing baths 5 batches Time (hours) for dyeing each batch in each bath:

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 3 2 1 . 2 2 2 2 3 . 1 2 . 1 3 1 5 . 1 2 1 1 3

The element (i,k) is the time for dyeing batch i in bath k .

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Writing the mathematical model

Preliminary observations

• Each batch must be dyed in the first bath, then in the second bath,….

The k-th bath must finish before that the (k+1)-th bath can start.

• The time for dyeing in the last bath must be bounded by the completion time

for last batch.

• Two different batches cannot be dyed in the same bath at the same time.
• If the bath i is to be dyed before batch j, then j cannot be dyed before i

(disjunction).

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Mathematical model

• Parameters

∀ i∈ L, k∈ V sik : time necessary to dye batch i in bath k M : upper bound to completion time of last bath T: completion time for last batch tik: starting time for dyeing batch i in bath k yijk = 1 if batch i is to be dyed before batch j in bath k,

0 otherwise.

• Variables
• Sets

L : set of fabric batches L={1,…,l} V : set of dyeing baths V ={1,...,v}

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Mathematical model

• Objective function

T min

we want to minimize the completion time of the last batch

• Constraints

( )

) 1 (

} { \ ,

+

≤ + ∈ ∈ ∀

k i ik ik

t s t v V k L i

( )

T s t L i

iv iv

≤ + ∈ ∀

( ) ( )

ijk jk ik ik

y M t s t j i V k L j i − + ≤ + ≠ ∈ ∈ ∀ 1 , , ,

( )

1 , , , = + ≠ ∈ ∈ ∀

jik ijk

y y j i V k L j i

• sequential:
• last bath:
• non overlapping:
• disjunction:

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AMPL model – Dyeing

param l >= 1; param v >= 1; set L := 1..l; set V := 1..v; set V0 := 1..v-1; param s{L,V} >= 0; param M default sum{i in L, k in V} s[i,k] ; var t{L,V} >= 0; var T >= 0; var y{L,L,V} binary; minimize makespan : T; subject to sequential{i in L, k in V0} : t[i,k] + s[i,k] <= t[i,k+1]; subject to lastbath{i in L} : t[i,v] + s[i,v] <= T; subject to nonoverlap{i in L, j in L, k in V : i != j} : t[i,k] + s[i,k] <= t[j,k] + M * (1 - y[i,j,k]); subject to disjunction{i in L, j in L, k in V : i != j} : y[i,j,k] + y[j,i,k] = 1;

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AMPL dat – Dyeing

param l := 5; param v := 3; param s : 1 2 3 := 1 3.0 1.0 1.0 2 2.0 1.5 1.0 3 3.0 1.2 1.3 4 2.0 2.0 2.0 5 2.1 2.0 3.0 ;

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AMPL run – Dyeing

model dyeing.mod; data dyeing.dat;

• ption solver cplex;

solve; display makespan; for {i in L} { printf "batch %d : ", i; for {k in V} { printf "[%f] ", t[i,k]; } printf "\n"; }

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Solution – Dyeing

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 14.1 1175 MIP simplex iterations 235 branch-and-bound nodes makespan = 14.1 batch 1 : [9.100000] [12.100000] [13.100000] batch 2 : [0.000000] [2.000000] [3.500000] batch 3 : [6.100000] [10.600000] [11.800000] batch 4 : [2.000000] [4.000000] [6.000000] batch 5 : [4.000000] [6.100000] [8.100000]