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24 Loading and Cutting
24.1 Wagon load balancing
% function Result = wagonloadbalancingEx(PriLev)
%
% Creates a TOMLAB LP problem for wagon loading. Minimizes the max weight
%
% WAGON LOAD BALANCING
%
% Three railway wagons with a carrying capacity of 100 quintals (1
% quintal = 100 kg) have been reserved to transport sixteen boxes.
% The weight of the boxes in quintals is given in the following
% table. How shall the boxes be assigned to the wagons in order to
% keep to the limits on the maximum carrying capacity and to minimize
% the heaviest wagon load?
%
% Weight of boxes
%
% +------+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
% |Box | 1| 2| 3| 4| 5| 6| 7| 8| 9|10|11|12|13|14|15|16|
% +------+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
% |Weight|34| 6| 8|17|16| 5|13|21|25|31|14|13|33| 9|25|25|
% +------+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
%
% Before implementing a Mathematical Programming solution, one may
% wish to try to see whether it is possible to solve this problem
% instance with the following simple heuristic: until all boxes are
% distributed to the wagons we choose in turn the heaviest unassigned
% box and put it onto the wagon with the least load.
%
%
% VARIABLES
%
% minload Minimal load accepted per wagon
% maxload Maximal load accepted per wagon
% weights Weight of each box
%
% RESULTS
%
% For an interpretation of the results, set PriLev > 1,
% Result = wagonloadbalancingEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev Print Level
%
% OUTPUT PARAMETERS
% Result Result structure.
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2006 by Tomlab Optimization Inc., $Release: 5.1.0$
% Written Oct 7, 2005. Last modified Feb 3, 2006.
function Result = wagonloadbalancingEx(PriLev)
if nargin < 1
PriLev = 1;
end
minload = [ 0; 0; 0;];
maxload = [100;100;100;];
weights = [34;6;8;17;16;5;13;21;25;31;14;13;33;9;25;25];
Prob = wagonloadbalancing(minload, maxload, weights);
Prob.SolverInf = 'cplex';
Result = infLinSolve(Prob, PriLev);
if PriLev > 1,
boxes = length(weights);
wagons = length(minload);
temp = reshape(Result.x_k,boxes,wagons);
for w = 1:wagons,
idx = find(temp(:,w) > 0.5);
disp(['wagon ' num2str(w) ...
' has a total load of ' num2str(sum(weights(idx))) ...
' by carrying the boxes: ' num2str(idx') ])
end
end
% MODIFICATION LOG
%
% 051007 med Created.
% 060111 per Added documentation.
% 060125 per Moved disp to end
% 060203 med Printing updated
24.2 Barge loading
% function Result = bargeloadingEx(PriLev)
%
% Creates a TOMLAB LP problem for barge loading, maximizing the profit
%
% BARGE LOADING
%
% A shipper on the river Rhine owns a barge of carrying capacity 1500
% m3. Over time he has specialized in the transport of wheat. He has
% seven regular customers who load and unload practically at the same
% places. The shipper knows his costs of transport from long
% experience and according to his personal preferences has concluded
% agreements with his clients for the price charged to them for the
% transport of their wheat. The following table summarizes the
% information about the seven clients. Every client wishes to
% transport a certain number of lots, deciding himself the size of
% his lots in m3. The table lists the price charged by the shipper
% for every transported lot. The last column of the table contains
% the cost incurred by the shipper per transported m3. This cost
% differs depending on the distance covered.
%
% Lots to transport
%
% +------+------------------+--------+----------+--------------+
% | |Available quantity|Lot size|Price per | Transport |
% |Client| (no. of lots) |(in m3) |lot (in $)|cost (in $/m3)|
% +------+------------------+--------+----------+--------------+
% | 1 | 12 | 10 | 1000 | 80 |
% | 2 | 31 | 8 | 600 | 70 |
% | 3 | 20 | 6 | 600 | 85 |
% | 4 | 25 | 9 | 800 | 80 |
% | 5 | 50 | 15 | 1200 | 73 |
% | 6 | 40 | 10 | 800 | 70 |
% | 7 | 60 | 12 | 1100 | 80 |
% +------+------------------+--------+----------+--------------+
%
% The objective of the shipper is to maximize his profit from
% transporting the wheat with lots that may be divided.
%
% Question 1:
% As a first step, assuming that every client has an unlimited
% quantity of wheat, which clients’ wheat should be transported?
%
% Question 2:
% If in addition the actual availability of wheat lots from the
% customers is taken into account, which strategy should the shipper
% adopt?
%
% Question 3:
% What happens if the lots cannot be divided?
%
% VARIABLES
%
% capacity = 1500;
% units = [12;31;20;25;50;40;60];
% sizes = [10;8;6;9;15;10;12];
% unitprice = [10;6;6;8;12;8;11]*100;
% cost = [80;70;85;80;73;70;80];
% var1 30 plus
% var2 30 minus
%
% RESULTS
%
% x_k is a long vector of values, and if we run
% Result = bargeloadingEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev Print Level
%
% OUTPUT PARAMETERS
% Result Result structure.
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., $Release: 5.0.0$
% Written Oct 7, 2005. Last modified Oct 7, 2005.
function Result = bargeloadingEx(PriLev)
if nargin < 1
PriLev = 1;
end
capacity = 1500;
units = [12;31;20;25;50;40;60];
sizes = [10;8;6;9;15;10;12];
unitprice = [10;6;6;8;12;8;11]*100;
cost = [80;70;85;80;73;70;80];
Prob = bargeloading(capacity, units, sizes, unitprice, cost);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1,
disp('Best buy')
for i = 1:length(units),
if Result.x_k(i) > 0,
disp([' ' num2str(Result.x_k(i)) ' unit(s) from ' num2str(i)])
end
end
end
% MODIFICATION LOG
%
% 051007 med Created.
% 060111 per Added documentation.
% 060125 per Moved disp to end
24.3 Tank loading
% function Result = tankloadingEx(PriLev)
%
% Creates a TOMLAB MIP problem for tank loading, maximizing the left over
% capacity
%
% TANK LOADING
%
% Five tanker ships have arrived at a chemical factory. They are
% carrying loads of liquid products that must not be mixed: 1200
% tonnes of Benzol, 700 tonnes of Butanol, 1000 tonnes of Propanol,
% 450 tonnes of Styrene, and 1200 tonnes of THF. Nine tanks of
% different capacities are available on site. Some of them are
% already partially filled with a liquid. The following table lists
% the characteristics of the tanks (in tonnes). Into which tanks
% should the ships be unloaded (question 1) to maximize the capacity
% of the tanks that remain unused, or (question 2) to maximize the
% number of tanks that remain free?
%
% Characteristics of tanks
%
% +---------------+---+------+---+---+---+---+---+---+---+
% |Tank | 1| 2 | 3| 4| 5| 6| 7| 8| 9|
% +---------------+---+------+---+---+---+---+---+---+---+
% |Capacity |500| 400 |400|600|600|900|800|800|800|
% |Current product| -|Benzol| -| -| -| -|THF| -| -|
% |Quantity | 0| 100 | 0| 0| 0| 0|300| 0| 0|
% +---------------+---+------+---+---+---+---+---+---+---+
%
% VARIABLES
%
% capacity Capacity of Empty tanks
% products Amount of incoming chemicals
%
% RESULTS
%
% In order to interpret the results, try the following:
% Result = tankloadingEx(2);
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev Print Level
%
% OUTPUT PARAMETERS
% Result Result structure.
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., $Release: 5.0.0$
% Written Oct 10, 2005. Last modified Dec 8, 2005.
function Result = tankloadingEx(PriLev)
if nargin < 1
PriLev = 1;
end
capacity = [500;0;400;600;600;900;0;800;800];
products = [1200-300;700;1000;450;1200-500];
Prob = tankloading(capacity, products);
Result = tomRun('cplex', Prob, PriLev);
Result.remaining = sum(capacity)-Result.f_k;
if PriLev > 1,
disp('NB: Tank 2 and tank 7 are filled before the optimization starts.')
temp = reshape(Result.x_k,9,5);
ships = length(products);
for ship = 1:ships,
tank = find(temp(:,ship));
disp(['Ship number ' num2str(ship) ' unloads in tank(s) ' num2str(tank')])
end
end
% MODIFICATION LOG
%
% 051010 med Created
% 051208 med Added remaining the return
% 060112 per Added documentation.
% 060124 per Interpretation of results upgraded.
% 060125 per Moved disp to end
24.4 Backing up files
% function Result = backingupfilesEx(PriLev)
%
% Creates a TOMLAB MIP problem for backing up files
%
% BACKING UP FILES
%
% Before leaving on holiday, you wish to backup your most important
% files onto floppy disks. You have got empty disks of 1.44Mb
% capacity. The sixteen files you would like to save have the
% following sizes: 46kb, 55kb, 62kb, 87kb, 108kb, 114kb, 137kb,
% 164kb, 253kb, 364kb, 372kb, 388kb, 406kb, 432kb, 461kb, and 851kb.
%
% Assuming that you do not have any program at hand to compress the
% files and that you have got a sufficient number of floppy disks to
% save everything, how should the files be distributed in order to
% minimize the number of floppy disks used?
%
% VARIABLES
%
% maxuse A maximal number of disks
% capacity Storage available on each disk
% sizes Filesizes
%
% RESULTS
%
% Results are explained by running:
% Result = backingupfilesEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev Print Level
%
% OUTPUT PARAMETERS
% Result Result structure.
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2006 by Tomlab Optimization Inc., $Release: 5.2.0$
% Written Oct 10, 2005. Last modified Mar 14, 2006.
function Result = backingupfilesEx(PriLev)
if nargin < 1
PriLev = 1;
end
maxuse = 10;
capacity = 1440;
sizes = [46;55;62;87;108;114;137;164;253;364;372;388;406;432;461;851];
Prob = backingupfiles(maxuse, capacity, sizes);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1,
files = length(sizes);
filepos = reshape(Result.x_k(1:files*maxuse),files,maxuse);
disks_in_use = sum(Result.x_k(files*maxuse+1:files*maxuse+maxuse));
disp(['a total of ' num2str(disks_in_use) ' disks are in use'])
for d = 1:maxuse,
files_on_disk = filepos(:,d); % what files on disk d
if (abs(sum(files_on_disk)) >= 0.5), % disk d is not empty?
file_ids = find(files_on_disk);
disp([' files on one of them: ' num2str(file_ids') ])
end
end
end
% MODIFICATION LOG
%
% 051010 med Created.
% 060112 per Added documentation
% 060125 per Moved disp to end
% 060314 med checking for empty disc changed
24.5 Cutting sheet metal
% function Result = cuttingsheetmetalEx(PriLev)
%
% Creates a TOMLAB MIP problem for cutting smaller parts from big ones.
%
% CUTTING SHEET METAL
%
% A sheet metal workshop cuts pieces of sheet metal from large
% rectangular sheets of 48 decimeters × 96 decimeters (dm). It has
% received an order for 8 rectangular pieces of 36 dm × 50 dm, 13
% sheets of 24 dm × 36 dm, 5 sheets of 20 dm × 60 dm, and 15 sheets
% of 18 dm × 30 dm. Theses pieces of sheet metal need to be cut from
% the available large pieces. How can this order by satisfied by
% using the least number of large sheets?
%
% VARIABLES
%
% patterns Each column represent a possible sheet.
% demand Wanted rectangles
%
% RESULTS
%
% To explain the results, run:
% Result = cuttingsheetmetalEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev Print Level
%
% OUTPUT PARAMETERS
% Result Result structure.
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., $Release: 5.0.0$
% Written Oct 10, 2005. Last modified Oct 10, 2005.
function Result = cuttingsheetmetalEx(PriLev)
if nargin < 1
PriLev = 1;
end
% Have to manually evaluate all the possible patterns.
patterns = [1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0;...
2 1 0 2 1 0 3 2 1 0 5 4 3 2 1 0;...
0 0 0 2 2 2 1 1 1 1 0 0 0 0 0 0;...
0 1 3 0 1 3 0 2 3 5 0 1 3 5 6 8];
demand = [8;13;5;15];
Prob = cuttingsheetmetal(demand, patterns);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1,
x = Result.x_k;
idx = find(x);
order = [x(idx) idx];
for i = 1:length(idx),
disp(['cut ' num2str([order(i,1)]) ' sheet(s) in pattern ' num2str([order(i,2)])])
end
end
% MODIFICATION LOG
%
% 051010 med Created.
% 060112 per Added documentation.
% 060125 per Moved disp to end
24.6 Cutting steel bars for desk legs
% function Result = cuttingsteelbarsfordesklegsEx(PriLev)
%
% Creates a TOMLAB MIP problem for cutting steel bars for desk legs.
%
% CUTTING STEEL BARS FOR DESK LEGS
%
% The company SchoolDesk produces desks of different sizes for
% kindergartens, primary and secondary schools, and colleges. The
% legs of the desks all have the same diameter, with different
% lengths: 40 cm for the smallest ones, 60 cm for medium height, and
% 70 cm for the largest ones. These legs are cut from steel bars of
% 1.5 or 2 meters. The company has received an order for 108 small,
% 125 medium and 100 large desks. How should this order be produced
% if the company wishes to minimize the trim loss?
%
% VARIABLES
%
% patterns The different cutting patterns
% demand Demand of the different lengths
% loss Loss for each pattern
% lengths The lengths
%
% RESULTS
%
% To interpret the results run this:
% Result = cuttingsteelbarsfordesklegsEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev Print Level
%
% OUTPUT PARAMETERS
% Result Result structure.
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., $Release: 5.0.0$
% Written Oct 10, 2005. Last modified Dec 8, 2005.
function Result = cuttingsteelbarsfordesklegsEx(PriLev)
if nargin < 1
PriLev = 1;
end
patterns = [0 0 2 0 2 3 0 0 1 3 0 5;...
0 1 0 2 1 0 1 2 0 0 3 0;...
2 1 1 0 0 0 2 1 2 1 0 0];
demand = [108;125;100]*4;
loss = [10 20 0 30 10 30 0 10 20 10 20 0]';
lengths = [150;200];
Prob = cuttingsteelbarsfordesklegs(demand, patterns, lengths);
Prob.fConstant = -75280; % Constant to deduct from objective
Result = tomRun('cplex', Prob, PriLev);
Result.loss = sum(Result.x_k.*loss);
if PriLev > 1,
x = Result.x_k;
idx = find(x);
order = [x(idx) idx];
disp(['a minimal loss of ' num2str(Result.loss) ' is found with this combination:' ])
for i = 1:length(idx),
disp([' cut ' num2str([order(i,1)]) ' bar(s) in pattern ' num2str([order(i,2)])])
end
end
% MODIFICATION LOG
%
% 051010 med Created
% 051208 med Loss added to Result
% 060112 per Added documentation.
% 060112 per Minor update.
% 060125 per Moved disp to end
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