Production of Alloys

Contents

Problem description

The company Steel has received an order for 500 tonnes of steel to be used in shipbuilding. This steel must have the following characteristics (‘grades’).

Characteristics of steel ordered

+---------------+-------+-------+
|Chemical       |Minimum|Maximum|
|Element        |Grade  |Grade  |
+---------------+-------+-------+
|Carbon (C)     |  2.0  | 3.0   |
|Copper (Cu)    |  0.4  | 0.6   |
|Manganese (Mn) |  1.2  | 1.65  |
+---------------+-------+-------+

The company has seven different raw materials in stock that may be used for the production of this steel. The table below lists the grades, available amounts and prices for all raw materials.

Raw material grades, availabilities, and prices

+----------------+---+----+----+------------+----+
|Raw material    |C %|Cu %|Mn %|Availability|Cost|
+----------------+---+----+----+------------+----+
|Iron alloy 1    |2.5| 0  |1.3 |     400    |200 |
|Iron alloy 2    | 3 | 0  |0.8 |     300    |250 |
|Iron alloy 3    | 0 | 0.3|0   |     600    |150 |
|Copper alloy 1  | 0 | 90 |0   |     500    |220 |
|Copper alloy 2  | 0 | 96 |4   |     200    |240 |
|Aluminum alloy 1| 0 | 0.4|1.2 |     300    |200 |
|Aluminum alloy 2| 0 | 0.6|0   |     250    |165 |
+----------------+---+----+----+------------+----+

The objective is to determine the composition of the steel that minimizes the production cost.

Variables

compsize             Amount of steel to produce
mincomp              Minimal contents of each component.
maxcomp              Maximal contents of each component.
rawcompmat           Contents of raw material.
rawavail             Amount available of raw material.
rawcost              Cost of raw material.

Reference

Applications of optimization... Gueret, Prins, Seveaux

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2009 by Tomlab Optimization Inc., $Release: 7.2.0$
% Written Oct 7, 2005.   Last modified Apr 8, 2009.

Problem setup

compsize   = 500;
mincomp    = [2;.4;1.2];
maxcomp    = [3;.6;1.65];
rawcompmat = [[2.5;  3;  0;  0;  0;  0;  0],...
    [  0;  0; .3; 90; 96; .4; .6],...
    [1.3; .8;  0;  0;  4;1.2;  0]];
rawavail   = [ 400;300;600;500;200;300;250];
rawcost    = [ 200;250;150;220;240;200;165];

comp = tom('comp',7,1);
objective = rawcost'*comp;
prodcons = {compsize == sum(comp)};
compminmax = {mincomp*compsize <= rawcompmat'*comp <= maxcomp*compsize};
availcons = {0 <= comp <= rawavail};
constraints = {prodcons, compminmax, availcons};
options = struct;
options.solver = 'cplex';
options.name = 'Production of Alloys';
sol = ezsolve(objective,constraints,[],options);

PriLev = 1;
if PriLev > 0
    names = [ 'Iron 1' ; 'Iron 2' ; 'Iron 3' ; 'Copp 1' ; ...
        'Copp 2' ; 'Alum 1' ; 'Alum 2'];
    disp('the optimal steel contains:');
    for i = 1:length(sol.comp),
        if sol.comp(i) > 0,
            disp(['   ' num2str(sol.comp(i)) ' tonnes of ' names(i,:) ])
        end
    end
end

% MODIFICATION LOG
%
% 051007 med   Created.
% 060110 per   Added documentation.
% 060125 per   Moved disp to end
% 090308 med   Converted to tomSym

Problem type appears to be: lp
===== * * * =================================================================== * * *
TOMLAB - Tomlab Optimization Inc. Development license  999001. Valid to 2010-02-05
=====================================================================================
Problem: ---  1: Production of Alloys           f_k   98121.635791681227000000
                                       sum(|constr|)      0.000000000000000568
                              f(x_k) + sum(|constr|)  98121.635791681227000000
                                              f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=1.
CPLEX Dual Simplex LP solver
Optimal solution found

FuncEv    5 Iter    5 
the optimal steel contains:
   400 tonnes of Iron 1
   39.7763 tonnes of Iron 3
   2.7613 tonnes of Copp 2
   57.4624 tonnes of Alum 1