max_volume

% Box volume maximization example.
% (a figure is generated if the tradeoff flag is turned on)
%
% This is an example taken directly from the paper:
%
%   A Tutorial on Geometric Programming (see pages 8 and 13)
%   by Boyd, Kim, Vandenberghe, and Hassibi.
%
% Maximizes volume of a box-shaped structure which has constraints
% on its total wall area, its total floor area, and which has lower
% and upper bounds on the aspect ratios. This leads to a GP:
%
%   maximize   h*w*d
%       s.t.   2(h*w + h*d) <= Awall, w*d <= Afloor
%              alpha <= h/w <= beta
%              gamma <= d/w <= delta
%
% where variables are the box height h, width w, and depth d.
%
% Almir Mutapcic 02/01/2006
clear all; close all;
PLOT_TRADEOFF = 1; % to disable tradeoff plot, set PLOT_TRADEOFF = 0

% problem constants
Awall  = 10000; Afloor = 1000;
alpha = 0.5; beta = 2; gamma = 0.5; delta = 2;

% GP variables
gpvar h w d

% objective function is the box volume
volume = h*w*d;

% set of constraints expressed as an array
constr = [ 2*(h*w + h*d) <= Awall; w*d <= Afloor;
           alpha <= h/w; h/w <= beta;
           gamma <= d/w; d/w <= delta;];

% solve the given GP problem
[max_volume solution status] = gpsolve(volume, constr, 'max');
assign(solution);

% display results
fprintf(1,'\nMaximum volume is %2.2f.\n', max_volume)
fprintf(1,['Optimal solution is height h = %3.4f, width w = %3.4f, '...
           'depth d = %3.4f.\n\n'], h, w, d);

%*********************************************************************
% tradeoff curve code
%*********************************************************************
if( PLOT_TRADEOFF )

% set the quiet flag (no solver reporting)
global QUIET; QUIET = 1;

% reintroduce the optimization variables
gpvar h w d

% varying parameters for an optimal trade-off curve
N = 10;
Aflr = logspace(1,3,N);
Awall = [100 1000 10000];
opt_volumes = zeros(length(Awall),N);

% setup various GP problems with varying parameters
for k = 1:length(Awall)
  for n = 1:N
    % change constraints with varying parameters
    constr(1) = 2*h*w + 2*h*d <= Awall(k);
    constr(2) = w*d <= Aflr(n);

    % solve the GP problem and compute the optimal volume
    [max_volume solution status] = gpsolve(volume, constr, 'max');
    opt_volumes(k,n) = max_volume;
  end
end

% enable solver reporting again
global QUIET; QUIET = 0;

% plot the tradeoff curve
loglog(Aflr,opt_volumes(1,:), Aflr,opt_volumes(2,:), Aflr,opt_volumes(3,:));
xlabel('Aflr'); ylabel('V');

end
% end tradeoff curve code

 
Iteration     primal obj.         gap        dual residual     previous step.
 
   1         1.00000e+00       1.20000e+01       2.73e+00               Inf
   2         1.13450e+00       5.09520e+00       3.18e-03       9.65862e-01
   3         7.84658e-01       4.08972e+00       1.25e-03       3.74018e-01
   4         4.86635e-01       3.25002e+00       3.24e-04       4.90589e-01
   5         2.84603e-01       2.69975e+00       4.39e-05       6.31798e-01
   6        -2.39995e-01       1.30243e+00       6.13e-13       1.00000e+00
 
Iteration     primal obj.         gap        dual residual     previous step.
 
   1         2.66126e+02       1.20000e+01       2.98e+00               Inf
   2         5.79616e+01       9.95483e+00       2.92e+00       9.58011e-03
   3         2.03729e+01       9.65217e+00       2.88e+00       7.37105e-03
   4         2.25731e+00       9.39954e+00       2.81e+00       1.18031e-02
   5        -1.01030e+01       8.91928e+00       2.64e+00       3.12500e-02
   6        -5.82528e+00       9.46552e+00       6.60e-01       5.00000e-01
   7        -9.57242e+00       6.85383e+00       1.65e-01       5.00000e-01
   8        -1.00194e+01       5.89507e+00       6.05e-06       1.00000e+00
   9        -1.04901e+01       2.94793e+00       8.35e-07       1.00000e+00
  10        -1.08661e+01       1.47474e+00       6.53e-07       1.00000e+00
  11        -1.10901e+01       7.37986e-01       1.67e-06       1.00000e+00
  12        -1.11902e+01       3.69166e-01       4.82e-06       1.00000e+00
  13        -1.12187e+01       1.84575e-01       1.35e-07       1.00000e+00
  14        -1.12350e+01       9.23682e-02       7.62e-07       1.00000e+00
  15        -1.12461e+01       4.62892e-02       1.05e-07       1.00000e+00
  16        -1.12517e+01       2.31810e-02       1.04e-08       1.00000e+00
  17        -1.12546e+01       1.16008e-02       7.80e-10       1.00000e+00
  18        -1.12561e+01       5.80299e-03       5.12e-11       1.00000e+00
  19        -1.12568e+01       2.90215e-03       3.23e-12       1.00000e+00
  20        -1.12571e+01       1.45124e-03       2.03e-13       1.00000e+00
  21        -1.12573e+01       7.25658e-04       1.27e-14       1.00000e+00
  22        -1.12574e+01       3.62839e-04       7.94e-16       1.00000e+00
  23        -1.12575e+01       1.81422e-04       4.96e-17       1.00000e+00
  24        -1.12575e+01       9.07117e-05       3.10e-18       1.00000e+00
  25        -1.12575e+01       4.53560e-05       1.94e-19       1.00000e+00
  26        -1.12575e+01       2.26780e-05       1.21e-20       1.00000e+00
  27        -1.12575e+01       1.13390e-05       7.58e-22       1.00000e+00
  28        -1.12575e+01       5.66952e-06       4.74e-23       1.00000e+00
  29        -1.12575e+01       2.83476e-06       2.96e-24       1.00000e+00
  30        -1.12575e+01       1.41738e-06       1.85e-25       1.00000e+00
  31        -1.12575e+01       7.08690e-07       1.15e-26       1.00000e+00
  32        -1.12575e+01       3.54345e-07       7.35e-28       1.00000e+00
  33        -1.12575e+01       1.77172e-07       5.39e-29       1.00000e+00
  34        -1.12575e+01       8.85862e-08       2.55e-30       1.00000e+00
  35        -1.12575e+01       4.42931e-08       6.98e-31       1.00000e+00
  36        -1.12575e+01       2.21466e-08       2.16e-31       1.00000e+00
  37        -1.12575e+01       1.10733e-08       2.75e-31       1.00000e+00
  38        -1.12575e+01       5.53664e-09       1.32e-31       1.00000e+00
Solved
Problem succesfully solved.

Maximum volume is 77459.67.
Optimal solution is height h = 77.4597, width w = 38.7298, depth d = 25.8199.

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Problem succesfully solved.