Animal Food Production

Contents

Problem description

The company CowFood produces food for farm animals that is sold in two forms: powder and granules. The raw materials used for the production of the food are: oat, maize and molasses. The raw materials (with the exception of molasses) first need to be ground, and then all raw materials that will form a product are blended. In the last step of the production process the product mix is either transformed to granules or sieved to obtain food in the form of powder.

                  Molasses -+
                            |
                            V       +---> Granulating --> Granules
Oat ----+                           |
        +-> Grinding --> Blending --+
Maize --+                           |
                                    +---> Sieving ------> Powder
Animal food production process

Every food product needs to fulfill certain nutritional requirements. The percentages of proteins, lipids and fibers contained in the raw materials and the required percentages in the final products are listed in the table below.

Contents of nutritional components in percent

+-----------------+--------+-------+-----+
|Raw material     |Proteins|Lipids |Fiber|
+-----------------+--------+-------+-----+
|Oat              | 13.6   |  7.1  | 7.0 |
|Maize            |  4.1   |  2.4  | 3.7 |
|Molasses         |  5.0   |  0.3  | 25  |
|Required contents| >=9.5  |  >=2  | <=6 |
+-----------------+--------+-------+-----+

There are limits on the availability of raw materials. The table below displays the amount of raw material that is available every day and the respective prices.

Raw material availabilities and prices

+------------+----------------------+------------+
|Raw material|Available amount in kg|Cost in $/kg|
+------------+----------------------+------------+
|Oat         | 11900                |   0.13     |
|Maize       | 23500                |   0.17     |
|Molasses    |   750                |   0.12     |
+------------+----------------------+------------+

The cost of the different production steps are given in the following table.

Production costs in $/kg

+--------+--------+-----------+-------+
|Grinding|Blending|Granulating|Sieving|
+--------+--------+-----------+-------+
|  0.25  | 0.05   |    0.42   |  0.17 |
+--------+--------+-----------+-------+

With a daily demand of nine tonnes of granules and twelve tonnes of powder, which quantities of raw materials are required and how should they be blended to minimize the total cost?

Variables

compsize                   Amount of different food to produce
mincomp                    Minimal content of nutrients
maxcomp                    Maximal content of nutrients
rawcompmat                 Content of nutrients in raw sources
rawavail                   Available raw sources
rawcost                    Cost of raw material
prodcostsraw               Cost to process raw material
prodcostsprod              Cost to produce products

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    = [9000;12000];
rawcompmat  = [[13.6;4.1;5],...
    [7.1;2.4;.3],...
    [7;3.7;25]];
rawavail     = [11900;23500;750];
rawcost      = [.13;.17;.12];

grindcost = 0.25;
blendcost = 0.05;
grancost = 0.42;
sievcost = 0.17;
use = tom('use',length(rawcost),length(compsize));
objective = sum(sum(repmat(rawcost,1,2).*use)) + ...
    sum(sum(grindcost*use(1:end-1,:))) + ...
    sum(sum(blendcost*use)) + ...
    sum(sum(grancost*use(:,1))) + ...
    sum(sum(sievcost*use(:,2)));

prodcons = {sum(use,1) == compsize'};
reqcons1 = {rawcompmat(:,1)'*use >= 9.5*compsize'};
reqcons2 = {rawcompmat(:,2)'*use >= 2*compsize'};
reqcons3 = {rawcompmat(:,3)'*use <= 6*compsize'};
availcons = {sum(use,2) <= rawavail};
bnds = {0 <= use};
constraints = {prodcons, reqcons1, reqcons2, reqcons3, availcons, bnds};
options = struct;
options.solver = 'cplex';
options.name   = 'Animal Food Production';
sol = ezsolve(objective,constraints,[],options);

PriLev = 1;
if PriLev > 0
    raws  = length(rawavail);
    prods = length(compsize);

    for prod = 1:prods,
        disp(['produce ' num2str(compsize(prod)) ...
            ' of product ' num2str(prod) ','])
        disp('   and use the following ingredients:')

        for raw = 1:raws,
            disp(['   ' num2str(sol.use(raw,prod)) ...
                ' units of ingredient ' num2str(raw) ])
        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: Animal Food Production         f_k   15086.795064700573000000
                                       sum(|constr|)      0.000000000000000606
                              f(x_k) + sum(|constr|)  15086.795064700573000000
                                              f(x_0)      0.000000000000000000

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

FuncEv   10 Iter   10 
produce 9000 of product 1,
   and use the following ingredients:
   5098.5555 units of ingredient 1
   3719.5305 units of ingredient 2
   181.9139 units of ingredient 3
produce 12000 of product 2,
   and use the following ingredients:
   6798.074 units of ingredient 1
   4959.3741 units of ingredient 2
   242.5519 units of ingredient 3