Planning the Production of Electronic Components

Contents

Problem description

To augment its competitiveness a small business wishes to improve the production of its best selling products. One of its main activities is the production of cards with microchips and electronic badges. The company also produces the components for these cards and badges. Good planning of the production of these components therefore constitutes a decisive success factor for the company. The demand for components is internal in this case and hence easy to anticipate.

For the next six months the production of four products with references X43-M1, X43-M2, Y54-N1, Y54-N2 is to be planned. The production of these components is sensitive to variations of the level of production, and every change leads to a non-negligible cost through controls and readjustments. The company therefore wishes to minimize the cost associated with these changes whilst also taking into account the production and storage costs.

The demand data per time period, the production and storage costs, and the initial and desired final stock levels for every product are listed in the following table. When the production level changes, readjustments of the machines and controls have to be carried out for the current month. The cost incurred is proportional to the increase or reduction of the production compared to the preceding month. The cost for an increase of the production is $ 1 per unit but only $ 0.50 for a decrease of the production level.

Data for the four products

+------------------------------------+------------------+-------------+
|           Demands                  |        Cost      |     Stock   |
+------+----+----+----+----+----+----+----------+-------+-------+-----+
| Month|  1 |  2 |  3 |  4 |  5 |  6 |Production|Storage|Initial|Final|
+------+----+----+----+----+----+----+----------+-------+-------+-----+
|X43-M1|1500|3000|2000|4000|2000|2500|    20    |  0.4  |  10   | 50  |
|X43-M2|1300| 800| 800|1000|1100| 900|    25    |  0.5  |   0   | 10  |
|Y54-N1|2200|1500|2900|1800|1200|2100|    10    |  0.3  |   0   | 10  |
|Y54-N2|1400|1600|1500|1000|1100|1200|    15    |  0.3  |   0   | 10  |
+------+----+----+----+----+----+----+----------+-------+-------+-----+

What is the production plan that minimizes the sum of costs incurred through changes of the production level, production and storage costs?

Variables

demand                      the demand for each component and month
prodcost                    Cost to produce a component
storagecost                 Cost to store a component
initialstock                Initial stock
finalstock                  Final stock
increasecost, decreasecost  Increase or decrease in cost
                            when producing more or less this month
                            than last month.

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

demand = [1500 3000 2000 4000 2000 2500;...
    1300  800  800 1000 1100  900;...
    2200 1500 2900 1800 1200 2100;...
    1400 1600 1500 1000 1100 1200];

prodcost        = [20;25;10;15];
storagecost     = [.4;.5;.3;.3];
initialstock    = [10;0;50;0];
finalstock      = [50;10;30;10];
increasecost    = 1;
decreasecost    = 0.5;

t = size(demand,2);
p = size(demand,1);

produce = tom('produce',p,t,'int');
store   = tom('store',p,t,'int');
add     = tom('add',t,1,'int');
reduce  = tom('reduce',t,1,'int');

% All slots are integers
bnds = {produce >= 0, store >= 0, add >= 0, reduce >= 0};
bnds1 = {store(:,end) >= finalstock};

% Production equilibrium constraint at start
con1 = {store(:,1) == initialstock + produce(:,1) - demand(:,1)};

% Production equilibrium in process
con2 = {store(:,2:end) == store(:,1:end-1) + produce(:,2:end) - demand(:,2:end)};

% Add/reduction constraint
con3 = {sum(produce(:,2:end),1)' - sum(produce(:,1:end-1),1)' == ...
    add(2:end) - reduce(2:end)};

% Objective
objective = sum(prodcost'*produce + storagecost'*store) + ...
    sum(increasecost*add(2:end)' + decreasecost*reduce(2:end)');

constraints = {bnds, bnds1, con1, con2, con3};
options = struct;
options.solver = 'cplex';
options.name   = 'Production of Electronic Components';
sol = ezsolve(objective,constraints,[],options);

PriLev = 1;
if PriLev > 0
    for month = 1:t,
        disp(['Solution for month ' num2str(month)])
        disp(['produce  ' num2str(sol.produce(:,month)')])
        disp(['stock    ' num2str(sol.store(:,month)')])
        disp(['increase ' num2str(sol.add(month))])
        disp(['decrease ' num2str(sol.reduce(month))])
        disp(' ')
    end
end

% MODIFICATION LOG
%
% 051018 med   Created.
% 060110 per   Added documentation.
% 060125 per   Moved disp to end
% 060203 med   Removed printing of temp
% 090407 med   Converted to tomSym
Problem type appears to be: mip
===== * * * =================================================================== * * *
TOMLAB - Tomlab Optimization Inc. Development license  999001. Valid to 2010-02-05
=====================================================================================
Problem: ---  1: Production of Electronic Components  f_k  683929.000000000000000000
                                             sum(|constr|)      0.000000000000056843
                                    f(x_k) + sum(|constr|) 683929.000000000000000000
                                                    f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=101.
CPLEX Branch-and-Cut MIP solver
Optimal integer solution found

FuncEv   37 
CPU time: 0.015625 sec. Elapsed time: 0.016000 sec. 
Solution for month 1
produce  1490  1300  2870  1400
stock    0    0  720    0
increase 0
decrease 0
 
Solution for month 2
produce  3000   800   780  2480
stock    0    0    0  880
increase 0
decrease 0
 
Solution for month 3
produce  2000   800  2900  1360
stock    0    0    0  740
increase 0
decrease 0
 
Solution for month 4
produce  4000           1000           1800            260
stock    0  0  0  0
increase 0
decrease 0
 
Solution for month 5
produce  2000  1100  1900  1100
stock    0    0  700    0
increase 0
decrease 960
 
Solution for month 6
produce  2550   910  1430  1210
stock    50  10  30  10
increase 0
decrease 0